1,001 Calculus Practice Problems

Part II

The Answers

 

In this part …

Here you get answers and explanations for all 1,001 problems. As you read the solutions, you may realize that you need a little more help. Lucky for you, the For Dummies series (published by Wiley) offers several excellent resources. I recommend checking out the following titles, depending on your needs:

·        Calculus For Dummies, Calculus Workbook For Dummies, and Calculus Essentials For Dummies, all by Mark Ryan

·        Pre-Calculus For Dummies, by Yang Kuang and Elleyne Kase, and Pre-Calculus Workbook For Dummies, by Yang Kuang and Michelle Rose Gilman

·        Trigonometry For Dummies and Trigonometry Workbook For Dummies, by Mary Jane Sterling

When you're ready to step up to more advanced calculus courses, you'll find the help you need in Calculus II For Dummies, by Mark Zegarelli.

Visit www.dummies.com for more information.

Chapter 16

Answers and Explanations

Here are the answer explanations for all 1,001 practice problems.

1. 9781118496718-eq16001.png

Get a common denominator of 12 and then perform the arithmetic in the numerator:

9781118496718-eq16002.png

2. 9781118496718-eq16003.png

Get a common denominator of 60 and then perform the arithmetic in the numerator:

9781118496718-eq16004.png

3. 11

Write each factor as a fraction and then multiply across the numerator and denominator (cancel common factors as well!):

9781118496718-eq16005.png

4. 9781118496718-eq16006.png

Recall that to divide by a fraction, you multiply by its reciprocal: 9781118496718-eq16007.tif. After rewriting the fraction, cancel common factors before multiplying:

9781118496718-eq16008.png

5. 9781118496718-eq16009.png

To add the fractions, you need a common denominator. The least common multiple of the denominators is x2y, so multiply each fraction accordingly to get this denominator for each term. Then write the answer as a single fraction:

9781118496718-eq16010.png

6. 9781118496718-eq16011.png

You need a common denominator so you can subtract the fractions. The least common multiple of the denominators is (x – 1)(x + 1) = x2 – 1, so multiply each fraction accordingly to get this denominator. Then perform the arithmetic in the numerator:

9781118496718-eq16012.png

7. 9781118496718-eq16013.png

Begin by factoring the expressions completely. Cancel any common factors and then simplify to get the answer:

9781118496718-eq16014.png

8. 9781118496718-eq16015.png

Recall that to divide by a fraction, you multiply by its reciprocal: 9781118496718-eq16016.tif. After rewriting the fraction, factor and cancel the common factors:

9781118496718-eq16017.png

9. 9781118496718-eq16018.png

Write each factor using positive exponents and then simplify. To start, notice that 9781118496718-eq16019.tif in the denominator becomes 9781118496718-eq16020.tif when it moves to the numerator:

9781118496718-eq16021.png

10. 9781118496718-eq16022.png

To simplify, cancel the common factors:

9781118496718-eq16023.png

11. 9781118496718-eq16024.png

Because the entire expression is being raised to the power of zero, the answer is 1:

9781118496718-eq16025.png

There's no point in simplifying the expression inside the parentheses.

12. 9781118496718-eq16026.png

Begin by factoring the numerator and then cancel the common factors:

9781118496718-eq16027.png

13. 9781118496718-eq16028.png

For this problem, you need to know the properties of exponents. The important properties to recall here are 9781118496718-eq16029.tif9781118496718-eq16030.tif9781118496718-eq16031.tif9781118496718-eq16032.tif9781118496718-eq16033.tif, and 9781118496718-eq16034.tif.

Write all factors using positive exponents and then use the properties of exponents to simplify the expression:

9781118496718-eq16035.png

14. 9781118496718-eq16036.png

Factor the number underneath the radical and then rewrite the radical using 9781118496718-eq16037.tif to simplify:

9781118496718-eq16038.png

15. 9781118496718-eq16039.png

Start by rewriting the expression so it has only one square root sign. Then reduce the fraction and use properties of radicals to simplify:

9781118496718-eq16040.png

16. 9781118496718-eq16041.png

Begin by writing the expression using a single square root sign. Then combine like bases and simplify:

9781118496718-eq16042.png

17. 9781118496718-eq16043.png

Begin by writing the expression using a single cube root sign. Combine the factors with like bases and then simplify:

9781118496718-eq16044.png

18. 9781118496718-eq16045.png

Simplify each root individually and then use properties of exponents to simplify further:

9781118496718-eq16046.png

19. 9781118496718-eq16047.png

Recall that 9781118496718-eq16048.tif. The expression becomes

9781118496718-eq16049.png

20. 9781118496718-eq16050.png

Recall that 9781118496718-eq16051.tif. Therefore, the expression becomes

9781118496718-eq16052.png

21. y = 3x + 8

Recall that a function is one-to-one if it satisfies the requirement that if x1 ≠ x2, then f (x1) ≠ f (x2); that is, no two x coordinates have the same y value. If you consider the graph of such a function, no horizontal line would cross the graph in more than one place.

Of the given functions, only the linear function y = 3x + 8 passes the horizontal line test. This function is therefore one-to-one and has an inverse.

9781118496718-un1601.tif

22. y = x2 – 4, x ≥ 0

When determining whether a function has an inverse, consider the domain of the given function. Some functions don't pass the horizontal line test — and therefore don't have an inverse — unless the domain is restricted. The function y = x2 – 4 doesn't pass the horizontal line test; however, if you restrict the domain to x ≥ 0, then y = x2 – 4 does pass the horizontal line test (and therefore has an inverse), because you're getting only the right half of the parabola.

9781118496718-un1602.tif

23. y = tan−1 x

Of the given functions, only the function y = tan−1 x passes the horizontal line test. Therefore, this function is one-to-one and has an inverse.

9781118496718-un1603.tif

24. 9781118496718-eq16053.png

First replace f (x) with y:

9781118496718-eq16054.png

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

9781118496718-eq16055.png

Therefore, the inverse of f (x) = 4 – 5x is

9781118496718-eq16056.png

25. 9781118496718-eq16057.png

First replace f (x) with y:

9781118496718-eq16058.png

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse. You need to complete the square so you can isolate y. (To complete the square, consider the quadratic expression on the right side of the equal sign; take one-half of the coefficient of the term involving y, square it, and add that value to both sides of the equation.) Then factor the right side and use square roots to solve for y:

9781118496718-eq16059.png

Notice that for the domain of the inverse to match the range of f (x), you have to keep the positive root. Therefore, the inverse of f (x) = x2 – 4x, where x ≥ 2, is

9781118496718-eq16060.png

26. 9781118496718-eq16061.png9781118496718-eq16062.png

First replace f (x) with y:

9781118496718-eq16063.png

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

9781118496718-eq16064.png

Therefore, the inverse of 9781118496718-eq16065.tif is

9781118496718-eq16066.png

The domain of the inverse is x ≥ 0 because the range of f (x) is y ≥ 0 and the domain of 9781118496718-eq16067.tif is equal to the range of f (x).

27. 9781118496718-eq16068.png

First replace f (x) with y:

9781118496718-eq16069.png

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

9781118496718-eq16070.png

Therefore, the inverse of f (x) = 3x5 + 7 is

9781118496718-eq16071.png

28. 9781118496718-eq16072.png

First replace f (x) with y:

9781118496718-eq16073.png

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

9781118496718-eq16074.png

Note that the range of 9781118496718-eq16075.tif is (–1, 1], so the domain of the inverse function is (–1, 1]. Therefore, the inverse of 9781118496718-eq16076.tif is

9781118496718-eq16077.png

29. 9781118496718-eq16078.png

First replace f (x) with y:

9781118496718-eq16079.png

Then replace y with x and each x with y. After making the replacements, solve for y to find the inverse:

9781118496718-eq16080.png

Therefore, the inverse of 9781118496718-eq16081.tif is

9781118496718-eq16082.png

30. {(1, 0), (4, 3), (–6, 5)}

If the point (a, b) is on the graph of a one-to-one function, the point (b, a) is on the graph of the inverse function. Therefore, just switch the x and y coordinates to get that the set of points {(1, 0), (4, 3), (–6, 5)} belongs to the graph of f −1(x).

31. domain: (–1, 2); range: [–2, 4)

For a one-to-one function (which by definition has an inverse), the domain of f (x) becomes the range of f −1(x), and the range of f (x) becomes the domain of f −1(x). Therefore, f −1(x) has domain (–1, 2) and range [–2, 4).

32. g −1(x) = f −1(x) – c

Replacing x with (x + c) shifts the graph c units to the left, assuming that c > 0. If the point (a, b) belongs to the graph of f (x), then the point (a – c, b) belongs to the graph of f (x + c).

Now consider the inverse. The point (b, a) belongs to the graph of f −1(x), so the point (b, a – c) belongs to the graph of g −1(x). Therefore, g −1(x) is the graph of f −1(x) shifted down c units so that g −1(x) = f −1(x) – c.

The same argument applies if c < 0.

33. x = 2

Put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x to get the solution:

9781118496718-eq16083.png

34. x = –4

Distribute to remove the parentheses. Then put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x to get the solution:

9781118496718-eq16084.png

35. 9781118496718-eq16085.png

Distribute to remove the parentheses. Then put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x to get the solution:

9781118496718-eq16086.png

36. 9781118496718-eq16087.png

Put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x (that is, multiply both sides by the reciprocal of the coefficient) to get the solution:

9781118496718-eq16088.png

37. 9781118496718-eq16089.png

Distribute to remove the parentheses. Then put all terms involving x on one side of the equation and all constants on the other side, combining all like terms. Finally, divide by the coefficient of x to get the solution:

9781118496718-eq16090.png

38. x = –3, 7

This quadratic factors without too much trouble using trial and error:

9781118496718-eq16091.png

Setting each factor equal to zero gives you x – 7 = 0 so that x = 7 is a solution and x + 3 = 0 so that x = –3 is a solution.

39. 9781118496718-eq16092.png

Complete the square to solve the quadratic equation. (To complete the square, consider the quadratic expression on the left side of the equal sign; take one-half of the coefficient of the term involving x, square it, and add that value to both sides of the equation.) Then factor the left side and use square roots to solve for x:

9781118496718-eq16093.png

40. 9781118496718-eq16094.png

Complete the square to solve the quadratic equation. Begin by moving the constant to the right side of the equal sign and then dividing both sides of the equation by 2:

9781118496718-eq16095.png

Next, consider the quadratic expression on the left side of the equal sign; take one-half of the coefficient of the term involving x, square it, and add that value to both sides of the equation. Then factor the left side and use square roots to solve for x:

9781118496718-eq16096.png

41. 9781118496718-eq16097.png

Factoring by trial and error gives you

9781118496718-eq16098.png

Setting each factor equal to zero gives you 2x – 1 = 0 so that 9781118496718-eq16099.tif is a solution and 3x + 4 = 0 so that 9781118496718-eq16100.tif is a solution.

42. 9781118496718-eq16101.png

The equation 3x2 + 4x – 2 = 0 doesn't factor nicely, so use the quadratic equation to solve for x. Here, a = 3, b = 4, and c = –2:

9781118496718-eq16102.png

43. 9781118496718-eq16103.png

The equation x10 + 7x5 + 10 = 0 isn't quadratic, but by using a substitution, you can produce a quadratic equation:

9781118496718-eq16104.png

Now use the substitution y = x5 to form a quadratic equation that you can easily factor:

9781118496718-eq16105.png

From y + 2 = 0, you have the solution y = –2, and from y + 5 = 0, you have the solution y = –5.

Now replace y using the original substitution and solve for x to get the final answer: x5 = –2 gives you 9781118496718-eq16106.tif, and x5= –5 gives you 9781118496718-eq16107.tif.

44. x = 0, 9781118496718-eq16108.tif, 1

Begin by factoring out the greatest common factor, x2. Then factor the remaining quadratic expression:

9781118496718-eq16109.png

Next, set each factor equal to zero and solve for x: x2 = 0 has the solution x = 0, 3x + 5 = 0 has the solution 9781118496718-eq16110.tif, and x – 1 = 0 has the solution x = 1.

45. no real solutions

Factoring this polynomial by trial and error gives you the following:

9781118496718-eq16111.png

Setting the first factor equal to zero gives you x4 + 5 = 0 so that x4 = –5, which has no real solutions. Setting the second factor equal to zero gives you x4 + 7 = 0 so that x4 = –7, which also has no real solutions.

46. x = –1, 1

Factor the polynomial repeatedly to get the following:

9781118496718-eq16112.png

Now set each factor equal to zero and solve for x. Setting the first factor equal to zero gives you x – 1 = 0 so that x = 1. The second factor gives you x + 1 = 0 so that x = –1. And the last factor gives you x2 + 4 = 0 so that x2 = –4, which has no real solutions.

47. x = –3, 3

Factor the polynomial repeatedly to get the following:

9781118496718-eq16113.png

Now set each factor equal to zero and solve for x. Setting the first factor equal to zero gives you x – 3 = 0 so that x = 3. Setting the second factor equal to zero gives you x + 3 = 0 so that x = –3. The last factor gives you x2 + 9 = 0, or x2 = –9, which has no real solutions.

48. x = 1, 9781118496718-eq16114.tif

To solve an absolute value equation of the form 9781118496718-eq16115.tif, where b > 0, you must solve the two equations a = b and a = –b. So for the equation 9781118496718-eq16116.tif, you have to solve 5x – 7 = 2:

9781118496718-eq16117.png

You also have to solve 5x – 7 = –2:

9781118496718-eq16118.png

Therefore, the solutions are 9781118496718-eq16119.tif and x = 1.

49. no real solutions

To solve an absolute value equation of the form 9781118496718-eq16120.tif, where b > 0, you must solve the two equations a = b and a = –b. However, in this case, you have 9781118496718-eq16121.tif, which gives you 9781118496718-eq16122.tif. Because the absolute value of a number can't be negative, this equation has no solutions.

50. x = –3, 9

To solve an absolute value equation of the form 9781118496718-eq16123.tif, where b > 0, you must solve the two equations a = b and a = –b. So for the equation 9781118496718-eq16124.tif, you must solve x2 – 6x = 27 and x2 – 6x = –27.

To solve x2 – 6x = 27, set the equation equal to zero and factor using trial and error:

9781118496718-eq16125.png

Setting each factor equal to zero gives you x – 9 = 0, which has the solution x = 9, and x + 3 = 0, which has the solution x = –3.

Then set x2 – 6x = –27 equal to zero, giving you x2 – 6x + 27 = 0. This equation doesn't factor nicely, so use the quadratic equation, with a = 1, b = –6, and c = 27:

9781118496718-eq16126.png

The number beneath the radical is negative, so this part of the absolute value has no real solutions.

Therefore, the only real solutions to 9781118496718-eq16127.tif are x = –3 and x = 9.

51. x = –3, 2

To solve an absolute value equation of the form 9781118496718-eq16128.tif, you must solve the two equations a = b and a = –b. So for the equation 9781118496718-eq16129.tif, you have to solve 15x – 5 = 35 – 5x and 15x – 5 = –(35 – 5x).

For the first equation, you have

9781118496718-eq16130.png

And for the second equation, you have

9781118496718-eq16131.png

Therefore, the solutions are x = –3 and x = 2.

52. x = –1

To solve a rational equation of the form 9781118496718-eq16132.tif, you need to solve the equation p(x) = 0. Therefore, to solve 9781118496718-eq16133.tif, you simply set the numerator equal to zero and solve for x:

9781118496718-eq16134.png

53. x = 9781118496718-eq16135.tif9781118496718-eq16136.tif

To solve the equation 9781118496718-eq16137.tif, first remove all fractions by multiplying both sides of the equation by the least common multiple of the denominators:

9781118496718-eq16138.png

In this case, multiplying leaves you with a quadratic equation to solve:

9781118496718-eq16139.png

Check the solutions to see whether they're extraneous (incorrect) answers. In this case, you can verify that both 9781118496718-eq16140.tif and 9781118496718-eq16141.tif satisfy the original equation by substituting these values into 9781118496718-eq16142.tif and checking that you get 1 on the left side of the equation.

54. –14

To solve an equation of the form 9781118496718-eq16143.tif, cross-multiply to produce the equation ad = bc:

9781118496718-eq16144.png

After cross-multiplying, you're left with a quadratic equation, which then reduces to a linear equation:

9781118496718-eq16145.png

You can verify that –14 is a solution of the original rational equation by substituting x = –14 into the equation 9781118496718-eq16146.tif and checking that you get the same value on both sides of the equation.

55. no real solutions

Begin by factoring all denominators. Then multiply each side of the equation by the least common multiple of the denominators to remove all fractions:

9781118496718-eq16147.png

Then simplify:

9781118496718-eq16148.png

Because x = 2 makes the original equation undefined, there are no real solutions.

56. (–4, 8)

To solve the inequality x2 – 4x – 32 < 0, begin by solving the corresponding equation x2 – 4x – 32 = 0. Then pick a point from each interval (determined by the solutions) to test in the original inequality.

Factoring x2 – 4x – 32 = 0 gives you (x + 4)(x – 8) = 0. Setting the first factor equal to zero gives you x + 4 = 0, which has the solution x = –4, and setting the second factor equal to zero gives you x – 8 = 0, which has the solution x = 8.

Therefore, you need to pick a point from each of the intervals, (–∞, –4), (–4, 8), and (8, ∞), to test in the original inequality. Substitute each test number into the expression x2 – 4x – 32 to see whether the answer is less than or greater than zero.

Using x = –10 to check the interval (–∞, –4) gives you

9781118496718-eq16149.png

which is not less than zero and so doesn't satisfy the inequality.

Using x = 0 to check the interval (–4, 8) gives you

9781118496718-eq16150.png

which is less than zero and does satisfy the inequality.

Using x = 10 to check the interval (8, ∞) gives you

9781118496718-eq16151.png

which is not less than zero and so doesn't satisfy the inequality.

Therefore, the solution set is the interval (–4, 8).

57. 9781118496718-eq16152.png

To solve the inequality 2x4 + 2x3 ≥ 12x2, begin by setting one side equal to zero. Solve the corresponding equation and then pick a point from each interval (determined by the solutions) to test in the inequality.

So from 2x4 + 2x3 ≥ 12x2, you have 2x4 + 2x3 – 12x2 ≥ 0. Factoring the corresponding equation so you can solve for x gives you

9781118496718-eq16153.png

Set each factor equal to zero and solve for x, giving you the solutions x = 0,x = –3, and x = 2. These values are also solutions to the inequality.

Next, pick a test point from each of the intervals, (–∞, –3), (–3, 0), (0, 2), and (2, ∞), to see whether the answer is positive or negative. Using x = –10 to check the interval (–∞, –3) gives you

9781118496718-eq16154.png

which is greater than zero.

Using x = –1 to check the interval (–3, 0) gives you

9781118496718-eq16155.png

which is less than zero.

Using x = 1 to check the interval (0, 2) gives you

9781118496718-eq16156.png

which is less than zero.

Finally, using x = 10 to check the interval (2, ∞) gives you

9781118496718-eq16157.png

which is greater than zero.

Therefore, the solution set is 9781118496718-eq16158.tif.

Note that you could've divided the original inequality by 2 to simplify the initial inequality.

58. 9781118496718-eq16159.png

To solve the rational inequality 9781118496718-eq16160.tif, first determine which values will make the numerator equal to zero and which values will make the denominator equal to zero. This helps you identify zeros (where the graph crosses the x-axis) and points where the function is not continuous.

From the numerator, you have x + 1 = 0, which has the solution x = –1. You also have x – 2 = 0, which has the solution x = 2. From the denominator, you have x + 3 = 0, which has the solution x = –3. Take a point from each interval determined by these solutions and test it in the original inequality.

You need to test a point from each of the intervals, (–∞, –3), (–3, –1), (–1, 2), and (2, ∞). Pick a point from each interval and substitute into the expression 9781118496718-eq16161.tif to see whether you get a value less than or greater than zero.

Using x = –10 to test the interval (–∞, –3) gives you

9781118496718-eq16162.png

which is less than zero and so satisfies the inequality.

Using x = –2 to test the interval (–3, –1) gives you

9781118496718-eq16163.png

which is greater than zero and doesn't satisfy the inequality.

Using x = 0 to test the interval (–1, 2) gives you

9781118496718-eq16164.png

which is less than zero and so satisfies the inequality.

Using x = 3 to test the interval (2, ∞) gives you

9781118496718-eq16165.png

which is greater than zero and so doesn't satisfy the inequality.

Therefore, the solution set is the interval (–∞, –3) and (–1, 2).

59. 9781118496718-eq16166.png(1, 3)

Begin by putting the rational inequality into the form 9781118496718-eq16167.tif by putting all terms on one side of the inequality and then getting common denominators:

9781118496718-eq16168.png

Next, set the numerator equal to zero and solve for x. Setting the numerator equal to zero gives you a quadratic equation that you can factor by trial and error:

9781118496718-eq16169.png

The solutions are 9781118496718-eq16170.tif and x = 3.

Also set each factor from the denominator equal to zero and solve for x. This gives you both (x – 1) = 0, which has the solution x = 1, and (x + 1) = 0, which has the solution x = –1.

Next, take a point from each of the intervals, (–∞, –1), 9781118496718-eq16171.tif9781118496718-eq16172.tif, (1, 3), and (3, ∞), and test it in the expression 9781118496718-eq16173.tif to see whether the answeris positive or negative; or equivalently, you can use the expression 9781118496718-eq16174.tif.

Using x = –10 to test the interval (–∞, 1) gives you

9781118496718-eq16175.png

which is less than zero and so doesn't satisfy the inequality.

Using 9781118496718-eq16176.tif to test the interval 9781118496718-eq16177.tif gives you

9781118496718-eq16178.png

which is greater than zero and so satisfies the inequality.

Using x = 0 to test the interval 9781118496718-eq16179.tif gives you

9781118496718-eq16180.png

which is less than zero and so doesn't satisfy the inequality.

Using x = 2 to test the interval (1, 3) gives you

9781118496718-eq16181.png

which is greater than zero and so satisfies the inequality.

And using x = 10 to test the interval (3, ∞) gives you

9781118496718-eq16182.png

which is less than zero and so doesn't satisfy the inequality.

Therefore, the solution set is the intervals 9781118496718-eq16183.tif and (1, 3).

60. 9781118496718-eq16184.png

To solve an absolute inequality of the form 9781118496718-eq16185.tif, where b > 0, solve the compound inequality –b < a < b. So for the inequality 9781118496718-eq16186.tif, you have

9781118496718-eq16187.png

The solution set is 9781118496718-eq16188.tif.

61. (–∞, 1) 9781118496718-eq16189.tif

To solve an absolute value inequality of the form 9781118496718-eq16190.tif, where b > 0, you have to solve the corresponding inequalities a > b and a < –b. Therefore, for the inequality 9781118496718-eq16191.tif, you have to solve 5x – 7 > 2 and 5x – 7 < –2. For the first inequality, you have

9781118496718-eq16192.png

And for the second inequality, you have

9781118496718-eq16193.png

Therefore, the solution set is the intervals (–∞, 1) and 9781118496718-eq16194.tif.

62. 9781118496718-eq16195.png

To solve an absolute inequality of the form 9781118496718-eq16196.tif, where b > 0, solve the compound inequality –b ≤ a ≤ b. For the inequality 9781118496718-eq16197.tif, you have

9781118496718-eq16198.png

Therefore, the solution is the interval 9781118496718-eq16199.tif.

63. 9781118496718-eq16200.png

The slope of a line that goes through the points (x1y1) and (x2y2) is given by 9781118496718-eq16201.tif. Therefore, the slope of the line that passes through (1, 2) and (5, 9) is

9781118496718-eq16202.png

64. y = 4x + 5

The graph of y = mx + b is a line with slope of m and a y-intercept at (0, b). Therefore, using m = 4 and b = 5 gives you the equation y = 4x + 5.

65. 9781118496718-eq16203.png

The equation of a line that goes through the point (x1y1) and has a slope of m is given by the point-slope formula: y – y1 = m(x – x1).

The slope of the line that goes through (–2, 3) and (4, 8) is

9781118496718-eq16204.png

Now use the slope and the point (4, 8) in the point-slope formula:

9781118496718-eq16205.png

66. 9781118496718-eq16206.png

The equation of a line that goes through the point (x1y1) and has a slope of m is given by the point-slope formula: y – y1 = m(x – x1).

The two lines are parallel, so the slopes are the same. Therefore, the slope of the line you're trying to find is 9781118496718-eq16207.tif.

Using the slope and the point (1, 5) in the point-slope formula gives you

9781118496718-eq16208.png

67. 9781118496718-eq16209.png

The equation of a line that goes through the point (x1y1) and has a slope of m is given by the point-slope formula: y – y1 = m(x – x1).

The slope of the line that goes through the points (–6, 2) and (3, –4) is

9781118496718-eq16210.png

The slopes of perpendicular lines are opposite reciprocals of each other, so the slope of the line you want to find is 9781118496718-eq16211.tif (you flip the fraction and change the sign).

Using the slope and the point (3, –4) in the point-slope formula gives you

9781118496718-eq16212.png

Here's what the perpendicular lines look like:

9781118496718-un1604.tif

68. 9781118496718-eq16213.png

Stretching the graph of 9781118496718-eq16214.tif vertically by a factor of 2 produces the equation 9781118496718-eq16215.tif. To shift the graph of 9781118496718-eq16216.tif to the right 3 units, replace x with (x – 3) to get 9781118496718-eq16217.tif. Last, to move the graph of 9781118496718-eq16218.tif up 4 units, add 4 to the right side to get 9781118496718-eq16219.tif.

69. 9781118496718-eq16220.png

The vertex form of a parabola is given by y = a(x – h)2 + k, with the vertex at the point (h, k). Using the vertex (–2, –4) gives you the following equation:

9781118496718-eq16221.png

Now use the point (0, 2) to solve for a:

9781118496718-eq16222.png

Therefore, the equation of the parabola is

9781118496718-eq16223.png

70. y = –(x + 1)2 + 6

The vertex form of a parabola is given by y = a(x – h)2 + k, with the vertex at the point (h, k). Using the vertex (–1, 6), you have the following equation:

9781118496718-eq16224.png

Now use the point (1, 2) to solve for a:

9781118496718-eq16225.png

Therefore, the vertex form of the parabola is

9781118496718-eq16226.png

Here's the graph of y = –(x + 1)2 + 6:

9781118496718-un1605.tif

71. y = 3(x – 5)2 + 2

To translate the parabola 6 units to the right, you replace x with the quantity (x – 6). And to translate the parabola 2 units down, you subtract 2 from the original expression for y:

9781118496718-eq16227.png

Another option is simply to count and determine that the new vertex is at (5, 2); then use the vertex form y = a(x – h)2 + k, where the vertex is at the point (h, k).

72. y = e−2x – 5

To compress the graph of y = ex horizontally by a factor of 2, replace x with 2x to get y = e2x. To reflect the graph of y = e2x across the y-axis, replace x with –x to get y = e2(–x), or y = e−2x. Last, to shift the graph of y = e−2x down 5 units, subtract 5 from the right side of the equation to get y = e−2x – 5.

Here's the graph of y = e−2x – 5:

9781118496718-un1606.tif

73. 9781118496718-eq16228.png

To stretch the graph of 9781118496718-eq16229.tif horizontally by a factor of 5, replace x with 9781118496718-eq16230.tif to get 9781118496718-eq16231.tif. Next, to reflect the graph of 9781118496718-eq16232.tif across the x-axis, multiply the right side of the equation 9781118496718-eq16233.tif by –1 to get 9781118496718-eq16234.tif. Last, to shift the graph of 9781118496718-eq16235.tif up 3 units, add 3 to the right side of the equation to get

9781118496718-eq16236.png

74. 9781118496718-eq16237.png

The polynomial has x intercepts at x = –4, x = –2, and x = 1, so you know that the polynomial has the factors (x + 4), (x + 2), and (x – 1). Therefore, you can write

9781118496718-eq16238.png

Use the point (0, 3) to solve for a:

9781118496718-eq16239.png

Now you can enter the value of a in the equation of the polynomial and simplify:

9781118496718-eq16240.png

Here's the graph of 9781118496718-eq16241.tif:

9781118496718-un1607.tif

75. 9781118496718-eq16242.png

The polynomial is fourth-degree and has the x-intercepts at x = –1, 2, and 3, where 3 is a repeated root. Therefore, the polynomial has the factors (x + 1), (x – 2), and (x – 3)2, so you can write

9781118496718-eq16243.png

Use the point (1, 4) to solve for a:

9781118496718-eq16244.png

Now you can enter the value of a in the equation of the polynomial and simplify:

9781118496718-eq16245.png

76. 9781118496718-eq16246.png

The parabola has x-intercepts at x = –4 and x = 6, so you know that (x + 4) and (x – 6) are factors of the parabola. Therefore, you can write

9781118496718-eq16247.png

Use the point (0, 8) to solve for a:

9781118496718-eq16248.png

Now you can enter the value of a in the equation of the parabola and simplify:

9781118496718-eq16249.png

77. 9781118496718-eq16250.png

The parabola has x-intercepts at x = –8 and x = –2, so you know that (x + 8) and (x + 2) are factors of the parabola. Therefore, you can write

9781118496718-eq16251.png

Use the point (–4, –12) to solve for a:

9781118496718-eq16252.png

Now you can enter the value of a in the equation of the parabola and simplify:

9781118496718-eq16253.png

78. domain: [–2, ∞); range: [1, ∞)

The function is continuous on its domain.

The lowest x coordinate occurs at the point (–2, 3), and then the graph extends indefinitely to the right; therefore, the domain is [–2, ∞). The lowest y coordinate occurs at the point (–1, 1), and then the graph extends indefinitely upward; therefore, the range is [1, ∞).

Note that the brackets in the interval notation indicate that the value –2 is included in the domain and that 1 is included in the range.

79. domain: [–5, 3]; range: [–2, 2]

Notice that the function is continuous on its domain.

The smallest x coordinate occurs at the point (–5, 0), and the largest x coordinate occurs at the point (3, 0); therefore, the domain is [–5, 3]. The smallest y coordinate occurs at the point (1, –2), and the largest y coordinate occurs at the point (–3, 2); therefore, the range is [–2, 2].

Note that the brackets in the interval notation indicate that the values –5 and 3 are included in the domain and that the values –2 and 2 are included in the range.

80. domain: 9781118496718-eq16254.tif; range: 9781118496718-eq16255.tif

Notice that the function is continuous on its domain.

For the portion of the graph that's to the left of the y-axis, there's no largest x value due to the hole at (–1, –1); instead, the x values get arbitrarily close to –1. The graph then extends indefinitely to the left so that the domain for the left side of the graph is (–∞, –1). There's also no largest y value due to the hole at (–1, –1); the y values get arbitrarily close to –1. The graph then decreases as it approaches the horizontal asymptote at y = –3, so the range for this portion of the graph is (–3, –1).

For the portion of the graph that's to the right of the y-axis, the smallest x value is at the point (2, 2). The graph then extends indefinitely to the right, so the domain for the right side of the graph is [2, ∞). The smallest y value also occurs at the point (2, 2), and then the graph increases as it approaches the horizontal asymptote at y = 4; therefore, the range of this portion of the graph is [2, 4).

Putting the two parts together, the domain of the function is 9781118496718-eq16256.tif, and the range is 9781118496718-eq16257.tif.

81. 9781118496718-eq16258.tif9781118496718-eq16259.tif

To determine the end behavior of a polynomial, you just have to determine the end behavior of the highest-powered term.

As x approaches –∞, you have

9781118496718-eq16260.png

Note that even though the limit is approaching negative infinity, the limit is still positive due to the even exponent.

As x approaches ∞, you have

9781118496718-eq16261.png

82. 9781118496718-eq16262.tif9781118496718-eq16263.tif

To determine the end behavior of a polynomial, determine the end behavior of the highest-powered term.

As x approaches –∞, you have

9781118496718-eq16264.png

Note that the limit is positive because a negative number raised to an odd power is negative, but after multiplying by –7, the answer becomes positive.

As x approaches ∞, you have

9781118496718-eq16265.png

Likewise, the limit here is negative because a positive number raised to an odd power (or any power for that matter!) is positive, but after multiplying by –7, the answer becomes negative.

83. 3x + 12

Remove the parentheses and combine like terms:

9781118496718-eq16266.png

84. 3x – 2

Remove the parentheses and combine like terms:

9781118496718-eq16267.png

85. x3 – x2 + 2x + 14

Remove the parentheses and combine like terms:

9781118496718-eq16268.png

86. –2x4 + 3x + 8

Remove the parentheses and combine like terms:

9781118496718-eq16269.png

87. x4 + 5x3 – 3x2

Remove the parentheses and combine like terms:

9781118496718-eq16270.png

88. 3x – 7

Begin by removing the parentheses, being careful to distribute the –1 to the second term. Then collect like terms to simplify:

9781118496718-eq16271.png

89. 6x2 – 5x + 5

Begin by removing the parentheses, being careful to distribute the –1 to the second term. Then collect like terms to simplify:

9781118496718-eq16272.png

90. 4x3 + 5x2 – 8x + 14

Begin by removing the parentheses, being careful to distribute the –1 to the second term. Then collect like terms to simplify:

9781118496718-eq16273.png

91. 2x2 + 3x + 1

Begin by removing the parentheses, being careful to distribute the –1 to the second and third terms. Then collect like terms to simplify:

9781118496718-eq16274.png

92. 10x4 – 7x3 – 9x2 – 8x + 10

Begin by removing the parentheses, being careful to distribute the –1 to the second term. Then collect like terms to simplify:

9781118496718-eq16275.png

93. 5x3 – 15x2

Distribute to get

9781118496718-eq16276.png

94. 3x2 + 7x – 20

Distribute each term in the first factor to each term in the second factor. Then collect like terms:

9781118496718-eq16277.png

95. x2y – xy2 + 6xy

Multiply each term in the first factor by xy:

9781118496718-eq16278.png

96. 2x3 – 3x2 + 9x – 4

Distribute each term in the first factor to each term in the second factor and then collect like terms:

9781118496718-eq16279.png

97. 9781118496718-eq16280.png

Begin by distributing –x to each term in the second factor:

9781118496718-eq16281.png

Next, multiply each term in the first factor by each term in the second factor:

9781118496718-eq16282.png

98. 9781118496718-eq16283.png

Using polynomial long division gives you

9781118496718-un1608.tif

Remember to put the remainder over the divisor when writing the answer:

9781118496718-eq16284.png

99. 9781118496718-eq16285.png

Using polynomial long division gives you

9781118496718-un1609.tif

When writing the answer, put the remainder over the divisor:

9781118496718-eq16286.png

100. 9781118496718-eq16287.png

First add a placeholder for the missing x2 term in the numerator. Rewriting x3 – 2x + 6 as x3 + 0x2 – 2x + 6 will make all the like terms line up when you do the long division, making the subtraction a bit easier to follow.

Then use polynomial long division:

9781118496718-un1610.tif

When you write the answer, put the remainder over the divisor:

9781118496718-eq16288.png

101. 9781118496718-eq16289.png

Begin by filling in all the missing terms so that everything will line up when you perform the long division. Here, add 0x3 and 0x as placeholders in the numerator and put 0x in the denominator:

9781118496718-eq16290.png

Then use polynomial long division:

9781118496718-un1611.tif

Put the remainder over the divisor when writing your answer:

9781118496718-eq16291.png

102. 9781118496718-eq16292.png

Begin by filling in all the missing terms so that everything will line up when you perform the long division:

9781118496718-eq16293.png

Using polynomial long division gives you

9781118496718-un1612.tif

Then write your answer, putting the remainder over the divisor:

9781118496718-eq16294.png

103. 9781118496718-eq16295.tif9781118496718-eq16296.tif9781118496718-eq16297.tif

When considering the sides of the right triangle, the values of the trigonometric functions are given by 9781118496718-eq16298.tif9781118496718-eq16299.tif, and 9781118496718-eq16300.tif; therefore, 9781118496718-eq16301.tif9781118496718-eq16302.tif, and 9781118496718-eq16303.tif.

104. 9781118496718-eq16304.tif9781118496718-eq16305.tif9781118496718-eq16306.tif

When considering the sides of the right triangle, the values of the trigonometric functions are given by 9781118496718-eq16307.tif9781118496718-eq16308.tif, and 9781118496718-eq16309.tif; therefore, 9781118496718-eq16310.tif9781118496718-eq16311.tif, and 9781118496718-eq16312.tif.

105. 9781118496718-eq16313.png

You know the value of 9781118496718-eq16314.tif, so if you can find the value of 9781118496718-eq16315.tif, you can evaluate 9781118496718-eq16316.tif using 9781118496718-eq16317.tif. To find 9781118496718-eq16318.tif, use the identity 9781118496718-eq16319.tif:

9781118496718-eq16320.png

Next, take the square root of both sides, keeping the negative solution for cosine because 9781118496718-eq16321.tif:

9781118496718-eq16322.png

Therefore, using 9781118496718-eq16323.tif, you have

9781118496718-eq16324.png

106. 9781118496718-eq16325.png

To find the value of 9781118496718-eq16326.tif, you can find the value of 9781118496718-eq16327.tif and then use 9781118496718-eq16328.tif. Use the identity 9781118496718-eq16329.tif to solve for 9781118496718-eq16330.tif:

9781118496718-eq16331.png

Next, take the square root of both sides, keeping the negative solution for sine because 9781118496718-eq16332.tif:

9781118496718-eq16333.png

Therefore, using 9781118496718-eq16334.tif, you have

9781118496718-eq16335.png

107. 9781118496718-eq16336.png

To find the value of 9781118496718-eq16337.tif, you can use the identity 9781118496718-eq16338.tif. Notice that because 9781118496718-eq16339.tif and 9781118496718-eq16340.tif, angle 9781118496718-eq16341.tif must be in the second quadrant. Using 9781118496718-eq16342.tif, you can make a right triangle and find the missing side using the Pythagorean theorem. When making the triangle, you can neglect the negative sign:

9781118496718-un1613.tif

9781118496718-eq16343.png

Because 9781118496718-eq16344.tif is in the second quadrant, you have 9781118496718-eq16345.tif and 9781118496718-eq16346.tif. Enter these values in the identity 9781118496718-eq16347.tif and solve:

9781118496718-eq16348.png

108. 9781118496718-eq16349.png

To find the value of 9781118496718-eq16350.tif, you can use the identity 9781118496718-eq16351.tif. Using 9781118496718-eq16352.tif you can make a right triangle and find the missing side using the Pythagorean theorem. When making the triangle, you can neglect the negative sign:

9781118496718-un1614.tif

9781118496718-eq16353.png

Now use the identity 9781118496718-eq16354.tif. The sine is negative, so you have 9781118496718-eq16355.tif:

9781118496718-eq16356.png

109. 9781118496718-eq16357.png rad

Because 180° = π rad, you have 9781118496718-eq16358.tif, so multiply the number of degrees by this value:

9781118496718-eq16359.png

110. 9781118496718-eq16360.png rad

Because 180° = π rad, you have 9781118496718-eq16361.tif, so multiply the number of degrees by this value:

9781118496718-eq16362.png

111. 9781118496718-eq16363.tif rad

Because 180° = π rad, you have 9781118496718-eq16364.tif, so multiply the number of degrees by this value:

9781118496718-eq16365.png

112. 9781118496718-eq16366.tif rad

Because 180° = π rad, you have 9781118496718-eq16367.tif, so multiply the number of degrees by this value:

9781118496718-eq16368.png

113. 210°

Because π rad = 180°, you have 9781118496718-eq16369.tif, so multiply the number of radians by this value:

9781118496718-eq16370.png

114. 165°

Because π rad = 180°, you have 9781118496718-eq16371.tif, so multiply the number of radians by this value:

9781118496718-eq16372.png

115. –108°

Because π rad = 180°, you have 9781118496718-eq16373.tif, so multiply the number of radians by this value:

9781118496718-eq16374.png

116. –630°

Because π rad = 180°, you have 9781118496718-eq16375.tif, so multiply the number of radians by this value:

9781118496718-eq16376.png

117. 9781118496718-eq16377.png

Because 9781118496718-eq16378.tif is in Quadrant II and the angle is measured counterclockwise, you have 9781118496718-eq16379.tif. Therefore, 9781118496718-eq16380.tif is the most appropriate choice.

118. 9781118496718-eq16381.png

Because 9781118496718-eq16382.tif is in Quadrant III and the angle is measured clockwise, you have 9781118496718-eq16383.tif. Therefore, 9781118496718-eq16384.tif is the most appropriate choice.

119. 9781118496718-eq16385.png

Because 9781118496718-eq16386.tif is in Quadrant IV and the angle is measured counterclockwise, you have 9781118496718-eq16387.tif. Therefore, 9781118496718-eq16388.tif is the most appropriate choice.

120. 9781118496718-eq16389.tif9781118496718-eq16390.tif9781118496718-eq16391.tif

9781118496718-un1615.tif

The given angle measure is 9781118496718-eq16391a.tif. Using the first quadrant of the unit circle, you have the following:

9781118496718-eq16392.png

121. 9781118496718-eq16393.tif9781118496718-eq16394.tif9781118496718-eq16395.tif

9781118496718-un1616.tif

The given angle measure is 9781118496718-eq16391b.tif. Using the second quadrant of the unit circle, you have the following:

9781118496718-eq16396.png

122. 9781118496718-eq16397.tif9781118496718-eq16398.tif9781118496718-eq16399.tif

9781118496718-un1617.tif

Use the third quadrant of the unit circle. Note that because the angle 9781118496718-eq16400.tif touches the unit circle in the same place as the angle 9781118496718-eq16401.tif, you have the following:

9781118496718-eq16402.png

123. 9781118496718-eq16403.tif9781118496718-eq16404.tif9781118496718-eq16405.png

9781118496718-un1618.tif

Use the third quadrant of the unit circle. The angle 9781118496718-eq16406.tif touches the unit circle in the same place as the angle 9781118496718-eq16407.tif, so you have the following:

9781118496718-eq16408.png

124. 9781118496718-eq16409.tif9781118496718-eq16410.tif9781118496718-eq16411.tif

Because the angle 9781118496718-eq16412.tif intersects the unit circle at the point (–1, 0), you have the following:

9781118496718-eq16413.png

125. 9781118496718-eq16414.png

Simply rewrite cotangent and simplify:

9781118496718-eq16415.png

126. sin x tan x

Begin by writing sec x as 9781118496718-eq16416.tif. Then get common denominators and simplify:

9781118496718-eq16417.png

127. 1 + sin 2x

Expand the given expression and use the identity sinx + cosx = 1 along with 2 sin x cos x = sin(2x):

9781118496718-eq16418.png

128. sin x

Use the identity sin(A – B) = sin A cos B – cos A sin B on the expression sin(π – x):

9781118496718-eq16419.png

129. cos x

Use the identity cos A cos B + sin A sin B = cos(A – B) to get

9781118496718-eq16420.png

Note that you can also use the following:

9781118496718-eq16421.png

130. 9781118496718-eq16422.png

First get common denominators:

9781118496718-eq16423.png

Then use the identity sinx = 1 – cosx in the denominator:

9781118496718-eq16424.png

131. csc x + cot x

Begin by multiplying the numerator and denominator by the conjugate of the denominator, 1 + cos x. Then start to simplify:

9781118496718-eq16425.png

Continue, using the identity 1 – cosx = sinx to simplify the denominator:

9781118496718-eq16426.png

132. 9781118496718-eq16427.png

Use the identity cos(A + B) = cos A cos B – sin A sin B:

9781118496718-eq16428.png

Then use the identities 9781118496718-eq16429.tif and 9781118496718-eq16430.tif and simplify:

9781118496718-eq16431.png

133. 9781118496718-eq16432.tif9781118496718-eq16433.tif

Solve for sin x:

9781118496718-eq16434.png

The solutions are x = 9781118496718-eq16435.tif and 9781118496718-eq16436.tif in the interval [0, 2π].

134. 0, π, 2π

Make one side of the equation zero, use the identity 9781118496718-eq16437.tif, and then factor:

9781118496718-eq16438.png

Setting the first factor equal to zero gives you sin x = 0, which has the solutions x = 0, π, and 2π. Setting the second factor equal to zero gives you 9781118496718-eq16439.tif so that cos x = 1, which has the solutions x = 0 and 2π.

135. 9781118496718-eq16440.tif, π, 9781118496718-eq16441.tif

Factor the equation:

9781118496718-eq16442.png

Setting the first factor equal to zero gives you 2 cos x – 1 = 0 so that 9781118496718-eq16443.tif, which has the solutions 9781118496718-eq16444.tif and 9781118496718-eq16445.png. Setting the second factor equal to zero gives you cos x + 1 = 0 so that cos x = –1, which has the solution x = π.

136. 9781118496718-eq16446.tif9781118496718-eq16447.tif9781118496718-eq16448.tif9781118496718-eq16449.tif

To solve the equation 9781118496718-eq16450.tif, you must find the solutions to two equations: tan x = 1 and tan x = –1. For the equation tan x = 1, you have the solutions 9781118496718-eq16451.tif and 9781118496718-eq16452.tif. For the equation tan x = –1, you have the solutions 9781118496718-eq16453.tif and 9781118496718-eq16454.tif.

137. 9781118496718-eq16455.tif9781118496718-eq16456.tif

Factor the equation 2 sinx – 5 sin x – 3 = 0 to get

9781118496718-eq16457.png

Setting the first factor equal to zero gives you 2 sin x + 1 = 0 so that 9781118496718-eq16458.tif, which has the solutions 9781118496718-eq16459.tif and 9781118496718-eq16460.tif. Setting the second factor equal to zero gives you sin x – 3 = 0 so that sin x = 3. Because 3 is outside the range of the sine function, this equation from the second factor has no solution.

138. 9781118496718-eq16461.tif9781118496718-eq16462.tif

Begin by making one side of the equation equal to zero. Then use the identity 9781118496718-eq16463.tif and factor:

9781118496718-eq16464.png

Setting the first factor equal to zero gives you cos x = 0, which has the solutions x = 9781118496718-eq16465.tif and 9781118496718-eq16466.tif. Setting the second factor equal to zero gives you 9781118496718-eq16467.tif so that sin x = 1, which has the solution 9781118496718-eq16468.tif.

139. 9781118496718-eq16469.tif9781118496718-eq16470.tif9781118496718-eq16471.tif9781118496718-eq16472.tif

Begin by making the substitution 2x = y. Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 2x ≤ 4π and that 0 ≤ y ≤ 4π.

Solving the equation 9781118496718-eq16473.tif gives you the solutions 9781118496718-eq16474.tif9781118496718-eq16475.tif9781118496718-eq16476.tif, and 9781118496718-eq16477.tif in the interval [0, 4π].

Next, take each solution, set it equal to 2x, and solve for x: 9781118496718-eq16478.tif so that 9781118496718-eq16479.tif9781118496718-eq16480.tif so that 9781118496718-eq16481.tif9781118496718-eq16482.tif so that 9781118496718-eq16483.tif; and 9781118496718-eq16484.tif so that 9781118496718-eq16485.tif.

140. 9781118496718-eq16486.tif9781118496718-eq16487.tif9781118496718-eq16488.tif9781118496718-eq16489.tif

Begin by making one side of the equation zero. Then use the identity sin 2x = 2 sin x cos x and factor:

9781118496718-eq16490.png

Setting the first factor equal to zero gives you cos x = 0, which has the solutions 9781118496718-eq16491.tif and 9781118496718-eq16492.tif. Setting the second factor equal to zero gives you 2 sin x – 1 = 0 so that 9781118496718-eq16493.tif, which has the solutions 9781118496718-eq16494.tif and 9781118496718-eq16495.tif.

141. 9781118496718-eq16496.tif9781118496718-eq16497.tif

Begin by using the identity 9781118496718-eq16498.tif and then factor:

9781118496718-eq16499.png

Set each factor equal to zero and solve for x. The equation cos x = 0 has the solutions 9781118496718-eq16500.tif and 9781118496718-eq16501.tif, and the equation 1 + sin x = 0 gives you sin x = –1, which has the solution 9781118496718-eq16502.tif.

142. 9781118496718-eq16503.tif9781118496718-eq16504.tif9781118496718-eq16505.tif

Use the identity cos 2x = 2 cosx – 1, make one side of the equation zero, and factor:

9781118496718-eq16506.png

Set each factor equal to zero and solve for x. Setting the first factor equal to zerogives you 2 cos x + 1 = 0 so that 9781118496718-eq16507.tif, which has the solutions 9781118496718-eq16508.tif and 9781118496718-eq16509.tif. Setting the second factor equal to zero gives you cos x + 1 = 0 so that cos x = –1, which has the solution x = π.

143. 9781118496718-eq16510.tif9781118496718-eq16511.tif9781118496718-eq16512.tif9781118496718-eq16513.tif9781118496718-eq16514.tif9781118496718-eq16515.tif

Begin by making the substitution y = 3x. Because 0 ≤ x ≤ 2π, you have 0 ≤ 3x ≤ 6π so that 0 ≤ y ≤ 6π. Finding all solutions of the equation tan y = –1 in the interval [0, 6π] gives you 9781118496718-eq16516.tif9781118496718-eq16517.tif9781118496718-eq16518.tif9781118496718-eq16519.tif9781118496718-eq16520.tif, and 9781118496718-eq16521.tif.

Next, take each solution, set it equal to 3x, and solve for x: 9781118496718-eq16522.tif so that 9781118496718-eq16523.tif9781118496718-eq16524.tif so that 9781118496718-eq16525.tif9781118496718-eq16526.tif so that 9781118496718-eq16527.tif9781118496718-eq16528.tif so that 9781118496718-eq16529.tif9781118496718-eq16530.tif so that 9781118496718-eq16531.tif; and 9781118496718-eq16532.tif so that 9781118496718-eq16533.tif.

144. 9781118496718-eq16534.tif9781118496718-eq16535.tif9781118496718-eq16536.tif9781118496718-eq16537.tif

Begin by making one side of the equation equal to zero and then factor:

9781118496718-eq16538.png

Next, make the substitution y = 2x:

9781118496718-eq16539.png

Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 2x ≤ 4π so that 0 ≤ y ≤ 4π.

Setting the first factor equal to zero gives you cos y = 0, which has the solutions 9781118496718-eq16540.tif9781118496718-eq16541.tif9781118496718-eq16542.tif, and 9781118496718-eq16543.tif. Take each solution, set it equal to 2x, and solve for x: 9781118496718-eq16544.tif so that 9781118496718-eq16545.tif9781118496718-eq16546.tif so that 9781118496718-eq16547.tif9781118496718-eq16548.tif so that 9781118496718-eq16549.tif; and 9781118496718-eq16550.tif so that 9781118496718-eq16551.tif.

Proceed in a similar manner for the second factor: 9781118496718-eq16552.tif so that sin y = 1, which has the solutions 9781118496718-eq16553.tif and 9781118496718-eq16554.tif. Now take each solution, set it equal to 2x, and solve for x: 9781118496718-eq16555.tif so that 9781118496718-eq16556.tif, and 9781118496718-eq16557.tif so that 9781118496718-eq16558.tif.

145. amplitude: 9781118496718-eq16559.tif; period: 2π; phase shift: 9781118496718-eq16560.tif; midline y = 0

For the function 9781118496718-eq16561.tif, the amplitude is 9781118496718-eq16562.tif, the period is 9781118496718-eq16563.png, the phase shift is 9781118496718-eq16564.tif, and the midline is y = D. Writing 9781118496718-eq16565.tif gives you the amplitude as 9781118496718-eq16566.tif, the period as 9781118496718-eq16567.tif, the phase shift as 9781118496718-eq16568.tif, and the midline as y = 0.

146. amplitude: 9781118496718-eq16569.tif; period: 2; phase shift: 9781118496718-eq16570.tif; midline y = 0

For the function 9781118496718-eq16571.tif, the amplitude is 9781118496718-eq16572.tif, the period is 9781118496718-eq16573.png, the phase shift is 9781118496718-eq16574.tif, and the midline is y = D. Writing 9781118496718-eq16575.tif gives you the amplitude as 9781118496718-eq16576.tif, the period as 9781118496718-eq16577.tif, the phase shift as 9781118496718-eq16578.tif, and the midline as y = 0.

147. amplitude: 3; period: 2; phase shift: 9781118496718-eq16579.tif; midline y = 2

For the function 9781118496718-eq16580.tif, the amplitude is 9781118496718-eq16581.tif, the period is 9781118496718-eq16582.tif, the phase shift is 9781118496718-eq16583.tif, and the midline is y = D. Writing 9781118496718-eq16584.tif gives you the amplitude as 9781118496718-eq16585.tif, the period as 9781118496718-eq16586.tif, the phase shift as 9781118496718-eq16587.tif, and the midline as y = 2.

148. amplitude: 1; period: 4π; phase shift: –π; midline 9781118496718-eq16588.tif

For the function 9781118496718-eq16589.tif, the amplitude is 9781118496718-eq16590.tif, the period is 9781118496718-eq16591.png, the phase shift is 9781118496718-eq16592.tif, and the midline is y = D. Writing 9781118496718-eq16593.tif gives you the amplitude as 9781118496718-eq16594.tif, the period as 9781118496718-eq16595.tif, the phase shift as –π, and the midline as 9781118496718-eq16596.tif.

149. f (x) = 2 sin(2x)

For the function 9781118496718-eq16597.tif, the amplitude is 9781118496718-eq16598.tif, the period is 9781118496718-eq16599.tif, the phase shift is 9781118496718-eq16600.tif, and the midline is y = D. The function has a period of π, so one possible value of B is B = 2. If you use a sine function to describe the graph, there's no phase shift, so you can use C = 0. The line y = 0 is the midline of the function so that D = 0. Last, the amplitude is 2, so you can use A = 2 because the function increases for values of x that are slightly larger than zero. Therefore, the function 9781118496718-eq16601.tif describes the given graph. Note that you can also use other functions to describe this graph.

150. f (x) = 2 cos(2x)

For the function 9781118496718-eq16602.tif, the amplitude is 9781118496718-eq16603.tif, the period is 9781118496718-eq16604.tif, the phase shift is 9781118496718-eq16605.tif, and the midline is y = D. The function has a period of π, so one possible value of B is B = 2. If you use a cosine function to describe the graph, there's no phase shift; therefore, you can use C = 0. The line y = 0 is the midline of the function, so D = 0. Last, the amplitude is 2; you can use A = 2 because the function decreases for values of x that are slightly larger than zero. Therefore, the function f (x) = 2 cos(2x) describes the given graph. Note that you can also use other functions to describe this graph.

151. 9781118496718-eq16606.tif

For the function 9781118496718-eq16607.tif, the amplitude is 9781118496718-eq16608.tif, the period is 9781118496718-eq16609.tif, the phase shift is 9781118496718-eq16610.tif, and the midline is y = D. The function has a period of 4π, so one possible value of B is 9781118496718-eq16611.tif. If you use a cosine function to describe the graph, there's no phase shift, so you can use C = 0. The line y = 0 is the midline of the function, so D = 0. Last, the amplitude is 2, so you can use A = –2 because the function increases for values of x that are slightly larger than zero. Therefore, the function 9781118496718-eq16612.tif describes the given graph. Note that you can also use other functions to describe this graph.

152. f (x) = 2 cos(2x) + 1

For the function 9781118496718-eq16613.tif, the amplitude is 9781118496718-eq16614.tif, the period is 9781118496718-eq16615.tif, the phase shift is 9781118496718-eq16616.tif, and the midline is y = D. The function has a period of π, so one possible value of B is B = 2. If you use a cosine function to describe the graph, there's no phase shift, so you can use C = 0. The line y = 1 is the midline of the function, so D = 1. Last, the amplitude is 2, so you can use A = 2 because the function decreases for values of x that are slightly larger than zero. Therefore, the function f (x) = 2 cos(2x) + 1 describes the given graph. Note that you can also use other functions to describe this graph.

153. 9781118496718-eq16617.png For the function 9781118496718-eq16618.tif, the amplitude is 9781118496718-eq16619.tif, the period is 9781118496718-eq16620.tif, the phase shift is 9781118496718-eq16621.tif, and the midline is y = D. The function has a period of 4π, so one possible value of B is 9781118496718-eq16622.tif. Think of the graph as a cosine graph that's flipped about the x-axis and shifted to the right; that means there's a phase shift of 9781118496718-eq16623.tif to the right, so 9781118496718-eq16624.tif. The line y = 0 is the midline of the function, so D = 0. Last, the amplitude is 2, so you can use A = –2 because the function increases for values of x that are slightly larger than 9781118496718-eq16625.tif. Therefore, the following function describes the graph:

9781118496718-eq16626.png

Note that you can also use other functions to describe this graph.

154. 9781118496718-eq16627.png

For the function 9781118496718-eq16628.tif, the amplitude is 9781118496718-eq16629.tif, the period is 9781118496718-eq16630.png, the phase shift is 9781118496718-eq16631.tif, and the midline is y = D. The function has a period of 4π, so one possible value of B is 9781118496718-eq16632.tif. Think of the graph as a cosine graph that's flipped about the x-axis and shifted to the right and down; there's a phase shift of 9781118496718-eq16633.tif to the right so that 9781118496718-eq16634.tif. The line y = –1 is the midline of the function, so D = –1. Last, the amplitude is 2, so you can use A = –2 because the function increases for values of x that are slightly larger than 9781118496718-eq16635.tif. Therefore, the following function describes the given graph.

9781118496718-eq16636.png

Note that you can also use other functions to describe this graph.

155. 9781118496718-eq16637.png

To evaluate 9781118496718-eq16638.tif, find the solution of 9781118496718-eq16639.tif, where 9781118496718-eq16640.png. Because 9781118496718-eq16641.tif, you have 9781118496718-eq16642.png.

156. 9781118496718-eq16643.png

To evaluate arctan(–1) = x, find the solution of –1 = tan x, where 9781118496718-eq16644.tif. Because 9781118496718-eq16645.tif, you have 9781118496718-eq16646.tif.

157. 9781118496718-eq16647.png

To evaluate 9781118496718-eq16648.tif, first find the value of 9781118496718-eq16649.tif. To evaluate 9781118496718-eq16650.png, where 9781118496718-eq16651.tif, find the solution of 9781118496718-eq16652.png. Because 9781118496718-eq16653.tif, you have 9781118496718-eq16654.tif. Therefore, 9781118496718-eq16655.tif.

158. 9781118496718-eq16656.png

To evaluate 9781118496718-eq16657.tif, first find the value of 9781118496718-eq16658.tif. To evaluate 9781118496718-eq16659.png, where 9781118496718-eq16660.tif, find the solution of 9781118496718-eq16661.png. Because 9781118496718-eq16662.tif, you have 9781118496718-eq16663.tif. Therefore, 9781118496718-eq16664.tif.

159. 9781118496718-eq16665.png

The value of 9781118496718-eq16666.tif probably isn't something you've memorized, so to evaluate 9781118496718-eq16667.tif, you can create a right triangle.

Let 9781118496718-eq16668.tif so that 9781118496718-eq16669.tif. Using 9781118496718-eq16670.tif, create the right triangle; then use the Pythagorean theorem to find the missing side:

9781118496718-un1619.tif

By the substitution, you have 9781118496718-eq16671.tif, and from the right triangle, you have 9781118496718-eq16672.tif.

160. 9781118496718-eq16673.png

To evaluate 9781118496718-eq16674.tif, first create two right triangles using the substitutions tan−1(2) = α and tan−1(3) = β.

To make the first right triangle, use tan−1(2) = α. You know that 2 = tan α, and you can find the missing side of the first right triangle using the Pythagorean theorem:

9781118496718-un1620.tif

As for the second right triangle, because tan−1(3) = β, you know that 3 = tan β. Again, you can use the Pythagorean theorem to find the missing side of the right triangle:

9781118496718-un1621.tif

The substitutions give you 9781118496718-eq16675.tif. Using a trigonometric identity, you know that sin(α + β) = sin α cos β + cos α sin β. From the right triangles, you can read off each of the values to get the following:

9781118496718-eq16676.png

161. x = 0.412, π – 0.412

To solve sin x = 0.4 for x over the interval [0, 2π], begin by taking the inverse sine of both sides:

9781118496718-eq16677.png

The other solution belongs to Quadrant II because the sine function also has positive values there:

9781118496718-eq16678.png

162. x = 2.465, 2π – 2.456

To solve cos x = –0.78 for x over the interval [0, 2π], begin by taking the inverse cosine of both sides:

9781118496718-eq16679.png

The other solution belongs to Quadrant III because the cosine function also has negative values there:

9781118496718-eq16680.png

163. x = 0.322, 9781118496718-eq16681.tif, π + 0.322, 9781118496718-eq16682.tif

To solve 5 sin(2x) + 1 = 4 for x over the interval [0, 2π], begin by isolating the term involving sine:

9781118496718-eq16683.png

You can also use the substitution y = 2x to help simplify. Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 2x ≤ 4π so that 0 ≤ y ≤ 4π. Use the substitution and take the inverse sine of both sides:

9781118496718-eq16684.png

It follows that in the interval [0, 4π], 2π + 0.644 is also a solution because when you add 2π, the resulting angle lies at the same place on the unit circle.

Likewise, there's a solution to the equation 9781118496718-eq16685.tif in the second quadrant because the function also has positive values there, namely π – 0.644; adding 2π gives you the other solution, 3π – 0.644. Therefore, y = 0.644, π – 0.644, 2π + 0.644, and 3π – 0.644 are all solutions.

Last, substitute 2x into the equations and divide to get the solutions for x:

9781118496718-eq16686.png

9781118496718-eq16687.png

9781118496718-eq16688.png

9781118496718-eq16689.png

The solutions are x = 0.322, 9781118496718-eq16690.tif, π + 0.322, and 9781118496718-eq16691.tif.

164. x = 0.321, 9781118496718-eq16692.tif9781118496718-eq16693.tif9781118496718-eq16694.tif9781118496718-eq16695.tif, 2π – 0.321

To solve 7 cos(3x) – 1 = 3 for x over the interval [0, 2π], first isolate the term involving cosine:

9781118496718-eq16696.png

You can also use the substitution y = 3x to simplify the equation. Because 0 ≤ x ≤ 2π, it follows that 0 ≤ 3x ≤ 6π so that 0 ≤ y ≤ 6π. Use the substitution and take the inverse cosine of both sides:

9781118496718-eq16697.png

It follows that in the interval [0, 6π], y = 2π + 0.963 and y = 4π + 0.963 are also solutions because adding multiples of 2π makes the resulting angles fall at the same places on the unit circle.

Likewise, there's a solution to the equation 9781118496718-eq16698.tif in the fourth quadrant because cosine also has positive values there, namely y = 2π – 0.963. Because y = 2π – 0.963 is a solution, it follows that y = 4π – 0.963 and y = 6π – 0.963 are also solutions.

Therefore, you have y = 0.963, 2π – 0.963, 2π + 0.963, 4π – 0.963, 4π + 0.963, and 6π – 0.963 as solutions to 9781118496718-eq16699.tif. Last, substitute 3x into the equations and divide to solve for x:

9781118496718-eq16700.png

9781118496718-eq16701.png

9781118496718-eq16702.png

9781118496718-eq16703.png

9781118496718-eq16704.png

9781118496718-eq16705.png

Therefore, the solutions are x = 0.321, 9781118496718-eq16706.tif9781118496718-eq16707.tif9781118496718-eq16708.tif9781118496718-eq16709.tif, and 2π – 0.321.

165. x = π + 0.887, 2π – 0.887

To solve 2 sinx + 8 sin x + 5 = 0 for x over the interval [0, 2π], first use the quadratic formula:

9781118496718-eq16710.png

Simplifying gives you 9781118496718-eq16711.tif and 9781118496718-eq16712.tif.

You now need to find solutions to sin x = –0.775 and sin x = –3.225. Notice that sin x = –3.225 has no solutions because –3.225 is outside the range of the sine function.

To solve sin x = –0.775, take the inverse sine of both sides:

9781118496718-eq16713.png

Note that this solution isn't in the desired interval, [0, 2π]. The solutions in the given interval belong to Quadrants III and IV, respectively, because in those quadrants, the sine function has negative values; those solutions are x = π + 0.887 and x = 2π – 0.887.

9781118496718-un1622.tif

166. x = π – 0.322, 2π – 0.322, 9781118496718-eq16714.tif9781118496718-eq16715.tif

To solve 3 secx + 4 tan x = 2 for x over the interval [0, 2π], begin by using the identity secx = 1 + tanx and make one side of the equation equal to zero:

9781118496718-eq16716.png

Next, use the quadratic formula to find tan x:

9781118496718-eq16717.png

Therefore, tan x = –1 and 9781118496718-eq16718.tif.

The solutions to the equation tan x = –1 are 9781118496718-eq16719.tif and 9781118496718-eq16720.tif. To solve 9781118496718-eq16721.tif, take the inverse tangent of both sides:

9781118496718-eq16722.png

Note that this solution isn't in the given interval. The solutions that are in the given interval and belong to Quadrants II and IV (where the tangent function is negative) are x = π – 0.322 and x = 2π – 0.322.

167. 1

As x approaches 3 from the left, the y values approach 1 so that 9781118496718-eq16723.tif.

168. 3

As x approaches 3 from the right, the y values approach 3 so that 9781118496718-eq16724.tif.

169. 2

As x approaches –3 both from the left and from the right, the y values approach 2 so that 9781118496718-eq16725.tif.

Note that the actual value of f (–3) doesn't matter when you're finding the limit.

170. 3

As x approaches 1 from the left, the y values approach 3 so that 9781118496718-eq16726.tif.

171. 3

As x approaches 1 from the right, the y values approach 3 so that 9781118496718-eq16727.tif.

172. 5

As x approaches –2 both from the left and from the right, the y values approach 5 so that 9781118496718-eq16728.tif.

173. 4

Note that substituting the limiting value, 3, into the function 9781118496718-eq16729.tif gives you the indeterminate form 9781118496718-eq16730.tif.

To find the limit, first factor the numerator and simplify:

9781118496718-eq16731.tif

Then substitute 3 for x:

9781118496718-eq16732.png

9781118496718-un16726.tif

174. 9781118496718-eq16733.png

Note that substituting the limiting value, 2, into the function 9781118496718-eq16734.tif gives you the indeterminate form 9781118496718-eq16735.tif.

To find the limit, first factor the numerator and denominator and simplify:

9781118496718-eq16736.png

Then substitute 2 for x:

9781118496718-eq16737.png

175. 9781118496718-eq16738.png

Note that substituting the limiting value, –5, into the function 9781118496718-eq16739.tif gives you the indeterminate form 9781118496718-eq16740.tif.

To find the limit, first factor the numerator and denominator and simplify:

9781118496718-eq16741.png

Then substitute –5 for x:

9781118496718-eq16742.png

9781118496718-un16726.tif

176. 4

Note that substituting the limiting value, 4, into the function 9781118496718-eq16743.tif gives you the indeterminate form 9781118496718-eq16744.tif.

To find the limit, first factor the numerator and simplify:

9781118496718-eq16745.png

Then substitute 4 for x:

9781118496718-eq16746.png

177. 1

Note that substituting in the limiting value, 0, gives you an indeterminate form. For example, as x approaches 0 from the right, you have the indeterminate form ∞ – ∞, and as x approaches 0 from the left, you have the indeterminate form –∞ + ∞.

To find the limit, first get common denominators and simplify:

9781118496718-eq16747.png

Then substitute 0 for x:

9781118496718-eq16748.png

178. 0

The given function is continuous everywhere, so you can simply substitute in the limiting value:

9781118496718-eq16749.png

179. 3

Note that substituting the limiting value, –1, into the function 9781118496718-eq16750.tif gives you the indeterminate form 9781118496718-eq16751.tif.

To find the limit, first factor the numerator and simplify:

9781118496718-eq16752.png

Then substitute –1 for x:

9781118496718-eq16753.png

180. 9781118496718-eq16754.png

Note that substituting the limiting value, 0, into the function 9781118496718-eq16755.tif gives you the indeterminate form 9781118496718-eq16756.tif.

To find the limit, first multiply the numerator and denominator by the conjugate of the numerator and then simplify:

9781118496718-eq16757.png

Then substitute 0 for h:

9781118496718-eq16758.png

181. 108

Note that substituting the limiting value, 3, into the function 9781118496718-eq16759.png gives you the indeterminate form 9781118496718-eq16760.png.

To find the limit, first factor the numerator and simplify:

9781118496718-eq16761.png

Then substitute 3 for x:

9781118496718-eq16762.png

182. ∞

Note that substituting in the limiting value gives you an indeterminate form.

Because x is approaching 0 from the right, you have x > 0 so that 9781118496718-eq16763.tif. Therefore, the limit becomes

9781118496718-eq16764.png

As x → 0+, you have (2 + 2x) → 2 and x2 → 0+ so that 9781118496718-eq16765.tif. The limit is positive infinity because dividing 2 by a very small positive number close to zero gives you a very large positive number.

183. ∞

Note that substituting in the limiting value gives you the indeterminate form ∞ – ∞.

Because x is approaching 0 from the left, you have x < 0 so that 9781118496718-eq16766.tif. Therefore, the limit becomes

9781118496718-eq16767.png

Now consider the numerator and denominator as x → 0. As x → 0, you have (2 + 2x) → 2 and x2 → 0+ so that 9781118496718-eq16768.tif. The limit is positive infinity because dividing 2 by a very small positive number close to zero gives you a very large positive number.

184. limit does not exist

Note that substituting in the limiting value gives you the indeterminate form 9781118496718-eq16769.tif.

Examine both the left-hand limit and right-hand limit to determine whether the limits are equal. To find the left-hand limit, consider values that are slightly smaller than 9781118496718-eq16770.tif and substitute into the limit. Notice that to simplify the absolute value in the denominator of the fraction, you replace the absolute values bars with parentheses and add a negative sign, because substituting in a value less than 9781118496718-eq16771.tif will make the number in the parentheses negative; the extra negative sign will make the value positive again.

9781118496718-eq16772.png

You deal with the right-hand limit similarly. Here, when removing the absolute value bars, you simply replace them with parentheses; you don't need the negative sign because the value in the parentheses is positive when you're substituting in a value larger than 9781118496718-eq16773.tif.

9781118496718-eq16774.png

Because the right-hand limit doesn't equal the left-hand limit, the limit doesn't exist.

185. 9781118496718-eq16775.png

Note that substituting the limiting value, 5, into the function 9781118496718-eq16776.tif gives you the indeterminate form 9781118496718-eq16777.tif.

To find the limit, first factor the numerator and denominator and simplify:

9781118496718-eq16778.png

Then substitute 5 for x:

9781118496718-eq16779.png

186. 9781118496718-eq16780.png

Note that substituting in the limiting value gives you the indeterminate form 9781118496718-eq16781.tif.

Begin by multiplying the numerator and denominator of the fraction by the conjugate of the numerator:

9781118496718-eq16782.png

Next, factor –1 from the numerator and continue simplifying:

9781118496718-eq16783.png

To find the limit, substitute 3 for x:

9781118496718-eq16784.png

187. 12

Note that substituting the limiting value, 0, into the function 9781118496718-eq16785.tif gives you the indeterminate form 9781118496718-eq16786.tif.

Expand the numerator and simplify:

9781118496718-eq16787.png

To find the limit, substitute 0 for h:

9781118496718-eq16788.png

188. 1

Note that substituting in the limiting value, 4, gives you the indeterminate form 9781118496718-eq16789.tif.

Because x is approaching 4 from the right, you have x > 4 so that 9781118496718-eq16790.tif. Therefore, the limit becomes

9781118496718-eq16791.png

9781118496718-un16726.tif

189. –1

Note that substituting in the limiting value gives you the indeterminate form 9781118496718-eq16792.tif.

Because x is approaching 5 from the left, you have x < 5 so that 9781118496718-eq16793.tif. Therefore, the limit becomes

9781118496718-eq16794.png

190. 9781118496718-eq16795.png

Note that substituting in the limiting value, –5, into the function 9781118496718-eq16796.tif gives you the indeterminate form 9781118496718-eq16797.tif.

Begin by writing the two fractions in the numerator as a single fraction by getting common denominators. Then simplify:

9781118496718-eq16798.png

To find the limit, substitute –5 for x:

9781118496718-eq16799.png

191. 9781118496718-eq16800.png

Note that substituting in the limiting value gives you an indeterminate form. For example, as x approaches 0 from the right, you have the indeterminate form ∞ – ∞, and as x approaches 0 from the left, you have the indeterminate form –∞ + ∞.

Begin by getting common denominators:

9781118496718-eq16801.png

Next, multiply the numerator and denominator of the fraction by the conjugate of the numerator and simplify:

9781118496718-eq16802.png

To find the limit, substitute 0 for x:

9781118496718-eq16803.png

192. 9781118496718-eq16804.png

Note that substituting in the limiting value, 0, gives you the indeterminate form 9781118496718-eq16805.tif.

Begin by rewriting the two terms in the numerator using positive exponents. After that, get common denominators in the numerator and simplify:

9781118496718-eq16806.png

To find the limit, substitute 0 for h:

9781118496718-eq16807.png

193. 5

The squeeze theorem states that if f (x) ≤ g(x) ≤ h(x) when x is near a (except possibly at a) and if 9781118496718-eq16808.tif, then 9781118496718-eq16809.tif.

Note that 9781118496718-eq16810.tif and that 9781118496718-eq16811.tif. Therefore, by the squeeze theorem, 9781118496718-eq16812.tif.

194. 4

Note that 9781118496718-eq16815.tif and that 9781118496718-eq16816.tif. Therefore, by the squeeze theorem, 9781118496718-eq16817.tif.

195. 2

Note that 9781118496718-eq16820.tif and that 9781118496718-eq16821.tif. Therefore, by the squeeze theorem, 9781118496718-eq16822.tif.

196. 0

Notice that for all values of x except for x = 0, you have 9781118496718-eq16823.tif due to the range of the cosine function. So for all values of x except for x = 0, you have

9781118496718-eq16824.png

Because x is approaching 0 (but isn't equal to 0), you can apply the squeeze theorem:

9781118496718-eq16825.png

Therefore, you can conclude that 9781118496718-eq16826.tif.

197. 0

Notice that for all values of x > 0, you have 9781118496718-eq16827.tif due to the range of the sine function. So for all values of x > 0, you have

9781118496718-eq16828.png

Because x is approaching 0 (but isn't equal to 0), you can apply the squeeze theorem:

9781118496718-eq16829.png

Therefore, you can conclude that 9781118496718-eq16830.tif.

198. 0

Notice that for all values of x except for x = 0, you have 9781118496718-eq16831.tif due to the range of the sine function. So for all values of x except for x = 0, you have the following (after multiplying by –1):

9781118496718-eq16832.png

You need to make the center expression match the given one, 9781118496718-eq16833.tif, so do a little algebra. Adding 3 gives you

9781118496718-eq16834.png

Now multiply by 9781118496718-eq16835.tif:

9781118496718-eq16836.png

Note that 9781118496718-eq16837.tif for values of x greater than 0, so you don't have to flip the inequalities.

Because the limit is approaching 0 from the right (but isn't equal to 0), you can apply the squeeze theorem to get

9781118496718-eq16838.png

Therefore, you can conclude that 9781118496718-eq16839.tif.

199. 5

To use 9781118496718-eq16840.tif, you need the denominator of the function to match the argument of the sine. Begin by multiplying the numerator and denominator by 5. Then simplify:

9781118496718-eq16841.png

200. 0

Factor the 2 from the numerator and then multiply the numerator and denominator by 9781118496718-eq16842.tif. Then simplify:

9781118496718-eq16843.png

201. 9781118496718-eq16844.png

Begin by using a trigonometric identity in the numerator and then factor:

9781118496718-eq16845.png

Next, factor out a –1 from the numerator and simplify:

9781118496718-eq16846.png

202. 9781118496718-eq16847.png

You want to rewrite the expression so you can use 9781118496718-eq16848.tif. Begin by multiplying the numerator and denominator by 9781118496718-eq16849.tif and rewrite the expression as the product of two fractions:

9781118496718-eq16850.png

Now you want a 5 in the denominator of the first fraction and a 9 in the numerator of the second fraction. Multiplying by 9781118496718-eq16851.tif and also by 9781118496718-eq16852.tif gives you

9781118496718-eq16853.png

203. 9781118496718-eq16854.png

Begin by rewriting tan(7x) to get

9781118496718-eq16855.png

Now you want 7x in the denominator of the first fraction and 3x in the numerator of the third fraction. Therefore, multiply by 9781118496718-eq16856.tif9781118496718-eq16857.tif, and 9781118496718-eq16858.tif to get

9781118496718-eq16859.png

204. 8

Begin by breaking up the fraction as

9781118496718-eq16860.png

Next, you want each fraction to have a denominator of 2x, so multiply by 9781118496718-eq16861.tif9781118496718-eq16862.tif, and 9781118496718-eq16863.tif — or equivalently, by 9781118496718-eq16864.tif — to get

9781118496718-eq16865.png

205. 9781118496718-eq16866.png

Begin by factoring the denominator and rewriting the limit:

9781118496718-eq16867.png

Notice that you can rewrite the first limit using the substitution 9781118496718-eq16868.tif so that as x → 2, you have 9781118496718-eq16869.tif; this step isn't necessary, but it clarifies how to use 9781118496718-eq16870.tif in this problem.

Replacing (x – 2) with 9781118496718-eq16871.tif and replacing x → 2 with 9781118496718-eq16872.tif in the limit 9781118496718-eq16873.tif gives you the following:

9781118496718-eq16874.png

206. 9781118496718-eq16875.png

Begin by rewriting tan x and then multiply the numerator and denominator by 9781118496718-eq16876.tif. Then simplify:

9781118496718-eq16877.png

Then substitute 0 into the equation and solve:

9781118496718-eq16878.png

207. ∞

As x approaches 3 from the left, the y values approach ∞ so that 9781118496718-eq16879.tif.

208. –∞

As x approaches 3 from the right, the y values approach –∞ so that 9781118496718-eq16880.tif.

209. ∞

As x approaches 5 from the left, the y values approach ∞ so that 9781118496718-eq16881.tif.

210. –∞

As x approaches 5 from the right, the y values approach –∞ so that 9781118496718-eq16882.tif.

211. limit does not exist

As x approaches 5 from the left, the y values approach ∞. However, as x approaches 5 from the right, the y values approach –∞. Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

212. ∞

As x approaches 1 from the right, you have (x – 1) → 0+ so that

9781118496718-eq16883.png

Note that the limit is positive infinity because dividing 3 by a small positive number close to zero gives you a large positive number.

9781118496718-eq16883.png

213. –∞

As x approaches 1 from the left, you have (x – 1) → 0 so that

9781118496718-eq16884.png

Note that the limit is negative infinity because dividing 3 by a small negative number close to zero gives you a negative number whose absolute value is large.

214. –∞

Begin by writing the limit as 9781118496718-eq16885.tif. Then consider what happens in the numerator and the denominator as x approaches 9781118496718-eq16886.tif from the right. As 9781118496718-eq16887.tif, you have sin x → 1 and cos x → 0. Therefore, as 9781118496718-eq16888.tif, it follows that

9781118496718-eq16889.png

215. ∞

Consider what happens in the numerator and denominator as x approaches π from the left. As x → π –, you have x2 → π2 and sin x → 0+. Therefore, as x → π –, it follows that

9781118496718-eq16890.png

216. –∞

Consider what happens in the numerator and denominator as x approaches 5 from the left. As x → 5, you have (x + 3) → 8 and (x – 5) → 0. Therefore, as x → 5, it follows that

9781118496718-eq16891.png

217. –∞

Consider what happens in the numerator and denominator as x approaches 0 from the left. As x → 0, you have (1 – x) → 1 and (ex – 1) → 0. Therefore, as x → 0, it follows that

9781118496718-eq16892.png

218. –∞

Begin by writing 9781118496718-eq16893.tif. Then consider what happens in the numerator and denominator as x approaches 0 from the left. As x → 0, you have cos x → 1 and sin x → 0. Therefore, as x → 0, it follows that

9781118496718-eq16894.png

219. ∞

Consider what happens in the numerator and denominator as x approaches 2. As x → 2, you have 4ex → 4e2 and 9781118496718-eq16895.tif. Therefore, as x → 2, it follows that

9781118496718-eq16896.png

220. –∞

Consider what happens to each factor in the numerator and denominator as x approaches 9781118496718-eq16897.tif from the right. As 9781118496718-eq16898.tif, you have 9781118496718-eq16899.tif and 9781118496718-eq16900.tif. Therefore, as 9781118496718-eq16901.tif, it follows that

9781118496718-eq16902.png

221. –∞

Consider what happens in the numerator and denominator as x approaches 5. As x → 5, you have sin x → sin 5, and because π < 5 < 2π, it follows that sin 5 < 0. And as x → 5, you also have (5 – x)4 → 0+. Therefore, as x → 5, it follows that

9781118496718-eq16904.png

Note that the limit is negative infinity because sin(5) is negative, and dividing sin(5) by a small positive number close to zero gives you a negative number whose absolute value is large.

222. –∞

Consider what happens to each factor in the numerator and denominator as x approaches 0. You need to examine both the left-hand limit and the right-hand limit.

For the left-hand limit, as x → 0, you have (x + 5) → 5, x4 → 0+, and (x – 6) → –6. Therefore, as x → 0, it follows that

9781118496718-eq16905.png

For the right-hand limit, you have (x + 5) → 5, x4 → 0+, and (x – 6) → –6. Therefore, as x → 0+, it follows that

9781118496718-eq16906.png

Because the left-hand limit is equal to the right-hand limit, 9781118496718-eq16907.tif.

223. limit does not exist

Begin by examining the left-hand limit and the right-hand limit to determine whether they're equal.

To find the limits, consider what happens in the numerator and denominator as x approaches 1. Using the left-hand limit, as x → 1, you have (3x) → 3 and (ex – e) → 0. Therefore, as x → 1, it follows that

9781118496718-eq16908.png

Using the right-hand limit, as x → 1+, you have (3x) → 3 and (ex – e) → 0+. So as x → 1+, it follows that

9781118496718-eq16909.png

Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

224. –∞

Consider what happens to each factor in the numerator and denominator as x approaches 0 from the right. You have (x – 1) → –1, x2 → 0+, and (x + 2) → 2. Therefore, as x → 0+, you get the following:

9781118496718-eq16910.png

225. –∞

Consider what happens in the numerator and denominator as x approaches e from the left. As x → e, you have x3 → e3 and (ln x – 1) → (ln e – 1) → 0. Therefore, as x → e, it follows that

9781118496718-eq16911.png

226. limit does not exist

Begin by examining both the left-hand limit and the right-hand limit to determine whether they're equal.

To find the limits, consider what happens in the numerator and denominator as x approaches e2. For the left-hand limit, as x → e2–, you have –x → –e2 and (ln x – 2) → 0. Therefore, as x → e2–, it follows that

9781118496718-eq16912.png

Note that the limit is positive infinity because dividing –e2 by a small negative number close to zero gives you a large positive number.

For the right-hand limit, as x → e2+, you have –x → –e2 and (ln x – 2) → 0+. Therefore, as x → e2+, it follows that

9781118496718-eq16913.png

Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

227. limit does not exist

Begin by examining the left-hand limit and the right-hand limit to determine whether they're equal.

To find the limits, consider what happens in the numerator and denominator as x approaches 2. For the left-hand limit, as x → 2, you have (x + 2) → 4 and (x2 – 4) → 0. Therefore, as x → 2, it follows that

9781118496718-eq16914.png

For the right-hand limit, as x → 2+, you have (x + 2) → 4 and (x2 – 4) → 0+. Therefore, as x → 2+, it follows that

9781118496718-eq16915.png

Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

228. limit does not exist

Begin by examining the left-hand limit and the right-hand limit to determine whether they're equal.

To find the limits, consider what happens in the numerator and denominator as x approaches 25. For the left-hand limit, as x → 25, you have 9781118496718-eq16916.tif and (x – 25) → 0. Therefore, as x → 25, it follows that

9781118496718-eq16917.png

For the right-hand limit, as x → 25+, you have 9781118496718-eq16918.tif and (x – 25) → 0+. Therefore, as x → 25+, you have

9781118496718-eq16919.png

Because the left-hand limit doesn't equal the right-hand limit, the limit doesn't exist.

9781118496718-un16726.tif

229. –∞

Consider what happens to each factor in the numerator and denominator as x approaches 0. As x → 0, you have (x2 + 4) → 4, x2 → 0+, and (x – 1) → –1. Therefore, as x → 0, it follows that

9781118496718-eq16920.png

230. –∞

Consider what happens to each factor in the numerator and denominator as x approaches 0 from the left. You have (x – 1) → –1, x2 → 0+, and (x + 2) → 2. Therefore, as x → 0, you get the following:

9781118496718-eq16921.png

231. ∞

Consider what happens in the numerator and denominator as x approaches 3. As x → 3, you have (3 + x) → 6 and 9781118496718-eq16922.tif. Therefore, as x → 3, it follows that

9781118496718-eq16923.png

232. 9781118496718-eq16924.png

As the x values approach –∞, the y values approach 9781118496718-eq16925.tif so that 9781118496718-eq16926.tif.

233. 9781118496718-eq16927.png

As the x values approach ∞, the y values approach 9781118496718-eq16928.tif so that 9781118496718-eq16929.tif.

234. 2

As the x values approach –∞, the y values approach 2 so that 9781118496718-eq16930.tif.

235. 2

As the x values approach ∞, the y values approach 2 so that 9781118496718-eq16931.tif.

236. 9781118496718-eq16932.png

Divide the numerator and denominator by the highest power of x that appears in the denominator, x1, and simplify:

9781118496718-eq16933.png

Then apply the limit:

9781118496718-eq16934.png

9781118496718-un16726.tif

237. –∞

Begin by multiplying to get

9781118496718-eq16935.png

For a polynomial function, the end behavior is determined by the term containing the largest power of x, so you have

9781118496718-eq16936.png

238. 3

Divide the numerator and denominator by the highest power of x that appears in the denominator, x1, and simplify:

9781118496718-eq16937.png

Then apply the limit:

9781118496718-eq16938.png

239. limit does not exist

Because cos x doesn't approach a specific value as x approaches ∞, the limit doesn't exist.

240. 0

Begin by expanding the denominator:

9781118496718-eq16939.png

Next, divide the numerator and denominator by the highest power of x that appears in the denominator, x5, and simplify:

9781118496718-eq16940.png

Then apply the limit:

9781118496718-eq16941.png

241. 9781118496718-eq16942.png

Begin by multiplying the numerator and denominator by 9781118496718-eq16943.tif so that you can simplify the expression underneath the square root:

9781118496718-eq16944.png

Because x is approaching –∞, you know that x < 0. So as you take the limit, you need to use the substitution 9781118496718-eq16945.tif in the denominator. Therefore, you have

9781118496718-eq16946.png

Now apply the limit:

9781118496718-eq16947.png

242. ∞

Divide the numerator and denominator by the highest power of x that appears in the denominator, x2, and simplify:

9781118496718-eq16948.png

Then apply the limit. Consider what happens in the numerator and the denominator asx → –∞. In the numerator, 9781118496718-eq16949.tif, and in the denominator, 9781118496718-eq16950.tif. Therefore, you have

9781118496718-eq16951.png

243. 9781118496718-eq16952.png

To simplify the expression underneath the radical, begin by multiplying the numerator and denominator by 9781118496718-eq16953.tif:

9781118496718-eq16954.png

Because x is approaching –∞, you know that x < 0. So as you take the limit, you need to use the substitution 9781118496718-eq16955.tif in the numerator:

9781118496718-eq16956.png

Now apply the limit:

9781118496718-eq16957.png

244. 9781118496718-eq16958.png

To simplify the expression underneath the radical, begin by multiplying the numerator and denominator by 9781118496718-eq16959.tif:

9781118496718-eq16960.png

Now apply the limit:

9781118496718-eq16961.png

245. –∞

Notice that if you consider the limit immediately, you have the indeterminate form ∞ – ∞.

Begin by factoring to get

9781118496718-eq16962.png

246. 9781118496718-eq16963.png

Notice that if you consider the limit immediately, you have the indeterminate form ∞ – ∞.

Create a fraction and multiply the numerator and denominator by the conjugate of the expression 9781118496718-eq16964.tif. The conjugate is 9781118496718-eq16965.tif, so you have the following:

9781118496718-eq16966.png

Next, multiply the numerator and denominator by 9781118496718-eq16967.tif so you can simplify the expression underneath the radical:

9781118496718-eq16968.png

Now apply the limit:

9781118496718-eq16969.png

247. 9781118496718-eq16970.png

Notice that if you consider the limit immediately, you have the indeterminate form –∞ + ∞.

Create a fraction and multiply the numerator and denominator by the conjugate of the expression 9781118496718-eq16971.tif. The conjugate is 9781118496718-eq16972.tif, so you have

9781118496718-eq16973.png

Next, multiply the numerator and denominator by 9781118496718-eq16974.tif so that you can simplify the expression underneath the square root:

9781118496718-eq16975.png

Because x is approaching –∞, you know that x < 0. So as you take the limit, you need to use the substitution 9781118496718-eq16976.tif in the denominator. Therefore, you have

9781118496718-eq16977.png

Now apply the limit:

9781118496718-eq16978.png

248. 9781118496718-eq16979.png

To find any horizontal asymptotes of the function 9781118496718-eq16980.tif, you need to consider the limit of the function as x → ∞ and as x → –∞. For the limit as x → ∞, begin by multiplying the numerator and denominator by 9781118496718-eq16981.tif:

9781118496718-eq16982.png

To find the limit as x → –∞, you can proceed in the same way in order to simplify:

9781118496718-eq16983.png

Therefore, the only horizontal asymptote is 9781118496718-eq16984.tif.

249. y = –1

To find any horizontal asymptotes of the function 9781118496718-eq16985.tif, you need to consider the limit of the function as x → ∞ and as x → –∞. For the limit as x → ∞, begin by multiplying the numerator and denominator by 9781118496718-eq16986.tif:

9781118496718-eq16987.png

To find the limit as x → –∞, proceed in the same way:

9781118496718-eq16988.png

Therefore, the only horizontal asymptote is y = –1.

250. 9781118496718-eq16989.png

In order to find any horizontal asymptotes of the function 9781118496718-eq16990.tif, you need to consider the limit of the function as x → ∞ and as x → –∞. For the limit as x → ∞, begin by multiplying the numerator and denominator by 9781118496718-eq16991.tif:

9781118496718-eq16992.png

In order to find the limit as x → –∞, proceed in the same way, noting that as x → –∞, you still use 9781118496718-eq16993.tif:

9781118496718-eq16994.png

Therefore, the only horizontal asymptote is 9781118496718-eq16995.tif.

9781118496718-un16726.tif

251. y = 1 and y = –1

To find any horizontal asymptotes of the function 9781118496718-eq16996.tif, you need to consider the limit of the function as x → ∞ and as x → –∞. For the limit as x → ∞, begin by multiplying the numerator and denominator by 9781118496718-eq16997.tif:

9781118496718-eq16998.png

To find the limit as x → –∞, proceed in the same way, noting that as 9781118496718-eq16999.tif, you need to use the substitution 9781118496718-eq161000.tif:

9781118496718-eq161001.png

Therefore, the function has the horizontal asymptotes y = 1 and y = –1.

252. removable discontinuity at x = –3, jump discontinuity at x = 3

The limit exists at x = –3 but isn't equal to f (–3), which corresponds to a removable discontinuity.

At x = 3, the left-hand limit doesn't equal the right-hand limit (both limits exist as finite values); this corresponds to a jump discontinuity.

253. removable discontinuity at x = 1, infinite discontinuity at x = 5

The limit exists at x = 1, but f (1) is undefined, which corresponds to a removable discontinuity.

At x = 5, the left-hand limit is ∞ and the right-hand limit is –∞, so an infinite discontinuity exists at x = 5.

254. jump discontinuity at x = –2, jump discontinuity at x = 3

At x = –2, the left-hand limit doesn't equal the right-hand limit (both limits exist as finite values), which corresponds to a jump discontinuity.

At x = 3, the left-hand limit again doesn't equal the right-hand limit (both limits exist as finite values), so this also corresponds to a jump discontinuity.

255. removable discontinuity at x = –1, jump discontinuity at x = 4, infinite discontinuity at x = 6

At x = –1, the left-hand limit equals the right-hand limit, but the limit doesn't equal f (–1), which is undefined. Therefore, a removable discontinuity is at x = –1.

At x = 4, the left-hand limit doesn't equal the right-hand limit (both limits exist as finite values), so a jump discontinuity is at x = 4.

At x = 6, both the left and right-hand limits equal ∞, so an infinite discontinuity is at x = 6.

256. not continuous, infinite discontinuity

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161002.tif. The left-hand limit at a is given by 9781118496718-eq161003.tif. As x → 2, you have (x – 2) → 0 so that 9781118496718-eq161004.tif. Because the discontinuity is infinite, you don't need to examine the right-hand limit; you can conclude that the function is not continuous.

257. continuous, f (a) = 2

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161005.tif. The left-hand limit at a is 9781118496718-eq161006.tif, and the right-hand limit at a is 9781118496718-eq161007.tif. The left-hand and right-hand limits match, so the limit at a exists and is equal to 2.

The value at a is 9781118496718-eq161008.tif. Because the function satisfies the definition of continuity, you can conclude that that function is continuous at a = 1.

258. continuous, f (a) = 5

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161009.tif. The left-hand limit at a is

9781118496718-eq161010.png

And the right-hand limit at a is

9781118496718-eq161011.png

Note that in this case, you could have simply evaluated the limit as x approaches 3 instead of examining the left-hand limit and right-hand limit separately.

The left-hand and right-hand limits match, so the limit exists and is equal to 5. Because f (a) = f (3) = 5, the definition of continuity is satisfied and the function is continuous.

259. continuous, 9781118496718-eq161012.tif

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161013.tif. The left-hand limit at a is

9781118496718-eq161014.png

The right-hand limit at a is

9781118496718-eq161015.png

The left-hand and right-hand limits match, so the limit exists.

Note that in this case, you could have simply evaluated the limit as x approaches 16 instead of examining the left-hand limit and right-hand limit separately.

The value at a is given by 9781118496718-eq161016.tif, which matches the limit. Because the definition of continuity is satisfied, you can conclude that that function is continuous at a = 16.

260. not continuous, jump discontinuity

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161017.tif. The left-hand limit at a is

9781118496718-eq161018.png

The right-hand limit at a is

9781118496718-eq161019.png

Because the left- and right-hand limits exist but aren't equal to each other, there's a jump discontinuity at a = –6.

261. not continuous, removable discontinuity

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161020.tif. The left-hand limit at a is

9781118496718-eq161021.png

The right-hand limit at a is

9781118496718-eq161022.png

The left-hand and right-hand limits match, so the limit exists and is equal to 3.

Note that in this case, you could have simply evaluated the limit as x approaches –1 instead of examining the left-hand limit and right-hand limit separately.

However, because f (a) = f (–1) = 2, the function is not continuous; it has a removable discontinuity.

262. continuous at a = 0 and at a = π

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161023.tif. To determine whether the function is continuous at a = 0, see whether it satisfies the equation 9781118496718-eq161024.tif. The left-hand limit at a = 0 is 9781118496718-eq161025.tif, and the right-hand limit at a = 0 is 9781118496718-eq161026.tif. Because f (0) = 2 cos(0) = 2, the function is continuous at a = 0.

Likewise, decide whether the function satisfies the equation 9781118496718-eq161027.tif. The left-hand limit at a = π is 9781118496718-eq161028.tif, and the right-hand limit at a = π is 9781118496718-eq161029.tif. Because 9781118496718-eq161030.tif, the function is also continuous at a = π.

9781118496718-un16726.tif

263. jump discontinuity at a = 1 and at a = 3

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161031.tif. To determine whether the function is continuous at a = 1, see whether it satisfies the equation 9781118496718-eq161032.tif. The left-hand limit at a = 1 is 9781118496718-eq161033.tif, and the right-hand limit at a= 1 is 9781118496718-eq161034.tif. The limits differ, so there's a jump discontinuity.

Likewise, decide whether the function satisfies the equation 9781118496718-eq161035.tif. The left-hand limit at a = 3 is 9781118496718-eq161036.tif, and the right-hand limit at a = 3 is 9781118496718-eq161037.tif, so there's another jump discontinuity.

264. continuous at a = 2, jump discontinuity at a = 3

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161038.tif. To determine whether the function is continuous at a = 2, see whether it satisfies the equation 9781118496718-eq161039.tif. The left-hand limit at a = 2 is 9781118496718-eq161040.tif, and the right-hand limit ata = 2 is 9781118496718-eq161041.tif. Because 9781118496718-eq161042.tif, the func-tion is continuous at 9781118496718-eq161043.tif.

Likewise, decide whether the function satisfies the equation 9781118496718-eq161044.tif. The left-hand limit at a = 3 is 9781118496718-eq161045.tif, and the right-hand limit at a = 3 is 9781118496718-eq161046.tif. The limits don't match, so there's a jump discontinuity at a = 3.

265. infinite discontinuity at a = 0, continuous at a = 4

A function f (x) is continuous at x = a if it satisfies the equation 9781118496718-eq161047.tif. To determine whether the function is continuous at a = 0, see whether it satisfies the equation 9781118496718-eq161048.tif. The left-hand limit at a = 0 is 9781118496718-eq161049.tif, and the right-hand limit at a= 0 is 9781118496718-eq161050.tif, so there's an infinite discontinuity at a = 0.

Likewise, decide whether the function satisfies the equation 9781118496718-eq161051.tif. The left-hand limit at a = 4 is 9781118496718-eq161052.tif, and the right-hand limit at a = 4 is 9781118496718-eq161053.tif. Because 9781118496718-eq161054.tif, the function is continuous at a = 4.

266. 9781118496718-eq161055.png

To determine the value of c, you must satisfy the definition of a continuous function: 9781118496718-eq161056.tif.

The left-hand limit at x = 2 is given by 9781118496718-eq161057.tif, and the right-hand limit is given by 9781118496718-eq161058.tif. Also note that f (2) = 2c – 2.

The left-hand limit must equal the right-hand limit, so set them equal to each other and solve for c:

9781118496718-eq161059.png

Therefore, 9781118496718-eq161060.tif is the solution.

267. c = 2

To determine the value of c, you must satisfy the definition of a continuous function: 9781118496718-eq161061.tif.

The left-hand limit at x = 4 is given by 9781118496718-eq161062.tif, and the right-hand limit is given by 9781118496718-eq161063.tif. Also note that 9781118496718-eq161064.tif.

The left-hand limit must equal the right-hand limit, so set them equal to each other:

9781118496718-eq161065.png

Therefore, c – 2 = 0, which gives you the solution c = 2.

268. [1, 2]

Recall the intermediate value theorem: Suppose that f is continuous on the closed interval [a, b], and let N be any number between f (a) and f (b), where f (a) ≠ f (b). Then a number c exists in (a, b) such that f (c) = N.

Notice that 9781118496718-eq161066.tif is a polynomial that's continuous everywhere, so the interme-diate value theorem applies. Checking the endpoints of the interval [1, 2] gives you

9781118496718-eq161067.png

Because the function changes signs on this interval, there's at least one root in the interval by the intermediate value theorem.

9781118496718-un16726.tif

269. [16, 25]

Notice that 9781118496718-eq161068.tif is a continuous function for x ≥ 0, so the intermediate value theorem applies. Checking the endpoints of the interval [16, 25] gives you

9781118496718-eq161069.png

Because the function changes signs on this interval, there's at least one root in the interval by the intermediate value theorem.

270. [2, 3]

The function f (x) = 2(3x) + x2 – 4 is continuous everywhere, so the intermediate value theorem applies. Checking the endpoints of the interval [2, 3] gives you

9781118496718-eq161070.png

The number 32 is between 18 and 59, so by the intermediate value theorem, there exists at least one point c in the interval [2, 3] such that f (c) = 32.

271. [3, 4]

The function 9781118496718-eq161071.tif is continuous everywhere, so the intermediate value theorem applies. Checking the endpoints of the interval [3, 4] gives you

9781118496718-eq161072.png

The number 22 is between 17 and 25, so by the intermediate value theorem, there exists at least one point c in the interval [3, 4] such that f (c) = 22.

272. x = 1

The graph has a point of discontinuity at x = 1, so the graph isn't differentiable there. However, the rest of the graph is smooth and continuous, so the derivative exists for all other points.

273. differentiable everywhere

Because the graph is continuous and smooth everywhere, the function is differentiable everywhere.

274. x = –2, x = 0, x = 2

The graph of the function has sharp corners at x = –2, x = 0, and x = 2, so the function isn't differentiable there.

You can also note that for each of the points x = –2, x = 0, and x = 2, the slopes of the tangent lines jump from 1 to –1 or from –1 to 1. This jump in slopes is another way to recognize values of x where the function is not differentiable.

275. 9781118496718-eq17001.tif, where n is an integer

Because y = tan x has points of discontinuity at 9781118496718-eq17002.tif, where n is an integer, the function isn't differentiable at those points.

276. x = 0

The tangent line would be vertical at x = 0, so the function isn't differentiable there.

277. 2

Use the derivative definition with f (x) = 2x – 1 and f (x + h) = 2(x + h) – 1 = 2x + 2h – 1:
9781118496718-eq17003.eps

278. 2 x

Use the derivative definition with f (x) = x2 and f (x + h) = (x + h)2 = (x + h)(x + h) = x2 + 2xh + h2:
9781118496718-eq17004.eps

279. 1

Use the derivative definition with f (x) = 2 + x and f (x + h) = 2 + (x + h):
9781118496718-eq17005.eps

280. 9781118496718-eq17006.eps

Use the derivative definition with
9781118496718-eq17007.tif and with
9781118496718-eq17008.tif to get the following:
9781118496718-eq17009.eps

281. 3 x2 – 2x

Using the derivative definition with f (x) = x3 – x2 and with
9781118496718-eq17010.eps

gives you the following:
9781118496718-eq17011.eps

282. 6 x + 4

Using the derivative definition with f (x) = 3x2 + 4x and with
9781118496718-eq17012.eps

gives you the following:
9781118496718-eq17013.eps

283. 9781118496718-eq17014.eps

Using the derivative definition with 9781118496718-eq17015.tif and with 9781118496718-eq17016.tif gives you the following:
9781118496718-eq17017.eps

284. 9781118496718-eq17018.eps

Using the derivative definition with 9781118496718-eq17019.tif and with 9781118496718-eq17020.tif gives you the following:
9781118496718-eq17021.eps

285. 9781118496718-eq17022.eps

Use the derivative definition with 9781118496718-eq17023.tif and with 9781118496718-eq17024.tif:
9781118496718-eq17025.eps

286. 3 x2 + 3

Use the derivative definition with f (x) = x3 + 3x and with f (x + h) = (x + h)3 + 3(x + h) = x3 + 3x2h + 3xh2 + h3 + 3x + 3h:

9781118496718-eq17026.eps

287. 9781118496718-eq17027.eps

Using the derivative definition with 9781118496718-eq17028.tif and with 9781118496718-eq17029.tif gives you the following:
9781118496718-eq17030.eps

288. 9781118496718-eq17031.eps

Using the derivative definition with 9781118496718-eq17032.tif and with 9781118496718-eq17033.tif gives you the following:
9781118496718-eq17034.eps

289. 9781118496718-eq17035.eps

Using the derivative definition with 9781118496718-eq17036.tif and with 9781118496718-eq17037.tif gives you the following:
9781118496718-eq17038.eps

290. 9781118496718-eq17039.eps

Using the derivative definition with 9781118496718-eq17040.tif and with 9781118496718-eq17041.tif gives you 9781118496718-eq17042.eps

To rationalize the numerator, consider the formula a3 – b3 = (a – b)(a2 + ab + b2). If you let a = (2x + 2h + 1)1/3 and b = (2x + 1)1/3, you have (a – b) in the numerator; that means you can rationalize the numerator by multiplying by (a2 + ab + b2):
9781118496718-eq17043.eps

291. 0

The tangent line at x = 3 is horizontal, so the slope is zero. Therefore, f'(3) = 0.

292. –1

The slope of the tangent line at x = –1 is equal to –1, so f'(–1) = –1.

293. 1

The slope of the tangent line at x = –3 is equal to 1, so f'(–3) =1.

294. 3

The slope of the tangent line at any point on the graph of y = 3x + 4 is equal to 3, so f'(–22π3) = 3.

295. f'(1) < f'(–2) < f'(–3)

The tangent line at x = –3 has a positive slope, the slope of the tangent line at x = –2 is equal to zero, and the slope of the tangent line at x = 1 is negative. Therefore, f'(1) < f'(–2) < f'(–3).

296. f'(1) < f'(2) < 0.1 < f'(5)

The slope of the tangent line at x = 1 is negative, the slope of the tangent line at x = 2 is equal to zero, and the slope of the tangent line at x = 5 is clearly larger than 0.1. Therefore, f'(1) < f'(2) < 0.1 < f'(5).

297. 5

Use basic derivative rules to get f'(x) = 5.

298. 2 x + 3

Apply the power rule to each term, recalling that the derivative of a constant is zero: f'(x) = 2x + 3.

299. 4 x + 7

Begin by multiplying the two factors together:
9781118496718-eq17044.eps

Then apply the power rule to get the derivative:
9781118496718-eq17045.eps

300. 0

Because π3 is constant, f'(x) = 0.

301. 9781118496718-eq17046.eps

Split up the radical and rewrite the power on the variable using exponential notation:
9781118496718-eq17047.eps

Then apply the power rule to find the derivative:
9781118496718-eq17048.eps

302. 9781118496718-eq17049.eps

Begin by breaking up the fraction:
9781118496718-eq17050.eps

Then apply the power rule to find the derivative:
9781118496718-eq17051.eps

303. 9781118496718-eq17052.eps

Begin by multiplying the factors together:
9781118496718-eq17053.eps

Then apply the power rule to find the derivative:
9781118496718-eq17054.eps

304. 9781118496718-eq17055.eps

Begin by rewriting the function using a negative exponent:
9781118496718-eq17056.eps

Then apply the power rule to each term to get the derivative:
9781118496718-eq17057.eps

305. 9781118496718-eq17058.eps

Apply the power rule to the first two terms of 9781118496718-eq17507.tif, recalling that the derivative of a constant is zero:
9781118496718-eq17059.eps

306. 9781118496718-eq17060.eps

Multiply the factors and rewrite the function using exponential notation:
9781118496718-eq17061.eps

Next, apply the power rule to find the derivative:
9781118496718-eq17062.eps

307. 9781118496718-eq17063.eps

Begin by multiplying the factors together:
9781118496718-eq17064.eps

Then find the derivative using the power rule:
9781118496718-eq17065.eps

308. 16 x3 – 2x + 8

Apply the power rule to each term, recalling that the derivative of a constant is zero: f'(x) = 16x3 – 2x + 8.

309. 9781118496718-eq17066.eps

Rewrite the function using exponential notation:
9781118496718-eq17067.eps

Then apply the power rule to each term to find the derivative:
9781118496718-eq17068.eps

310. 9781118496718-eq17069.eps, 1

Begin by finding the derivative of the function f (x) = x3 – x2 – x + 1:
9781118496718-eq17070.eps

A horizontal tangent line has a slope of zero, so set the derivative equal to zero, factor, and solve for x:
9781118496718-eq17071.eps

Setting each factor equal to zero gives you 3x + 1 = 0, which has the solution 9781118496718-eq17072.tif, and gives you x – 1 = 0, which has the solution x = 1.
9781118496718-un1701.eps

311. 9781118496718-eq17073.eps

Begin by finding the derivative of the function:
9781118496718-eq17074.eps

Next, set the derivative equal to 6 and solve for x:
9781118496718-eq17075.eps

312. 16 x7 + 5x4 – 8x3 – 1

Recall that the product rule states 9781118496718-eq17076.eps

You can multiply out the expression f (x) = (2x3 + 1)(x5 – 1) first and then avoid using the product rule, but here's how to find the derivative using the product rule:
9781118496718-eq17077.eps

313. x(2 sin x + x cos x)

Applying the product rule to f (x) = xsin x gives you
9781118496718-eq17079.eps

314. 9781118496718-eq17080.eps

Apply the product rule to f (x) = sec x tan x:
9781118496718-eq17082.eps

315. (sec x)(1 + x tan x)

Begin by rewriting the original expression as 9781118496718-eq17084.tif and then apply the product rule:
9781118496718-eq17085.eps

316. 4 (csc x)(1 – x cot x)

Apply the product rule to f (x) = 4x csc x to get
9781118496718-eq17087.eps

317. 12

According to the product rule, 9781118496718-eq17088.eps

To find (fg)'(4), enter the numbers and solve:
9781118496718-eq17089.eps

318. 9781118496718-eq17090.eps

Recall that the product rule states 9781118496718-eq17091.eps

You can apply the quotient rule directly, or you can rewrite the original function as 9781118496718-eq17092.tif and then apply the product rule to get the following:
9781118496718-eq17093.eps

319. x sec x tan x + 2 secx

Apply the product rule to f (x) = sec x(x + tan x) as follows:
9781118496718-eq17095.eps

320. 9781118496718-eq17096.eps

Apply the product rule to f (x) = (x2 + x)csc x to get

9781118496718-eq17098.eps

321. 4 x(sec x)(3 + x tan x)

Applying the product rule to f (x) = 4x3 sec x gives you

9781118496718-eq17100.eps

322. 9781118496718-eq17101.eps

You can apply the quotient rule directly, or you can rewrite the original function as 9781118496718-eq17103.tif and then apply the product rule as follows:
9781118496718-eq17104.eps

323. 9781118496718-eq17105.eps

Using the product rule on f (x) = x2g (x) and then factoring gives you the derivative as follows:
9781118496718-eq17107.eps

324. 9781118496718-eq17108.eps

Begin by breaking up the fraction and simplifying:
9781118496718-eq17110.eps

Next, apply the power rule to the first term and the product rule to the second term:
9781118496718-eq17111.eps

325. 46

The product rule tells you that 9781118496718-eq17112.eps

To find (fg)'(3), enter the numbers and solve:
9781118496718-eq17113.eps

326. 2 x cos x sin x – xsin2 x + xcosx

Recall that the product rule states 9781118496718-eq17114.eps

You can group the factors however you want and then apply the product rule within the product rule. If you group together the trigonometric functions and apply the product rule, you have f (x) = x2(cos x sin x) so that 9781118496718-eq17115.eps

327. 9781118496718-eq17116.eps

Begin by simplifying the given expression:
9781118496718-eq17118.eps

Then apply the product rule to the second term and the power rule to the first term:
9781118496718-eq17119.eps

328. 9781118496718-eq17120.eps

Rewrite the original expression:
9781118496718-eq17122.eps

Then apply the product rule to get the derivative:
9781118496718-eq17123.eps

329. 9781118496718-eq17124.eps

First simplify the given function:
9781118496718-eq17126.eps

Then apply the product rule to find the derivative:
9781118496718-eq17127.eps

330. 9781118496718-eq17128.eps

You can use the quotient rule directly, or you can rewrite the original expression as 9781118496718-eq17130.tif and then apply the product rule as follows:
9781118496718-eq17131.eps

331. 9781118496718-eq17132.eps

To find the derivative of 9781118496718-eq17134.tif, use the product rule within the product rule:
9781118496718-eq17135.eps

332. 9781118496718-eq17136.eps

Recall that the quotient rule states 9781118496718-eq17137.eps

Apply the quotient rule to 9781118496718-eq17138.tif:
9781118496718-eq17139.eps

333. 9781118496718-eq17140.eps

Apply the quotient rule to 9781118496718-eq17142.tif to get the derivative:
9781118496718-eq17143.eps

334. 9781118496718-eq17144.eps

Apply the quotient rule to 9781118496718-eq17146.tif:

9781118496718-eq17147.eps

335. 9781118496718-eq17148.eps

Apply the quotient rule to 9781118496718-eq17150.tif to get the following:
9781118496718-eq17151.eps

336. 9781118496718-eq17152.eps

The quotient rule gives you 9781118496718-eq17153.eps

To find 9781118496718-eq17154.tif, enter the numbers and solve:
9781118496718-eq17155.eps

337. 9781118496718-eq17156.eps

Recall that the quotient rule states 9781118496718-eq17157.eps

Apply the quotient rule to 9781118496718-eq17158.tif:
9781118496718-eq17159.eps

338. 9781118496718-eq17160.eps

Apply the quotient rule to 9781118496718-eq17162.tif to find the derivative:
9781118496718-eq17163.eps

339. 9781118496718-eq17164.eps

Apply the quotient rule to 9781118496718-eq17166.tif to get the derivative:
9781118496718-eq17167.eps

340. 9781118496718-eq17168.eps

First, simplify the numerator and denominator:
9781118496718-eq17170.eps

Then apply the quotient rule to get 9781118496718-eq17171.eps

341. 9781118496718-eq17172.eps

Apply the quotient rule to 9781118496718-eq17174.tif to get the following:
9781118496718-eq17175.eps

342. 9781118496718-eq17176.eps

Apply the quotient rule to 9781118496718-eq17178.tif:
9781118496718-eq17179.eps

343. 9781118496718-eq17180.eps

The quotient rule says that 9781118496718-eq17181.eps

To find 9781118496718-eq17182.tif, enter the numbers and solve:
9781118496718-eq17183.eps

344. 9781118496718-eq17184.eps

Recall that the quotient rule states 9781118496718-eq17185.eps

Apply the quotient rule to 9781118496718-eq17186.tif to get
9781118496718-eq17187.eps

345. 9781118496718-eq17188.eps

Apply the quotient rule to 9781118496718-eq17190.tif to get the following:
9781118496718-eq17191.eps

346. 9781118496718-eq17192.eps

Apply the quotient rule to 9781118496718-eq17194.tif:
9781118496718-eq17195.eps

347. 9781118496718-eq17196.eps

Apply the quotient rule to 9781118496718-eq17198.tif to get 9781118496718-eq17199.eps

348. 9781118496718-eq17200.eps

Apply the quotient rule to 9781118496718-eq17202.tif as follows:
9781118496718-eq17203.eps

Note that the identity secx – tanx = 1 was used to simplify the final expression.

349. 9781118496718-eq17204.eps

First multiply the numerator and denominator by x:
9781118496718-eq17206.eps

Then use the quotient rule to find the derivative:
9781118496718-eq17207.eps

350. 9781118496718-eq17208.eps

Applying the quotient rule to 9781118496718-eq17210.tif, followed by factoring and simplifying, gives you the following:
9781118496718-eq17211.eps

351. 9781118496718-eq17212.eps

Applying the quotient rule to 9781118496718-eq17214.tif gives you 9781118496718-eq17215.eps

352. 9781118496718-eq17216.eps

Recall that the chain rule states 9781118496718-eq17217.eps

Applying the chain rule to 9781118496718-eq17218.tif gives you 9781118496718-eq17219.eps

353. 4 cos(4x)

Apply the chain rule to f (x) = sin(4x):
9781118496718-eq17221.eps

354. 9781118496718-eq17222.eps

Rewrite the function using exponential notation:
9781118496718-eq17224.eps

Then apply the chain rule:
9781118496718-eq17225.eps

355. 9781118496718-eq17226.eps

Rewrite the function as 9781118496718-eq17228.eps

Then apply the chain rule to get the derivative:
9781118496718-eq17229.eps

356. 9781118496718-eq17230.eps

First rewrite the function:
9781118496718-eq17232.eps

Then apply the chain rule:
9781118496718-eq17233.eps

357. 9781118496718-eq17234.eps

Apply the chain rule to 9781118496718-eq17236.tif to get 9781118496718-eq17237.eps

358. 9781118496718-eq17238.eps

Recall that the chain rule states 9781118496718-eq17239.eps

and that the quotient rule states 9781118496718-eq17240.eps

To find the derivative of 9781118496718-eq17241.tif, apply the quotient rule along with the chain rule:
9781118496718-eq17242.eps

359. 9781118496718-eq17243.eps

Recall that the chain rule states 9781118496718-eq17244.eps

and that the product rule states 9781118496718-eq17245.eps

To find the derivative of f (x) = cos(x sin x), apply the chain rule and the product rule:
9781118496718-eq17246.eps

360. 9781118496718-eq17247.eps

Rewrite the function using exponential notation:
9781118496718-eq17250.eps

Now apply the product rule, being careful to use the chain rule when taking the derivative of the second factor:
9781118496718-eq17251.eps

361. 9781118496718-eq17252.eps

Recall that the chain rule states 9781118496718-eq17253.eps

and that the quotient rule states 9781118496718-eq17254.eps

Rewrite the function using exponential notation:
9781118496718-eq17255.eps

Next, apply the quotient rule, making sure to use the chain rule when taking the derivative of the denominator:
9781118496718-eq17256.eps

362. 4 secx tan x

First rewrite the function:
9781118496718-eq17258.eps

Applying the chain rule gives you the derivative:
9781118496718-eq17259.eps

363. 9781118496718-eq17260.eps

Recall that the chain rule states 9781118496718-eq17261.eps

and that the quotient rule states 9781118496718-eq17262.eps

Rewrite the function using exponential notation:
9781118496718-eq17263.eps

Apply the quotient rule and the chain rule to get the derivative:
9781118496718-eq17264.eps

364. 9781118496718-eq17265.eps

Recall that the chain rule states 9781118496718-eq17266.eps

and that the product rule states 9781118496718-eq17267.eps

Rewrite the function using exponential notation:
9781118496718-eq17268.eps

Then apply the product rule and also use the chain rule on the second factor to get the derivative:
9781118496718-eq17269.eps

365. 9781118496718-eq17270.eps

To find the derivative of 9781118496718-eq17273.tif, use the product rule along with the chain rule and then factor:
9781118496718-eq17274.eps

366. 9781118496718-eq17275.eps

Recall that the chain rule states 9781118496718-eq17276.eps

and that the quotient rule states 9781118496718-eq17277.eps

Applying the quotient rule and the chain rule to 9781118496718-eq17278.tif gives you the following derivative:
9781118496718-eq17279.eps

367. 9781118496718-eq17280.eps

First rewrite the function using exponential notation:
9781118496718-eq17283.eps

Then apply both the chain rule and the quotient rule:
9781118496718-eq17284.eps

368. 9781118496718-eq17285.eps

To find the derivative of 9781118496718-eq17287.tif, apply the chain rule repeatedly:
9781118496718-eq17288.eps

369. 9781118496718-eq17289.eps

Rewrite the function using exponential notation:
9781118496718-eq17291.eps

Then apply the chain rule repeatedly to get the derivative:
9781118496718-eq17292.eps

370. 9781118496718-eq17293.eps

Recall that the chain rule states 9781118496718-eq17294.eps

and that the product rule states 9781118496718-eq17295.eps

Applying the product rule and chain rule to f (x) = (1 + 5x)4(2 + x – x2)7 and then factoring (factoring can be the tricky part!) gives you the following:
9781118496718-eq17296.eps

371. x = 0, π, 2π

Recall that the chain rule states 9781118496718-eq17297.eps

Begin by finding the derivative of the function f (x) = 2 cos x + sinx:
9781118496718-eq17298.eps

Then set the derivative equal to zero to find the x values where the function has a horizontal tangent line:
9781118496718-eq17299.eps

Next, set each factor equal to zero and solve for x: sin x = 0 has the solutions x = 0, π, 2π, and cos x – 1 = 0, or cos x = 1, has the solutions x = 0, 2π. The slope of the tangent line is zero at each of these x values, so the solutions are x = 0, π, 2π.

372. 9781118496718-eq17300.eps

To find the derivative of 9781118496718-eq17302.tif, use the chain rule:
9781118496718-eq17303.eps

373. 28

Because the chain rule gives you 9781118496718-eq17304.tif, it follows that
9781118496718-eq17305.eps

374. –40

Because the chain rule gives you 9781118496718-eq17306.tif, it follows that
9781118496718-eq17307.eps

375. 9781118496718-eq17308.eps

Recall that the chain rule states 9781118496718-eq17309.eps

Using the chain rule on f (x) = H(x3) gives you the following:
9781118496718-eq17310.eps

Because 9781118496718-eq17311.tif, you know that that 9781118496718-eq17312.tif, so the derivative becomes 9781118496718-eq17313.eps

376. 80

Because the chain rule gives you 9781118496718-eq17314.tif, it follows that
9781118496718-eq17315.eps

377. 9781118496718-eq17316.eps

Use properties of logarithms to rewrite the function:
9781118496718-eq17317.eps

Then take the derivative:
9781118496718-eq17318.eps

378. 9781118496718-eq17319.eps

To find the derivative of f (x) = (ln x)4, apply the chain rule:
9781118496718-eq17320.eps

379. 9781118496718-eq17321.eps

Begin by using properties of logarithms to rewrite the function:
9781118496718-eq17322.eps

Then apply the chain rule to find the derivative:
9781118496718-eq17323.eps

380. 9781118496718-eq17324.eps

Begin by writing the function using exponential notation:
9781118496718-eq17325.eps

Then use the chain rule to find the derivative:
9781118496718-eq17326.eps

381. 9781118496718-eq17327.eps

To make the derivative easier to find, use properties of logarithms to break up the function:
9781118496718-eq17328.eps

Now apply the chain rule to each term to get the derivative:
9781118496718-eq17329.eps

382. 9781118496718-eq17330.eps

Begin by rewriting the function using properties of logarithms:
9781118496718-eq17331.eps

The derivative becomes
9781118496718-eq17332.eps

Note that log75 is a constant, so its derivative is equal to zero.

You can further simplify by using the change of base formula to write 9781118496718-eq17333.tif:
9781118496718-eq17334.eps

383. sec x

Applying the chain rule to 9781118496718-eq17335.tif gives you
9781118496718-eq17336.eps

384. 9781118496718-eq17337.eps

To find the derivative of 9781118496718-eq17338.tif, apply the quotient rule and chain rule to the first term and apply the chain rule to the second term:
9781118496718-eq17339.eps

385. 9781118496718-eq17340.eps

Begin by using properties of logarithms to rewrite the function:
9781118496718-eq17341.eps

Then apply the quotient rule and the chain rule:
9781118496718-eq17342.eps

You can now further simplify by using the change of base formula to write 9781118496718-eq17343.tif so that you have
9781118496718-eq17344.eps

386. 9781118496718-eq17345.eps

Begin by rewriting the function and taking the natural logarithm of each side:
9781118496718-eq17346.eps

Take the derivative of each side with respect to x:
9781118496718-eq17347.eps

Then multiply both sides by y:
9781118496718-eq17348.eps

Finally, replacing y with xtan x gives you the answer:
9781118496718-eq17349.eps

387. 9781118496718-eq17350.eps

Begin by rewriting the function and taking the natural logarithm of each side:
9781118496718-eq17351.eps

Then take the derivative of each side with respect to x:
9781118496718-eq17352.eps

Multiplying both sides by y produces
9781118496718-eq17353.eps

And replacing y with 9781118496718-eq17354.tif gives you the answer:
9781118496718-eq17355.eps

388. 9781118496718-eq17356.eps

Begin by rewriting the function and taking the natural logarithm of each side:
9781118496718-eq17357.eps

Using properties of logarithms to expand gives you
9781118496718-eq17358.eps

Next, take the derivative of each side with respect to x:
9781118496718-eq17359.eps

Multiplying both sides by y produces the following:
9781118496718-eq17360.eps

Replacing y with 9781118496718-eq17361.tif and simplifying gives you the solution:
9781118496718-eq17362.eps

Note that you can find the derivative without logarithmic differentiation, but using it makes the math easier.

389. 9781118496718-eq17363.eps

Begin by rewriting the function and taking the natural logarithm of each side:
9781118496718-eq17364.eps

Use properties of logarithms to expand:
9781118496718-eq17365.eps

Then take the derivative of each side with respect to x:
9781118496718-eq17366.eps

Multiplying both sides by y produces
9781118496718-eq17367.eps

Replacing with y with 9781118496718-eq17368.tif gives you the answer:
9781118496718-eq17369.eps

Note that you can find the derivative without logarithmic differentiation, but using this technique makes the calculations easier.

390. 5 e5x

Apply the chain rule to find the derivative of f (x) = e5x:
9781118496718-eq17370.eps

391. 9781118496718-eq17371.eps

Applying the chain rule to 9781118496718-eq17372.tif gives you
9781118496718-eq17373.eps

392. 9781118496718-eq17374.eps

Applying the product rule to 9781118496718-eq17375.tif gives you the derivative as follows:
9781118496718-eq17376.eps

393. 2 x

Use properties of logarithms to simplify the function:
9781118496718-eq17377.eps

The derivative is simply equal to 9781118496718-eq17378.eps

394. 9781118496718-eq17379.eps

To find the derivative of 9781118496718-eq17380.tif, apply the product rule while also applying the chain rule to the first factor:
9781118496718-eq17381.eps

395. 9781118496718-eq17382.eps

Put the radical in exponential form:
9781118496718-eq17383.eps

Then apply the product rule as well as the chain rule to the first factor to get the derivative:
9781118496718-eq17384.eps

396. 9781118496718-eq17385.eps

Applying the chain rule to 9781118496718-eq17386.tif gives you the derivative as follows:
9781118496718-eq17387.eps

397. 9781118496718-eq17388.eps

Use properties of logarithms to break up the function:
9781118496718-eq17389.eps

Then apply the chain rule to each term to get the derivative:
9781118496718-eq17390.eps

398. 9781118496718-eq17391.eps

Apply the quotient rule to find the derivative of 9781118496718-eq17392.tif:
9781118496718-eq17393.eps

399. 9781118496718-eq17394.eps

First rewrite the function:
9781118496718-eq17395.eps

Applying the chain rule gives you the derivative as follows:
9781118496718-eq17396.eps

400. 9781118496718-eq17397.eps

To find the derivative of 9781118496718-eq17398.tif, apply the quotient rule while applying the chain rule to the numerator:
9781118496718-eq17399.eps

401. 9781118496718-eq17400.eps

To find the derivative of 9781118496718-eq17401.tif, apply the product rule while applying the chain rule to the second factor:
9781118496718-eq17402.eps

402. y = πx + 3

You can begin by finding the y value, because it isn't given:
9781118496718-eq17403.eps

Next, find the derivative of the function:
9781118496718-eq17404.eps

Substitute in the given x value to find the slope of the tangent line:
9781118496718-eq17405.eps

Now use the point-slope formula for a line to get the tangent line at x = 0:
9781118496718-eq17406.eps

403. y = x + 1

Begin by finding the derivative of the function f (x) = x2 – x + 2:
9781118496718-eq17407.eps

Substitute in the given x value to find the slope of the tangent line:
9781118496718-eq17408.eps

Now use the point-slope formula for a line to get the tangent line at (1, 2):
9781118496718-eq17409.eps

9781118496718-un1702.eps

404. 9781118496718-eq17410.eps

You can begin by finding the y value, because it isn't given:
9781118496718-eq17411.eps

Next, find the derivative of the function:
9781118496718-eq17412.eps

Substitute in the given x value to find the slope of the tangent line:
9781118496718-eq17413.eps

Now use the point-slope formula for a line to get the tangent line at x = 2:
9781118496718-eq17414.eps

405. 9781118496718-eq17415.eps

The normal line is perpendicular to the tangent line. Begin by finding the derivative of the function f (x) = 3x2 + x – 2:
9781118496718-eq17416.eps

Then substitute in the given x value to find the slope of the tangent line:
9781118496718-eq17417.eps

To find the slope of the normal line, take the opposite reciprocal of the slope of the tangent line to get 9781118496718-eq17418.tif.

Now use the point-slope formula for a line to get the normal line at (3, 28):
9781118496718-eq17419.eps

406. 9781118496718-eq17420.eps

The normal line is perpendicular to the tangent line. You can begin by finding the y value at 9781118496718-eq17421.tif, because it isn't given:
9781118496718-eq17422.eps

Next, find the derivative of the function:
9781118496718-eq17423.eps

Then substitute in the given x value to find the slope of the tangent line:
9781118496718-eq17424.eps

To find the slope of the normal line, take the opposite reciprocal of the slope of the tangent line to get –1.

Now use the point-slope formula for a line to get the normal line at 9781118496718-eq17425.tif:
9781118496718-eq17426.eps

9781118496718-un1702.eps

407. 9781118496718-eq17427.eps

The normal line is perpendicular to the tangent line. You can begin by finding the y value at x = e2, because it isn't given:
9781118496718-eq17428.eps

Next, find the derivative of the function:
9781118496718-eq17429.eps

Then substitute in the given x value to find the slope of the tangent line:
9781118496718-eq17430.eps

To find the slope of the normal line, take the opposite reciprocal of the slope of the tangent line to get 9781118496718-eq17431.tif.

Now use the point-slope formula for a line to get the normal line at x = e2:
9781118496718-eq17432.eps

408. 9781118496718-eq17433.eps

Taking the derivative of both sides of x2 + y2 = 9 and solving for 9781118496718-eq17434.tif gives you the following:
9781118496718-eq17435.eps

409. 9781118496718-eq17436.eps

Take the derivative of both sides of y5 + x2y3 = 2 + x2y and solve for 9781118496718-eq17437.tif:
9781118496718-eq17438.eps

410. 9781118496718-eq17439.eps

Take the derivative of both sides of x3y3 + x cos(y) = 7 and solve for 9781118496718-eq17440.tif:
9781118496718-eq17441.eps

411. 9781118496718-eq17442.eps

Take the derivative of both sides of 9781118496718-eq17443.tif and solve for 9781118496718-eq17444.tif:
9781118496718-eq17445.eps

412. 9781118496718-eq17446.eps

Take the derivative of both sides of 9781118496718-eq17447.tif and solve for 9781118496718-eq17448.tif:

9781118496718-eq17449.eps

413. 9781118496718-eq17450.eps

Take the derivative of both sides of 9781118496718-eq17451.tif and solve for 9781118496718-eq17452.tif:
9781118496718-eq17453.eps

414. 9781118496718-eq17454.eps

Begin by finding the first derivative of 8x2 + y2 = 8:
9781118496718-eq17455.eps

Next, find the second derivative:
9781118496718-eq17456.eps

Substituting in the value of 9781118496718-eq17457.tif gives you
9781118496718-eq17458.eps

And now using 8x2 + y2 = 8 gives you the answer:
9781118496718-eq17459.eps

415. 9781118496718-eq17460.eps

Begin by finding the first derivative of x5 + y5 = 1:
9781118496718-eq17461.eps

Next, find the second derivative:
9781118496718-eq17462.eps

Substituting in the value of 9781118496718-eq17463.tif gives you
9781118496718-eq17464.eps

And now using x5 + y5 = 1 gives you the answer:
9781118496718-eq17465.eps

416. 9781118496718-eq17466.eps

Begin by finding the first derivative of x3 + y3 = 5:
9781118496718-eq17467.eps

Next, find the second derivative:
9781118496718-eq17468.eps

Substituting in the value of 9781118496718-eq17469.tif gives you
9781118496718-eq17470.eps

And now using x3 + y3 = 5 gives you the answer:
9781118496718-eq17471.eps

417. 9781118496718-eq17472.eps

Begin by finding the first derivative of 9781118496718-eq17473.tif:
9781118496718-eq17474.eps

Next, find the second derivative:
9781118496718-eq17475.eps

Substituting in the value of 9781118496718-eq17476.tif gives you
9781118496718-eq17477.eps

And now using 9781118496718-eq17478.tif gives you the answer:
9781118496718-eq17479.eps

418. y = –x + 2

You know a point on the tangent line, so you just need to find the slope. Begin by finding the derivative of x2 + xy + y2 = 3:
9781118496718-eq17480.eps

Next, enter the values x = 1 and y = 1 and solve for the slope 9781118496718-eq17481.tif:
9781118496718-eq17482.eps

The tangent line has a slope of –1 and passes through (1, 1), so its equation is
9781118496718-eq17483.eps

9781118496718-eq17483.eps

419. 9781118496718-eq17484.eps

You know a point on the tangent line, so you just need to find the slope. Begin by finding the derivative of 3(x2 + y2)2 = 25(x2 – y2):
9781118496718-eq17485.eps

Then enter the values x = 2 and y = 1 and solve for the slope 9781118496718-eq17486.eps:
9781118496718-eq17487.eps

The tangent line has a slope of 9781118496718-eq17488.tif and passes through (2, 1), so its equation is
9781118496718-eq17489.eps

420. y = –x + 1

You know a point on the tangent line, so you just need to find the slope. Begin by taking the derivative of x2 + 2xy + y2:
9781118496718-eq17490.eps

Then use the values x = 0 and y = 1 and solve for the slope 9781118496718-eq17491.tif:
9781118496718-eq17492.eps

The tangent line has a slope of –1 and passes through (0, 1), so its equation is
9781118496718-eq17493.eps

421. 9781118496718-eq17494.eps

You know a point on the tangent line, so you just need to find the slope. Begin by taking the derivative of cos(xy) + x2 = sin y: 9781118496718-eq17495.eps

Then use the values x = 1 and 9781118496718-eq17496.tif and solve for the slope 9781118496718-eq17497.tif:
9781118496718-eq17498.eps

The tangent line has a slope of 9781118496718-eq17499.tif and passes through 9781118496718-eq17500.tif, so its equation is 9781118496718-eq17501.eps

422. y = 1

You know a point on the tangent line, so you just need to find the slope. Begin by simplifying the left side of the equation. 9781118496718-eq17502.eps

Next, find the derivative:
9781118496718-eq17503.eps

Use the values x = 0 and y = 1 and solve for the slope 9781118496718-eq17504.tif:
9781118496718-eq17505.eps

The tangent line has a slope of 0 and passes through (0, 1), so its equation is
9781118496718-eq17506.eps

423. 9781118496718-eq180001.png

Use the formula for the differential, dy = f '(x) dx:

9781118496718-eq180002.png

So for the given values, x = 3 and 9781118496718-eq180003.tif, find dy:

9781118496718-eq180004.png

424. 9781118496718-eq180005.png

Use the formula for the differential, dy = f '(x) dx:

9781118496718-eq180006.png

So for the given values, x = 1 and dx = –0.1, find dy:

9781118496718-eq180007.png

425. 9781118496718-eq180008.png

Use the formula for the differential, dy = f '(x) dx:

9781118496718-eq180009.png

So for the given values, 9781118496718-eq180010.tif and dx = 0.02, find dy:

9781118496718-eq180011.png

426. L(x) = 6x – 3

9781118496718-un1801.tif

Finding the linearization L(x) is the same as finding the equation of the tangent line. First find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

9781118496718-eq180012.png

Next, find the derivative of the function to get the slope of the line:

9781118496718-eq180013.png

Using the point-slope formula with the slope m = 6 and the point (1, 3) gives you

9781118496718-eq180014.png

Replace y with L(x) to get the solution: L(x) = 6x – 3.

427. 9781118496718-eq180015.png

Finding the linearization L(x) is the same as finding the equation of the tangent line. First find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

9781118496718-eq180016.png

Next, find the derivative of the function to get the slope of the line:

9781118496718-eq180017.png

Using the point-slope formula with the slope m = –1 and the point 9781118496718-eq180018.png gives you

9781118496718-eq180019.png

Replace y with L(x) to get the solution: 9781118496718-eq180020.png.

428. 9781118496718-eq180021.png

Finding the linearization L(x) is the same as finding the equation of the tangent line. First find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

9781118496718-eq180022.png

Next, find the derivative of the function to get the slope of the line:

9781118496718-eq180023.png

Using the point-slope formula with the slope 9781118496718-eq180024.tif and the point (2, 61/3) gives you

9781118496718-eq180025.png

Replace y with L(x) to get the solution:

9781118496718-eq180026.png.

429. 3.987

9781118496718-un1802.tif

To estimate the value, you can find a linearization by using f (x) = x2/3 and a = 8 and then substitute 7.96 into the linearization.

Find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

9781118496718-eq180027.png

Next, find the derivative of the function to get the slope of the line:

9781118496718-eq180028.png

Using the point-slope formula with the slope 9781118496718-eq180029.tif and the point (8, 4) gives you

9781118496718-eq180030.png

Replacing y with L(x) gives you the linearization

9781118496718-eq180031.png.

Finally, substitute in the value x = 7.96 to find the estimate:

9781118496718-eq180032.png

430. 9781118496718-eq180033.png

To estimate the value, you can find a linearization by using 9781118496718-eq180034.tif and a = 100 and then substitute 102 into the linearization.

Find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

9781118496718-eq180035.png

Next, find the derivative of the function to get the slope of the line:

9781118496718-eq180036.png

Using the point-slope formula with slope 9781118496718-eq180037.tif and point (100, 10) gives you

9781118496718-eq180038.png

Replacing y with L(x) gives you the linearization 9781118496718-eq180039.tif.

Finally, substitute in the value x = 102 to find the estimate:

9781118496718-eq180040.png

431. 9781118496718-eq180041.png

To estimate the value of tan 46°, you can find a linearization using f (x) = tan x and 9781118496718-eq180042.tif and then substitute the value 9781118496718-eq180043.tif into the linearization.

Find a point on the line by entering the value of a in the function. For the given value of a, the corresponding y value is

9781118496718-eq180044.png

Next, find the derivative of the function to get the slope of the line:

9781118496718-eq180045.png

Using the point-slope formula with slope m = 2 and point 9781118496718-eq180046.tif gives you

9781118496718-eq180047.png

Replacing y with L(x) gives you the linearization 9781118496718-eq180048.tif.

Finally, substitute in the value of x = 9781118496718-eq180049.tif radians (that is, 46°) to find the estimate of tan 46°:

9781118496718-eq180050.png

432. 9781118496718-eq180051.png

The volume of a sphere is 9781118496718-eq180052.tif. To find the rate of expansion, take the derivative of each side with respect to time:

9781118496718-eq180053.png

433. 8π m2/s

The area of a circle is A = πr2. To find the rate of increase, take the derivative of each side with respect to time:

9781118496718-eq180054.png

The circle increases at a rate of 9781118496718-eq180055.tif meter per second, so when radius r = 4 meters, the area increases at the following rate:

9781118496718-eq180056.png

434. 508

Take the derivative of both sides with respect to t:

9781118496718-eq180057.png

Then substitute in the given values, x = 3 and 9781118496718-eq180058.tif:

9781118496718-eq180059.png

435. 9781118496718-eq180060.png

Take the derivative of both sides with respect to t:

9781118496718-eq180061.png

To find the value of z, enter x = 4 and y = 1 in the original equation:

9781118496718-eq180062.png

Then substitute in the given values along with the value of z and solve for 9781118496718-eq180063.tif:

9781118496718-eq180064.png

436. 2.49 m2/s

9781118496718-un1803.tif

To find the area of the triangle, you need the base and the height. If you let the length of the base equal 8 meters, then one side of the triangle equals 6 meters. Therefore, if the height equals h, then 9781118496718-eq180065.tif so that 9781118496718-eq180066.tif.

Because the area of a triangle is 9781118496718-eq180067.tif, you can write the area as

9781118496718-eq180068.png

This equation now involves 9781118496718-eq180069.tif, so you can take the derivative of both sides with respect to t to get the rate of increase:

9781118496718-eq180070.png

Substitute in the given values, 9781118496718-eq180071.tif and 9781118496718-eq180072.tif radians/second, and simplify:

9781118496718-eq180073.png

When the angle between the sides is 9781118496718-eq180074.tif, the area is increasing at a rate of about 2.49 square meters per second.

437. 9781118496718-eq180075.png rad/s

9781118496718-un1804.tif

The problem tells you how quickly the bottom of the ladder slides away from the wall 9781118496718-eq180076.tif, so first write an expression for x, the distance from the wall. Assuming that the ground and wall meet at a right angle, you can write

9781118496718-eq180077.png

Taking the derivative with respect to time, t, gives you the rate at which the angle is changing:

9781118496718-eq180078.png

Substitute in the given information, where 9781118496718-eq180079.tif and 9781118496718-eq180080.tif feet per second:

9781118496718-eq180081.png

When the angle is 9781118496718-eq180082.tif, the angle is increasing at a rate of 9781118496718-eq180083.tif radians per second.

438. 72 cm2/min

The area of a triangle is 9781118496718-eq180084.tif, and the problem tells you how quickly the base and height are changing. To find the rate of the change in area, take the derivative of both sides of the equation with respect to time, t:

9781118496718-eq180085.png

Note that you have to use the product rule to find the derivative of the right side of the equation.

Substitute in the given information, 9781118496718-eq180086.tif centimeters per minute, h = 32 centimeters, b = 20 centimeters, and 9781118496718-eq180087.tif centimeters per minute:

9781118496718-eq180088.png

439. 13.56 km/h

9781118496718-un1805.tif

Let x be the distance sailed by Ship A and let y be the distance sailed by Ship B. Using the Pythagorean theorem, the distance between the ships is

9781118496718-eq180089.png

From the given information, you have 9781118496718-eq180090.tif kilometers per hour and 9781118496718-eq180091.tif kilometers per hour. Taking the derivative of both sides of the equation with respect to time gives you

9781118496718-eq180092.png

Notice that after 3 hours have elapsed, x = 60 and y = 105. Therefore, using the Pythagorean theorem, you can deduce that 9781118496718-eq180093.tif kilometers.

Substitute in all these values and solve for 9781118496718-eq180094.tif:

9781118496718-eq180095.png

At 3 p.m., the ships are moving apart at a rate of about 13.56 kilometers per hour.

440. 4.83 cm/s

From the Pythagorean theorem, the distance from the origin is D2 = x2 + y2. Using the particle's path, 9781118496718-eq180096.tif, you get

9781118496718-eq180097.png

Taking the derivative of both sides with respect to time gives you

9781118496718-eq180098.png

From the given values x = 8 centimeters and y = 3 centimeters, you find that the distance from the origin is 9781118496718-eq180099.tif centimeters. Use these values along with 9781118496718-eq180100.tif centimeters per second and solve for 9781118496718-eq180101.tif:

9781118496718-eq180102.png

441. 2.95 mi/h

9781118496718-un1806.tif

Let x be the distance that the person who is walking west has traveled, let y be the distance that the person traveling southwest has traveled, and let z be the distance between them.

Use the law of cosines to relate x, y, and z:

9781118496718-eq180103.png

Take the derivative of both sides of the equation with respect to time:

9781118496718-eq180104.png

After 40 minutes, or 9781118496718-eq180105.tif hour, 9781118496718-eq180106.tif and 9781118496718-eq180107.tif. Using the initial equation, you find that z ≈ 1.96 miles at this time. Substitute these values into the derivative and solve for 9781118496718-eq180108.tif:

9781118496718-eq180109.png

After 40 minutes, the distance between these people is increasing at a rate of about 2.95 miles per hour.

442. 9781118496718-eq180110.png

9781118496718-un1807.tif

The water in the trough has a volume of 9781118496718-eq180111.tif. As the water level rises, the base and height of the triangle shapes increase. By similar triangles, you have 9781118496718-eq180112.tif so that 9781118496718-eq180113.tif. Therefore, the volume is

9781118496718-eq180114.png

Taking the derivative with respect to time gives you

9781118496718-eq180115.png

Use h = 1 foot and 9781118496718-eq180116.tif cubic feet per minute to find how quickly the water level is rising:

9781118496718-eq180117.png

When the water is 1 foot deep, the water level is rising at a rate of 9781118496718-eq180118.tif feet per minute.

443. 698.86 km/h

9781118496718-un1810.tif

Let x be the distance that the jet travels, and let y be the distance between the jet and the radar station.

Use the law of cosines to relate x and y:

9781118496718-eq180119.png

Taking the derivative of both sides of the equation with respect to time gives you

9781118496718-eq180120.png

After 2 minutes, the jet has traveled 9781118496718-eq180121.tif kilometers, so the distance from the radar station is

9781118496718-eq180122.png

Enter the value of y in the 9781118496718-eq180123.tif equation and solve:

9781118496718-eq180124.tif

After 2 minutes, the jet is moving away from the radar station at a rate of about 698.86 kilometers per hour.

444. 9781118496718-eq180125.png

9781118496718-un1809.tif

Let x be the distance from the Point P to the spot on the shore where the light is shining. From the diagram, you have 9781118496718-eq180126.tif, or 9781118496718-eq180127.tif. Taking the derivative of both sides of the equation with respect to time gives you

9781118496718-eq180128.png

When x = 2, 9781118496718-eq180129.tif. Using a trigonometric identity, you have

9781118496718-eq180130.png

From the given information, you know that 9781118496718-eq180131.tif revolutions per minute. Because 2π radians are in one revolution, you can convert as follows: 9781118496718-eq180132.tif radians per minute:

9781118496718-eq180133.png

445. 9781118496718-eq180134.png

9781118496718-un1810.tif

The volume of a cone is 9781118496718-eq180135.tif. The problem tells you that 9781118496718-eq180136.tif cubic feet per minute and that d = 2h. Because the diameter is twice the radius, you have 2r = d = 2h so that r = h; therefore, the volume becomes

9781118496718-eq180137.png

Take the derivative with respect to time:

9781118496718-eq180138.png

Substitute in the given information and solve for 9781118496718-eq180139.tif:

9781118496718-eq180140.png

When the pile is 12 feet high, the height of the pile is increasing at a rate of 9781118496718-eq180141.tif (about 0.04) feet per minute.

446. no absolute maximum; absolute minimum: y = –1; no local maxima; local minimum: (1, –1)

There's no absolute maximum because the graph doesn't attain a largest y value. There are also no local maxima. The absolute minimum is y = –1. The point (1, –1) is a local minimum.

447. absolute maximum: y = 4; absolute minimum: y = 0; local maximum: (5, 4); local minima: (1, 0), (7, 2)

The absolute maximum value is 4, which the graph attains at the point (5, 4). The absolute minimum is 0, which is attained at the point (1, 0). The point (5, 4) also corresponds to a local maximum, and the local minima occur at (1, 0) and (7, 2).

448. absolute maximum: y = 3; no absolute minimum; local maxima: (3, 3), (5, 3); local minimum: (4, 1)

The absolute maximum value is 3, which the graph attains at the points (3, 3) and (5, 3). The function approaches the x-axis but doesn't cross or touch it, so there's no absolute minimum. The points (3, 3) and (5, 3) also correspond to local maxima, and the local minimum occurs at (4, 1).

449. no maxima or minima

The graph has no absolute maximum or minimum because the graph approaches ∞ on the left and –∞ on the right. Likewise, there are no local maxima or minima because (2, 1) and (5, 3) are points of discontinuity.

450. I and II

Because Point A satisfies the definition of both a local maximum and a local minimum, it's both. The graph decreases to negative infinity on the left side, so Point A is not an absolute minimum.

451. absolute maximum: 5; absolute minimum: –7

Begin by finding the derivative of the function; then find any critical numbers on the given interval by determining where the derivative equals zero or is undefined. Note that by finding critical numbers, you're finding potential turning points or cusp points of the graph.

The derivative of f (x) = 3x2 – 12x + 5 is

9781118496718-eq180142.png

Setting the derivative equal to zero and solving gives you the only critical number, x = 2. Next, substitute the endpoints of the interval and the critical number into the original function and pick the largest and smallest values:

9781118496718-eq180143.png

Therefore, the absolute maximum is 5, and the absolute minimum is –7.

452. absolute maximum: 67; absolute minimum: 3

Begin by finding the derivative of the function; then find any critical numbers on the given interval. The derivative of f (x) = x4 – 2x2 + 4 is

9781118496718-eq180144.png

Setting the derivative equal to zero and solving gives you the critical numbers x = 0 and x = ±1, all of which fall within the given interval. Next, substitute the endpoints of the interval along with the critical numbers into the original function and pick the largest and smallest values:

9781118496718-eq180145.png

Therefore, the absolute maximum is 67, and the absolute minimum is 3.

453. absolute maximum: 9781118496718-eq180146.tif; absolute minimum: 0

Begin by finding the derivative of the function; then find any critical numbers on the given interval. The derivative of 9781118496718-eq180147.tif is

9781118496718-eq180148.png

Next, find the critical numbers by setting the numerator equal to zero and solving for x (note that the denominator will never be zero):

9781118496718-eq180149.png

Only x = 1 is in the given interval, so don't use x = –1. Next, substitute the endpoints of the interval along with the critical number into the original function and pick the largest and smallest values:

9781118496718-eq180150.png

Therefore, the absolute maximum is 9781118496718-eq180151.tif, and the absolute minimum is 0.

454. absolute maximum: 2; absolute minimum: 9781118496718-eq180152.tif

Begin by finding the derivative of the function; then find any critical numbers on the given interval. The derivative of 9781118496718-eq180153.tif is

9781118496718-eq180154.png

Next, find any critical numbers of the function by setting the numerator and denominator of the derivative equal to zero and solving for t.

From the numerator of the derivative, you have 4 – 2t2 = 0 so that 4 = 2t2, or 2 = t2, which has the solutions 9781118496718-eq180155.tif. From the denominator, you have 4 – t2 = 0, or 4 = t2, which has the solutions t = ±2.

Substitute the endpoints along with the critical points that fall within the interval into the original function and pick the largest and smallest values:

9781118496718-eq180156.png

Therefore, the absolute maximum is 2, and the absolute minimum is 9781118496718-eq180157.tif.

455. absolute maximum: π + 2; absolute minimum: 9781118496718-eq180158.tif

Begin by finding the derivative of the function; then find any critical numbers on the given interval. The derivative of f (x) = x – 2 cos x is

9781118496718-eq180159.png

Setting the derivative equal to zero in order to find the critical numbers gives you 1+ 2 sin x = 0, or 9781118496718-eq180160.tif, which has the solutions 9781118496718-eq180161.tif and 9781118496718-eq180162.tif in the given interval of [–π, π].

Substitute these critical numbers along with the endpoints of the interval into the original function and pick the largest and smallest values:

9781118496718-eq180163.png

Therefore, the absolute maximum is π + 2, and the absolute minimum is 9781118496718-eq180164.tif.

456. increasing on (–∞, –2) and (2, ∞); decreasing on (–2, 2)

Begin by finding the derivative of the function; then find any critical numbers on the given interval by determining where the derivative equals zero or is undefined. Note that by finding critical numbers, you're finding potential turning points or cusp points of the graph.

The derivative of the function f (x) = 2x3 – 24x + 1 is

9781118496718-eq180165.png

Setting the derivative equal to zero in order to find the critical points gives you x2 – 4 = 0, or x2 = 4, so that x = ±2.

To determine where the function is increasing or decreasing, substitute a test point inside each interval into the derivative to see whether the derivative is positive or negative.

To test (–∞, –2), you can use x = –3. In that case, f '(–3) = 6(–3)2 – 24 = 30. The derivative is positive, so the function is increasing on (–∞, –2). Proceed in a similar manner for the intervals (–2, 2) and (2, ∞). When x = 0, you have f '(0) = 6(0)2 – 24 = –24, so the function is decreasing on (–2, 2). And if x = 3, then f '(3) = 6(3)2 – 24 = 30, so the function is increasing on (2, ∞).

457. increasing on (–2, ∞); decreasing on (–3, –2)

Begin by finding the derivative of the function. The derivative of 9781118496718-eq180166.tif is

9781118496718-eq180167.png

Next, find any critical numbers that fall inside the interval (–3, ∞) by setting the derivative equal to zero and solving for x. (You only need to set the numerator equal to zero, because the denominator will never equal zero.) This gives you 3x + 6 = 0, or x = –2.

Next, pick a value in (–3, –2) and determine whether the derivative is positive or negative. So if x = –2.5, then 9781118496718-eq180168.tif; therefore, the function is decreasing on (–3, –2). Likewise, you can show that f '(x) > 0 on (–2, ∞), which means that f (x) is increasing on (–2, ∞).

458. increasing on 9781118496718-eq180169.tif9781118496718-eq180170.tif; decreasing on 9781118496718-eq180171.tif9781118496718-eq180172.tif9781118496718-eq180173.tif

Begin by finding the derivative of the function. The derivative of f (x) = cosx – sin x is

9781118496718-eq180174.png

Next, find any critical numbers that fall inside the specified interval by setting the derivative equal to zero and solving: –2 cos x sin x = 0, or (–cos x)(2 sin x + 1) = 0. Solving –cos x = 0 gives you the solutions 9781118496718-eq180175.tif and 9781118496718-eq180176.tif, and solving 2 sin + 1 = 0 gives you 9781118496718-eq180177.tif, which has the solutions 9781118496718-eq180178.tif and 9781118496718-eq180179.tif.

To determine where the original function is increasing or decreasing, substitute a test point from inside each interval into the derivative to see whether the derivative is positive or negative. So for the interval 9781118496718-eq180180.tif, you can use 9781118496718-eq180181.tif. In that case, 9781118496718-eq180182.tif, so the function is decreasing on 9781118496718-eq180183.tif. Proceeding in a similar manner, you can show that for a point inside the interval 9781118496718-eq180184.tiff ' > 0; that for a point inside the interval 9781118496718-eq180185.tiff '> 0; that for a point inside 9781118496718-eq180186.tiff ' > 0; and that for a point inside 9781118496718-eq180187.tiff ' > 0. Therefore, f (x) is increasing on 9781118496718-eq180188.tif and 9781118496718-eq180189.tif, and f (x) is decreasing on 9781118496718-eq180190.tif9781118496718-eq180191.tif, and 9781118496718-eq180192.tif.

459. increasing on 9781118496718-eq180193.tif9781118496718-eq180194.tif; decreasing on 9781118496718-eq180195.tif9781118496718-eq180196.tif

Begin by finding the derivative of the function. The derivative of f (x) = 2 cos x – cos 2x is

9781118496718-eq180197.png

Now find the critical numbers by setting each factor equal to zero and solving for x. The equation sin x = 0 gives you the solutions x = 0, x = π, and x = 2π. Solving 2 cos x – 1 = 0 gives you 9781118496718-eq180198.tif, which has solutions 9781118496718-eq180199.tif and 9781118496718-eq180200.tif.

To determine whether the function is increasing or decreasing on 9781118496718-eq180201.tif, pick a point in the interval and substitute it into the derivative. So if you use 9781118496718-eq180202.tif, then you have 9781118496718-eq180203.tif, so the function is increasing on 9781118496718-eq180204.tif. Likewise, if you use the point 9781118496718-eq180205.tif in the interval 9781118496718-eq180206.tif, you have 9781118496718-eq180207.tif. If you use 9781118496718-eq180208.tif in the interval 9781118496718-eq180209.tif, then 9781118496718-eq180210.tif. And if you use 9781118496718-eq180211.tif in the interval 9781118496718-eq180212.tif, you have 9781118496718-eq180213.tif. Therefore, f (x) is increasing on the intervals 9781118496718-eq180214.tif and 9781118496718-eq180215.tif, and f (x) is decreasing on the intervals 9781118496718-eq180216.tif and 9781118496718-eq180217.tif.

460. increasing on (0, 1); decreasing on (1, ∞)

Begin by finding the derivative of the function. The derivative of f (x) = 4 ln x – 2x2 is

9781118496718-eq180218.png

Now find the critical numbers. Setting the numerator equal to zero gives you (1 – x)(1 + x) = 0 so that x = 1 or –1. Setting the denominator of the derivative equal to zero gives you x = 0. Notice that neither x = 0 nor x = –1 is in the domain of the original function.

To determine whether the function is increasing or decreasing on (0, 1), take a point in the interval and substitute it into the derivative. If 9781118496718-eq180219.tif, then 9781118496718-eq180220.tif, so the function is increasing on (0, 1). Likewise, if you use x = 2, then 9781118496718-eq180221.tif, so the function is decreasing on (1, ∞).

461. local maximum at (–1, 7); local minimum at (2, –20)

Begin by finding the derivative of the function f (x) = 2x3 – 3x2 – 12x:

9781118496718-eq180222.png

Then set this derivative equal to zero and solve for x to find the critical numbers:

9781118496718-eq180223.png

Next, determine whether the function is increasing or decreasing on the intervals (–∞, –1), (–1, 2), and (2, ∞) by picking a point inside each interval and substituting it into the derivative. Using the values x = –2, x = 0, and x = 3 gives you

9781118496718-eq180224.png

Therefore, f (x) is increasing on (–∞, –1), decreasing on (–1, 2), and increasing again on (2, ∞). That means there's a local maximum at x = –1 and a local minimum at x = 2.

Now enter these values in the original function to find the coordinates of the local maximum and minimum. Because f (–1) = 2(–1)3 – 3(–1)2 – 12(–1) = 7, the local maximum occurs at (–1, 7). Because f (2) = 2(2)3 – 3(2)2 – 12(2) = –20, the local minimum occurs at (2, –20).

462. no local maxima; local minimum at (16, –16)

Begin by finding the derivative of the function 9781118496718-eq180225.tif:

9781118496718-eq180226.png

Then find the critical numbers. Setting the numerator equal to zero gives you 9781118496718-eq180227.tif so that 9781118496718-eq180228.tif, or x = 16. Setting the denominator equal to zero gives you x = 0. Next, determine whether the function is increasing or decreasing on the intervals (0, 16) and (16, ∞) by taking a point in each interval and substituting it into the derivative. So if x = 1, you have 9781118496718-eq180229.tif so that f (x) is decreasing on (0, 16). And if x = 25, you have 9781118496718-eq180230.tif, so f (x) is increasing on (16, ∞). Therefore, the function has a local minimum at x = 16. Finish by finding the coordinates: 9781118496718-eq180231.tif, so the local minimum is at (16, –16).

463. local maximum at (64, 32); local minimum at (0, 0)

Begin by finding the derivative of the function f (x) = 6x2/3 – x:

9781118496718-eq180232.png

Then find the critical numbers. Setting the numerator equal to zero gives you 9781118496718-eq180233.tif so that 4 = x1/3, or 64 = x. Setting the denominator equal to zero gives you x = 0 as a solution.

Next, determine whether the function is increasing or decreasing on the intervals (–∞, 0), (0, 64), and (64, ∞) by taking a point in each interval and substituting it into the derivative. Using the test points x = –1, x = 1, and x = 125 gives you the following:

9781118496718-eq180234.png

Therefore, f (x) is decreasing on (–∞, 0), increasing on (0, 64), and decreasing on (64, ∞). That means f (x) has a local minimum at x = 0; f (0) = 0, so a local minimum occurs at (0, 0). Also, f (x) has a local maximum at x = 64; f (64) = 6(64)2/3 – 64 = 32, so the local maximum occurs at (64, 32).

464. local maximum at 9781118496718-eq180235.tif; local minimum at 9781118496718-eq180236.tif

Begin by finding the derivative of the function f (x) = 2 sin x – sin 2x:

9781118496718-eq180237.png

Next, find the critical numbers by setting the derivative equal to zero:

9781118496718-eq180238.png

Setting the first factor equal to zero gives you 9781118496718-eq180239.tif, which has the solutions 9781118496718-eq180240.tif and 9781118496718-eq180241.tif. Setting the second factor equal to zero gives you 9781118496718-eq180242.tif, which has the solutions x = 0 and 2π.

Next, determine whether the function is increasing or decreasing on the intervals 9781118496718-eq180243.tif9781118496718-eq180244.tif, and 9781118496718-eq180245.tif by taking a point inside each interval and substituting it into the derivative to see whether it's positive or negative. Using the points 9781118496718-eq180246.tif, π, and 9781118496718-eq180247.tif gives you

9781118496718-eq180248.png

9781118496718-eq180249.png

9781118496718-eq180250.png

So f (x) has a local maximum when 9781118496718-eq180251.tif because the function changes from increasing to decreasing at this value, and f (x) has a local minimum when 9781118496718-eq180252.tif because the function changes from decreasing to increasing at this value.

To find the points on the original function, substitute in these values. If 9781118496718-eq180253.tif, then

9781118496718-eq180254.png

So the local maximum occurs at 9781118496718-eq180255.tif. If 9781118496718-eq180256.tif, then

9781118496718-eq180257.png

Therefore, the local minimum occurs at 9781118496718-eq180258.tif.

465. local maxima at 9781118496718-eq180259.tif9781118496718-eq180260.tif; local minima at 9781118496718-eq180261.tif9781118496718-eq180262.tif

Begin by finding the derivative of the function f (x) = x + 2 cos x:

9781118496718-eq180263.png

Next, find the critical numbers by setting the function equal to zero and solving for x on the given interval:

9781118496718-eq180264.png

Then determine whether the function is increasing or decreasing on the intervals 9781118496718-eq180265.tif9781118496718-eq180266.tif9781118496718-eq180267.tif9781118496718-eq180268.tif, and 9781118496718-eq180269.tif by taking a point inside each interval and substituting it into the derivative. Using the value x = –6.27, which is slightly larger than –2π, gives you f '(–6.27) ≈ f '(–2π) = 1 > 0. Likewise, using the points 9781118496718-eq180270.tif, 0, 9781118496718-eq180271.tif, and 9781118496718-eq180272.tif from each of the remaining four intervals gives you the following:

9781118496718-eq180273.png

Therefore, the function has local maxima when 9781118496718-eq180274.tif and 9781118496718-eq180275.tif and local minima when 9781118496718-eq180276.tif and 9781118496718-eq180277.tif. To find the points on the original function, substitute these x values into the original function:

9781118496718-eq180278.png

9781118496718-eq180279.png

9781118496718-eq180280.png

9781118496718-eq180281.png

Therefore, the local maxima are 9781118496718-eq180282.tif and 9781118496718-eq180283.tif, and the local minima are 9781118496718-eq180284.tif and 9781118496718-eq180285.tif.

466. concave up on (1, ∞); concave down on (–∞, 1)

To determine concavity, examine the second derivative. Because a derivative measures a “rate of change” and the first derivative of a function gives the slopes of tangent lines, the derivative of the derivative (the second derivative) measures the rate of change of the slopes of the tangent lines. If the second derivative is positive on an interval, the slopes of the tangent lines are increasing, so the function is bending upward, or is concave up. Likewise, if the second derivative is negative on an interval, the slopes of the tangent lines are decreasing, so the function is bending downward, or is concave down.

Begin by finding the first and second derivatives of the function f (x) = x3 – 3x2 + 4:

9781118496718-eq180286.png

Setting the second derivative equal to zero gives you 6x – 6 = 0 so that x = 1.

To determine the concavity on the intervals (–∞, 1) and (1, ∞), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = 0 and x = 2, you have f "(0) = 6(0) – 6 > 0 and f "(2) = 6(2) – 6 > 0. Therefore, f (x) is concave up on the interval (1, ∞) and concave down on the interval (–∞, 1).

467. concave up nowhere; concave down on (–∞, 0), (0, ∞)

Begin by finding the first and second derivatives of the function f (x) = 9x2/3 – x:

9781118496718-eq180287.png

9781118496718-eq180288.png

Next, set the denominator of the second derivative equal to zero to get x4/3 = 0 so that x = 0.

To determine the concavity on the intervals (–∞, 0) and (0, ∞), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –1 and x = 1, you have the following:

9781118496718-eq180289.png

Therefore, the f (x) is concave down on the intervals (–∞, 0) and (0, ∞).

468. concave up on (–∞, 0), 9781118496718-eq180290.tif; concave down on 9781118496718-eq180291.tif

Begin by finding the first derivative of the function f (x) = x1/3(x + 1) = x4/3 + x1/3:

9781118496718-eq180292.png

Next, use the power rule to find the second derivative:

9781118496718-eq180293.png

Then find where the second derivative is equal to zero or undefined. Setting the numerator equal to zero gives you 2x – 1 = 0 so that 9781118496718-eq180294.tif, and setting the denominator equal to zero gives you 9x5/3 = 0 so that x = 0.

To determine the concavity on the intervals (–∞, 0), 9781118496718-eq180295.tif, and 9781118496718-eq180296.tif, pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –1, 9781118496718-eq180297.tif, and x = 1 gives you the following:

9781118496718-eq180298.png

Therefore, f (x) is concave up on the intervals (–∞, 0) and 9781118496718-eq180299.tif, and f (x) is concave down on the interval 9781118496718-eq180300.tif.

469. concave up on (–∞, –2), 9781118496718-eq180301.tif, (2, ∞); concave down on 9781118496718-eq180302.tif9781118496718-eq180303.tif

Begin by finding the first and second derivatives of the function f (x) = (x2 – 4)3:

9781118496718-eq180304.png

9781118496718-eq180305.png

Next, set each factor in the second derivative equal to zero and solve for x: x2 – 4 = 0 gives you the solutions x = 2 and x = –2, and 5x2 – 4 = 0, or 9781118496718-eq180306.tif, has the solutions 9781118496718-eq180307.tif and 9781118496718-eq180308.tif.

To determine the concavity on the intervals (–∞, –2), 9781118496718-eq180309.tif9781118496718-eq180310.tif9781118496718-eq180311.tif, and (2, ∞), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –3, x = –1, x = 0, x = 1, and x = 3, you have the following:

9781118496718-eq180312.png

Therefore, f (x) is concave up on the intervals (–∞, –2), 9781118496718-eq180313.tif, and (2, ∞), and f (x) is concave down on the intervals 9781118496718-eq180314.tif and 9781118496718-eq180315.tif.

470. concave up on 9781118496718-eq180316.tif9781118496718-eq180317.tif; concave down on the intervals (0, 0.253), 9781118496718-eq180318.tif9781118496718-eq180319.tif

Begin by finding the first and second derivatives of the function f (x) = 2 cos x – sin(2x):

9781118496718-eq180320.png

9781118496718-eq180321.png

Next, set each factor of the second derivative equal to zero and solve: cos x = 0 has the solutions 9781118496718-eq180322.tif and 9781118496718-eq180323.tif, and 4 sin x – 1 = 0, or 9781118496718-eq180324.tif, has the solutions 9781118496718-eq180325.tif and x = π – 0.253.

To determine the concavity on the intervals (0, 0.25), 9781118496718-eq180326.tif9781118496718-eq180327.tif9781118496718-eq180328.tif, and 9781118496718-eq180329.tif, pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = 0.1, 9781118496718-eq180330.tif9781118496718-eq180331.tifx = π, and 9781118496718-eq180332.tif gives you the following:

9781118496718-eq180333.png

Therefore, f (x) is concave up on the intervals 9781118496718-eq180334.tif and 9781118496718-eq180335.tif, and f (x) is concave down on the intervals (0, 0.253), 9781118496718-eq180336.tif, and 9781118496718-eq180337.tif.

471. no inflection points

Inflection points are points where the function changes concavity. To determine concavity, examine the second derivative. Because a derivative measures a “rate of change” and the first derivative of a function gives the slopes of tangent lines, the derivative of the derivative (the second derivative) measures the rate of change of the slopes of the tangent lines. If the second derivative is positive on an interval, the slopes of the tangent lines are increasing, so the function is bending upward, or is concave up. Likewise, if the second derivative is negative on an interval, the slopes of the tangent lines are decreasing, so the function is bending downward, or is concave down.

Begin by finding the first and second derivatives of the function 9781118496718-eq180338.tif. The first derivative is

9781118496718-eq180339.png

Now find the second derivative:

9781118496718-eq180340.png

Notice that the numerator of the second derivative never equals zero and that the denominator equals zero when x = ±3; however, x = ±3 isn't in the domain of the original function. Therefore, there can be no inflection points.

472. 9781118496718-eq180341.png

Begin by finding the first and second derivatives of the function f (x) = 2x3 + x2. The first derivative is

9781118496718-eq180342.png

And the second derivative is

9781118496718-eq180343.png

Setting the second derivative equal to zero gives you 12x + 2 = 0, which has the solution 9781118496718-eq180344.tif.

To determine the concavity on the intervals 9781118496718-eq180345.tif and 9781118496718-eq180346.tif, pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –1 and x = 0 gives you the following:

9781118496718-eq180347.png

Therefore, the function is concave down on the interval 9781118496718-eq180348.tif and concave up on the interval 9781118496718-eq180349.tif, so 9781118496718-eq180350.tif is an inflection point. The corresponding y value on f (x) is

9781118496718-eq180351.png

Therefore, the inflection point is 9781118496718-eq180352.tif.

473. no inflection points

Being by finding the first and second derivatives of the function 9781118496718-eq180353.tif. The first derivative is

9781118496718-eq180354.png

And the second derivative is

9781118496718-eq180355.png

Next, find where the second derivative is equal to zero or undefined. Setting the numerator equal to zero gives you sin x = 0, which has solutions x = 0, x = π, and x = 2π. Setting the denominator equal to zero gives you 1 + cos x = 0, or cos x = –1, which has the solution x = π.

To determine the concavity on the intervals (0, π) and (π, 2π), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values 9781118496718-eq180356.tif and 9781118496718-eq180357.tif gives you the following:

9781118496718-eq180358.png

However, note that at x = π, the original function is undefined, so there are no inflection points. In fact, if you noticed this at the beginning, there's really no need to determine the concavity on the intervals!

474. (π, 0)

Begin by finding the first and second derivatives of the function f (x) = 3 sin x – sinx. The first derivative is

9781118496718-eq180359.png

And the second derivative is

9781118496718-eq180360.png

Next, find where the second derivative is equal to zero by setting each factor equal to zero: cos x has the solutions 9781118496718-eq180361.tif and 9781118496718-eq180362.tif, and sin x = 0 has the solutions x = 0, x = π, and x = 2π.

To determine the concavity on the intervals 9781118496718-eq180363.tif9781118496718-eq180364.tif9781118496718-eq180365.tif, and 9781118496718-eq180366.tif, pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values 9781118496718-eq180367.tif9781118496718-eq180368.tif9781118496718-eq180369.tif, and 9781118496718-eq180370.tif, you have the following:

9781118496718-eq180371.png

Therefore, the concavity changes when x = π. The corresponding y value on f (x) is f (π) = 3 sin π – (sin π)3 = 0, so the inflection point is (π, 0).

475. (–1, –6)

Begin by finding the first and second derivatives of the function f (x) = x5/3 – 5x2/3. The first derivative is

9781118496718-eq180372.png

And the second derivative is

9781118496718-eq180373.png

Next, find where the second derivative is equal to zero or undefined by setting the numerator and the denominator equal to zero. For the numerator, you have 10x + 10 = 0, which has the solution x = –1. For the denominator, you have 9x4/3 = 0, which has the solution x = 0.

To determine the concavity on the intervals (–∞, –1), (1, 0), and (0, ∞), pick a point from each interval and substitute it into the second derivative to see whether it's positive or negative. Using the values x = –2, 9781118496718-eq180374.tif, and x = 1 gives you

9781118496718-eq180375.png

Because the concavity changes at x = –1, the original function has an inflection point there. The corresponding y value on f (x) is

9781118496718-eq180376.png

Therefore, the inflection point is (–1, –6).

476. no local maxima; local minimum at (0, 1)

Begin by finding the first derivative of the function 9781118496718-eq180377.tif:

9781118496718-eq180378.png

Next, find any critical numbers of the function. Setting the numerator of the derivative equal to zero gives you the solution x = 0. Note that no real values can make the denominator equal to zero.

Then find the second derivative:

9781118496718-eq180379.png

To see whether the original function is concave up or concave down at the critical number, substitute x = 0 into the second derivative:

9781118496718-eq180380.png

The second derivative is positive, so the original function is concave up at the critical point; therefore, a local minimum is at x = 0. The corresponding y value on the original function is 9781118496718-eq180381.tif, so the local minimum is at (0, 1).

477. local maximum at (0, 1); local minima at 9781118496718-eq180382.tif9781118496718-eq180383.tif

Begin by finding the first derivative of the function f (x) = x4 – 4x2 + 1:

9781118496718-eq180384.png

Set the first derivative equal to zero and solve for x to find the critical numbers of f. The critical numbers are x = 0, 9781118496718-eq180385.tif, and 9781118496718-eq180386.tif.

Next, find the second derivative:

9781118496718-eq180387.png

Substitute each of the critical numbers into the second derivative to see whether the second derivative is positive or negative at those values:

9781118496718-eq180388.png

Therefore, the original function has a local maximum when x = 0 and local minima when 9781118496718-eq180389.tif and 9781118496718-eq180390.tif. The corresponding y values on the original function are 9781118496718-eq180391.tiff (0) = 04 – 4(0)2 + 1 = 1, and 9781118496718-eq180392.tif. Therefore, the local maximum occurs at (0, 1), and the local minima occur at 9781118496718-eq180393.tif and 9781118496718-eq180394.tif.

478. local maxima at 9781118496718-eq180395.tif9781118496718-eq180396.tif; local minimum at (0, 0)

Begin by finding the first derivative of the function f (x) = 2x2(1 – x2) = 2x2 – 2x4:

9781118496718-eq180397.png

Set the first derivative equal to zero and solve for x to find the critical numbers. Solving 4x(1 – 2x2) = 0 gives you x = 0, 9781118496718-eq180398.tif, and 9781118496718-eq180399.tif as the critical numbers.

Next, find the second derivative:

9781118496718-eq180400.png

Substitute the critical numbers into the second derivative to see whether the second derivative is positive or negative at those values:

9781118496718-eq180401.png

So the original function has local maxima when 9781118496718-eq180402.tif and 9781118496718-eq180403.tif and a local minimum when x = 0. Finding the corresponding y values on the original function gives you the following:

9781118496718-eq180404.png

Therefore, the local maxima are at 9781118496718-eq180405.tif and 9781118496718-eq180406.tif, and the local minimum is at (0, 0).

479. local maximum at 9781118496718-eq180407.tif; local minimum at 9781118496718-eq180408.tif

Begin by finding the first derivative of the function 9781118496718-eq180409.tif:

9781118496718-eq180410.png

Then set the numerator and denominator of the first derivative equal to zero to find the critical numbers. Setting the numerator equal to zero gives you –x2 + 4 = 0, which has the solutions x = 2 and x = –2. Notice that the denominator of the derivative doesn't equal zero for any value of x.

Next, find the second derivative of the function:

9781118496718-eq180411.png

Substitute the critical numbers into the second derivative to see whether the second derivative is positive or negative at those values:

9781118496718-eq180412.png

Because f (x) is concave up when x = –2 and concave down when x = 2, a local minimum occurs at x = –2, and a local maximum occurs when x = 2. Because 9781118496718-eq180413.tif and 9781118496718-eq180414.tif, the local minimum occurs at 9781118496718-eq180415.tif, and the local maximum occurs at 9781118496718-eq180416.tif.

480. local maximum at 9781118496718-eq180417.tif; local minimum at 9781118496718-eq180418.tif

Begin by finding the first derivative of the function f (x) = 2 sin x – x:

9781118496718-eq180419.png

Set the first derivative equal to zero to find the critical numbers. From the equation 2 cos x – 1 = 0, you get 9781118496718-eq180420.tif, so the critical numbers are 9781118496718-eq180421.tif and 9781118496718-eq180422.tif.

Next, find the second derivative:

9781118496718-eq180423.png

Substitute the critical numbers into the second derivative to see whether the second derivative is positive or negative at these values:

9781118496718-eq180424.png

Because the function is concave down at 9781118496718-eq180425.tif, it has a local maximum there, and because the function is concave up at 9781118496718-eq180426.tif, it has a local minimum there. To find the corresponding y value on f (x), substitute the critical numbers into the original function:

9781118496718-eq180427.png

9781118496718-eq180428.png

Therefore, the local maximum is at 9781118496718-eq180429.tif, and the local minimum is at 9781118496718-eq180430.tif.

481. 9781118496718-eq180431.png

Recall Rolle's theorem: If f is a function that satisfies the following three hypotheses:

·        f is continuous on the closed interval [a, b]

·        f is differentiable on the open interval (a, b)

·        f (a) = f (b)

then there is a number c in (a, b) such that f '(c= 0.

Notice that the given function is differentiable everywhere (and therefore continuous everywhere) because it's a polynomial. Then verify that f (0) = f (6) so that Rolle's theorem can be applied:

9781118496718-eq180432.png

Next, find the derivative of the function, set it equal to zero, and solve for c:

9781118496718-eq180433.png

9781118496718-un1811.tif

482. 9781118496718-eq180434.png

Notice that the given function is differentiable on (–8, 0). Then verify that f (–8) = f (0) so that Rolle's theorem can be applied:

9781118496718-eq180435.png

Next, find the derivative of the function, set it equal to zero, and solve. Here's the derivative:

9781118496718-eq180436.png

Setting the numerator equal to zero and solving for c gives you

9781118496718-eq180437.png

483. 0, 9781118496718-eq180438.tif, ±1

Notice that the given function is differentiable everywhere. Then verify that f (–1) = f (1) so that Rolle's theorem can be applied:

9781118496718-eq180439.png

Next, find the derivative of the function, set it equal to zero, and solve for c. Here's the derivative:

9781118496718-eq180440.png

You need to find the solutions of sin(2πc) = 0 on the interval –1 ≤ c ≤ 1 so that –2π ≤ 2πc ≤ 2π. Notice that the solutions occur when 2πc = –2π, –π, 0, π, and 2π. Solving each of these equations for c gives you the solutions c = –1, 9781118496718-eq180441.tifc = 0, 9781118496718-eq180442.tif, and c = 1.

484. 9781118496718-eq180443.png

Recall the mean value theorem: If f is a function that satisfies the following hypotheses:

·        f is continuous on the closed interval [a, b]

·        f is differentiable on the open interval (a, b)

then there is a number c in (a, b) such that 9781118496718-eq180444.tif.

Notice that the given function is differentiable everywhere, so you can apply the mean value theorem. Next, find the value of 9781118496718-eq180445.tif on the given interval:

9781118496718-eq180446.png

Now find the derivative of f (x) = x3 + 3x – 1, the given function, and set it equal to 7. The derivative is f ' = 3x2 + 3, so

9781118496718-eq180447.png

The negative root falls outside the given interval, so keeping only the positive root gives you the solution 9781118496718-eq180448.tif.

9781118496718-un1812.tif

485. 9781118496718-eq180449.png

Notice that the given function is differentiable on (0, 1) and continuous on [0, 1], so you can apply the mean value theorem. Next, find the value of 9781118496718-eq180451.tif on the given interval:

9781118496718-eq180452.png

Now find the derivative of 9781118496718-eq180453.tif, the given function, and set it equal to 2. The derivative is 9781118496718-eq180454.tif, so you have

9781118496718-eq180455.png

486. 9781118496718-eq180456.png

Notice that the given function is differentiable everywhere, so you can apply the mean value theorem. Next, find the value of 9781118496718-eq180458.tif on the given interval:

9781118496718-eq180459.png

Now find the derivative of 9781118496718-eq180460.tif, the given function, and set it equal to 9781118496718-eq180461.tif. The derivative is 9781118496718-eq180462.tif, so you have

9781118496718-eq180463.png

Keeping only the positive root, you have 9781118496718-eq180464.tif, which has the solution 9781118496718-eq180465.tif. Note that the negative root would give you 9781118496718-eq180466.tif, which is outside of the interval.

487. 24

By the mean value theorem, you have the following equation for some c on the interval [1, 5]:

9781118496718-eq180468.png

Next, solve for f (5) using the fact that f (1) = 12 and f '(x) ≥ 3 (and so f '(c) ≥ 3):

9781118496718-eq180469.png

488. 8 ≤ f (8) – f (4) ≤ 24

If 2 ≤ f '(x) ≤ 6, then by the mean value theorem, you have the following equation for some c in the interval [4, 8]:

9781118496718-eq180471.png

Using 2 ≤ f '(x) ≤ 6, you can bound the value of 4f '(c) because 2 ≤ f '(c) ≤ 6:

9781118496718-eq180472.png

Because f (8) – f (4) = 4f '(c), it follows that

9781118496718-eq180473.png

489. 9781118496718-eq180474.png

The given function 9781118496718-eq180476.tif is continuous and differentiable on the given interval. By the mean value theorem, there exists a number c in the interval [8, 9] such that

9781118496718-eq180477.png

But because 9781118496718-eq180478.tif, you have 9781118496718-eq180479.tif. Replacing 9781118496718-eq180480.tif in the equation 9781118496718-eq180481.tif gives you

9781118496718-eq180482.png

Because 9781118496718-eq180483.tif, you have 9781118496718-eq180484.tif, so

9781118496718-eq180485.png

Notice also that f ' is decreasing (f " will be negative on the given interval), so the following is true:

9781118496718-eq180486.png

Because 9781118496718-eq180487.tif, you have 9781118496718-eq180488.tif. Therefore, it follows that

9781118496718-eq180489.png

490. velocity: 2; acceleration: 2

Begin by taking the derivative of the position function s(t) to find the velocity function:

9781118496718-eq180490.png

Substituting in the value of t = 5, find the velocity:

9781118496718-eq180491.png

Next, take the derivative of the velocity function to find the acceleration function:

9781118496718-eq180492.png

Because the acceleration is a constant at t = 5, the acceleration is a(5) = 2.

491. velocity: 1; acceleration: –2

Begin by taking the derivative of the position function s(t) to find the velocity function:

9781118496718-eq180493.png

Substituting in the value 9781118496718-eq180494.tif gives you the velocity:

9781118496718-eq180495.png

Next, take the derivative of the velocity function to find the acceleration function:

9781118496718-eq180496.png

Substituting in the value of 9781118496718-eq180497.tif gives you the acceleration:

9781118496718-eq180498.png

492. velocity: 0; acceleration: –1

Begin by taking the derivative of the position function s(t) to find the velocity function:

9781118496718-eq180499.png

Substituting in the value t = 1 gives you the velocity:

9781118496718-eq180500.png

Next, take the derivative of the velocity function to find the acceleration function:

9781118496718-eq180501.png

Substituting in the value of t = 1 gives you the acceleration:

9781118496718-eq180502.png

493. compressing; speed: 8 ft/s

Begin by taking the derivative of the equation of motion to find the velocity function:

9781118496718-eq180503.png

At 9781118496718-eq180504.tif, you have

9781118496718-eq180505.png

The velocity is negative, so the spring is compressing. Because the speed is the absolute value of the velocity, the speed is 8 feet per second.

494. maximum height: 25 ft; upward velocity: 9781118496718-eq180506.tif ft/s; downward velocity: 9781118496718-eq180507.tif ft/s

One way to find the maximum height of the stone is to determine when the velocity of the stone is zero, because that's when the stone stops going upward and begins falling. (The stone follows a parabolic path, so you can also use the formula to find the vertex, but try the calculus way instead.)

Finding the derivative of the position function gives you the velocity function:

9781118496718-eq180508.png

Set the velocity function equal to zero and solve for time t:

9781118496718-eq180509.png

Substituting this time value into the position equation gives you

9781118496718-eq180510.png

Therefore, the maximum height of the stone is 25 feet.

To answer the second part of the question, use algebra to find at what points in time the stone is at a height of 20 feet:

9781118496718-eq180511.png

Using the quadratic formula with a = 4, b = –10, and c = 5, you get the solutions 9781118496718-eq180512.tif and 9781118496718-eq180513.tif. Substituting these values into the derivative gives you the velocity at these times:

9781118496718-eq180514.png

9781118496718-eq180515.png

9781118496718-un1813.tif

The velocities are approximately equal to 17.89 feet per second and –17.89 feet per second. The signs on the answers reflect that the stone is moving up at the first time and down at the second time.

495. maximum height: 6.25 ft; velocity on impact: –20 ft/s

One way to find the height of the stone is to determine when the velocity of the stone is zero, because that's when the stone stops going upward and begins falling. (The stone follows a parabolic path, so you can also use the formula to find the vertex, but try the calculus way instead.)

Finding the derivative of the position function gives you the velocity function:

9781118496718-eq180516.png

Set the velocity function equal to zero and solve for time t:

9781118496718-eq180517.png

Substitute in this value to find the height:

9781118496718-eq180518.png

To find the velocity of the stone when it hits the ground, first determine when the stone hits the ground by setting the position equation equal to zero and solving for t:

9781118496718-eq180519.png

Because t = 0 corresponds to when the stone is first released, substitute 9781118496718-eq180520.tif into the velocity function to get the final velocity:

9781118496718-eq180521.png

496. downward on 9781118496718-eq180522.tif; upward on 9781118496718-eq180523.tif

Note that because the velocity is the rate of change in position with respect to time, you want to find when the velocity is positive and when the velocity is negative. Assume that when the velocity is positive, the particle is moving upward, and when the velocity is negative, the particle is moving downward.

Begin by taking the derivative of the position function to find the velocity function:

9781118496718-eq180524.png

Set the velocity function equal to zero and solve for time t:

9781118496718-eq180525.png

Use only the positive solution based on the given interval for t.

Now take a point from the interval 9781118496718-eq180526.tif and a point from 9781118496718-eq180527.tif and substitute them into the velocity function to see whether the velocity is positive or negative on those intervals. Using t = 1, you have v(1) = 3(1)2 – 4 > 0, and using t = 10, you have v(10) = 3(10)2 – 4 > 0; therefore, the particle is moving downward on the interval 9781118496718-eq180528.tif and upward on the interval 9781118496718-eq180529.tif.

497. downward on 9781118496718-eq180530.tif; upward on 9781118496718-eq180531.tif

Note that because the velocity is the rate of change in position with respect to time, you want to find when the velocity is positive and when the velocity is negative. Assume that when the velocity is positive, the particle is moving upward, and when the velocity is negative, the particle is moving downward.

Begin by taking the derivative of the position function to find the velocity function:

9781118496718-eq180532.png

Then find when the velocity is equal to zero:

9781118496718-eq180533.png

Now take a point from the interval 9781118496718-eq180534.tif and a point from 9781118496718-eq180535.tif and substitute them into the velocity function to see whether the velocity is positive or negative on those intervals. If you use 9781118496718-eq180536.tif, you have 9781118496718-eq180537.tif, and if t = 1, you have v(1) = 8(1) – 6 > 0; therefore, the particle is moving downward on the interval 9781118496718-eq180538.tifand upward on the interval 9781118496718-eq180539.tif.

498. –25, 25

Let x and y be the two numbers so that x – y = 50. The function that you want to minimize, the product, is given by

9781118496718-eq180540.png

Solve for x so you can write the product in terms of one variable:

9781118496718-eq180541.png

Then substitute the value of x into P:

9781118496718-eq180542.png

Next, find the derivative, set it equal to zero, and solve for y:

9781118496718-eq180543.png

Using y = –25 and x = 50 + y gives you x = 25, so the product is

9781118496718-eq180544.png

You can verify that y = –25 gives you a minimum by using the first derivative test.

499. 20, 20

Let x and y be the two positive numbers so that xy = 400. The function that you want to minimize, the sum, is given by

9781118496718-eq180545.png

You can write the product in terms of one variable:

9781118496718-eq180546.png

Then substitute the value of y into the sum function:

9781118496718-eq180547.png

Next, find the derivative of the function:

9781118496718-eq180548.png

Setting the derivative equal to zero gives you x2 – 400 = 0, and keeping only the positive solution, you have x = 20. You can verify that this value gives you a minimum by using the first derivative test.

Using x = 20 and xy = 400 gives you y = 20.

500. 15 m × 15 m

If you let x and y be the side lengths of the rectangle, you have the following equation for the perimeter:

9781118496718-eq180549.png

The function that you want to maximize, the area, is given by

9781118496718-eq180550.png

You can write this as a function of one variable by using 2x + 2y = 60 to get y = 30 – x. Substituting the value of y into the area equation gives you

9781118496718-eq180551.png

Take the derivative and set it equal to zero to find x:

9781118496718-eq180552.png

You can verify that this gives you a maximum by using the first derivative test.

Using 2x + 2y = 60 and x = 15 meters, you get y = 15 meters.

501. 56,250 ft2

Let x and y be the lengths of the sides of the pen as in the following diagram:

9781118496718-un1814.tif

From the diagram, you have the following perimeter formula:

9781118496718-eq180553.png

The function you want to maximize, the area, is given by

9781118496718-eq180554.png

You can write this as a function of one variable by using the equation 5x + 2y = 1,500 to get 9781118496718-eq180555.tif. Substituting the value of y into the area equation gives you

9781118496718-eq180556.png

Take the derivative of this function, set it equal to zero, and solve for x:

9781118496718-eq180557.png

You can verify that this value gives you a maximum by using the first derivative test. Using 5x + 2y = 1,500 and x = 150 gives you y = 375; therefore, the area is

9781118496718-eq180558.png

502. 16 ft3

9781118496718-un1815.tif

If a square with side x is removed from each corner of the piece of cardboard, you have the following volume, where 0 > x > 3:

9781118496718-eq180559.png

By distributing, you have

9781118496718-eq180560.png

You want the maximum volume, so take the derivative of this function:

9781118496718-eq180561.png

Then set the derivative equal to zero and solve for x:

9781118496718-eq180562.png

Because x = 3 makes the volume zero, you know that x = 1 gives you a maximum,which you can verify by using the first derivative test.

Using V = (6 – 2x)(6 – 2x)x and x = 1 gives you the volume:

9781118496718-eq180563.png

503. 9781118496718-eq180564.tif cm × 9781118496718-eq180565.tif cm × 9781118496718-eq180566.tif cm

9781118496718-un1816.tif

If you let x be a length of one side of the base and let y be the height of the box,the volume is

9781118496718-eq180567.png

The square base has an area of x2, and the box has four sides, each with an areaof xy, so the surface area is given by

9781118496718-eq180568.png

You can write this as a function of one variable by using x2y = 16,000 to get 9781118496718-eq180569.tif. Substituting the value of y into the surface area formula gives you

9781118496718-eq180570.png

You want the minimum surface area, so take the derivative of this function:

9781118496718-eq180571.png

Then set the derivative equal to zero and solve for x:

9781118496718-eq180572.png

Use 9781118496718-eq180573.tif to find y:

9781118496718-eq180574.png

Therefore, the dimensions are 9781118496718-eq180575.tif cm × 9781118496718-eq180576.tif cm × 9781118496718-eq180577.tif cm (approximately 31.75 cm × 31.75 cm × 15.87 cm).

Note that you can easily use the second derivative test to verify that 9781118496718-eq180578.tif gives you a minimum because S" > 0 for all x > 0.

504. 9781118496718-eq180579.tif9781118496718-eq180580.tif

You want to find the maximum distance. The distance from a point (x, y) to the point (1, 0) is given by

9781118496718-eq180581.png

Rewrite the equation of the ellipse as y2 = 8 – 8x2 and substitute the value of y2 into the distance equation:

9781118496718-eq180582.png

Tip: You can take the derivative of this function and use the first derivative test to find a maximum, but it's easier to use the square of the distance, which gets rid of the radical. For a function that satisfies f (x) ≥ 0, its local maxima and minima occur at the same x values as the local maxima and minima of its square, [f (x)]2. Obviously, the corresponding y values would change, but that doesn't matter here!

The square of the distance is

9781118496718-eq180583.png

And the derivative of this function is

9781118496718-eq180584.png

Setting the derivative equal to zero and solving gives you –2 – 14x = 0, or –2 = 14x, which has the solution 9781118496718-eq180585.tif. You can verify that 9781118496718-eq180586.tif gives you a maximum by using the first derivative test.

Using 9781118496718-eq180587.tif and y2 = 8 – 8x2, find the y coordinate:

9781118496718-eq180588.png

Therefore, the two points that are farthest from (1, 0) are the points 9781118496718-eq180589.tif and 9781118496718-eq180590.tif.

505. 9781118496718-eq180591.png

You want to find the minimum distance. The distance from a point (x, y) to the origin is

9781118496718-eq180592.png

Using y = 4x + 6, the distance is

9781118496718-eq180593.png

Tip: You can take the derivative of this function and use the first derivative test to find a minimum, but it's easier to use the square of the distance, which gets rid of the radical. For a function that satisfies f (x) ≥ 0, its local maxima and minima occur at the same x values as the local maxima and minima of its square, [f (x)]2. Obviously, the corresponding y values would change, but that doesn't matter here!

Using the square of the distance, you have

9781118496718-eq180594.png

The derivative is

9781118496718-eq180595.png

Setting the derivative equal to zero gives you 34x + 48 = 0, which has the solution 9781118496718-eq180596.tif. You can verify that this value gives you a minimum by using the first derivative test.

Using 9781118496718-eq180597.tif and y = 4x + 6, find the y coordinate:

9781118496718-eq180598.png

Therefore, the point closest to the origin is 9781118496718-eq180599.tif.

506. 9781118496718-eq180600.tif in. × 9781118496718-eq180601.tif in.

9781118496718-un1817.tif

You want to maximize the printed area. Let x and y be the length and width of the poster. Because the total area must be 90 square inches, you have A = xy = 90.

The poster has a 1-inch margin at the bottom and sides and a 3-inch margin at the top, so the printed area of the poster is

9781118496718-eq180602.png

Using 9781118496718-eq180603.tif, the printed area is

9781118496718-eq180604.png

Find the derivative of the function for the printed area:

9781118496718-eq180605.png

Set the derivative equal to zero and solve for x: –4x2 + 180 = 0 so that x2 = 45. Keeping the positive solution, you have 9781118496718-eq180606.tif.

Using the first derivative test, you can verify that 9781118496718-eq180607.tif gives you a maximum. From the equation 9781118496718-eq180608.tif, you get the y value:

9781118496718-eq180609.png

Therefore, the dimensions are 9781118496718-eq180610.tif inches × 9781118496718-eq180611.tif inches (approximately 6.7 inches ×13.4 inches).

507. 9781118496718-eq180612.tif and 9781118496718-eq180613.tif

You want to locate the maximum slope of f (x) = 2 + 20x3 – 4x5. The slope of the tangent line is given by the derivative

9781118496718-eq180614.png

At a point p, the slope is

9781118496718-eq180615.png

You want to maximize this function, so take another derivative:

9781118496718-eq180616.png

To find the critical numbers, set this function equal to zero and solve for p:

9781118496718-eq180617.png

Take a point from each interval and test it in s'(p) to see whether the answer is positive or negative. By taking a point from the interval 9781118496718-eq180618.tif, you have s'(p) > 0; by using a point from 9781118496718-eq180619.tif, you have s'(p) > 0; in 9781118496718-eq180620.tif, you have s'(p) > 0; and in 9781118496718-eq180621.tif, you haves'(p) > 0. Because s(p) approaches –∞ as x approaches ±∞, the maximum value must occur at one (or both) of 9781118496718-eq180622.tif.

Substituting these values into the slope equation gives you

9781118496718-eq180623.png

Therefore, the maximum slope occurs when 9781118496718-eq180624.tif and when 9781118496718-eq180625.tif.

508. $396.23

If you let x be the length of the base and let y be the height, the volume is

9781118496718-eq180626.png

The area of the base is 2x2, and the box has four sides, each with an area of xy. With the base material at $20 per square meter and the side material at $12 per square meter, the total cost is

9781118496718-eq180627.png

You can write this as a function of one variable by using the volume equation to get 9781118496718-eq180628.tif. The cost becomes

9781118496718-eq180629.png

You want the minimum cost, so find the derivative of this function:

9781118496718-eq180630.png

Setting this equal to zero, you get 80x3 – 480 = 0, which has the solution 9781118496718-eq180631.tif meters. Substituting this value into the cost function C(x) = 40x2 + 480x−1 gives you the minimum cost:

9781118496718-eq180632.png

509. 20 m

9781118496718-un1818.tif

You want to maximize the total area. If x is the length of wire used for the square, each side of the square has a length of 9781118496718-eq180633.tif meters, so the area of the square is 9781118496718-eq180634.tif. You'll have (20 – x) meters of wire left for the triangle, so each side of the triangle is 9781118496718-eq180635.tif meters. Because the triangle is equilateral, its height is 9781118496718-eq180636.tif meters, so the area of the triangle is 9781118496718-eq180637.tif. Therefore, the total area of the square and the triangle together is

9781118496718-eq180638.png

where 0 ≤ x ≤ 20.

Find the derivative of this function:

9781118496718-eq180639.png

Then set the derivative equal to zero and solve for x:

9781118496718-eq180640.png

It's reasonable to let x = 0 (the wire is used entirely to make the triangle) or x = 20 (the wire is used entirely to make the square), so check those values as well. Here are the areas:

9781118496718-eq180641.png

The maximum occurs when x = 20.

510. 9781118496718-eq180642.png

9781118496718-un1819.tif

You want to minimize the total area. If x is the length of wire used for the square, each side of the square has a length of 9781118496718-eq180643.tif meters, so the area of the square is 9781118496718-eq180644.tif. You'll have (20 – x) meters of wire left for the triangle, so each side of the triangle is 9781118496718-eq180645.tif meters. Because the triangle is equilateral, its height is 9781118496718-eq180646.tif meters, so the area of the triangle is 9781118496718-eq180647.tif. Therefore, the total area of the square and the triangle together is

9781118496718-eq180648.png

where 0 ≤ x ≤ 20.

Find the derivative of this function:

9781118496718-eq180649.png

Set the derivative equal to zero and solve for x:

9781118496718-eq180650.png

It's reasonable to let x = 0 (the wire is used entirely to make the triangle) or x = 20 (the wire is used entirely to make the square), so check those values as well. Here are the areas:

9781118496718-eq180651.png

Therefore, the minimum occurs when

9781118496718-eq180652.png

511. 9781118496718-eq180653.tif feet from the bright light source

9781118496718-un1820.tif

Let x be the distance of the object from the brighter light source and k be the strength of the weaker light source. The illumination of each light source is directly proportional to the strength of the light source (5k and k) and inversely proportional to the square of the distance from the source (x and 20 – x), so the total illumination is

9781118496718-eq180654.png

where 0 > x > 20.

Taking the derivative of this function gives you

9781118496718-eq180655.png

Set this derivative equal to zero and simplify:

9781118496718-eq180656.png

Then solve for x. Taking the cube root of both sides and solving gives you

9781118496718-eq180657.png

Note that you can use the first derivative test to verify that this value does in fact give a minimum.

512. 12

9781118496718-un1821.tif

You want to maximize the area of the rectangle. From the diagram, the area of the rectangle is

9781118496718-eq180658.png

Using the equation of the ellipse, solve for x:

9781118496718-eq180659.png

Keeping only the positive solution, you have 9781118496718-eq180660.tif (although you can just as easily work with the negative solution). Therefore, the area in terms of y becomes

9781118496718-eq180661.png

Take the derivative by using the product rule and the chain rule:

9781118496718-eq180662.png

Then set the derivative equal to zero and solve for y:

9781118496718-eq180663.png

Keeping the positive solution (either works), you have 9781118496718-eq180664.tif, which gives you the following x value:

9781118496718-eq180665.png

Therefore, the area of the rectangle is

9781118496718-eq180666.png

Note that you can verify that the value 9781118496718-eq180667.tif does indeed give a maximum by using the first derivative test.

513. 0.84771

Use the formula 9781118496718-eq180668.tif with f (x) = x3 + 4x – 4, f '(x) = 3x2 + 4, and x1 = 1. Note that the formula gives you 9781118496718-eq180669.tif9781118496718-eq180670.tif, and so on. Therefore, you have

9781118496718-eq180671.png

514. 2.0597671

Use the formula 9781118496718-eq180672.tif with f (x) = x4 – 18, f '(x) = 4x3, and x1 = 2. Note that the formula gives you 9781118496718-eq180673.tif9781118496718-eq180674.tif, and so on. Therefore, you have

9781118496718-eq180675.png

515. –1.2457342

Use the formula 9781118496718-eq180676.tif with f (x) = x5 + 3, f '(x) = 5x4, and x1 = –1. Note that the formula gives you 9781118496718-eq180677.tif9781118496718-eq180678.tif, and so on. Therefore, you have

9781118496718-eq180679.png

516. 0.73909

Newton's method begins by making one side of the equation zero, labeling the other side f (x), and then picking a value of x1 that's “close” to a root of f (x). There's certainly a bit of trial and error involved; in this case, you could graph both y = cos x and y = x and look for a point of intersection to get a rough idea of what the root may be.

Use the formula 9781118496718-eq180680.tif with f (x) = cos x – x and f '(x) = –sin x – 1. Note that the formula gives you 9781118496718-eq180681.tif9781118496718-eq180682.tif, and so on. You also have to decide on a value for x1. Notice that f (0) = 1 – 0 = 1, which is close to the desired value of 0, so you can start with the value x1 = 0. Therefore, you have

9781118496718-eq180683.png

The approximation is no longer changing, so the solution is 0.73909.

517. 1.69562

Newton's method begins by making one side of the equation zero, labeling the other side f (x), and then picking a value of x1 that's “close” to a root of f (x). In general, a bit of trial and error is involved, but in this case, you're given a closed interval to work with, which considerably narrows down the choices!

Use the formula 9781118496718-eq180684.tif with f (x) = x3 – x2 – 2 and f '(x) = 3x2– 2x. Note that the formula gives you 9781118496718-eq180685.tif9781118496718-eq180686.tif, and so on. You can choose any value in the interval [1, 2] for x1, but x1 = 1 makes the computations a bit easier for the first step, so use that value:

9781118496718-eq180687.png

The approximation is no longer changing, so the solution is 1.69562.

518. 1.22074

Newton's method begins by making one side of the equation zero, labeling the other side f (x), and then picking a value of x1 that's “close” to a root of f (x). There's certainly a bit of trial and error involved; in this case, you could graph both 9781118496718-eq180688.tif and y = x2 and look for a point of intersection to get a rough idea of what the root may be.

Use the formula 9781118496718-eq180689.tif with 9781118496718-eq180690.tif and 9781118496718-eq180691.tif. Note that the formula gives you 9781118496718-eq180692.tif9781118496718-eq180693.tif, and so on. Notice that 9781118496718-eq180694.tif, which is close to the desired value of 0, so you can start withx1 = 1:

9781118496718-eq180695.png

The approximation is no longer changing, so the solution is 1.22074.

519. 9781118496718-eq180696.png

9781118496718-un1822.tif

Because the given function has values that are strictly greater than or equal to zero on the given interval, you can interpret the Riemann sum as approximating the area that's underneath the curve and bounded below by the x-axis.

You want to use four rectangles of equal width to estimate the area under f (x) = 2 + x2 over the interval 0 ≤ x ≤ 2. Begin by dividing the length of the interval by 4 to find the width of each rectangle:

9781118496718-eq180697.png

Then divide the interval [0, 2] into 4 equal pieces, each with a width of 9781118496718-eq180698.tif, to get the intervals 9781118496718-eq180699.tif9781118496718-eq180700.tif9781118496718-eq180701.tif, and 9781118496718-eq180702.tif. Using the left endpoint of each interval to calculate the heights of the rectangles gives you the values x = 0, 9781118496718-eq180703.tifx = 1, and 9781118496718-eq180704.tif. The height of each rectangle is f (x). To approximate the area under the curve, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180705.png

520. 10.43

Because the given function has values that are strictly greater than or equal to zero on the given interval, you can interpret the Riemann sum as approximating the area that's underneath the curve and bounded below by the x-axis.

You want to use five rectangles of equal width to estimate the area under 9781118496718-eq180706.tif over the interval 1 ≤ x ≤ 4. Begin by dividing the length of the interval by 5 to find the width of each rectangle:

9781118496718-eq180707.png

Then divide the interval [1, 4] into 5 equal pieces, each with a width of 9781118496718-eq180708.tif, to get the intervals 9781118496718-eq180709.tif9781118496718-eq180710.tif9781118496718-eq180711.tif9781118496718-eq180712.tif, and 9781118496718-eq180713.tif. Using the left endpoint of each interval to calculate the heights of the rectangles gives you the values x = 1, 9781118496718-eq180714.tif9781118496718-eq180715.tif9781118496718-eq180716.tif, and 9781118496718-eq180717.tif. The height of each rectangle is f (x). To approximate the area under the curve, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180718.png

521. 22.66

Because the given function has values that are strictly greater than or equal to zero on the given interval, you can interpret the Riemann sum as approximating the area that's underneath the curve and bounded below by the x-axis.

You want to use seven rectangles of equal width to estimate the area under f (x) = 4 ln x + 2x over the interval 1 ≤ x ≤ 4. Begin by dividing the length of the interval by 7 to find the width of each rectangle:

9781118496718-eq180719.png

Then divide the interval [1, 4] into 7 equal pieces, each with a width of 9781118496718-eq180720.tif, to get the intervals 9781118496718-eq180721.tif9781118496718-eq180722.tif9781118496718-eq180723.tif9781118496718-eq180724.tif9781118496718-eq180725.tif9781118496718-eq180726.tif, and 9781118496718-eq180727.tif. Using the left endpoint of each interval to calculate the heights of the rectangles gives you the values x = 1, 9781118496718-eq180728.tif9781118496718-eq180729.tif9781118496718-eq180730.tif9781118496718-eq180731.tif9781118496718-eq180732.tif, and 9781118496718-eq180733.tif. The height of each rectangle is f (x). To approximate the area under the curve, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180734.png

522. 2.788 × 1010

Because the given function has values that are strictly greater than or equal to zero on the given interval, you can interpret the Riemann sum as approximating the area that's underneath the curve and bounded below by the x-axis.

You want to use eight rectangles of equal width to estimate the area under f (x) = e3x + 4 over the interval 1 ≤ x ≤ 9. Begin by dividing the length of the interval by 8 to find the width of each rectangle:

9781118496718-eq180735.png

Then divide the interval [1, 9] into 8 equal pieces, each with a width of 1, to get the intervals [1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], and [8, 9]. Using the left endpoint of each interval to calculate the heights of the rectangles gives you the values x = 1, x = 2, x = 3, x = 4, x = 5,x = 6, x = 7, and x = 8. The height of each rectangle is f (x). To approximate the area under the curve, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180736.png

523. 24

Because the given function has values that are strictly greater than or equal to zero on the given interval, you can interpret the Riemann sum as approximating the area that's underneath the curve and bounded below by the x-axis.

You want to use four rectangles of equal width to estimate the area under f (x) = 1 + 2x over the interval 0 ≤ x ≤ 4. Begin by dividing the length of the interval by 4 to find the width of each rectangle:

9781118496718-eq180737.png

Then divide the interval [0, 4] into 4 equal pieces, each with a width of 1, to get the intervals [0, 1], [1, 2], [2, 3], and [3, 4]. Using the right endpoint of each interval to calculate the heights of the rectangles gives you the values x = 1, x = 2, x = 3, and x = 4.The height of each rectangle is f (x). To approximate the area under the curve, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180738.png

524. –8.89

In this example, the given function has values that are both positive and negative on the given interval, so don't make the mistake of thinking the Riemann sum approximates the area between the curve and the x-axis. In this case, the Riemann sum approximates the value of

9781118496718-eq180739.png

You want to use five rectangles of equal width over the interval 2 ≤ x ≤ 6. Begin by dividing the length of the interval by 5 to find the width of each rectangle:

9781118496718-eq180740.png

Then divide the interval [2, 6] into 5 equal pieces, each with a width of 9781118496718-eq180741.tif, to get the intervals 9781118496718-eq180742.tif9781118496718-eq180743.tif9781118496718-eq180744.tif9781118496718-eq180745.tif, and 9781118496718-eq180746.tif. Using the right endpoint of each interval to calculate the heights of the rectangles gives you the values 9781118496718-eq180747.tif9781118496718-eq180748.tif9781118496718-eq180749.tif9781118496718-eq180750.tif, and x = 6. The height of each rectangle is f (x). (Note that a “negative height” indicates that the rectangle is below the x-axis.) To find the Riemann sum, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180751.png

525. 3.24

In this example, the given function has values that are both positive and negative on the given interval, so the Riemann sum approximates the value of

9781118496718-eq180752.png

You want to use six rectangles of equal width over the interval 0 ≤ x ≤ 5. Begin by dividing the length of the interval by 6 to find the width of each rectangle:

9781118496718-eq180753.png

Then divide the interval [0, 5] into 6 equal pieces, each with a width of 9781118496718-eq180754.tif, to get the intervals 9781118496718-eq180755.tif9781118496718-eq180756.tif9781118496718-eq180757.tif9781118496718-eq180758.tif9781118496718-eq180759.tif, and 9781118496718-eq180760.tif. Using the right endpoint of each interval to calculate the height of the rectangles gives you the values 9781118496718-eq180761.tif9781118496718-eq180762.tif9781118496718-eq180763.tif9781118496718-eq180764.tif9781118496718-eq180765.tif, and x = 5. The height of each rectangle is f (x). (Note that a “negative height” indicates that the rectangle is below the x-axis.) To find the Riemann sum, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180766.png

526. 1.34

9781118496718-un1823.tif

Because the given function has values that are strictly greater than or equal to zero on the given interval, you can interpret the Riemann sum as approximating the area that's underneath the curve and bounded below by the x-axis.

You want to use eight rectangles of equal width to estimate the area under 9781118496718-eq180767.tif over the interval 1 ≤ x ≤ 3. Begin by dividing the length of the interval by 8 to find the width of each rectangle:

9781118496718-eq180768.png

Then divide the interval [1, 3] into eight equal pieces, each with a width of 9781118496718-eq180769.tif, to get the intervals 9781118496718-eq180770.tif9781118496718-eq180771.tif9781118496718-eq180772.tif9781118496718-eq180773.tif9781118496718-eq180774.tif9781118496718-eq180775.tif9781118496718-eq180776.tif, and 9781118496718-eq180777.tif. Using the right endpoint of each interval to calculate the height of the rectangles gives you the values 9781118496718-eq180778.tif9781118496718-eq180779.tif9781118496718-eq180780.tifx = 2, 9781118496718-eq180781.tif9781118496718-eq180782.tif9781118496718-eq180783.tif, and x = 3. The height of each rectangle is f (x). To approximate the area under the curve, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180784.png

527. 0.29

In this example, the given function has values that are both positive and negative on the given interval, so don't make the mistake of thinking the Riemann sum approximates the area between the curve and the x-axis. In this case, the Riemann sum approximates the value of

9781118496718-eq180785.png

You want to use four rectangles of equal width over the interval 0 ≤ x ≤ 3. Begin by dividing the length of the interval by 4 to find the width of each rectangle:

9781118496718-eq180786.png

Then divide the interval [0, 3] into 4 equal pieces, each with a width of 9781118496718-eq180787.tif, to get the intervals 9781118496718-eq180788.tif9781118496718-eq180789.tif9781118496718-eq180790.tif, and 9781118496718-eq180791.tif. Recall that to find the midpoint of an interval, you simply add the left and the right endpoints together and then divide by 2 (that is, you average the two values). In this case, the midpoints are 9781118496718-eq180792.tif9781118496718-eq180793.tif9781118496718-eq180794.tif, and 9781118496718-eq180795.tif. The height of each rectangle is f (x). (Note that a “negative height” indicates that the rectangle is below the x-axis.) To find the Riemann sum, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180796.png

528. 0.32

In this example, the given function has values that are both positive and negative on the given interval, so the Riemann sum approximates the value of

9781118496718-eq180797.png

You want to use five rectangles of equal width over the interval 1 ≤ x ≤ 5. Begin by dividing the length of the interval by 5 to find the width of each rectangle:

9781118496718-eq180798.png

Divide the interval [1, 5] into 5 equal pieces, each with a width of 9781118496718-eq180799.tif, to get the intervals 9781118496718-eq180800.tif9781118496718-eq180801.tif9781118496718-eq180802.tif9781118496718-eq180803.tif, and 9781118496718-eq180804.tif. Recall that to find the midpoint of an interval, you simply add the left and the right endpoints together and then divide by 2 (that is, you average the two values). In this case, the midpoints are 9781118496718-eq180805.tif9781118496718-eq180806.tif9781118496718-eq180807.tif9781118496718-eq180808.tif, and 9781118496718-eq180809.tif. The height of each rectangle is f (x). (Note that a “negative height” indicates that the rectangle is below the x-axis.) To find the Riemann sum, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180810.png

529. 160.03

Because the given function has values that are strictly greater than or equal to zero on the given interval, you can interpret the Riemann sum as approximating the area that's underneath the curve and bounded below by the x-axis.

You want to use six rectangles of equal width over the interval 1 ≤ x ≤ 3. Begin by dividing the length of the interval by 6 to find the width of each rectangle:

9781118496718-eq180811.png

Then divide the interval [1, 4] into 6 equal pieces, each with a width of 9781118496718-eq180812.tif, to get the intervals 9781118496718-eq180813.tif9781118496718-eq180814.tif9781118496718-eq180815.tif9781118496718-eq180816.tif9781118496718-eq180817.tif, and 9781118496718-eq180818.tif. Recall that to find the midpoint of an interval, you simply add the left and the right endpoints together and then divide by 2 (that is, you average the two values). In this case, the midpoints are 9781118496718-eq180819.tif9781118496718-eq180820.tif9781118496718-eq180821.tif9781118496718-eq180822.tif9781118496718-eq180823.tif, and 9781118496718-eq180824.tif. The height of each rectangle is f (x). To find the Riemann sum, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180825.png

530. 18.79

9781118496718-un1824.tif

Because the given function has values that are strictly greater than or equal to zero on the given interval, you can interpret the Riemann sum as approximating the area that's underneath the curve and bounded below by the x-axis.

You want to use eight rectangles of equal width over the interval 1 ≤ x ≤ 5. Begin by dividing the length of the interval by 8 to find the width of each rectangle:

9781118496718-eq180826.png

Then divide the interval [1, 5] into 8 equal pieces, each with a width of 9781118496718-eq180827.tif, to get the intervals 9781118496718-eq180828.tif9781118496718-eq180829.tif9781118496718-eq180830.tif9781118496718-eq180831.tif9781118496718-eq180832.tif9781118496718-eq180833.tif9781118496718-eq180834.tif, and 9781118496718-eq180835.tif. Recall that to find the midpoint of an interval, you simply add the left and the right endpoints together and then divide by 2 (that is, you average the two values). In this case, the midpoints are 9781118496718-eq180836.tif9781118496718-eq180837.tif9781118496718-eq180838.tif9781118496718-eq180839.tif9781118496718-eq180840.tif9781118496718-eq180841.tif9781118496718-eq180842.tif, and 9781118496718-eq180843.tif. The height of each rectangle is f (x). To approximate the area under the curve, multiply the width of each rectangle by f (x) and add the areas:

9781118496718-eq180844.png

531. 9781118496718-eq180845.png

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180846.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180847.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180848.png

Substituting those values into the definition of the definite integral gives you

9781118496718-eq180849.png

532. 9781118496718-eq180850.png

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180851.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180852.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180853.png

Substituting those values into the definition of the definite integral gives you

9781118496718-eq180854.png

533. 9781118496718-eq180855.png

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180856.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180857.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180858.png

Substituting those values into the definition of the definite integral gives you

9781118496718-eq180859.png

534. 9781118496718-eq180860.png

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180861.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180862.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180863.png

Substituting those values into the definition of the definite integral gives you

9781118496718-eq180864.png

535. 9781118496718-eq180865.png

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180866.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180867.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180868.png

Substituting those values into the definition of the definite integral gives you

9781118496718-eq180869.png

536. 9781118496718-eq180870.png

Recall that in the limit representation of a definite integral, you divide the interval over which you're integrating into n pieces of equal width. You also have to select a point from each interval; the formula a + (Δx)i lets you select the right endpoint of each interval.

Begin by looking for the factor that would represent Δx. In this example, 9781118496718-eq180871.tif; if you ignore the n in this expression, you're left with the length of the interval over which the definite integral is being evaluated. In this case, the length of the interval is 3. Notice that if you consider 9781118496718-eq180872.tif, this is of the form a + (Δx)i, where a = 4.

You now have a value for a and know the length of the interval, so you can conclude that you're integrating over the interval [4, 7]. To produce the function, replace each expression of the form 9781118496718-eq180873.tif that appears in the summation with the variable x. In this example, you replace 9781118496718-eq180874.tifwith x6 so that f (x) = x6. Therefore, the Riemann sum could represent the definite integral 9781118496718-eq180875.tif.

537. 9781118496718-eq180876.png

Recall that in the limit representation of a definite integral, you divide the interval over which you're integrating into n pieces of equal width. You also have to select a point from each interval; the formula a + (Δx)i lets you select the right endpoint of each interval.

Begin by looking for the factor that would represent Δx. In this example, 9781118496718-eq180877.tif; if you ignore the n in this expression, you're left with the length of the interval over which the definite integral is being evaluated. In this case, the length of the interval is 9781118496718-eq180878.tif. Notice that if you consider 9781118496718-eq180879.tif, this is of the form a + (Δx)i, where a = 0.

You now have a value for a and know the length of the interval, so you can conclude that you're integrating over the interval 9781118496718-eq180880.tif. To produce the function, replace each expression of the form 9781118496718-eq180881.tif that appears in the summation with the variable x. In this example, you replace 9781118496718-eq180882.tif with sec x so that f (x) = sec x. Therefore, the Riemann sum could represent the definite integral 9781118496718-eq180883.tif.

538. 9781118496718-eq180884.png

Recall that in the limit representation of a definite integral, you divide the interval over which you're integrating into n pieces of equal width. You also have to select a point from each interval; the formula a + (Δx)i lets you select the right endpoint of each interval.

Begin by looking for the factor that would represent Δx. In this example, 9781118496718-eq180885.tif; if you ignore the n in this expression, you're left with the length of the interval over which the definite integral is being evaluated. In this case, the length of the interval is 1. Notice that if you consider 9781118496718-eq180886.tif, this is of the form a + (Δx)i, where a = 6.

You now have a value for a and know the length of the interval, so you can conclude that you're integrating over the interval [6, 7]. To produce the function, replace each expression of the form 9781118496718-eq180887.tif that appears in the summation with the variable x. In this example, you replace 9781118496718-eq180888.tif with x so that f (x) = x. Therefore, the Riemann sum could represent the definite integral 9781118496718-eq180889.tif.

539. 9781118496718-eq180890.png

Recall that in the limit representation of a definite integral, you divide the interval over which you're integrating into n pieces of equal width. You also have to select a point from each interval; the formula a + (Δx)i lets you select the right endpoint of each interval.

Begin by looking for the factor that would represent Δx. In this example, 9781118496718-eq180891.tif; if you ignore the n in this expression, you're left with the length of the interval over which the definite integral is being evaluated. In this case, the length of the interval is 9781118496718-eq180892.tif. Notice that if you consider 9781118496718-eq180893.tif, this is of the form a + (Δx)i, where a = 0.

You now have a value for a and know the length of the interval, so you can conclude that you're integrating over the interval 9781118496718-eq180894.tif. To produce the function, replace each expression of the form 9781118496718-eq180895.tif that appears in the summation with the variable x. In this example, you replace 9781118496718-eq180896.tif with cos x + sin x so that f (x) = cos x + sin x. Therefore, the Riemann sum could represent the definite integral 9781118496718-eq180897.tif.

540. 9781118496718-eq180898.png

Recall that in the limit representation of a definite integral, you divide the interval over which you're integrating into n pieces of equal width. You also have to select a point from each interval; the formula a + (Δx)i lets you select the right endpoint of each interval.

Begin by looking for the factor that would represent Δx. In this example, 9781118496718-eq180899.tif; if you ignore the n in this expression, you're left with the length of the interval over which the definite integral is being evaluated. In this case, the length of the interval is 5. Notice that if you consider 9781118496718-eq180900.tif, this is of the form a + (Δx)i, where a = 0.

You now have a value for a and know the length of the interval, so you can conclude that you're integrating over the interval [0, 5]. To produce the function, replace each expression of the form 9781118496718-eq180901.tif that appears in the summation with the variable x. Notice that you can rewrite the summation as 9781118496718-eq180902.tif; that means you can replace 9781118496718-eq180903.tif with 9781118496718-eq180904.tif so that 9781118496718-eq180905.tif. Therefore, the Riemann sum could represent the definite integral 9781118496718-eq180906.tif.

541. 6

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180907.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180908.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180909.png

Substituting those values into the definition of the integral gives you the following:

9781118496718-eq180910.png

Note that formulas 9781118496718-eq180911.tif and 9781118496718-eq180912.tif were used to simplify the summations.

542. 196

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180913.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180914.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180915.png

Substituting those values into the definition of the integral gives you the following:

9781118496718-eq180916.png

Note that the formulas 9781118496718-eq180917.tif and 9781118496718-eq180918.tifwere used to simplify the summations.

543. 9781118496718-eq180919.png

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180920.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180921.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180922.png

Substituting those values into the definition of the integral gives you the following:

9781118496718-eq180923.png

Note that the formulas 9781118496718-eq180924.tif and 9781118496718-eq180925.tif were used to simplify the summations.

544. 9781118496718-eq180926.png

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180927.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180928.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180929.png

Substituting those values into the definition of the integral gives you the following:

9781118496718-eq180930.png

Note that the formulas 9781118496718-eq180931.tif9781118496718-eq180932.tif, and 9781118496718-eq180933.tif were used to simplify the summations.

545. 9781118496718-eq180934.png

To begin, you split the interval into n pieces of equal width using the formula 9781118496718-eq180935.tif, where a is the lower limit of integration and b is the upper limit of integration. In this case, you have

9781118496718-eq180936.png

You also want to select a point from each interval. The formula xi = a + (Δx)i gives you the right endpoint from each interval. Here, you have

9781118496718-eq180937.png

Substituting those values into the definition of the integral gives you the following:

9781118496718-eq180938.png

Note that the formulas 9781118496718-eq180939.tif9781118496718-eq180940.tif, and 9781118496718-eq180941.tif were used to simplify the summations.

546. 9781118496718-eq180942.png

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq180943.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

To find the derivative of the function 9781118496718-eq180944.tif, simply substitute the upper limit of integration, x, into the integrand:

9781118496718-eq180945.png

547. 9781118496718-eq180946.png

To find the derivative of the function 9781118496718-eq180948.tif, simply substitute the upper limit of integration, x, into the integrand:

9781118496718-eq180949.png

548. x3 cos x

To find the derivative of the function 9781118496718-eq180951.tif, simply substitute the upper limit of integration, x, into the integrand:

9781118496718-eq180952.png

549. 9781118496718-eq180953.png

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq180954.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

To use the fundamental theorem of calculus, you need to have the variable in the upper limit of integration. Therefore, to find the derivative of the function 9781118496718-eq180955.tif, first flip the limits of integration and change the sign:

9781118496718-eq180956.png

Then substitute the upper limit of integration into the integrand to find the derivative:

9781118496718-eq180957.png

550. –cosx

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq180958.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

To use the fundamental theorem of calculus, you need to have the variable in the upper limit of integration. Therefore, to find the derivative of the function 9781118496718-eq180959.tif, first flip the limits of integration and change the sign:

9781118496718-eq180960.png

Note also that to find 9781118496718-eq180961.tif, you can use the substitution u = h(x) and then apply the chain rule as follows:

9781118496718-eq180962.png

All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of 9781118496718-eq180963.tif is

9781118496718-eq180964.png

Note that the identity 1 – sinx = cos2 x was used to simplify the derivative.

551. –2x

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq180965.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

To use the fundamental theorem of calculus, you need to have the variable in the upper limit of integration. Therefore, to find the derivative of the function 9781118496718-eq180966.tif, flip the limits of integration and change the sign of the integral:

9781118496718-eq180967.png

Note also that to find 9781118496718-eq180968.tif, you can use the substitution u = h(x) and then apply the chain rule as follows:

9781118496718-eq180969.png

All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of 9781118496718-eq180970.tif is

9781118496718-eq180971.png

552. 9781118496718-eq180972.png

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq180973.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

To use the fundamental theorem of calculus, you need to have the variable in the upper limit of integration. Therefore, to find the derivative of the function 9781118496718-eq180974.tif, first flip the limits of integration and change the sign:

9781118496718-eq180975.png

Note also that to find 9781118496718-eq180976.tif, you can use the substitution

u = h(x) and then apply the chain rule as follows:

9781118496718-eq180977.png

All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of 9781118496718-eq180978.tif is

9781118496718-eq180979.png

553. 9781118496718-eq180980.png

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq180981.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

To use the fundamental theorem of calculus, you need to have the variable in the upper limit of integration. Therefore, to find the derivative of the function 9781118496718-eq180982.tif, first flip the limits of integration and change the sign:

9781118496718-eq180983.png

Note also that to find 9781118496718-eq180984.tif, you can use the substitution u = h(x) and then apply the chain rule as follows:

9781118496718-eq180985.png

All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of 9781118496718-eq180986.tif is

9781118496718-eq180987.png

554. 9781118496718-eq180988.png

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq180989.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

Note also that to find 9781118496718-eq180990.tif, you can use the substitution u = h(x) and then apply the chain rule as follows:

9781118496718-eq180991.png

All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of 9781118496718-eq180992.tif is

9781118496718-eq180993.png

555. 9781118496718-eq180994.png

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq180995.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

To use the fundamental theorem of calculus, you need to have the variable in the upper limit of integration. Therefore, to find the derivative of the function 9781118496718-eq180996.tif, first split the integral into two separate integrals:

9781118496718-eq180997.png

Then flip the limits of integration and change the sign of the first integral:

9781118496718-eq180998.png

Note also that to find 9781118496718-eq180999.tif, you can use the substitution u = h(x) and then apply the chain rule as follows:

9781118496718-eq181000.png

All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of 9781118496718-eq181001.tif is

9781118496718-eq181002.png

556. 9781118496718-eq181003.png

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq181004.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

To use the fundamental theorem of calculus, you need to have the variable in the upper limit of integration. Therefore, to find the derivative of the function 9781118496718-eq181005.tif, first split the integral into two separate integrals:

9781118496718-eq181006.png

Then flip the limits of integration and change the sign of the first integral:

9781118496718-eq181007.png

Note also that to find 9781118496718-eq181008.tif, you can use the substitution u = h(x) and then apply the chain rule as follows:

9781118496718-eq181009.png

All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of 9781118496718-eq181010.tif is

9781118496718-eq181011.png

557. 9781118496718-eq181012.png

Part of the fundamental theorem of calculus states that if the function g is continuous on [a, b], then the function f defined by

9781118496718-eq181013.png

is continuous on [a, b] and is differentiable on (a, b). Furthermore, f'(x= g(x).

To use the fundamental theorem of calculus, you need to have the variable in the upper limit of integration. Therefore, to find the derivative of the function 9781118496718-eq181014.tif, first split the integral into two separate integrals:

9781118496718-eq181015.png

Then flip the limits of integration and change the sign of the first integral:

9781118496718-eq181016.png

Note also that to find 9781118496718-eq181017.tif, you can use the substitution u = h(x) and then apply the chain rule as follows:

9781118496718-eq181018.png

All this tells you to substitute the upper limit of integration into the integrand and multiply by the derivative of the upper limit of integration. Therefore, the derivative of 9781118496718-eq181019.tif is

9781118496718-eq181020.png

558. 10

Using elementary antiderivative formulas gives you

9781118496718-eq181021.png

559. 9781118496718-eq181022.png

Using elementary antiderivative formulas gives you

9781118496718-eq181023.png

560. 1

Using elementary antiderivative formulas gives you

9781118496718-eq181024.png

561. 1

Using elementary antiderivative formulas gives you

9781118496718-eq181025.png

562. 9781118496718-eq181026.png

Using the basic antiderivative formula for an exponential function gives you

9781118496718-eq181027.png

563. 9781118496718-eq181028.png

Using elementary antiderivative formulas gives you

9781118496718-eq181029.png

564. 9781118496718-eq181030.png

Using elementary antiderivative formulas gives you

9781118496718-eq181031.png

565. 9781118496718-eq181032.png

Rewrite the radical using exponential notation:

9781118496718-eq181033.png

Then use elementary antiderivative formulas:

9781118496718-eq181034.png

566. 9781118496718-eq181035.png

Using elementary antiderivative formulas gives you

9781118496718-eq181036.png

567. 10

Using elementary antiderivative formulas gives you

9781118496718-eq181037.png

568. 9781118496718-eq181038.png

Begin by rewriting the square root using an exponent:

9781118496718-eq181039.png

Then use elementary antiderivative formulas:

9781118496718-eq181040.png

569. 9781118496718-eq181041.png

Integrate each piece of the function over the given integral. Here's the piecewise function:

9781118496718-eq181042.png

And here are the calculations:

9781118496718-eq181043.png

570. 9781118496718-eq181044.png

The given integral is

9781118496718-eq181045.png

Begin by splitting up the integral by using the definition of the absolute value function:

9781118496718-eq181046.png

So to evaluate 9781118496718-eq181047.tif, use the function –(x – 2) over the interval [–3, 2] and use the function (x – 2) over the interval [2, 4].

Therefore, the integral becomes

9781118496718-eq181048.png

571. 9781118496718-eq181049.png

Add 1 to the exponent and divide by that new exponent to get the antiderivative:

9781118496718-eq181050.png

572. 9781118496718-eq181051.png

First move the variable to the numerator:

9781118496718-eq181052.png

Then integrate by adding 1 to the exponent while dividing by that new exponent:

9781118496718-eq181053.png

573. 9781118496718-eq181054.png

Find the antiderivative of each term:

9781118496718-eq181055.png

574. 3 sin x + 4 cos x + C

Use elementary antiderivative formulas:

9781118496718-eq181056.png

575. 4x + C

This problem can be a bit difficult, unless you remember the trigonometric identities! In this problem, simply factor out the 4 and use a trigonometric identity:

9781118496718-eq181057.png

576. tan x – x + C

Use a trigonometric identity followed by elementary antiderivatives:

9781118496718-eq181058.png

577. x3 + x2 + x + C

Find the antiderivative of each term:

9781118496718-eq181059.png

578. 9781118496718-eq181060.png

Use elementary antiderivative formulas:

9781118496718-eq181061.png

579. 9781118496718-eq181062.png

Begin by using a trigonometric identity to replace tanx. Then simplify:

9781118496718-eq181063.png

580. 9781118496718-eq181064.png

Begin by rewriting the radicals using exponential notation. Then use an elementary antiderivative formula and simplify:

9781118496718-eq181065.png

581. 9781118496718-eq181066.png

Rewrite the integrand using exponential notation and factor out the constant. Then use elementary antiderivative formulas:

9781118496718-eq181067.png

582. 9781118496718-eq181068.png

Begin by writing the radical using exponential notation:

9781118496718-eq181069.png

Then use an elementary antiderivative formula:

9781118496718-eq181070.png

583. 9781118496718-eq181071.png

Use a trigonometric identity in the numerator of the integrand and simplify:

9781118496718-eq181072.png

584. 9781118496718-eq181073.png

Begin by multiplying out the integrand. Then integrate each term:

9781118496718-eq181074.png

585. 9781118496718-eq181075.png

Begin by breaking up the fraction and simplifying each term:

9781118496718-eq181076.png

Then find the antiderivative of each term:

9781118496718-eq181077.png

586. 9781118496718-eq181078.png

Expand the integrand and use elementary antiderivative formulas:

9781118496718-eq181079.png

587. 3x + cot x + C

Use a trigonometric identity on the integrand and simplify:

9781118496718-eq181080.png

588. 9781118496718-eq181081.png

Begin by splitting up the fraction and writing the radical using exponential notation:

9781118496718-eq181082.png

Then use elementary antiderivative formulas and simplify:

9781118496718-eq181083.png

589. 9781118496718-eq181084.png

Begin by writing the cube root using exponential notation and multiplying out the integrand:

9781118496718-eq181085.png

Then use elementary antiderivative formulas:

9781118496718-eq181086.png

590. 9781118496718-eq181087.png

Begin by rewriting the radicals using exponential notation and then distribute:

9781118496718-eq181088.png

Next, use elementary antiderivative formulas for each term:

9781118496718-eq181089.png

591. –csc x + C

Rewrite the integrand. Then use trigonometric identities followed by using a basic antiderivative formula:

9781118496718-eq181090.png

592. 9781118496718-eq181091.png

Begin by splitting up the fraction and simplifying the integrand:

9781118496718-eq181092.png

Then use elementary antiderivative formulas:

9781118496718-eq181093.png

593. 9781118496718-eq181094.png

Begin by splitting up the fraction and simplifying the integrand:

9781118496718-eq181095.png

Then use elementary antiderivative formulas:

9781118496718-eq181096.png

594. –cot x + x + C

Begin by splitting up the fraction and using trigonometric identities to rewrite the integrand:

9781118496718-eq181097.png

Now apply elementary antiderivative formulas:

9781118496718-eq181098.png

595. 9781118496718-eq181099.png

Multiply out the integrand and then use elementary antiderivative formulas on each term:

9781118496718-eq181100.png

596. tan x – x + C

Begin by multiplying out the integrand. Then simplify and use elementary antiderivative formulas:

9781118496718-eq181101.png

597. 9781118496718-eq181102.png

Factor the numerator out of the integrand. Then simplify and use elementary antiderivative formulas:

9781118496718-eq181103.png

598. 9781118496718-eq181104.png

Begin by factoring the numerator. Then simplify and use elementary antiderivative formulas:

9781118496718-eq181105.png

599. 9781118496718-eq181106.png

When finding the antiderivative, don't be thrown off by the decimal exponents; simply add 1 to the exponent of each term and divide by that new exponent for each term to get the solution:

9781118496718-eq181107.png

600. 9781118496718-eq181108.png

Begin by splitting up the fraction and simplifying:

9781118496718-eq181109.png

Then use elementary antiderivative formulas. To get the final answer, turn the decimals in the denominators into fractions and simplify:

9781118496718-eq181110.png

601. 9781118496718-eq181111.png

Start by bringing all variables to the numerator. Then use elementary antiderivative formulas:

9781118496718-eq181112.png

602. 9781118496718-eq181113.png

Use a trigonometric identity and simplify:

9781118496718-eq181114.png

603. x + C

Start by using a trigonometric identity for 1 + cotx, followed by rewriting cscx as 9781118496718-eq181115.tif. Then simplify:

9781118496718-eq181116.png

604. 9781118496718-eq181117.png

Begin by splitting the fraction into three terms:

9781118496718-eq181118.png

Then use elementary antiderivative formulas and simplify:

9781118496718-eq181119.png

605. 7x + C

Begin by factoring the 3 out of the first two terms of the integrand. Then use a trigonometric identity:

9781118496718-eq181120.png

606. 9781118496718-eq181121.png

Begin by factoring the numerator. Then simplify and use elementary antiderivative formulas:

9781118496718-eq181122.png

607. tan x + C

Begin by writing tanx as 9781118496718-eq181123.tif. Then simplify:

9781118496718-eq181124.png

608. sin x + cos x + C

Use a trigonometric identity on the numerator of the integrand, factor, and simplify:

9781118496718-eq181125.png

609. –cot x – 2x + C

Begin by using a trigonometric identity on the numerator of the integrand. Then split up the fraction and rewrite both fractions using trigonometric identities:

9781118496718-eq181126.png

Notice that the first term of the integrand does not have an elementary antiderivative, so you can use a trigonometric identity again to simplify and then integrate:

9781118496718-eq181127.png

610. sin x + C

Begin by factoring cos x from the numerator and use a trigonometric identity to simplify:

9781118496718-eq181128.png

611. the net change in the baby's weight in pounds during the first 2 weeks of life

The net change theorem states that the integral of a rate of change is the net change:

9781118496718-eq181129.png

Because w'(t) is the rate of the baby's growth in pounds per week, w(t) represents the child's weight in pounds at age t. From the net change theorem, you have

9781118496718-eq181130.png

Therefore, the integral represents the increase in the child's weight (in pounds) from birth to the start of the second week.

612. the total amount of oil, in gallons, that leaks from the tanker for the first 180 minutes after the leak begins.

Because r(t) is the rate at which oil leaks from a tanker in gallons per minute, you can write r(t) = –V'(t), where V(t) is the volume of the oil at time t. Notice that you need the minus sign because the volume is decreasing, so V'(t) is negative but r(t) is positive. By the net change theorem, you have

9781118496718-eq181132.png

This is the number of gallons of oil that leaks from the tank in the first 180 minutes after the leak begins.

613. the total bird population 6 months after they're placed in the refuge

Because p'(t) is the rate of growth of the bird population per month, p(t) represents the bird population after t months. From the net change theorem, you have

9781118496718-eq181134.png

This represents the change in the bird population in 6 months. Therefore, 9781118496718-eq181135.tif represents the total bird population after 6 months.

614. the net change of the particle's position, or displacement, during the first 10 seconds

Because s'(t) = v(t), where s(t) is the position function of the particle after t seconds, you have

9781118496718-eq181137.png

This represents the net change of the particle's position, or displacement, during the first 10 seconds. Notice that because the velocity is in meters per second, the displacement units are in meters.

615. the net change in the car's velocity, in meters per second, from the end of the third second to the end of the fifth second

Because v'(t) = a(t), where v(t) is the velocity function of the car after t seconds, you have

9781118496718-eq181139.png

This represents the net change in the car's velocity, in meters per second, from the end of the third second to the end of the fifth second.

616. the total number of solar panels produced from the end of the second week to the end of the fourth week

Because P'(t) is the rate of production per week, P(t) represents the total number of solar panels produced after t weeks. By the net change theorem, you have

9781118496718-eq181141.png

This represents the total number of solar panels produced from the end of the second week to the end of the fourth week.

617. the net change in the charge from time t1 to time t2

I(t) is defined as the derivative of the charge, Q'(t) = I(t), so by the net change theorem, you have

9781118496718-eq181143.png

This represents the net change in the charge from time t1 to time t2.

618. the net change in income in dollars during the first 10 years at the job

I(t= I'(t), where I(t) represents your total income after t years, so by the net change theorem, you have

9781118496718-eq181145.png

This represents the net change in income in dollars during the first 10 years at the job.

619. the net change in the amount of water in the pool from the end of the 60th minute to the end of the 120th minute

Because w'(t) is the rate at which water enters the pool in gallons per minute, w(t) represents the amount of water in the pool at time t. By the net change theorem, you have

9781118496718-eq181147.png

This represents the net change in the amount of water in the pool from the end of the 60th minute to the end of the 120th minute.

620. 5

Note that displacement, or change in position, can be positive or negative or zero. You can think of the particle moving to the left if the displacement is negative and moving to the right if the displacement is positive.

Velocity is the rate of change in displacement with respect to time, so if you integrate the velocity function over an interval where the velocity is negative, you're finding how far the particle travels to the left over that time interval (the value of the integral is negative to indicate that the displacement is to the left). Likewise, if you integrate the velocity function over an interval where the velocity is positive, you're finding the distance that the particle travels to the right over that time interval (in this case, the value of the integral is positive). By combining these two values — that is, by integrating the velocity function over the given time interval — you find the net displacement.

To find the displacement, simply integrate the velocity function over the given interval:

9781118496718-eq181148.png

621. 9781118496718-eq181149.png

To find the displacement, simply integrate the velocity function over the given interval:

9781118496718-eq181150.png

622. 0

To find the displacement, simply integrate the velocity function over the given interval:

9781118496718-eq181151.png

623. 9781118496718-eq181152.png

To find the displacement, simply integrate the velocity function over the given interval:

9781118496718-eq181153.png

624. 9781118496718-eq181154.png

To find the displacement, simply integrate the velocity function over the given interval:

9781118496718-eq181155.png

625. 13

Note that displacement, or change in position, can be positive or negative or zero. You can think of the particle moving to the left if the displacement is negative and moving to the right if the displacement is positive.

Velocity is the rate of change in displacement with respect to time, so if you integrate the velocity function over an interval where the velocity is negative, you're finding how far the particle travels to the left over that time interval (the value of the integral is negative to indicate that the displacement is to the left). Likewise, if you integrate the velocity function over an interval where the velocity is positive, you're finding the distance that the particle travels to the right over that time interval (in this case, the value of the integral is positive). By combining these two values, you find the net displacement.

Unlike displacement, distance can't be negative. In order to find the distance traveled, you can integrate the absolute value of the velocity function because you're then integrating a function that's greater than or equal to zero on the given interval.

9781118496718-eq181156.png

Find any zeros of the function on the given interval so you can determine where the velocity function is positive or negative. In this case, you have 2t – 4 = 0 so that t = 2. Because the velocity function is negative on the interval (0, 2) and positive on the interval (2, 5), the distance traveled is

9781118496718-eq181157.png

9781118496718-un1868.tif

626. 5

Unlike displacement, distance can't be negative. In order to find the distance traveled, you can integrate the absolute value of the velocity function because you're then integrating a function that's greater than or equal to zero on the given interval.

9781118496718-eq181158.png

Find any zeros of the function on the given interval so you can determine where the velocity function is positive or negative. In this case, you have t2 – t – 6 = 0, or (t – 3)(t + 2) = 0, which has solutions t = 3 and t = –2. Because the velocity function is negative on the interval (2, 3) and positive on the interval (3, 4), the distance traveled is

9781118496718-eq181159.png

627. 4

Unlike displacement, distance can't be negative. In order to find the distance traveled, you can integrate the absolute value of the velocity function because you're then integrating a function that's greater than or equal to zero on the given interval.

9781118496718-eq181160.png

Find any zeros of the function on the given interval so you can determine where the velocity function is positive or negative. In this case, you have 2 cos t ≥ 0 on 9781118496718-eq181161.tif and 2 cos t ≤ 0 on 9781118496718-eq181162.tif. Therefore, the distance traveled is

9781118496718-eq181163.png

628. 9781118496718-eq181164.png

Unlike displacement, distance can't be negative. In order to find the distance traveled, you can integrate the absolute value of the velocity function because you're then integrating a function that's greater than or equal to zero on the given interval.

9781118496718-eq181165.png

Find any zeros of the function on the given interval so you can determine where the velocity function is positive or negative. In this case, you have sin t – cos t = 0 so that sin t = cos t, which has a solution 9781118496718-eq181166.tif on the given interval. Note that on the interval 9781118496718-eq181167.tif, you have sint – cos t < 0 and that on the interval 9781118496718-eq181168.tif, sin t – cos t > 0. Therefore, the distance traveled is

9781118496718-eq181169.png

629. 9781118496718-eq181170.png

Unlike displacement, distance can't be negative. In order to find the distance traveled, you can integrate the absolute value of the velocity function because you're then integrating a function that's greater than or equal to zero on the given interval.

9781118496718-eq181171.png

Find the any zeros of the function on the given interval so you can determine where the velocity function is positive or negative. In this case, you have 9781118496718-eq181172.tif so that 9781118496718-eq181173.tif, or t = 16. Notice that on the interval (1, 16), you have 9781118496718-eq181174.tif, whereas on the interval (16, 25), you have 9781118496718-eq181175.tif. Therefore, the distance traveled is

9781118496718-eq181176.png

630. 9781118496718-eq181177.png

Because the derivative of the position function is the rate of change in position with respect to time, the derivative of the position function gives you the velocity function. Likewise, the derivative of the velocity function is the rate of change in velocity with respect to time, so the derivative of the velocity function gives you the acceleration function. It follows that the antiderivative of the acceleration function is the velocity function and that the antiderivative of the velocity function is the position function.

First find the velocity function by evaluating the antiderivative of the acceleration function, a(t) = t + 2:

9781118496718-eq181178.png

Next, use the initial condition, v(0) = –6, to solve for the arbitrary constant of integration:

9781118496718-eq181179.png

Therefore, the velocity function is

9781118496718-eq181180.png

To find the displacement, simply integrate the velocity function over the given interval, 0 ≤ t ≤ 8:

9781118496718-eq181181.png

631. 9781118496718-eq181182.png

First find the velocity function by evaluating the antiderivative of the acceleration function, a (t) = 2t + 1:

9781118496718-eq181183.png

Next, use the initial condition, v (0) = –12, to solve for the arbitrary constant of integration:

9781118496718-eq181184.png

Therefore, the velocity function is

9781118496718-eq181185.png

Next, find the displacement by integrating the velocity function over the given interval, 0 ≤ t ≤ 5:

9781118496718-eq181186.png

632. 9781118496718-eq181187.png

First find the velocity function by evaluating the antiderivative of the acceleration function, a (t) = sin t + cos t:

9781118496718-eq181188.png

Next, use the initial condition, 9781118496718-eq181189.tif, to solve for the arbitrary constant of integration:

9781118496718-eq181190.png

Therefore, the velocity function is

9781118496718-eq181191.png

Next, find the displacement by integrating the velocity function over the given interval, 9781118496718-eq181192.tif:

9781118496718-eq181193.png

633. 9781118496718-eq181194.png

Because the derivative of the position function is the rate of change in position with respect to time, the derivative of the position function gives you the velocity function. Likewise, the derivative of the velocity function is the rate of change in velocity with respect to time, so the derivative of the velocity function gives you the acceleration function. It follows that the antiderivative of the acceleration function is the velocity function and that the antiderivative of the velocity function is the position function.

Note that displacement, or change in position, can be positive or negative or zero. You can think of the particle moving to the left if the displacement is negative and moving to the right if the displacement is positive. But unlike displacement, distance can't be negative. To find the distance traveled, you can integrate the absolute value of the velocity function because you're then integrating a function that's greater than or equal to zero on the given interval.

Begin by finding the velocity function by evaluating the antiderivative of the acceleration function, a(t) = t + 2:

9781118496718-eq181195.png

Next, use the initial condition, v(0) = –6, to solve for the arbitrary constant of integration:

9781118496718-eq181196.png

Therefore, the velocity function is

9781118496718-eq181197.png

To find the distance traveled, integrate the absolute value of the velocity function over the given interval:

9781118496718-eq181198.png

Find any zeros of the function on the given interval so that you can determine where the velocity function is positive or negative: 9781118496718-eq181199.tif, or t2 + 4t – 12 = 0, which factors as (t + 6)(t – 2) = 0 and has the solutions of t = –6, t = 2. Notice that on the interval (0, 2), the velocity function is negative and that on the interval (2, 8), the velocity function is positive. Therefore, the distance traveled is

9781118496718-eq181200.png

634. 9781118496718-eq181201.png

Begin by finding the velocity function by evaluating the antiderivative of the acceleration function, a (t) = 2t + 1:

9781118496718-eq181202.png

Next, use the initial condition, v (0) = –12, to solve for the arbitrary constant of integration:

9781118496718-eq181203.png

Therefore, the velocity function is

9781118496718-eq181204.png

To find the distance traveled, integrate the absolute value of the velocity function over the given interval:

9781118496718-eq181205.png

Find any zeros of the function on the given interval so you can determine where the velocity function is positive or negative: t2 + t – 12 = 0 factors as (t + 4)(t – 3) = 0 and has the solutions t = –4, t = 3. The velocity function is negative on the interval (0, 3) and positive on the interval (3, 5), so the distance traveled is

9781118496718-eq181206.png

635. 9781118496718-eq181207.png

Begin by finding the velocity function by evaluating the antiderivative of the acceleration function, a (t) = sin t + cos t:

9781118496718-eq181208.png

Next, use the initial condition, 9781118496718-eq181209.tif, to solve for the arbitrary constant of integration:

9781118496718-eq181210.png

Therefore, the velocity function is

9781118496718-eq181211.png

To find the distance traveled, integrate the absolute value of the velocity function over the given interval, 9781118496718-eq181212.tif:

9781118496718-eq181213.png

Find any zeros of the velocity function on the given interval so you can determine where the velocity function is positive or negative:

9781118496718-eq181214.png

The velocity function is negative on the interval 9781118496718-eq181215.tif and positive on the interval 9781118496718-eq181216.tif, so the total distance traveled is

9781118496718-eq181217.png

636. 9781118496718-eq181218.png

Recall that the area A of the region bounded by the curves y = f (x) and y = g(x) and by the lines x = a and x = b, where f and g are continuous and f (x) ≥ g(x) for all x in [a, b], is

9781118496718-eq181219.png

More generically, in terms of a graph, you can think of the formula as

9781118496718-eq181220.png

Note that lines x = a and x = b may not be given, so the limits of integration often correspond to points of intersection of the functions.

Begin by finding the points of intersection by setting the functions equal to each other and solving for x:

9781118496718-eq181221.png

Because x2 ≥ x4 on the interval [–1, 1], the integral to find the area is

9781118496718-eq181222.png

The integrand is an even function, so it's symmetric about the y-axis; therefore, you can instead integrate on the interval [0, 1] and multiply by 2:

9781118496718-eq181223.png

This gives you the following:

9781118496718-eq181224.png

The following figure shows the region bounded by the given curves:

9781118496718-un1825.tif

637. 9781118496718-eq181225.png

Begin by finding the points of intersection by setting the functions equal to each other and solving for x:

9781118496718-eq181228.png

To determine which function is larger on the interval (0, 1), take a point inside the interval and substitute it into each function. If you use 9781118496718-eq181229.tif, then the first curve gives you 9781118496718-eq181230.tif, and the second curve gives you 9781118496718-eq181231.tif. Therefore, 9781118496718-eq181232.tif on (0, 1). That means the integral for the area of the bounded region is

9781118496718-eq181233.png

638. 9781118496718-eq181234.png

Because cos x + 1 ≥ x on [0, 1], the integral to find the area is

9781118496718-eq181237.png

639. 9781118496718-eq181238.png

Recall that the area A of the region bounded by the curves x = f (y) and x = g(y) and by the lines y = a and y = b, where f and g are continuous and f (y) ≥ g(y) for all y in [a, b], is

9781118496718-eq181239.png

More generically, in terms of a graph, you can think of the formula as

9781118496718-eq181240.png

Note that lines y = a and y = b may not be given, so the limits of integration often correspond to points of intersection of the curves.

In this case, you could integrate with respect to x, but you'd have to solve each equation for y by completing the square, which would needlessly complicate the problem.

Begin by finding the points of intersection by setting the expressions equal to each other and solving for y:

9781118496718-eq181241.png

To determine which function has larger x values for y in the interval (0, 2), pick a point in the interval and substitute it into each function. So if y = 1, then x = 12 – 1 = 0 and x = 3(1) – 12 = 2. Therefore, the integral to find the area is

9781118496718-eq181242.png

640. 9781118496718-eq181243.png

In this case, you can easily solve the two equations for x, so integrating with respect to y makes sense. If you instead solved the equations for y, you'd have to use more than one integral when integrating with respect to x to set up the area.

Begin by solving the first equation for x to get x = y2 – 1. Because 9781118496718-eq181246.tif on the interval [0, 1], the integral to find the area is

9781118496718-eq181247.png

641. 9781118496718-eq181248.png

Notice that you can easily solve the two equations for x, so integrating with respect to y makes sense. If you were to solve the first equation for y in order to integrate with respect to x, you'd have to use more than one integral to set up the area.

Begin by solving the second equation for x to get x = y + 7. Then find the points of intersection by setting the expressions equal to each other and solving for y:

9781118496718-eq181251.png

To determine which curve has the larger x values for y on the interval (–2, 3), pick a point in the interval and substitute it into each equation. If y = 0, then x = 1 + 02 = 1 and x = 0 + 7 = 7. Therefore, the integral to find the area is

9781118496718-eq181252.png

The following figure shows the region bounded by the given curves:

9781118496718-un1826.tif

642. 9781118496718-eq181253.png

Begin by finding the point of intersection of the two curves by setting them equal to each other and solving for y:

9781118496718-eq181256.png

Notice that 3y – 2 > y2 for y in the interval (1, 2). Therefore, the integral to find the area of the region is

9781118496718-eq181257.png

643. 9781118496718-eq181258.png

In this case, integrating with respect to y makes sense. You could integrate with respect to x, but you'd have to solve x = 2y2 for y and then use two integrals to compute the area, because the “top function” isn't the same for the entire region.

Begin by isolating x in the second equation to get x = 1 – y. Then find the points of intersection by setting the expressions equal to each other and solving for y:

9781118496718-eq181261.png

To determine which curve has the larger x values for y in the interval 9781118496718-eq181262.tif, pick a point in the interval and substitute it into each equation. So if y = 0, then x = 2(0)2 = 0 and x + 0 = 1. Therefore, x = 1 – y is the rightmost curve and x = 2y2 is the leftmost curve, which means the integral to find the area is

9781118496718-eq181263.png

The following figure shows the region bounded by the given curves:

9781118496718-un1827.tif

644. 36

Begin by finding the points of intersection by setting the functions equal to each other and solving for x:

9781118496718-eq181266.png

To determine which function is larger on the interval (–4, 2), take a point inside the interval and substitute it into each function. If you let x = 0, then y = 2(0) = 0 and y = 8 – 02 = 8. Therefore, 8 – x2 > 2x on (–4, 2), so the integral for the area of the bounded region is

9781118496718-eq181267.png

The following figure shows the region bounded by the given curves:

9781118496718-un1828.tif

645. 9781118496718-eq181268.png

In this example, integrating with respect to y makes sense. You could integrate with respect to x, but you'd have to solve both x = 2 – y2 and x = y2 – 2 for y and then use two integrals to compute the area, because the “top function” and the “bottom function” aren't the same for the entire region. (You could actually reduce the region to a single integral by using symmetry, but that isn't possible in general.)

Begin by finding the points of intersection by setting the expressions equal to each other and solving for y:

9781118496718-eq181271.png

To determine which curve has the larger x values for y in the interval 9781118496718-eq181272.tif, take a point inside the interval and substitute it into each function. So if y = 0, then x = 2 – 02 = 2 and x = 02 – 2 = –2; therefore, 2 – y2 > y2 – 2 on 9781118496718-eq181273.tif. That means the integral to find the area is

9781118496718-eq181274.png

By symmetry, you can rewrite the integral as

9781118496718-eq181275.png

Now you can evaluate the integral:

9781118496718-eq181276.png

The following figure shows the region bounded by the given curves:

9781118496718-un1829.tif

646. 72

Begin by finding the points of intersection by setting the functions equal to each other and solving for x:

9781118496718-eq181279.png

To determine which function is larger on the interval (–3, 3), take a point inside the interval and substitute into each function. If x = 0, then y = 14 – 02 = 14 and y = 02 – 4 = –4. Therefore, 14 – x2 > x2 on the interval (–3, 3), so the integral for the area of the bounded region becomes

9781118496718-eq181280.png

The integrand is an even function, so it's symmetric about the y-axis; therefore, you can instead integrate on the interval [0, 3] and multiply by 2:

9781118496718-eq181281.png

647. 9781118496718-eq181282.png

In this example, integrating with respect to y makes sense. You could integrate with respect to x, but you'd have to solve 4x + y2 = –3 for y and then use two integrals to compute the area, because the “top function” isn't the same for the entire region.

Begin by solving the second equation for x to get 9781118496718-eq181285.tif. To find the points of intersection, set the expressions equal to each other and solve for y:

9781118496718-eq181286.png

To determine which curve has larger x values for y in the interval (–3, 1), pick a point inside the interval and substitute it into each function. So if y = –2, then x = –2 and 9781118496718-eq181287.tif. Therefore, the integral to find the area is

9781118496718-eq181288.png

648. 9781118496718-eq181289.png

Begin by finding the points of intersection of the two curves by setting them equal to each other and solving for y:

9781118496718-eq181292.png

To determine which curve has the larger x values for y on the interval (0, 9), pick a value in the interval and substitute it into each equation. So if y = 1, then 9781118496718-eq181293.tif and 9781118496718-eq181294.tif. Therefore, the integral to find the area is

9781118496718-eq181295.png

649. 9781118496718-eq181296.png

Notice that the functions 9781118496718-eq181299.tif and y = cos x intersect when 9781118496718-eq181300.tif. On the interval 9781118496718-eq181301.tif, you have 9781118496718-eq181302.tif, and on the interval 9781118496718-eq181303.tif, you have 9781118496718-eq181304.tif. Therefore, the integrals to find the area of the region are

9781118496718-eq181305.png

650. 9781118496718-eq181306.png

Begin by finding the points of intersection by setting the functions equal to each other and solving for x:

9781118496718-eq181309.png

To determine which function is larger on the interval 9781118496718-eq181310.tif, take a point in the interval and substitute it into each function to determine which is larger. If x = –1, then y = (–1)3 – (–1) = 0 and y = 2(–1) = –2; therefore, x3 – x > 2x on 9781118496718-eq181311.tif. In a similar manner, check which function is larger on the interval 9781118496718-eq181312.tif. By letting x = 1, you can show that 2x > x3 – x on 9781118496718-eq181313.tif. Therefore, the integral to find the area of the region is

9781118496718-eq181314.png

The following figure shows the region bounded by the given curves:

9781118496718-un1830.tif

651. 9781118496718-eq181315.png

Notice that you can easily solve the two equations for x, so integrating with respect to y makes sense. If you were to solve the second equation for y in order to integrate with respect to x, you'd have to use more than one integral to set up the area.

Solve the first equation for x to get x = –y. Then find the points of intersection by setting the equations equal to each other and solving for y:

9781118496718-eq181318.png

To determine which curve has larger x values on the interval (–5, 0), pick a point in the interval and substitute it into each equation. If y = –1, then x = –(–1) = 1 and x = (–1)2 + 4(–1) = –3. Therefore, the integral to find the area is

9781118496718-eq181319.png

652. 9781118496718-eq181320.png

Begin by setting the expressions equal to each other and solving for y to find the points of intersection:

9781118496718-eq181323.png

To determine which curve has larger x values for y on the interval (–3, 1), take a point in the interval and substitute it into each equation. So if y = –2, then 9781118496718-eq181324.tif and 9781118496718-eq181325.tif. Therefore, the integral to find the area of the region is

9781118496718-eq181326.png

653. 9781118496718-eq181327.png

Begin by finding the point of intersection on the interval 9781118496718-eq181330.tif by setting the functions equal to each other and solving for x: sin x = cos x has a solution when 9781118496718-eq181331.tif.

To determine which function is larger on each interval, pick a point in the interval and substitute it into each function. On the interval 9781118496718-eq181332.tif, you have cos x ≥ sin x, and on the interval 9781118496718-eq181333.tif, you have sin x ≥ cos x. Therefore, the integrals required to find the area of the region are

9781118496718-eq181334.png

The following figure shows the region bounded by the functions.

9781118496718-un1831.tif

654. 9781118496718-eq181335.png

Begin by finding the points of intersection of the two curves by setting the expressions equal to each other and solving for y:

9781118496718-eq181338.png

To determine which expression has larger x values for y on the interval (0, 1), pick a point in the interval and substitute it into each equation. So if 9781118496718-eq181339.tif, then 9781118496718-eq181340.tif and 9781118496718-eq181341.tif. For y on the interval (0, 1), you have 9781118496718-eq181342.tif. Similarly, you can show that 9781118496718-eq181343.tiffor y on the interval (1, 2). Therefore, the integrals to find the area of the region are

9781118496718-eq181344.png

655. 9781118496718-eq181345.png

Begin by finding the points of intersection of the two curves by setting the expressions equal to each other:

9781118496718-eq181348.png

To determine which curve has the larger x values for y on the interval (0, 5), take a point in the interval and substitute it into each equation. So if y = 1, then x = 12 – 1 = 0 and x = 4(1) = 4. Therefore, the integral to find the area of the region is

9781118496718-eq181349.png

656. 18

Notice that you can easily solve the two equations for x, so integrating with respect to y makes sense. If you were to solve the second equation for y in order to integrate with respect to x, you'd have to complete the square to solve for y, which would needlessly complicate the problem.

Begin by solving both equations for x to get 9781118496718-eq181352.tif and x = y + 1. Then set the expressions equal to each other and solve for y to find the points of intersection:

9781118496718-eq181353.png

To find which curve has the larger x values for y on the interval (–2, 4), take a point in the interval and substitute it into each equation. So if y = 0, then 9781118496718-eq181354.tif and x = 0 + 1 = 1. Therefore, the integral to find the area of the region is

9781118496718-eq181355.png

657. 9781118496718-eq181356.png

Begin by finding all the points of intersection of the three functions. To do so, solve each equation for y and then set the functions equal to each other. Solving the second equation for y gives you 9781118496718-eq181359.tif, and solving the third equation for y gives you y = 3 – 2x. Set these two functions equal to each other and solve for x:

9781118496718-eq181360.png

Likewise, finding the other points of intersection gives you 9781118496718-eq181361.tif so that x = 0 is a solution and gives you x = 3 – 2x so that x = 1 is a solution.

Notice that on the interval (0, 1), the region is bounded above by the function y = x and below by the function 9781118496718-eq181362.tif. On the interval (1, 2), the region is bounded above by the function y = 3 – 2x and below by the function 9781118496718-eq181363.tif. Therefore, the integrals to find the area of the region are

9781118496718-eq181364.png

658. 9781118496718-eq181365.png

Begin by finding the points of intersection by setting the functions equal to each other. Square both sides of the equation and factor to solve for x:

9781118496718-eq181368.png

To determine which function is larger on the interval (–4, 0), pick a point inside the interval and substitute it into each equation. So if x = 3, then 9781118496718-eq181369.tif and 9781118496718-eq181370.tif. Because 9781118496718-eq181371.tif on (–4, 0), the integral to find the area is

9781118496718-eq181372.png

659. 18

To find the points of intersection, begin by noting that 9781118496718-eq181375.tif on the interval [0, ∞). To find the point of intersection on [0, 8), set the functions equal to each other and solve for x:

9781118496718-eq181376.png

Because the interval under consideration is [0, 8), use the solution x = 3.

Likewise, on (–∞, 0), you have 9781118496718-eq181377.tif. Again, find the point of intersection by setting the functions equal to each other and solving for x:

9781118496718-eq181378.png

Because the interval under consideration is (–∞, 0), keep only the solution x = –3. (You could have also noted that because both 9781118496718-eq181379.tif and y = x2 – 3 are even functions, then if there's a point of intersection at x = 3, there must also be a point of intersection at x = –3.)

On the interval (–3, 3), you have 9781118496718-eq181380.tif. Therefore, the integral to find the area is

9781118496718-eq181381.png

By symmetry, you can rewrite the integral as

9781118496718-eq181382.png

Finally, simplify and evaluate this integral:

9781118496718-eq181383.png

The following figure shows the region bounded by the given curves:

9781118496718-un1832.tif

660. 9781118496718-eq181384.png

Begin by finding the points of intersection on the interval 9781118496718-eq181387.tif by setting the functions equal to each other:

9781118496718-eq181388.png

Use an identity on the right-hand side of the equation and factor:

9781118496718-eq181389.png

Now set each factor equal to zero and solve for x: cos x = 0 and 2 sin x – 1 = 0, or 9781118496718-eq181390.tif. On the interval 9781118496718-eq181391.tif, you have cos x = 0 if 9781118496718-eq181392.tif and 9781118496718-eq181393.tif if 9781118496718-eq181394.tif. On the interval 9781118496718-eq181395.tif, you have cos x > sin(2x), and on 9781118496718-eq181396.tif, you have sin(2x) > cos x. Therefore, the integrals to find the area are

9781118496718-eq181397.png

The following figure shows the region bounded by the given curves:

9781118496718-un1833.tif

661. 9781118496718-eq181398.png

Begin by finding the point of intersection by setting the functions y = 2e2x and y = 3 – 5ex equal to each other and solving for x:

9781118496718-eq181401.png

Because ex = –3 has no solution, the only solution is 9781118496718-eq181402.tif. Note that 9781118496718-eq181403.tif, so the line x = 0 bounds the region on the right so that the upper limit of integration is b = 0.

To determine which function is larger on the interval 9781118496718-eq181404.tif, pick a point inside the interval and substitute it into each equation. On the interval 9781118496718-eq181405.tif, notice that 2e2x > 5ex. Therefore, the integral to find the area is

9781118496718-eq181406.png

662. 9781118496718-eq181407.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181408.png

Here's the region that's being rotated. Because the cross-sectional slice is a circle, you have A(x) = π(y)2 = π(f (x))2.

9781118496718-un1834.tif

Note that when y = 0, you have 0 = x4 so that x = 0 is the lower limit of integration.

The integral to find the volume is

9781118496718-eq181409.png

663. 2π

Recall the definition of volume: Let S be a solid that lies between y = c and y = d. If A(y) is the cross-sectional area of S in the plane Py that goes through y and is perpendicular to the y-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181410.png

Here's the region that's being rotated. Because the cross-sectional slice is a circle, you have A(y) = π(x)2 = π(g(y))2.

9781118496718-un1835.tif

In this case, the limits of integration correspond to y = 0 and y = π. Because 9781118496718-eq181411.tif, the integral to find the area becomes

9781118496718-eq181412.png

664. 9781118496718-eq181413.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181414.png

Because the cross-sectional slice is a circle in this case, you have A(x) = π(y)2 = π(f (x))2. Here, the limits of integration correspond to the lines x = 3 and x = 5. Because 9781118496718-eq181415.tif, the integral to find the volume is

9781118496718-eq181416.png

665. π ln 3

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181417.png

Because the cross-sectional slice is a circle in this case, you have A(x) = π(y)2 = π(f (x))2. Here, the limits of integration correspond to the lines x = 1 and x = 3. Because 9781118496718-eq181418.tif, the integral to find the volume is

9781118496718-eq181419.png

666. π

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181420.png

Here's the region that's being rotated. Because the cross-sectional slice is a circle, you have A(x) = π(y)2 = π(f (x))2.

9781118496718-un1836.tif

In this case, the limits of integration correspond to the lines 9781118496718-eq181421.tif and 9781118496718-eq181422.tif. Because y = csc x, the integral to find the volume is

9781118496718-eq181423.png

667. 9781118496718-eq181424.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181425.png

Because the cross-sectional slice is a circle in this case, you have A(x) = π(y)2 = π(f (x))2.

Begin by isolating y in the first equation:

9781118496718-eq181426.png

If you let y = 0 in this equation, you get x = 4, which corresponds to the upper limit of integration. Therefore, you have the following integral:

9781118496718-eq181427.png

668. 9781118496718-eq181428.png

Recall the definition of volume: Let S be a solid that lies between y = c and y = d. If A(y) is the cross-sectional area of S in the plane Py that goes through y and is perpendicular to the y-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181429.png

Here's the region that's being rotated. Because the cross-sectional slice is a circle, you have A(y) = π(x)2 = π(g(y))2.

9781118496718-un1837.tif

Begin by setting the expressions equal to each other and solving for y in order to find the limits of integration:

9781118496718-eq181430.png

Therefore, the integral to find the volume is

9781118496718-eq181431.png

669. 8π

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181432.png

Because the cross-sectional slice is a circle in this case, you have A(x) = π(y)2 = π(f (x))2.

To get the lower limit of integration, find where the function 9781118496718-eq181433.tif intersects the line y = 0. Set the equations equal to each other and solve for x:

9781118496718-eq181434.png

Therefore, x = 1 and x = 5 are the limits of integration. Because 9781118496718-eq181435.tif, the integral becomes

9781118496718-eq181436.png

670. 9781118496718-eq181437.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181438.png

In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional slice is a washer, you can find A(x) by using

9781118496718-eq181439.png

where the outer radius, rout, is the distance of the function farther away from the line of rotation and the inner radius, rin, is the distance of the function closer to the line of rotation at a particular value of x.

First find the limits of integration by finding the points of intersection of the two curves. Set the functions equal to each other and solve for x:

9781118496718-eq181440.png

On the interval 9781118496718-eq181441.tif, you have 9781118496718-eq181442.tif so that 9781118496718-eq181443.tif and rin = 2. Therefore, the integral to find the volume is

9781118496718-eq181444.png

Notice that 9781118496718-eq181445.tif is an even function, so it's symmetric about the y-axis. Therefore, you can instead make zero the lower limit of integration and double the value of the integral:

9781118496718-eq181446.png

671. 9781118496718-eq181447.png

Recall the definition of volume: Let S be a solid that lies between y = c and y = d. If A(y) is the cross-sectional area of S in the plane Py that goes through y and is perpendicular to the y-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181448.png

Because the cross-sectional slice is a circle in this case, you have A(y) = π(x)2 = π(g(y))2.

Note that y = 8 gives you one of the limits of integration; find the other limit of integration by setting the functions equal to each other and solving for y:

9781118496718-eq181449.png

With the limits of integration y = 0 and y = 8 and with g(y) = y2/3, the volume of the region is

9781118496718-eq181450.png

672. 9781118496718-eq181451.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181452.png

Because the cross-sectional slice is a circle in this case, you have A(x) = π(y)2 = π(f (x))2.

Notice that 9781118496718-eq181453.tif intersects y = 0 when x = r and when x = –r, which corresponds to the limits of integration. Therefore, the integral to find the volume becomes

9781118496718-eq181454.png

Because y = r2 – x2 is an even function, you can change the lower limit of integration to zero and multiply by 2 to evaluate the integral:

9781118496718-eq181455.png

By rotating the semicircular region, you've found the formula for the volume of a sphere!

673. 9781118496718-eq181456.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181457.png

Here's the region that's being rotated:

9781118496718-un1838.tif

In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional slice is a washer, you can find A(x) by using

9781118496718-eq181458.png

where the outer radius, rout, is the distance of the function farther away from the line of rotation and the inner radius, rin, is the distance of the function closer to the line of rotation at a particular value of x.

Begin by finding the point of intersection of the two functions by setting them equal to each other: sin x = cos x, which has the solution 9781118496718-eq181459.tif on the interval 9781118496718-eq181460.tif. Notice that on the interval 9781118496718-eq181461.tif, you have cos x > sin x so that rout = cos x and rin = sin x, and on the interval 9781118496718-eq181462.tif, you have sin x > cos x so that rout = sin x and rin = sin x. Therefore, the integrals to find the volume are

9781118496718-eq181463.png

Because 9781118496718-eq181464.tif and 9781118496718-eq181465.tif are equal, you can compute the volume by evaluating the integral on the interval 9781118496718-eq181466.tif and doubling the answer:

9781118496718-eq181467.png

674. 9781118496718-eq181468.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181469.png

Here's the region that's being rotated. Because the cross-sectional slice is a circle, you have A(x) = π(y)2 = π(f (x))2.

9781118496718-un1839.tif

Because the function is bounded by x = 0 and x = 1, these are the limits of integration. With 9781118496718-eq181470.tif, the integral to find the volume becomes

9781118496718-eq181471.png

Now use the trigonometric substitution x = tan θ so that dx = secθ dθ. Find the new limits of integration by noting that if x = 1, then 1 = tan θ so that 9781118496718-eq181472.tif. Likewise, if x = 0, then 0 = tan θ so that 0 = θ. Using these new values, you produce the following integral:

9781118496718-eq181473.png

Now use the identity 9781118496718-eq181474.tif = 9781118496718-eq181475.tif:

9781118496718-eq181476.png

675. 9781118496718-eq181477.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181478.png

In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional slice is a washer, you can find A(x) by using

9781118496718-eq181479.png

where the outer radius, rout, is the distance of the function farther away from the line of rotation and the inner radius, rin, is the distance of the function closer to the line of rotation at a particular value of x.

Begin by finding the points of intersection in order to find the limits of integration. Solve the second equation for y, set the expressions equal to each other, and solve for x:

9781118496718-eq181480.png

To find out which function is greater on the interval (0, 3) without graphing (and is therefore farther away from the line of rotation), take a point in this interval and substitute that value into each function. So if x = 1, you have y = 3 + 2 – 12 = 4 and y = 3 – 1 = 2. Therefore, rout = 3 + 2x – x2 and rin = 3 – x, so the integral becomes

9781118496718-eq181481.png

676. 9781118496718-eq181482.png

Recall the definition of volume: Let S be a solid that lies between y = c and y = d. If A(y) is the cross-sectional area of S in the plane Py that goes through y and is perpendicular to the y-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181483.png

In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional slice is a washer, you can find A(y) by using

9781118496718-eq181484.png

where the outer radius, rout, is the distance of the expression farther away from the line of rotation and the inner radius, rin, is the distance of the expression closer to the line of rotation at a particular value of y.

The region is being rotated about the line the y-axis, so solve the first equation for x:

9781118496718-eq181485.png

Note that you keep the positive root because the region being rotated is in the first quadrant.

Find the limits of integration by setting the functions equal to each other and solving for y:

9781118496718-eq181486.png

For a value in the interval (0, 1), you have 9781118496718-eq181487.tif so that 9781118496718-eq181488.tif and 9781118496718-eq181489.tif. Therefore, the integral to find the volume becomes

9781118496718-eq181490.png

677. 9781118496718-eq181491.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181492.png

Here's the region being rotated:

9781118496718-un1840.tif

In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional slice is a washer, you can find A(x) by using

9781118496718-eq181493.png

where the outer radius, rout, is the distance of the function farther away from the line of rotation and the inner radius, rin, is the distance of the function closer to the line of rotation at a particular value of x.

Begin by finding the point of intersection by setting the functions equal to each other and solving for x:

9781118496718-eq181494.png

The limits of integration are therefore x = 0 (given) and x = 1. Note that rout = 2 – x2/3 and rin = 2 – 1 = 1 because you're rotating the region about the line y = 2. Therefore, the integral to find the volume is

9781118496718-eq181495.png

678. 9781118496718-eq181496.png

Recall the definition of volume: Let S be a solid that lies between y = c and y = d. If A(y) is the cross-sectional area of S in the plane Py that goes through y and is perpendicular to the y-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181497.png

In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional slice is a washer, you can find A(y) by using

9781118496718-eq181498.png

where the outer radius, rout, is the distance of the expression farther away from the line of rotation and the inner radius, rin, is the distance of the expression closer to the line of rotation at a particular value of y.

At x = 0, you have y = 02/3 = 0, which will be the lower limit of integration; y = 1 corresponds to the upper limit of integration.

Solving the equation y = x2/3 for x gives you x = y3/2. Note that because you're rotating the region about the line x = –1, you have rout = y3/2 + 1 and rin = 0 + 1 = 1. Therefore, the integral to find the volume is

9781118496718-eq181499.png

679. 9781118496718-eq181500.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181501.png

Here's the region being rotated:

9781118496718-un1841.tif

In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional slice is a washer, you can find A(x) by using

9781118496718-eq181502.png

where the outer radius, rout, is the distance of the function farther away from the line of rotation and the inner radius, rin, is the distance of the function closer to the line of rotation at a particular value of x.

On the interval 9781118496718-eq181503.tif, you have sec x < 4. Because you're rotating the region about the line y = 4, you have rout = 4 and rin = 4 – sec x. Therefore, the integral to find the volume is

9781118496718-eq181504.png

680. 9781118496718-eq181505.png

Recall the definition of volume: Let S be a solid that lies between y = c and y = d. If A(y) is the cross-sectional area of S in the plane Py that goes through y and is perpendicular to the y-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181506.png

In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional slice is a washer, you can find A(y) by using

9781118496718-eq181507.png

where the outer radius, rout, is the distance of the expression farther away from the line of rotation and the inner radius, rin, is the distance of the expression closer to the line of rotation at a particular value of y.

Begin by setting the expressions equal to each other to find the points of intersection, which correspond to the limits of integration:

9781118496718-eq181508.png

Because you're rotating the region about the line x = 5, you have rout = 5 – y2 and rin = 5 – 4 = 1. Therefore, the integral to find the volume is

9781118496718-eq181509.png

681. 9781118496718-eq181510.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181511.png

Here's the region that's being rotated:

9781118496718-un1842.tif

In this case, the cross-sectional slice has the shape of a washer. When the cross-sectional slice is a washer, you can find A(x) by using

9781118496718-eq181512.png

where the outer radius, rout, is the distance of the function farther away from the line of rotation and the inner radius, rin, is the distance of the function closer to the line of rotation at a particular value of x.

Note that the lines x = 0 and x = 1 correspond to the limits of integration. Because you're rotating the region about the line y = –1, you have rout = ex + 1 and rin = 0 + 1 = 1. Therefore, the integral to find the volume becomes

9781118496718-eq181513.png

682. 9781118496718-eq181514.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181515.png

Here, you want to find an expression for A(x).

The base is a circle that has a radius of 4 and is centered at the origin, so the equation of the circle is x2 + y2 = 16.

9781118496718-un1843.tif

If you consider a cross-sectional slice at point (x, y) on the circle, where y > 0, the base of the square equals 2y; therefore, the area of the cross-sectional slice is (2y)2 = 4y2.

9781118496718-un1844.tif

In order to integrate with respect to x, you can solve the equation of the circle for y2 and write the area as 4y2 = 4(16 – x2) = A(x). Because the base of the solid varies for x over the interval [–4, 4], the limits of integration are –4 and 4. Therefore, the integral to find the volume is

9781118496718-eq181516.png

683. 9781118496718-eq181517.png

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181518.png

Here, you want to find an expression for A(x).

The base is a circle that has a radius of 4 and is centered at the origin, so the equation of the circle is x2 + y2 = 16.

9781118496718-un1845.tif

If you consider a cross-sectional slice at point (x, y) on the circle, where y > 0, the base of the equilateral triangle equals 2y and has a height of 9781118496718-eq181519.tif; therefore, the area of the triangle is 9781118496718-eq181520.tif.

9781118496718-un1846.tif

In order to integrate with respect to x, you can solve the equation of the circle for y2 and write the area as 9781118496718-eq181521.tif = A(x). Because the base of the solid varies for x over the interval [–4, 4], the limits of integration are –4 and 4. Therefore, the integral to find the volume is

9781118496718-eq181522.png

684. 96

Recall the definition of volume: Let S be a solid that lies between y = c and y = d. If A(y) is the cross-sectional area of S in the plane Py that goes through y and is perpendicular to the y-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181523.png

Here, you want to find an expression for A(y).

9781118496718-un1847.tif

If you consider a cross-sectional slice that goes through a point (x, y) on the ellipse, where x > 0, then one side of the square has length 2x. The area of the square is

9781118496718-eq181524.png

9781118496718-un1848.tif

Because the base of the solid varies for y over the interval [–2, 2] the limits of integration are –2 and 2. Therefore, the integral to find the volume is

9781118496718-eq181525.png

Note that because 36 – 9y2 is an even function of y, the lower limit of integration was changed to zero and the resulting integral was multiplied by 2 to make the integral a bit easier to compute.

685. 9781118496718-eq181526.png

Recall the definition of volume: Let S be a solid that lies between y = c and y = d. If A(y) is the cross-sectional area of S in the plane Py that goes through y and is perpendicular to the y-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181527.png

Here, you want to find an expression for A(y).

9781118496718-un1849.tif

The base of S is the region bounded by the curves x = 0, y = 0, and 9781118496718-eq181528.tif. If you consider a cross-sectional slice that goes through a point (x, y) on the line 9781118496718-eq181529.tif, the cross-sectional slice is an isosceles triangle with height equal to the base, so the area is

9781118496718-eq181530.png

9781118496718-un1850.tif

Because the base of the solid varies for y over the interval [0, 4], the limits of integration are 0 and 4. Therefore, the integral to find the volume is

9781118496718-eq181531.png

686. 16

Recall the definition of volume: Let S be a solid that lies between x = a and x = b. If A(x) is the cross-sectional area of S in the plane Px that goes through x and is perpendicular to the x-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181532.png

Here, you want to find an expression for A(x).

9781118496718-un1851.tif

If you consider a cross-sectional slice that goes through a point (x, y) on the ellipse, where y > 0, then the hypotenuse of the right triangle has length 2y. If l is a leg of the isosceles right triangle, then by the Pythagorean theorem, l2 + l2 = (2y)2 so that l2 = 2y2. Solving the elliptical boundary function for y2 gives you 9781118496718-eq181533.tif. Therefore, the area of the triangle is

9781118496718-eq181534.png

9781118496718-un1852.tif

Because the base of the solid varies for x over the interval [–3, 3], the limits of integration are –3 and 3. Therefore, the integral to find the volume is

9781118496718-eq181535.png

Note that because y = 36 – 4x2 is an even function, the lower limit of integration was changed to zero and the resulting integral was multiplied by 2 to make the integral a bit easier to compute.

687. 9781118496718-eq181536.png

Recall the definition of volume: Let S be a solid that lies between y = c and y = d. If A(y) is the cross-sectional area of S in the plane Py that goes through y and is perpendicular to the y-axis, where A is a continuous function, then the volume of S is

9781118496718-eq181537.png

Here, you want to find an expression for A(y).

9781118496718-un1853.tif

The base of S is the region bounded by the curves x = 0, y = 0, and 9781118496718-eq181538.tif. If you consider a cross-sectional slice that goes through a point (x, y) on the line 9781118496718-eq181539.tif, that cross-sectional slice is a semicircle with a radius of 9781118496718-eq181540.tif. That means the area of the slice is

9781118496718-eq181541.png

9781118496718-un1854.tif

Because the base of the solid varies for y over the interval [0, 4], the limits of integration are 0 and 4. Therefore, the integral to find the volume is

9781118496718-eq181542.png

688. 4π

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181543.png

More generically, you can use the formula

9781118496718-eq181544.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Here, the limits of integration are given as x = 1 and x = 3 and the shell height is 9781118496718-eq181545.tif, so the integral becomes

9781118496718-eq181546.png

689. 8π

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181547.png

More generically, you can use the formula

9781118496718-eq181548.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f(x– g(x). Note that the limits of integration often correspond to points of intersection.

Find the other limit of integration by determining where y = x2 and y = 0 intersect: x2 = 0 gives you x = 0. The shell height is x2 – 0, so the integral becomes

9781118496718-eq181549.png

690. 9781118496718-eq181550.png

To find the volume of a solid obtained by rotating about the x-axis a region that's to the left of the curve x = f (y) and to the right of the y-axis from y = c to y = d with cylindrical shells, use the integral

9781118496718-eq181551.png

More generically, you can use the formula

9781118496718-eq181552.png

To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance from y to the line of rotation. To find the shell height, let f (y) be the curve that bounds the region on the right and let g(y) be the curve that bounds the region on the left; the shell height is given by f(y– g(y). Note that the limits of integration often correspond to points of intersection.

Because y = 1 corresponds to one of the limits of integration, find the other limit of integration by solving the equation 0 = y1/3, which gives you 0 = y. Then find the volume using cylindrical shells:

9781118496718-eq181553.png

691. 9781118496718-eq181554.png

The region is being rotated about the line x = –1, which is parallel to the y-axis. To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b using cylindrical shells, use the following integral:

9781118496718-eq181555.png

More generically, you can use the formula

9781118496718-eq181556.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

The line x = 2 gives you one of the limits of integration, so begin by finding the other limit of integration by determining where the function intersects the x-axis; 0 = x2 gives you 0 = x. Notice that the shell radius is (x + 1), so here's the integral to find the volume using cylindrical shells:

9781118496718-eq181557.png

692. 216π

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181558.png

More generically, you can use the formula

9781118496718-eq181559.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Here's the region being rotated about the y-axis:

9781118496718-un1855.tif

Begin by finding the limits of integration by setting the functions equal to each other and solving for x:

9781118496718-eq181560.png

Because 2x ≥ x2 – 4x on the interval [0, 6], the shell height is given by 2x – (x2 – 4x). Therefore, the integral to find the volume is

9781118496718-eq181561.png

693. 9781118496718-eq181562.png

To find the volume of a solid obtained by rotating about the x-axis a region to the left of the curve x = f (y) and to the right of the y-axis from y = c to y = d with cylindrical shells, use the integral

9781118496718-eq181563.png

More generically, you can use the formula

9781118496718-eq181564.png

To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance from y to the line of rotation. To find the shell height, let f (y) be the curve that bounds the region on the right and let g(y) be the curve that bounds the region on the left; the shell height is given by f(y– g(y). Note that the limits of integration often correspond to points of intersection.

Begin by isolating the y in the first equation to get y1/4 = x. Find the lower limit of integration by solving y1/4 = 0 to get y = 0. Then find the volume using cylindrical shells:

9781118496718-eq181565.png

694. 9781118496718-eq181566.png

To find the volume of a solid obtained by rotating about the x-axis a region to the left of the curve x = f (y) and to the right of the y-axis from y = c to y = d with cylindrical shells, use the integral

9781118496718-eq181567.png

More generically, you can use the formula

9781118496718-eq181568.png

To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance from y to the line of rotation. To find the shell height, let f (y) be the curve that bounds the region on the right and let g(y) be the curve that bounds the region on the left; the shell height is given by f(y– g(y). Note that the limits of integration often correspond to points of intersection.

Here's the region being rotated about the x-axis:

9781118496718-un1856.tif

Begin by finding the limits of integration. To do so, find the points of intersection of the two curves by setting the expressions equal to each other and solving for y:

9781118496718-eq181569.png

Note that 5y2 – y3 ≥ 0 on the interval [0, 5] so that the shell height is 5y2 – y3 – 0, or simply 5y2 – y3. Therefore, the integral to find the volume is

9781118496718-eq181570.png

695. 16π

The region is being rotated about the line x = 4, which is parallel to the y-axis. To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181571.png

More generically, you can use the formula

9781118496718-eq181572.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Begin by finding the points of intersection by setting the functions equal to each other and solving for x in order to find the limits of integration:

9781118496718-eq181573.png

Note that on the interval [0, 2], you have 4x – x2 ≥ x2 so that the shell height is (4x – x2) – (x2). Also note that the region is to the left of the line of rotation x = 4, so the shell radius is (4 – x). Therefore, the integral to find the volume is

9781118496718-eq181574.png

696. 9781118496718-eq181575.png

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181576.png

More generically, you can use the formula

9781118496718-eq181577.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Here, the limits of integration correspond to the lines x = 0 and x = 1, the shell height is (1 + x + x2), and the shell radius is x. Therefore, the integral to find the volume is

9781118496718-eq181578.png

697. 9781118496718-eq181579.png

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181580.png

More generically, you can use the formula

9781118496718-eq181581.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Set the functions equal to each other and solve for x in order to find the other limit of integration:

9781118496718-eq181582.png

Here, the shell height is given by 4 – (4x – x2) and the shell radius is x, so the integral to find the volume is

9781118496718-eq181583.png

698. 9781118496718-eq181584.png

To find the volume of a solid obtained by rotating about the x-axis a region to the left of the curve x = f (y) and to the right of the y-axis from y = c to y = d with cylindrical shells, use the integral

9781118496718-eq181585.png

More generically, you can use the formula

9781118496718-eq181586.png

To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance from y to the line of rotation. To find the shell height, let f (y) be the curve that bounds the region on the right and let g(y) be the curve that bounds the region on the left; the shell height is given by f(y– g(y). Note that the limits of integration often correspond to points of intersection.

Notice that if x = 0, then 9781118496718-eq181587.tif, so the limits of integration are y = 0 (given) and y = 3. Solving for x in the equation 9781118496718-eq181588.tif gives you x = 9 – y2 so that the shell height is (9 – y2). Therefore, the integral to find the volume is

9781118496718-eq181589.png

699. 9781118496718-eq181590.png

The region is being rotated about the line x = 2, which is parallel to the y-axis. To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181591.png

More generically, you can use the formula

9781118496718-eq181592.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Find the limits of integration by setting the functions equal to each other and solving for x:

9781118496718-eq181593.png

Because the line of rotation x = 2 is to the right of the region being rotated, the shell radius is (2 – x). Therefore, to find the volume of revolution using cylindrical shells, you use the following integral:

9781118496718-eq181594.png

700. 9781118496718-eq181595.png

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181596.png

More generically, you can use the formula

9781118496718-eq181597.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Here's the region being rotated about the y-axis:

9781118496718-un1857.tif

Begin by isolating y in the second equation to get y = 5 – 2x. Then set the expressions equal to each other and solve for x to find the x coordinates of the points of intersection:

9781118496718-eq181598.png

Because 5 + 3x – x2 ≥ 5 – 2x on the interval [0, 5], the shell height is given by (5 + 3x – x2) – (5 – 2x). Therefore, the integral to find the volume is

9781118496718-eq181599.png

701. 9781118496718-eq181600.png

The region is being rotated about the line x = –1, which is parallel to the y-axis. To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181601.png

More generically, you can use the formula

9781118496718-eq181602.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Here's the region being rotated about the line x = –1:

9781118496718-un1858.tif

Find the point of intersection using y = xx + y = 5, and substitution to get x + x = 5 so that 9781118496718-eq181603.tif. Notice that on the interval 9781118496718-eq181604.tif, the upper boundary of the region is x and the lower boundary is y = 0; on 9781118496718-eq181605.tif the upper boundary of the region is 5 – x and the lower boundary isy = 0. Because the line of rotation is x = –1, the shell radius is (x + 1). Therefore, to find the volume using cylindrical shells, you use two integrals:

9781118496718-eq181606.png

702. 2π

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181607.png

More generically, you can use the formula

9781118496718-eq181608.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

In this example, the limits of integration are given as x = 0 and 9781118496718-eq181609.tif, and the shell height is sin(x2) – 0, so the integral becomes

9781118496718-eq181610.png

To evaluate this integral, use the substitution u = x2 so that du = 2 dx, or 9781118496718-eq181611.tif. You can find the new limits of integration by noting that if 9781118496718-eq181612.tif, then 9781118496718-eq181613.tif and that if x = 0, then u = 02 = 0. Using these values gives you

9781118496718-eq181614.png

703. 2π(e2 + 1)

To find the volume of a solid obtained by rotating about the x-axis a region to the left of the curve x = f (y) and to the right of the y-axis from y = c to y = d with cylindrical shells, use the integral

9781118496718-eq181615.png

More generically, you can use the formula

9781118496718-eq181616.png

To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance from y to the line of rotation. To find the shell height, let f (y) be the curve that bounds the region on the right and let g(y) be the curve that bounds the region on the left; the shell height is given by f(y– g(y). Note that the limits of integration often correspond to points of intersection.

In this example, the limits of integration are given as y = 0 and y = 2, and the shell height is ey – 0. Therefore, to find the volume using cylindrical shells, you use

9781118496718-eq181617.png

To evaluate this integral, use integration by parts with u = y so that du = dy, and let dv = edy so that v = ey:

9781118496718-eq181618.png

704. 9781118496718-eq181619.png

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181620.png

More generically, you can use the formula

9781118496718-eq181621.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

In this example, the limits of integration correspond to the lines x = 0 and x = 3 and the shell height is given by 9781118496718-eq181622.tif. Therefore, to find the volume of revolution using cylindrical shells, use the following integral:

9781118496718-eq181623.png

To evaluate this integral, use the substitution u = –x2 so that du = –2x dx, or 9781118496718-eq181624.tif. Find the new limits of integration by noting that if x = 3, then u = –9, and if x = 0, then u = 0. With these new values, the integral becomes

9781118496718-eq181625.png

705. 9781118496718-eq181626.png

The region is being rotated about the line x = –1, which is parallel to the y-axis. To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181627.png

More generically, you can use the formula

9781118496718-eq181628.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Here's the region being rotated about the line x = –1:

9781118496718-un1859.tif

Begin by writing x = y3 as x1/3 = y. Then set the functions equal to each other to find the points of intersection, which give you the limits of integration. Cube both sides of the equation and factor to solve for x:

9781118496718-eq181629.png

Notice that x1/3 > x2 on the interval [0, 1], so the shell height is (x1/3 – x2). For a value in the interval [0, 1], the distance from x to the line of rotation x = –1 is (x + 1), so the shell radius is (x + 1). Therefore, the integral to find the volume of revolution using cylindrical shells is

9781118496718-eq181630.png

706. 4π(2 ln 3 – 1)

The region is being rotated about the line x = 4, which is parallel to the y-axis. To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181631.png

More generically, you can use the formula

9781118496718-eq181632.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Because the region being rotated is to the left of the line of rotation x = 4, the shell radius is (4 – x). To find the volume using cylindrical shells, use the following integral:

9781118496718-eq181633.png

707. 9781118496718-eq181634.png

The region is being rotated about the line y = 1, which is parallel to the x-axis. To find the volume of a solid obtained by rotating about the x-axis a region to the left of the curve x = f (y) and to the right of the y-axis from y = c to y = d with cylindrical shells, use the integral

9781118496718-eq181635.png

More generically, you can use the formula

9781118496718-eq181636.png

To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance from y to the line of rotation. To find the shell height, let f (y) be the curve that bounds the region on the right and let g(y) be the curve that bounds the region on the left; the shell height is given by f(y– g(y). Note that the limits of integration often correspond to points of intersection.

Here's the region being rotated about the line y = 1:

9781118496718-un1860.tif

Begin by isolating x in each equation to get y2 = x and y1/3 = x. Set these expressions equal to each other to find the points of intersection, which give you the limits of integration. Cube both sides of the equation and factor to solve for y:

9781118496718-eq181637.png

Note that y1/3 > y2 for a point in the interval (0, 1), so the shell height is y1/3 – y2. Also note that the region being rotated is below the line of rotation y = 1, so in the interval (0, 1), the distance from y to the line of rotation is (1 – y), making the shell radius (1 – y). Therefore, the integral to find the volume using cylindrical shells is

9781118496718-eq181638.png

708. 9781118496718-eq181639.png

To find the volume of a solid obtained by rotating about the x-axis a region to the left of the curve x = f (y) and to the right of the y-axis from y = c to y = d with cylindrical shells, use the integral

9781118496718-eq181640.png

More generically, you can use the formula

9781118496718-eq181641.png

To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance from y to the line of rotation. To find the shell height, let f (y) be the curve that bounds the region on the right and let g(y) be the curve that bounds the region on the left; the shell height is given by f(y– g(y). Note that the limits of integration often correspond to points of intersection.

One of the limits of integration, y = 0, is given. Begin by finding the points of intersection (which will help you find the other limit of integration) by setting the functions equal to each other. Square both sides and factor to solve for x:

9781118496718-eq181642.png

Notice that x = –1 is an extraneous solution. If x = 2, then 9781118496718-eq181643.tif, which gives you the other limit of integration.

Solving 9781118496718-eq181644.tif for x gives you x = y2 – 2. Note that y > y2 – 2 in the interval [0, 2], so the shell height is y – (y2 – 2). Therefore, the integral to find the volume using cylindrical shells is

9781118496718-eq181645.png

709. 9781118496718-eq181646.png

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181647.png

More generically, you can use the formula

9781118496718-eq181648.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Note that the lines x = 0 and x = 1 correspond to the limits of integration and that the shell height is 9781118496718-eq181649.tif, or simply 9781118496718-eq181650.tif. To find the volume using cylindrical shells, use the following integral:

9781118496718-eq181651.png

Now use the substitution 9781118496718-eq181652.tif so that du = –x dx, or –du = x dx. Find the new limits of integration by noting that if x = 1, then 9781118496718-eq181653.tif, and if x = 0, then u = 0. Using these new values gives you the following:

9781118496718-eq181654.png

710. 9781118496718-eq181655.png

To find the volume of a solid obtained by rotating about the y-axis a region that's under the curve y = f (x) and above the x-axis from x = a to x = b with cylindrical shells, use the integral

9781118496718-eq181656.png

More generically, you can use the formula

9781118496718-eq181657.png

To find the shell radius, let x be in the interval [a, b]; the shell radius is the distance from x to the line of rotation. To find the shell height, let f (x) be the function that bounds the region above and let g(x) be the function that bounds the region below; the shell height is given by f (x– g(x). Note that the limits of integration often correspond to points of intersection.

Notice that on the interval [1, 2], you have 9781118496718-eq181658.tif, so the shell height is 9781118496718-eq181659.tif. Therefore, the integral to find the volume is

9781118496718-eq181660.png

To evaluate 9781118496718-eq181661.tif, use integration by parts with u = ln x so that 9781118496718-eq181662.tif, and use dv = x dx so that 9781118496718-eq181663.tif. These substitutions give you

9781118496718-eq181664.png

Evaluating the integral 9781118496718-eq181665.png gives you

9781118496718-eq181666.png

711. π2 – 2π

To find the volume of a solid obtained by rotating about the x-axis a region to the left of the curve x = f (y) and to the right of the y-axis from y = c to y = d with cylindrical shells, use the integral

9781118496718-eq181667.png

More generically, you can use the formula

9781118496718-eq181668.png

To find the shell radius, let y be in the interval [c, d]; the shell radius is the distance from y to the line of rotation. To find the shell height, let f (y) be the curve that bounds the region on the right and let g(y) be the curve that bounds the region on the left; the shell height is given by f(y– g(y). Note that the limits of integration often correspond to points of intersection.

Here, the limits of integration correspond to the lines y = 0 and 9781118496718-eq181669.tif, and the shell height is cos y. To find the volume using cylindrical shells, use the integral

9781118496718-eq181670.png

To evaluate the integral, use integration by parts with u = y so that du = dy, and let dv = cos y dy so that v = sin y:

9781118496718-eq181671.png

712. 1,470 J

First find the force by multiplying mass by acceleration:

9781118496718-eq181672.png

The force is constant, so no integral is required to find the work:

9781118496718-eq181673.png

713. 9,600 J

The force is constant, so no integral is required to find the work. Enter the numbers in the work equation and solve:

9781118496718-eq181674.png

714. 337.5 ft·lb

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181675.tif be a sample point in the 9781118496718-eq181676.tif subinterval [xi–1xi].

The portion of the rope that is from xi–1 feet to xi feet below the top of the cliff weighs (0.75)Δx pounds and must be lifted approximately 9781118496718-eq181677.tif feet. Therefore, its contribution to the total work is approximately 9781118496718-eq181678.tif ft·lb. The total work is

9781118496718-eq181679.png

715. 253.125 ft·lb

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181680.tif be a sample point in the 9781118496718-eq181681.tifsubinterval [xi–1xi].

First consider the work required to pull the top half of the rope that is from xi–1 feet to xi feet below the top of the cliff, where x is in the interval [0, 15]. This part of the rope weighs (0.75)Δx pounds and must be lifted approximately 9781118496718-eq181682.tif feet, so its contribution to the total work is approximately 9781118496718-eq181683.tif foot-pounds. Therefore, the total work for the top half of the rope is

9781118496718-eq181684.png

The bottom half of the rope must be lifted 15 feet, so the work required to lift the bottom half is

9781118496718-eq181685.png

To find the total work required, add the two values: 84.375 + 168.75 = 253.125 ft·lb.

Note that you can find Wbottom without an integral because there's a constant force on this section of the rope!

716. 630,000 ft·lb

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181686.tif be a sample point in the 9781118496718-eq181687.tif subinterval [xi–1xi].

A section of cable that is from xi–1 feet to xi feet below the top of the building weighs 4Δx pounds and must be lifted approximately 9781118496718-eq181688.tif feet, so its contribution to the total work is approximately 9781118496718-eq181689.tif foot-pounds. Therefore, the work required to lift the cable is

9781118496718-eq181690.png

The work required to lift the piece of metal is (1, 500 lb)(300 ft) = 450,000 ft·lb. Therefore, the total work required is 180,000 + 450,000 = 630,000 ft·lb.

717. 22,500 ft·lb

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181691.tif be a sample point in the 9781118496718-eq181692.tifsubinterval [xi–1xi].

The cable weighs 9781118496718-eq181693.tif. The portion of the cable that is from xi–1 feet to xi feet below the top of the building weighs 2Δx pounds and must be lifted approximately 9781118496718-eq181694.tif feet, so its contribution to the total work is approximately 9781118496718-eq181695.tif foot-pounds. The total work is

9781118496718-eq181696.png

718. 39,200 J

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181697.tif be a sample point in the 9781118496718-eq181698.tifsubinterval [xi–1xi].

A horizontal slice of water that is Δx meters thick and is at a distance of 9781118496718-eq181699.tif meters from the top of the tank has a volume of ((4)(2)(Δx)) cubic meters and has a mass of

9781118496718-eq181700.png

Because mass is not a force, you must multiply it by the acceleration due to gravity, 9.8 meters per second squared, in order to find the weight of the slice in newtons:

9781118496718-eq181701.png

The work required to pump out this slice of water is approximately 9781118496718-eq181702.tif joules. Therefore, the total work is

9781118496718-eq181703.png

719. 7.875 ft·lb

Begin by finding the spring constant, k, using the information about the work required to stretch the spring 2 feet beyond its natural length:

9781118496718-eq181704.png

Next, find the work required to stretch the spring 18 inches, or 1.5 feet, beyond its natural length:

9781118496718-eq181705.png

720. 1.51 J

Begin by using the equation F(x) = kx to find the value of the spring constant. Make sure you convert the centimeters to meters so you can use the units newton-meters, or joules.

9781118496718-eq181706.png

Now compute the work, again making sure to convert the lengths into meters. Because you're stretching the spring from 12 centimeters beyond its natural length to 22 centimeters beyond its natural length, the limits of integration are from 9781118496718-eq181707.tif to 9781118496718-eq181708.tif; therefore, the integral to compute the work is

9781118496718-eq181709.png

721. 9781118496718-eq181710.png

Use an integral to find the work:

9781118496718-eq181711.png

722. 25.2 J

Begin by finding the spring constant, k, using the information about the initial work required to stretch the spring 10 centimeters, or 0.10 meters, from its natural length:

9781118496718-eq181712.png

Now find the work required to stretch the spring from 30 centimeters to 42 centimeters, or from 0.15 meters from its natural length to 0.27 meters from its natural length:

9781118496718-eq181713.png

723. 9781118496718-eq181714.png

Begin by using the equation F(x) = kx to find the value of k, the spring constant. The force is measured in foot-pounds, so make sure the length is in feet rather than inches.

9781118496718-eq181715.png

Therefore, the force equation is F (x) = 30x.

Because the spring starts at its natural length, x = 0 is the lower limit of integration. Because the spring is stretched 8 inches, or 9781118496718-eq181716.tif feet, beyond its natural length, the upper limit of integration is 9781118496718-eq181717.tif. Therefore, the integral to find the work is

9781118496718-eq181718.png

724. 2.25 J

Begin by using the equation F (x) = kx to find the value of the spring constant. Make sure you convert the centimeters to meters so you can use the units newton-meters, or joules. If x is the distance the spring is stretched beyond its natural length, then 9781118496718-eq181719.tif. Therefore, the spring constant is

9781118496718-eq181720.png

Now compute the work, making sure to convert the distances into meters. Because you're computing the work while moving the spring from 5 centimeters beyond its natural length to 10 centimeters beyond its natural length, the limits of integration are from 9781118496718-eq181721.tif to 9781118496718-eq181722.tif.

9781118496718-eq181723.png

725. 400 ft·lb

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181724.tif be a sample point in the ith subinterval [xi–1xi]. Let 9781118496718-eq181726.tif be the distance from the middle of the chain.

9781118496718-un1861.tif

Notice that when you're lifting the end of the chain to the top, only the bottom half of the chain moves. A section of chain that is Δx in length and that is 9781118496718-eq181727.tif feet from the middle of the chain weighs 4Δx pounds and will move approximately 9781118496718-eq181728.tif feet. Therefore, the work required is

9781118496718-un1861.tif

9781118496718-eq181729.png

726. 613 J

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181730.tif be a sample point in the ith subinterval [xi–1xi].

The part of the chain x meters from the lifted end is raised 5 – x meters if 0 ≤ x ≤ 5 and is lifted 0 meters otherwise. Therefore, the work required is

9781118496718-eq181732.png

727. 3,397,333 J

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181733.tif be a sample point in the ith subinterval [xi–1xi].

Consider a horizontal slice of water that is Δx meters thick and is at a height of 9781118496718-eq181735.tif meters from the bottom of the tank. The trough has a triangular face with a width and height of 4 meters, so by similar triangles, the width of the slice of water is the same as the height 9781118496718-eq181736.tif; therefore, the volume of the slice of water is 9781118496718-eq181737.tif cubic meters. To find the weight of the water, multiply the acceleration due to gravity (g = 9.8 m/s2) by the mass, which equals the density of water (1,000 kg/m3) multiplied by the volume:

9781118496718-eq181738.png

The water must travel a distance of 9781118496718-eq181739.tif meters to exit the tank. Therefore, the total work required is

9781118496718-eq181740.png

728. 169,646 ft·lb

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181741.tif be a sample point in the ith subinterval [xi–1xi].

9781118496718-un1862.tif

A horizontal slice of water that is Δx feet thick and is at a depth of 9781118496718-eq181743.tif feet from the top of the tank is cylindrical, so it has a volume of 9781118496718-eq181744.tif cubic feet. Because water weighs 62.5 pounds per cubic foot, this slice of water weighs 9781118496718-eq181745.tif pounds. The work required to pump out this slice of water is

9781118496718-eq181746.png

Therefore, the total work is

9781118496718-eq181747.png

729. 3,000 ft·lb

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181748.tif be a sample point in the ith subinterval [xi–1xi].

The work required to lift only the bucket is

9781118496718-eq181750.png

Lifting the bucket takes 50 seconds, so the bucket is losing 1 pound of water per second. At time t (in seconds), the bucket is 9781118496718-eq181751.tif feet above the original 100-foot depth, but it now holds only (50 – t) pounds of water. In terms of distance, the bucket holds 9781118496718-eq181752.tif pounds of water when it's 9781118496718-eq181753.tif feet above the original 100-foot depth. Moving this amount of water a distance of Δx requires 9781118496718-eq181754.tif foot-pounds of work. Therefore, the work required to lift the only the water is

9781118496718-eq181755.png

Adding the work to lift the bucket gives you the total work required: 500 + 2,500 = 3,000 foot-pounds.

730. 4 cm

Begin by using the information about how much work is required to stretch the spring from 8 to 10 centimeters to solve for the spring constant and the natural length of the spring; you do so by setting up a system of equations.

You don't know the natural length L for the integral involved in stretching the spring from 8 centimeters to 10 centimeters beyond its natural length, so the limits of integration for one of the integrals involving work are from 0.08 – L to 0.10 – L. Likewise, the other integral will have limits of integration from 0.10 – L to 0.12 – L. Therefore, you have the following two equations for work:

9781118496718-eq181756.png

9781118496718-eq181757.png

Expanding the first of the two integrals gives you

9781118496718-eq181758.png

Likewise, expanding the second integral gives you

9781118496718-eq181759.png

Now solve for the spring constant k. By subtracting 20 = k[0.0036 – 0.04L] from 28 = k[0.0044 – 0.04L], you're left with 8 = k(0.0008) so that k = 10,000.

Finally, find the natural length of the spring. Substituting the value of k into the equation 20 = k[0.0036 – 0.04L] gives you 20 = 10,000[0.0036 – 0.04L], and solving for L yields 0.04 m = L. Therefore, the natural length of the spring is 4 centimeters.

731. 5.29 ft

Let n be the number of subintervals of length Δx, and let 9781118496718-eq181760.tif be a sample point in the ith subinterval [xi–1xi].

A horizontal slice of water that is Δx feet thick and is at a distance of 9781118496718-eq181762.tif feet from the top of the tank is cylindrical and therefore has a volume of 9781118496718-eq181763.tif cubic feet. Because water weighs 62.5 pounds per cubic foot, this slice of water weighs 9781118496718-eq181764.tif pounds. The work required to pump out this slice of water is

9781118496718-eq181765.png

You don't know how many feet of water are pumped out, so you need to solve for the upper limit of integration in the equation:

9781118496718-eq181766.png

Because the water level in the tank was reduced, the water is now 5.29 feet from the top of the tank.

732. 16.67 cm

Begin by using the information about how much work is required to stretch the spring from 40 to 60 centimeters (0.40 to 0.60 meters) beyond its natural length to solve for the spring constant and the natural length of the spring; to do so, set up a system of equations.

Because you don't know the natural length L for the integral involved in stretching the spring from 0.40 meters to 0.60 meters beyond its natural length, the limits of integration for one of the integrals involving work are from 0.40 – L to 0.60 – L. Likewise, the other integral will have limits of integration from 0.60 – L to 0.80 – L. Therefore, you have the following two equations for work:

9781118496718-eq181767.png

9781118496718-eq181768.png

Expanding the first of the two integrals gives you

9781118496718-eq181769.png

Likewise, expanding the second integral gives you

9781118496718-eq181770.png

Now solve for the spring constant k. By subtracting 50 = k[0.2 – 0.4L] from 80 = k[0.28 – 0.4L], you're left with 30 = k(0.08) so that k = 375.

Finally, find the natural length of the spring. Substituting the value of k into the equation 50 = k[0.2 – 0.4L] gives you 50 = 375[0.2 – 0.4L]; solving for L yields 9781118496718-eq181771.tif meters, so the natural length of the spring is approximately 16.67 centimeters.

733. 9781118496718-eq181772.png

Let n be the number of subintervals of length Δx and let 9781118496718-eq181773.tif be a sample point in the ith subinterval.

Here's a view of the tank underground:

9781118496718-un1863.tif

And here's a cross-section of the tank:

9781118496718-un1864.tif

A horizontal slice of water that is Δx meters thick and is at a distance of 9781118496718-eq181774.tif meters down from the middle of the tank has a volume of 9781118496718-eq181775.tif cubic meters. To find the weight of the water, multiply the acceleration due to gravity (g = 9.8 m/s2) by the mass of the water, which equals the density of water (1,000 kg/m3) times the volume:

9781118496718-eq181776.png

The slice of water must travel a distance of 9781118496718-eq181777.tif meters to reach ground level. (Note that negative x values correspond to slices of water that are above the middle of the tank.) Therefore, the total work required is

9781118496718-eq181778.png

Notice that second integral is zero because 9781118496718-eq181779.tif is an odd function being integrated over an interval that's symmetric about the origin. To evaluate 9781118496718-eq181780.tif, notice that the integral represents the area of a semicircle with a radius of 1, so the work equals

9781118496718-eq181781.png

734. 20,944 ft·lb

Here's a view of the tank and its cross-section:

9781118496718-un1865.tif

Let n be the number of subintervals of length Δx and let 9781118496718-eq181782.tif be a sample point in the ith subinterval. Consider a horizontal slice of water that is Δx feet thick and is at a distance of 9781118496718-eq181783.tif feet from the top of the tank; let 9781118496718-eq181784.tif be the radius of the slice of water. Because the tank has a shape of a hemisphere of radius 4, you have the following relationship:

9781118496718-eq181785.png

Therefore, the slice has a volume of

9781118496718-eq181786.png

And the force on this slice is

9781118496718-eq181787.png

The slice of water must travel a distance of 9781118496718-eq181788.tif feet to exit the tank. Therefore, the total work required is

9781118496718-eq181789.png

735. 14,726 ft·lb

Here's a view of the tank and its cross-section:

9781118496718-un1866.tif

9781118496718-un1867.tif

Let n be the number of subintervals of length Δx and let 9781118496718-eq181790.tif be a sample point in the ith subinterval. Consider a horizontal slice of water that is Δx feet thick and is 9781118496718-eq181791.tif feet from the bottom of the tank with a width of 9781118496718-eq181792.tif. This slice has a volume of

9781118496718-eq181793.png

By similar triangles, you have 9781118496718-eq181794.tif, so 9781118496718-eq181795.tif; therefore, the volume becomes

9781118496718-eq181796.png

The force on this slice of water is

9781118496718-eq181797.png

The distance this slice must travel to exit the tank is 9781118496718-eq181798.tif feet. Therefore, the total work required is

9781118496718-eq181799.png

736. 9781118496718-eq181800.png

Using the average value formula with a = –1, b = 2, and f (x) = x3 gives you

9781118496718-eq181801.png

737. 9781118496718-eq181802.png

Using the average value formula with a = 0, 9781118496718-eq181803.tif, and f (x) = sin x gives you

9781118496718-eq181804.png

738. 9781118496718-eq181805.png

Using the average value formula with a = 0, 9781118496718-eq181806.tif, and f (x) = (sinx)(cos x) gives you

9781118496718-eq181807.png

Next, use the substitution u = sin x so that du = cos x dx. Find the new limits of integration by noting that if 9781118496718-eq181808.tif, then 9781118496718-eq181809.tif, and if x = 0, then u = sin 0 = 0. With these new values, you get the following:

9781118496718-eq181810.png

739. 9781118496718-eq181811.png

Using the average value formula with a = 0, b = 2, and 9781118496718-eq181812.tif gives you

9781118496718-eq181813.png

Begin by using the substitution u = 1 + x3 so that du = 3xdx, or 9781118496718-eq181814.tif. You can also find the new limits of integration by noting that if x = 2, then u = 1 + 23 = 9, and if x = 0, then u = 1 + 03 = 1. With these values, you have the following:

9781118496718-eq181815.png

740. 9781118496718-eq181816.png

Using the average value formula with a = 0, b = ln 3, and f (x) = sinh x cosh x, gives you

9781118496718-eq181817.png

Use the substitution u = cosh x so that du = sinh x. You can find the new limits of integration by noting that if x = ln 3, then 9781118496718-eq181818.tif, and if x = 0, then 9781118496718-eq181819.tif. With these new values, you have

9781118496718-eq181820.png

741. 9781118496718-eq181821.png

Using the average value formula with a = 1, b = 4, and 9781118496718-eq181822.tif gives you

9781118496718-eq181823.png

Next, use the substitution u = 1 + r so that du = dr. Find the new limits of integration by noting that if r = 4, then u = 1 + 4 = 5, and if r = 1, then u = 1 + 1 = 2. With these new values, you find that

9781118496718-eq181824.png

742. 9781118496718-eq181825.png

Using the average value formula with a = 0, b = 8, and 9781118496718-eq181826.tif gives you

9781118496718-eq181827.png

Next, use the substitution u = x + 2 so that du = dx. You can find the new limits of integration by noting that if x = 8, then u = 8 + 2 = 10, and if x = 0, then u = 0 + 2 = 2. With these new values, you find that

9781118496718-eq181828.png

743. d = 1

Using the average value formula with a = 0, b = d, and f (x) = 2 + 4x – 3x2 gives you

9781118496718-eq181829.png

Next, set this expression equal to 3 and solve for d:

9781118496718-eq181830.png

744. d = 4

Using the average value formula with a = 0, b = d, and f (x) = 3 + 6x – 9x2 gives you

9781118496718-eq181831.png

Next, set this expression equal to –33 and solve for d:

9781118496718-eq181832.png

Because the interval is of the form [0, d], the only solution is d = 4.

745. 9781118496718-eq181833.png

Begin by finding the average value of the function on the given interval by using the average value formula with a = 1, b = 3, and 9781118496718-eq181834.tif:

9781118496718-eq181835.png

Next, set the original function equal to the average value and solve for c:

9781118496718-eq181836.png

Because 9781118496718-eq181837.tif isn't in the specified interval, the solution is 9781118496718-eq181838.tif.

746. 9781118496718-eq181839.png

Begin by finding the average value of the function on the given interval by using the average value formula with a = 4, b = 9, and 9781118496718-eq181840.tif:

9781118496718-eq181841.png

Next, set the original function equal to the average value and solve for c:

9781118496718-eq181842.png

Because 9781118496718-eq181843.tif is in the given interval, you've found the solution.

747. 9781118496718-eq181844.png

Begin by finding the average value of the function on the given interval by using the average value formula with a = –2, b = 2, and f (x) = 5 – 3x2:

9781118496718-eq181845.png

Next, set the original function equal to the average value and solve for c:

9781118496718-eq181846.png

Both values fall in the given interval, so both are solutions.

748. 9781118496718-eq181847.png

Using the average value formula with a = 0, 9781118496718-eq181848.tif, and f (x) = x sin x gives you

9781118496718-eq181849.png

To integrate this function, you can use integration by parts. Notice that for the indefinite integral 9781118496718-eq181850.tif, you can let u = x so that du = dx and let dv = sin x dx so that v = –cos x dx. Using integration by parts formula gives you

9781118496718-eq181851.png

Therefore, to evaluate the definite integral 9781118496718-eq181852.tif, you have

9781118496718-eq181853.png

749. 9781118496718-eq181854.png

Using the average value formula with a = 0, 9781118496718-eq181855.tif, and 9781118496718-eq181856.tif gives you

9781118496718-eq181857.png

To evaluate this integral, begin by splitting up the fraction:

9781118496718-eq181858.png

For the first integral, use a substitution, letting u = x2 + 1 so that du = 2x dx, or 9781118496718-eq181859.tif. You can also find the new limits of integration for the first integral by noting that if 9781118496718-eq181860.tif, then 9781118496718-eq181861.tif, and if x = 0, then u = 02 + 1 = 1. With these new values, you have

9781118496718-eq181862.png

750. 9781118496718-eq190001.png

To find the derivative of 9781118496718-eq190002.tif, use the derivative formula for sin−1 x and the chain rule:

9781118496718-eq190003.png

751. 9781118496718-eq190004.png

To find the derivative of 9781118496718-eq190005.tif, you need to know the derivative formula for the inverse cosine function and also use the chain rule:

9781118496718-eq190006.png

752. 9781118496718-eq190007.png

First rewrite the radical using an exponent:

9781118496718-eq190008.png

Then use the derivative formula for tan−1 x and the chain rule to find the derivative:

9781118496718-eq190009.png

753. 9781118496718-eq190010.png

First rewrite the radical using an exponent:

9781118496718-eq190011.png

Find the derivative using the product rule and the derivative formula for sin−1 x:

9781118496718-eq190012.png

754. 9781118496718-eq190013.png

To find the derivative of 9781118496718-eq190014.tif, use the derivative formula for tan−1 x along with the chain rule:

9781118496718-eq190015.png

755. 9781118496718-eq190016.png

Find the derivative of 9781118496718-eq190017.tif using the chain rule and the derivative formula for sec−1 t:

9781118496718-eq190018.png

756. 9781118496718-eq190019.png

To find the derivative of y = csc−1 e2x, use the derivative formula for csc−1 x along with the chain rule:

9781118496718-eq190020.png

757. 9781118496718-eq190021.png

To find the derivative of 9781118496718-eq190022.tif, use the derivative formula for ex along with the chain rule and product rule:

9781118496718-eq190023.png

758. 9781118496718-eq190025.png

Rewrite the arccosine as the inverse cosine:

9781118496718-eq190026.png

Now use the chain rule along with the derivative formula for cos−1 x:

9781118496718-eq190027.png

Next, simplify the term underneath the square root. Begin by getting common denominators under the radical:

9781118496718-eq190028.png

Then split up the square root and simplify:

9781118496718-eq190029.png

759. 9781118496718-eq190030.png

First, rewrite the radical using an exponent:

9781118496718-eq190031.png

Use the product rule on the first term and the chain rule on the second term to get the derivative:

9781118496718-eq190032.png

760. 0

Rewrite 9781118496718-eq190033.tif in the given function using a negative exponent:

9781118496718-eq190034.png

Use the derivative formula for cot−1 x for the first term, and use the same derivative formula and the chain rule for the second term. Here are the calculations:

9781118496718-eq190035.png

761. 9781118496718-eq190036.png

Here's the given function:

9781118496718-eq190037.png

Begin by using the derivative formula for tan−1 x on the first term and using the quotient rule on the second term:

9781118496718-eq190038.png

Then get common denominators and simplify:

9781118496718-eq190039.png

762. 9781118496718-eq190040.png

Rewrite the radical in the given function using an exponent:

9781118496718-eq190041.png

Use the derivative formula for tan−1 x along with the chain rule:

9781118496718-eq190042.png

Now the fun algebra simplification begins:

9781118496718-eq190043.png

763. 9781118496718-eq190044.png

Here's the initial problem:

9781118496718-eq190045.png

Because you have the antiderivative formula 9781118496718-eq190046.tif, you can simply apply that here to find the solution:

9781118496718-eq190047.png

764. 9781118496718-eq190048.png

Here's the given problem:

9781118496718-eq190049.png

Because you have the antiderivative formula 9781118496718-eq190050.tif, you can simply apply that here to find the solution:

9781118496718-eq190051.png

765. 9781118496718-eq190052.png

You may want to write the second term underneath the radical as a quantity squared so you can more easily see which substitution to use:

9781118496718-eq190053.png

Now begin by using the substitution u = 3x so that du = 3 dx, or 9781118496718-eq190054.tif. This gives you the following:

9781118496718-eq190055.png

766. 9781118496718-eq190056.png

Here's the given problem:

9781118496718-eq190057.png

Begin by using the substitution u = sin−1 x so that 9781118496718-eq190058.tif. Find the new limits of integration by noting that if 9781118496718-eq190059.tif, then 9781118496718-eq190060.tif, and if x = 0, then u = sin−1 0 = 0. With these new values, you get the following answer:

9781118496718-eq190061.png

767. 9781118496718-eq190062.png

Here's the given problem:

9781118496718-eq190063.png

Begin with the substitution u = sin x so that du = cos x dx. Notice that if 9781118496718-eq190064.tif, then 9781118496718-eq190065.tif, and that if x = 0, then u = sin 0 = 0. With these new values, you get the following:

9781118496718-eq190066.png

768. 9781118496718-eq190067.png

Start with the given problem:

9781118496718-eq190068.png

Use the substitution u = ln x so that 9781118496718-eq190069.tif:

9781118496718-eq190070.png

769. 9781118496718-eq190071.png

Here's the given problem:

9781118496718-eq190072.png

Begin with the substitution 9781118496718-eq190073.tif so that u2 = x and 2u du = dx. Using these substitutions gives you the following:

9781118496718-eq190074.png

770. 9781118496718-eq190075.png

Here's the given problem:

9781118496718-eq190076.png

Begin by doing a bit of algebra to manipulate the denominator of the integrand:

9781118496718-eq190077.png

Now use the substitution 9781118496718-eq190078.tif so that 9781118496718-eq190079.tif, or 5du = dx:

9781118496718-eq190080.png

771. 9781118496718-eq190081.png

Here's the given problem:

9781118496718-eq190082.png

Begin by completing the square on the expression underneath the square root.

9781118496718-eq190083.png

This gives you

9781118496718-eq190084.png

Now use the substitution u = x – 1 so that du = dx:

9781118496718-eq190085.png

772. 9781118496718-eq190086.png

The given problem is

9781118496718-eq190087.png

You'd like to have “1 – (a quantity squared)” under the radical so that you can use the formula 9781118496718-eq190088.tif, so you may want to write the second term under theradical as a quantity squared and try a substitution to see whether it all works out. (Of course, you can use the substitution without rewriting, but rewriting may help you see which substitution to use.)

9781118496718-eq190089.png

Now you can use the substitution u = e3x so that du = 3e3dx, or 9781118496718-eq190090.tif:

9781118496718-eq190091.png

773. 9781118496718-eq190092.png

Begin by splitting the integral:

9781118496718-eq190093.png

For the first integral, you can use the substitution u = x2 + 4 so that du = 2x dx, or 9781118496718-eq190094.tif. For the second integral, simply use the tan−1 x formula for integration,which gives you the following:

9781118496718-eq190095.png

774. 9781118496718-eq190096.png

Here's the given problem:

9781118496718-eq190097.png

Begin by completing the square on the expression under the radical:

9781118496718-eq190098.png

This gives you

9781118496718-eq190099.png

Now use the substitution u = x – 2 so that u + 2 = x and du = dx. Notice you can find the new limits of integration by noting that if x = 3, then u = 3 – 2 = 1, and if x = 2, then u = 2 – 2 = 0.

9781118496718-eq190100.png

Next, use some algebra to manipulate the expression under the radical:

9781118496718-eq190101.png

Now you can use the substitution 9781118496718-eq190102.tif so that 2w = u and 2 dw = du. You can again find the new upper limit of integration by using the substitution 9781118496718-eq190103.tif and noting that when u = 1, you have 9781118496718-eq190104.tif. Likewise, you can find the new lower limit of integration by noting that when u= 0, you have 9781118496718-eq190105.tif.

9781118496718-eq190106.png

Now split the integral:

9781118496718-eq190107.png

In the first integral, use the substitution v = 1 – w2 so that dv = –2w dw and –2 dv = 4w dw. You can again update the limits of integration by using the substitution v = 1 – w2. To find the new upper limit of integration, note that when 9781118496718-eq190108.tif, you have 9781118496718-eq190109.tif. To find the new lower limit of integration, note that when w = 0, you have v = 1 – 02 = 1. Again, use this substitution only for the first integral:

9781118496718-eq190110.png

Reverse the bounds in the first integral and change the sign. Then simplify:

9781118496718-eq190111.png

Whew!

775. 0

Using the definition of hyperbolic sine gives you the following:

9781118496718-eq190112.png

776. 9781118496718-eq190113.png

Using the definition of hyperbolic cosine gives you the following:

9781118496718-eq190114.png

777. 9781118496718-eq190115.png

Using the definition of hyperbolic cotangent gives you the following:

9781118496718-eq190116.png

778. 9781118496718-eq190117.png

Using the definition of hyperbolic tangent gives you the following:

9781118496718-eq190118.png

779. 9781118496718-eq190119.png

Using the definition of hyperbolic cosine gives you the following:

9781118496718-eq190120.png

780. 2 cosh x sinh x

The given function is y = coshx, which equals (cosh x)2. Using the chain rule gives you the derivative y′ = 2 cosh x sinh x.

781. 9781118496718-eq190121.png

The given function is 9781118496718-eq190122.tif. Using the chain rule gives you

9781118496718-eq190123.png

782. 9781118496718-eq190124.png

Here's the given function:

9781118496718-eq190125.png

Use the chain rule to find the derivative:

9781118496718-eq190126.png

783. 9781118496718-eq190127.png

Here's the given function:

9781118496718-eq190128.png

Use the chain rule to find the derivative:

9781118496718-eq190129.png

784. 9781118496718-eq190130.png

The given function is

y = tanh(sinh x)

Use the chain rule to get the derivative:

9781118496718-eq190131.png

785. 9781118496718-eq190132.png

Here's the given function:

9781118496718-eq190133.png

Use the chain rule to find the derivative:

9781118496718-eq190134.png

786. 9781118496718-eq190135.png

Rewrite the given function:

9781118496718-eq190136.png

Then use the chain rule to find the derivative:

9781118496718-eq190137.png

787. 9781118496718-eq190138.png

Here's the given function:

9781118496718-eq190139.png

Use the chain rule to find the derivative:

9781118496718-eq190140.png

788. 2x3 – x

Before taking the derivative, simplify the function:

9781118496718-eq190141.png

Now simply take the derivative using the power rule:

9781118496718-eq190142.png

789. 9781118496718-eq190143.png

The given function is

9781118496718-eq190144.png

Use the chain rule to find the derivative:

9781118496718-eq190145.png

Rewrite and simplify the equation to get the answer:

9781118496718-eq190146.png

790. 9781118496718-eq190147.png

Here's the given problem:

9781118496718-eq190148.png

To find the antiderivative, use the substitution u = 1 – 3x so that du = –3 dx, or 9781118496718-eq190149.tif:

9781118496718-eq190150.png

791. 9781118496718-eq190151.png

Here's the given problem:

9781118496718-eq190152.png

To find the antiderivative, begin by using the substitution u = cosh (x – 3) so that du = sinh (x – 3) dx. With this substitution, you get the following:

9781118496718-eq190153.png

792. 9781118496718-eq190154.png

Rewrite the problem using the definition of hyperbolic cotangent:

9781118496718-eq190155.png

Begin finding the antiderivative by using the substitution u = sinh x so that du = cosh x dx. With these substitutions, you get the following:

9781118496718-eq190156.png

793. 9781118496718-eq190157.png

Here's the given problem:

9781118496718-eq190158.png

To find the antiderivative, use the substitution u = 3x – 2 to get du = 3 dx so that 9781118496718-eq190159.tif. With these substitutions, you get the following:

9781118496718-eq190160.png

794. 9781118496718-eq190161.png

The given problem is

9781118496718-eq190162.png

Begin by using the substitution u = 2 + tanh x so that du = sechx dx:

9781118496718-eq190163.png

795. 9781118496718-eq190164.png

Here's the given problem:

9781118496718-eq190165.png

To find the antiderivative, use integration by parts with u = x so that du = dx, and let dv = cosh(6x) dx so that 9781118496718-eq190166.tif. This gives you the following:

9781118496718-eq190167.png

796. 9781118496718-eq190168.png

Here's the given problem:

9781118496718-eq190169.png

To find the antiderivative, begin by using the substitution 9781118496718-eq190170.tif so that 9781118496718-eq190171.tif, or 9781118496718-eq190172.tif. With these substitutions, you get the following:

9781118496718-eq190173.png

797. 9781118496718-eq190174.png

Start with the given problem:

9781118496718-eq190175.png

To find the antiderivative, begin by using the substitution 9781118496718-eq190176.tif so that du = –2x3,or 9781118496718-eq190177.tif. With these substitutions, you get the following:

9781118496718-eq190178.png

798. 9781118496718-eq190179.png

The given problem is

9781118496718-eq190180.png

To find the antiderivative, begin by using the hyperbolic identity cosh2 x – 1 = sinh2 x:

9781118496718-eq190181.png

799. 9781118496718-eq190182.png

Here's the given problem:

9781118496718-eq190183.png

To find the antiderivative, begin by using the substitution u = sinh x so that du = cosh x dx. You can find the new limits of integration by noting that if x = ln 2, then 9781118496718-eq190184.tif, and if x = 0, then 9781118496718-eq190185.tif. With these substitutions, you produce the following:

9781118496718-eq190186.png

Now use a new substitution 9781118496718-eq190187.tif so that 9781118496718-eq190188.tif, or 3 dw = du. You can again find the new limits of integration by noting that if 9781118496718-eq190189.tif, then 9781118496718-eq190190.png, and if u = 0, then 9781118496718-eq190191.png. With these substitutions, you get the following:

9781118496718-eq190192.png

800. 3

The given problem is

9781118496718-eq190193.png

Notice that if you substitute in the value x = –1, you get the indeterminate form 9781118496718-eq190194.tif. Now use L'Hôpital's rule to find the limit:

9781118496718-eq190195.png

801. 9781118496718-eq190196.png

Here's the given problem:

9781118496718-eq190197.png

Notice that if you substitute in the value x = 1, you get the indeterminate form 9781118496718-eq190198.tif. Using L'Hôpital's rule gives you the limit:

9781118496718-eq190199.png

802. 9781118496718-eq190200.png

Start with the given problem:

9781118496718-eq190201.png

Notice that if you substitute in the value x = 2, you get the indeterminate form 9781118496718-eq190202.tif. Use L'Hôpital's rule to find the limit:

9781118496718-eq190203.png

803. ∞

The given problem is

9781118496718-eq190204.png

Notice that if you substitute in the value 9781118496718-eq190205.tif, you get the indeterminate form 9781118496718-eq190206.tif. Using L'Hôpital's rule gives you the limit:

9781118496718-eq190207.png

804. 0

Start with the given problem:

9781118496718-eq190208.png

Notice that if you substitute in the value x = 0, you get the indeterminate form 9781118496718-eq190209.tif. Use L'Hôpital's rule to find the limit:

9781118496718-eq190210.png

805. 0

Here's the given problem:

9781118496718-eq190211.png

Notice that if you take the limit, you get the indeterminate form 9781118496718-eq190212.tif. Using L'Hôpital's rule gives you the limit as follows:

9781118496718-eq190213.png

806. 9781118496718-eq190214.png

Here's the given problem:

9781118496718-eq190215.png

Notice that if you substitute in the value x = 1, you get the indeterminate form 9781118496718-eq190216.tif. Using L'Hôpital's rule gives you

9781118496718-eq190217.png

807. 1

The given problem is

9781118496718-eq190218.png

Notice that if you substitute in the value x = 0, you get the indeterminate form 9781118496718-eq190219.tif. Using L'Hôpital's rule gives you the limit:

9781118496718-eq190220.png

808. 9781118496718-eq190221.png

Here's the given problem:

9781118496718-eq190222.png

Applying the limit gives you the indeterminate form ∞ – ∞. Write the expression as a single fraction by getting common denominators and then apply L'Hôpital's rule:

9781118496718-eq190223.png

809. 9781118496718-eq190224.png

The given problem is

9781118496718-eq190225.png

Notice that if you substitute in the value x = 0, you get the indeterminate form 9781118496718-eq190226.tif. Apply L'Hôpital's rule to find the limit:

9781118496718-eq190227.png

810. 0

Here's the given problem:

9781118496718-eq190228.png

Applying the limit gives you the indeterminate form 0(–∞). Rewrite the product as a quotient and use L'Hôpital's rule to find the limit:

9781118496718-eq190229.png

811. 0

Here's the given problem:

9781118496718-eq190230.png

Applying the limit gives you the indeterminate form 9781118496718-eq190231.tif. Use properties of logarithms to rewrite the expression and then use L'Hôpital's rule to find the limit:

9781118496718-eq190232.png

812. 1

The given problem is

9781118496718-eq190233.png

Substituting the value x = 0 gives you the indeterminate form 9781118496718-eq190234.tif. Apply L'Hôpital's rule to find the limit:

9781118496718-eq190235.png

813. 0

Here's the initial problem:

9781118496718-eq190236.png

Make sure you have an indeterminate form before applying L'Hospital's rule, or you can get incorrect results! Notice that as x → ∞ in the numerator, 9781118496718-eq190237.tif → 9781118496718-eq190238.tif, and as x → ∞ in the denominator, (x – 1) → ∞. Therefore, the solution becomes

9781118496718-eq190239.png

814. –∞

Here's the given problem:

9781118496718-eq190240.png

Applying the limit gives you the indeterminate form 9781118496718-eq190241.tif. Applying L'Hôpital's rule gives you the following:

9781118496718-eq190242.png

This limit is still an indeterminate form, and you can continue using L'Hôpital's rule four more times to arrive at the solution. However, because the degree of the numerator is larger than the degree of the denominator, you can find the limit by noting that

9781118496718-eq190243.png

815. 0

The given problem is

9781118496718-eq190244.png

Applying the limit gives you the indeterminate form ∞ – ∞. Rewrite the difference as a quotient and use L'Hôpital's rule:

9781118496718-eq190245.png

816. 9781118496718-eq190246.png

Start with the given problem:

9781118496718-eq190247.png

Applying the limit gives you the indeterminate form (∞)(0). Rewrite the product as a quotient and use L'Hôpital's rule as follows:

9781118496718-eq190248.png

817. 0

Here's the initial problem:

9781118496718-eq190249.png

Taking the limit gives you the indeterminate form (–∞)(0). Rewrite the product as a quotient and use L'Hôpital's rule:

9781118496718-eq190250.png

Applying the limit again gives you the indeterminate form 9781118496718-eq190251.tif. Using L'Hôpital's rule again gives you the following:

9781118496718-eq190252.png

If you again take the limit, you get the indeterminate form 9781118496718-eq190253.tif. Use L'Hôpital's rule one more time to get the final answer:

9781118496718-eq190254.png

818. 1

The given problem is

9781118496718-eq190255.png

Applying the limit gives you the indeterminate form ∞ – ∞. Begin by factoring out x; then rewrite the expression as a fraction and use L'Hôpital's rule:

9781118496718-eq190256.png

819. ∞

The given problem is

9781118496718-eq190257.png

Substituting in the value x = ∞ gives you the indeterminate form 9781118496718-eq190258.tif, so use L'Hôpital's rule:

9781118496718-eq190259.png

If you again substitute in the value x = ∞, you still get the indeterminate form 9781118496718-eq190260.tif. Applying L'Hôpital's rule one more time gives you the limit:

9781118496718-eq190261.png

820. 9781118496718-eq190262.png

Here's the initial problem:

9781118496718-eq190263.png

Substituting in the value x = 1 gives you the indeterminate form 9781118496718-eq190264.tif, so use L'Hôpital's rule:

9781118496718-eq190265.png

If you again substitute in the value x = 1, you still get the indeterminate form 9781118496718-eq190266.tif. Using L'Hôpital's rule again gives you the limit:

9781118496718-eq190267.png

821. e−5

Here's the given problem:

9781118496718-eq190268.png

Substituting in the value x = 0 gives you the indeterminate form 1. Create a new limit by taking the natural logarithm of the original expression, 9781118496718-eq190269.tif, so that 9781118496718-eq190270.tif, or 9781118496718-eq190271.tif. Take the limit of the new expression:

9781118496718-eq190272.png

Then apply L'Hôpital's rule:

9781118496718-eq190273.png

Because 9781118496718-eq190274.tif and 9781118496718-eq190275.tif, it follows that

9781118496718-eq190276.png

822. 1

Here's the initial problem:

9781118496718-eq190277.png

Applying the limit gives you the indeterminate form ∞0. Create a new limit by taking the natural logarithm of the original expression, 9781118496718-eq190278.tif, so that 9781118496718-eq190279.tif, or 9781118496718-eq190280.tif. Then take the limit of this new expression:

9781118496718-eq190281.png

Apply L'Hôpital's rule:

9781118496718-eq190282.png

Because 9781118496718-eq190283.tif and 9781118496718-eq190284.tif, it follows that

9781118496718-eq190285.png

823. 1

Here's the given problem:

9781118496718-eq190286.png

Applying the limit gives you the indeterminate form 1. Create a new limit by taking the natural logarithm of the original expression, y = (cos x)2/x, so that ln y = ln(cos x)2/x, or 9781118496718-eq190287.tif. Then take the limit of this new expression:

9781118496718-eq190288.png

Apply L'Hôpital's rule:

9781118496718-eq190289.png

Because 9781118496718-eq190290.tif and 9781118496718-eq190291.tif, it follows that

9781118496718-eq190292.png

824. e−15/9

The given problem is

9781118496718-eq190293.png

Applying the limit gives you the indeterminate form 1. Create a new limit by taking the natural logarithm of the original expression, 9781118496718-eq190294.tif, so that 9781118496718-eq190295.tif, or 9781118496718-eq190296.tif. Next, rewrite the expression as a fraction and use properties of logarithms to expand the natural logarithm (this step will make finding the derivative a bit easier):

9781118496718-eq190297.png

Apply L'Hôpital's rule:

9781118496718-eq190298.png

Because 9781118496718-eq190299.tif and 9781118496718-eq190300.tif, it follows that

9781118496718-eq190301.png

825. 1

Here's the given problem:

9781118496718-eq190302.png

Substituting in the value x = 0 gives you the indeterminate form 00. Create a new limit by taking the natural logarithm of the original expression, 9781118496718-eq190303.tif, so that 9781118496718-eq190304.tif, or 9781118496718-eq190305.tif. Then take the limit of the new expression:

9781118496718-eq190306.png

Apply L'Hôpital's rule:

9781118496718-eq190307.png

Because 9781118496718-eq190308.tif and 9781118496718-eq190309.tif, it follows that

9781118496718-eq190310.png

826. 1

Here's the given problem:

9781118496718-eq190311.png

Substituting in the value x = 0 gives you the indeterminate form 00. Create a new limit by taking the natural logarithm of the original expression, 9781118496718-eq190312.tif, so that 9781118496718-eq190313.tif, or 9781118496718-eq190314.tif. Next, take the limit of the new expression:

9781118496718-eq190315.png

Apply L'Hôpital's rule:

9781118496718-eq190316.png

Rewrite the expression and simplify:

9781118496718-eq190317.png

Because 9781118496718-eq190318.tif and 9781118496718-eq190319.tif, it follows that

9781118496718-eq190320.png

827. 0

The given problem is

9781118496718-eq190321.png

Applying the limit gives you the indeterminate form 0(–∞). Rewrite the product as a quotient and use L'Hôpital's rule to get the limit:

9781118496718-eq190322.png

828. e

Here's the initial problem:

9781118496718-eq190323.png

Applying the limit gives you the indeterminate form 1. Create a new limit by taking the natural logarithm of the original expression, 9781118496718-eq190324.tif, so that 9781118496718-eq190325.tif, or 9781118496718-eq190326.tif. Noting that 9781118496718-eq190327.tif is an indeterminate product, you can rewrite this limit as

9781118496718-eq190328.png

Next, apply L'Hôpital's rule:

9781118496718-eq190329.png

Because 9781118496718-eq190330.tif and 9781118496718-eq190331.tif, it follows that

9781118496718-eq190332.png

829. e4

The given problem is

9781118496718-eq190333.png

Applying the limit gives you the indeterminate form 1. Create a new limit by taking the natural logarithm of the original expression, 9781118496718-eq190334.tif, so that 9781118496718-eq190335.tif, or 9781118496718-eq190336.tif. Then use L'Hôpital's rule on the new limit:

9781118496718-eq190337.png

Because 9781118496718-eq190338.tif and 9781118496718-eq190339.tif, it follows that

9781118496718-eq190340.png

830. –∞

Here's the given problem:

9781118496718-eq190341.png

Applying the limit gives you the indeterminate form ∞ – ∞, so write the expression as a single fraction by getting common denominators and then apply L'Hôpital's rule:

9781118496718-eq190342.png

Multiply the numerator and denominator by x:

9781118496718-eq190343.png

Notice that as x → 1+ in the numerator, 2x(ln x) – x2 → –1, and as x → 1+ in the denominator, x2 – 1 + 2xln x → 0+. Therefore, you get the following answer:

9781118496718-eq190344.png

831. 9781118496718-eq190345.png

Here's the given problem:

9781118496718-eq190346.png

Evaluating the limit gives you the indeterminate form ∞ – ∞, so write the expression as a single fraction by getting a common denominator and then use L'Hôpital's rule:

9781118496718-eq190347.png

Multiply the numerator and denominator by x:

9781118496718-eq190348.png

Substituting in the value x = 1 gives you the indeterminate form 9781118496718-eq190349.tif, so you can again use L'Hôpital's rule:

9781118496718-eq190350.png

832. 9781118496718-eq190351.png

Here's the given integral:

9781118496718-eq190352.png

If you let u = 5x, then du = 5dx, or 9781118496718-eq190353.tif. Substituting into the original integral, you get

9781118496718-eq190354.png

833. 9781118496718-eq190355.png

Here's the given integral:

9781118496718-eq190356.png

If you let u = x + 4, then du = dx. Substituting into the original integral gives you the following:

9781118496718-eq190357.png

834. 9781118496718-eq190358.png

The given integral is

9781118496718-eq190359.png

If you let u = x3 + 1, then du = 3x2 dx. Substitute into the original integral and solve:

9781118496718-eq190360.png

835. 9781118496718-eq190361.png

The given integral is

9781118496718-eq190362.png

If you let 9781118496718-eq190363.tif, then you get 9781118496718-eq190364.tif, or equivalently, 9781118496718-eq190365.tif. Substitute into the original integral:

9781118496718-eq190366.png

836. 9781118496718-eq190367.png

Here's the given integral:

9781118496718-eq190368.png

If you let u = 4 + 5x, then you get du = 5 dx, or 9781118496718-eq190369.tif. Substitute into the original integral:

9781118496718-eq190370.png

837. 9781118496718-eq190371.png

Start by rewriting the given integral:

9781118496718-eq190372.png

If you let 9781118496718-eq190373.tif, then 9781118496718-eq190374.tif. Substitute into the original integral to find the answer:

9781118496718-eq190375.png

838. 0

Recall that 9781118496718-eq190376.tif is an odd function; that is, 9781118496718-eq190377.tif. Likewise, 9781118496718-eq190378.tif so that 9781118496718-eq190379.tif is also an odd function. Because this function is symmetric about the origin and you're integrating over an interval of the form [–aa], you get 9781118496718-eq190380.tif.

Here's an alternate approach: Begin by saving a factor of the tangent and use an identity to allow you to split up the integral:

9781118496718-eq190381.png

To evaluate the first integral, rewrite the tangent:

9781118496718-eq190382.png

Then use the substitution 9781118496718-eq190383.tif so that 9781118496718-eq190384.tif, or 9781118496718-eq190385.tif. You can find the new limits of integration by noting that if 9781118496718-eq190386.tif or if 9781118496718-eq190387.tif, you have 9781118496718-eq190388.tif:

9781118496718-eq190389.png

Because the upper and lower limits of integration are the same, you have the following for the first integral:

9781118496718-eq190390.png

The second integral is

9781118496718-eq190391.png

To evaluate the second integral, use the substitution 9781118496718-eq190392.tif so that 9781118496718-eq190393.tif. Notice that you can find the new limits of integration by noting that if 9781118496718-eq190394.tif, then 9781118496718-eq190395.tif, and if 9781118496718-eq190396.tif, then 9781118496718-eq190397.tif. Therefore, the second integral becomes

9781118496718-eq190398.png

Combining the values of the first and second integrals gives you the answer:

9781118496718-eq190399.png

839. 4

The given integral is

9781118496718-eq190400.png

Start by using the substitution u = ln x so that 9781118496718-eq190401.tif. You can find the new limits of integration by noting that if x = e2, then u = ln e2 = 2, and if x = 1, then u = ln 1 = 0. Therefore, the integral becomes

9781118496718-eq190402.png

840. 9781118496718-eq190403.png

The given integral is

9781118496718-eq190404.png

Start by letting u = 4 – 3x so that du = –3dx, or 9781118496718-eq190405.tif. Substitute this back into the original integral:

9781118496718-eq190406.png

841. 9781118496718-eq190407.png

Here's the given integral:

9781118496718-eq190408.png

Let u = tan−1 x so that 9781118496718-eq190409.tif. Substituting these values into the original integral, you get

9781118496718-eq190410.png

842. 9781118496718-eq190411.png

Start by rewriting the integral:

9781118496718-eq190412.png

Now you can perform a u-substitution with u = cos x so that du = –sin x dx, or –du = sin x dx. Substituting this expression into the original integral gives you the following:

9781118496718-eq190413.png

You can now use the power property of logarithms to rewrite the integral:

9781118496718-eq190414.png

9781118496718-eq190415.png

843. e – e1/2

The given integral is

9781118496718-eq190416.png

Begin by using the substitution u = cos x so that du = –sin x dx, or –du = sin x dx. Notice you can find the new limits of integration by noting that if 9781118496718-eq190417.tif, then 9781118496718-eq190418.tif, and if x = 0, then u = cos 0 = 1. With these values, you have the following:

9781118496718-eq190419.png

844. 9781118496718-eq190420.png

Rewrite the given integral:

9781118496718-eq190421.png

Begin by using the substitution u = 1 + x3/2 so that 9781118496718-eq190422.tif, or 9781118496718-eq190423.tif. Notice you can find the new limits of integration by noting that if x = 1, then u = 1 + 13/2 = 2, and if x = 0, then u = 1 + 03/2 = 1. With these new values, you find that

9781118496718-eq190424.png

845. 9781118496718-eq190425.png

Rewrite the given integral:

9781118496718-eq190426.png

Then use the substitution u = tan x so that du = secx dx:

9781118496718-eq190427.png

846. 9781118496718-eq190428.png

Here's the given integral:

9781118496718-eq190429.png

Use the substitution 9781118496718-eq190430.tif so that 9781118496718-eq190431.tif, or 9781118496718-eq190432.tif:

9781118496718-eq190433.png

847. 9781118496718-eq190434.png

Here's the given integral:

9781118496718-eq190435.png

If you let u = 1 + 2x + 3x2, then you get du = (2 + 6x)dx, or 2du = (4 + 12x)dx. Substitute this value into the original integral:

9781118496718-eq190436.png

848. 9781118496718-eq190437.png

Here's the given integral:

9781118496718-eq190438.png

If you let u = cot x, then you get du = –cscx dx, or –du = cscx dx. Substitute this value into the original integral:

9781118496718-eq190439.png

849. 9781118496718-eq190440.png

Here's the given integral:

9781118496718-eq190441.png

If you let u = x2, then du = 2x dx, or 9781118496718-eq190442.tif. You can also find the new limits of integration by using the original limits of integration along with the u-substitution. When 9781118496718-eq190443.tif9781118496718-eq190444.tif. Likewise, if x = 0, then u = (0)2 = 0.

Now substitute these values into the original integral:

9781118496718-eq190445.png

850. 9781118496718-eq190446.png

The given integral is

9781118496718-eq190447.png

Begin by breaking the integral into two separate integrals:

9781118496718-eq190448.png

Notice that 9781118496718-eq190449.tif.

All that's left is to integrate the second integral; you can do this with a u-substitution. For 9781118496718-eq190450.tif, let u = 1 + x2 so that du = 2x dx, or 9781118496718-eq190451.tif. Substituting these values into the second integral gives you the following:

9781118496718-eq190452.png

The expression 1 + x2 is always positive, so you can drop the absolute value sign:

9781118496718-eq190453.png

Combining the solutions gives you

9781118496718-eq190454.png

851. 9781118496718-eq190455.png

The given integral is

9781118496718-eq190456.png

Begin with the u-substitution u = ln x so that 9781118496718-eq190457.tif. You can also compute the new limits of integration by noting that if x = e, then u = ln e = 1, and if x = e2, then u = ln e2 = 2. Now you can integrate:

9781118496718-eq190458.png

852. 9781118496718-eq190459.png

Here's the given integral:

9781118496718-eq190460.png

Start with the substitution 9781118496718-eq190461.tif so that 9781118496718-eq190462.tif. Substituting these values into the original integral, you get the following:

9781118496718-eq190463.png

853. 9781118496718-eq190464.png

The given integral is

9781118496718-eq190465.png

Begin by letting u = –x3 so that du = –3xdx, or 9781118496718-eq190466.tif. Substitute these values into the original integral:

9781118496718-eq190467.png

854. 9781118496718-eq190468.png

Here's the given integral:

9781118496718-eq190469.png

Begin by using the substitution u = x + 1 so that u – 1 = x and du = dx. Notice you can find the new limits of integration by noting that if x = 3, then u = 3 + 1 = 4, and if x = 0, then u = 0 + 1 = 1. With these new values, you have the following:

9781118496718-eq190470.png

855. 9781118496718-eq190471.png

At first, you may have trouble seeing how a substitution will work for this problem. But note that you can split x5 into x3x2:

9781118496718-eq190472.png

Now you can use the substitution u = x3 + 1 so that u – 1 = x3 and du = 3xdx, or 9781118496718-eq190473.tif. Notice that you can find the new limits of integration by noting that if x = 1, then u = 13 + 1 = 2, and if x = 0, then u = 03 + 1 = 1. With these values, you get the following integral:

9781118496718-eq190474.png

856. 9781118496718-eq190475.png

Here's the given integral:

9781118496718-eq190476.png

If you let u = x + 3, then du = dx and u – 3 = x. Substituting into the original integral, you get the following:

9781118496718-eq190477.png

857. 9781118496718-eq190478.png

At first glance, substitution may not seem to work for this problem. However, you can rewrite the integral as follows:

9781118496718-eq190479.png

If you let u = x4 + 1, then you get du = 4xdx, or 9781118496718-eq190480.tif. Notice also from the u-substitution that u – 1 = x4. Substitute into the original integral and simplify:

9781118496718-eq190481.png

858. 9781118496718-eq190483.png

The given integral is

9781118496718-eq190484.png

Use integration by parts with u = x so that du = dx and let dv = cos(4x) dx so that 9781118496718-eq190485.tif. Using the integration by parts formula gives you the following:

9781118496718-eq190486.png

859. 9781118496718-eq190487.png

Here's the given integral:

9781118496718-eq190488.png

Use integration by parts with u = x so that du = dx and let dv = edx so that v = ex. Using the integration by parts formula gives you the following:

9781118496718-eq190489.png

860. 9781118496718-eq190490.png

Here's the given integral:

9781118496718-eq190491.png

Use integration by parts with u = x so that du = dx and let dv = sinh(2x) dx so that 9781118496718-eq190492.tif. Using the integration by parts formula gives you the following:

9781118496718-eq190493.png

861. 9781118496718-eq190494.png

The given integral is

9781118496718-eq190495.png

Use integration by parts with u = x so that du = dx and let dv = 6dx so that 9781118496718-eq190496.tif. Using the integration by parts formula gives you the following:

9781118496718-eq190497.png

862. 9781118496718-eq190498.png

The given integral is

9781118496718-eq190499.png

Use integration by parts with u = x so that du = dx and let dv = sec x tan xdx so that v = sec x:

9781118496718-eq190500.png

863. 9781118496718-eq190501.png

The given integral is

9781118496718-eq190502.png

Begin by using integration by parts with u = x so that du = dx and let dv = sin(5x) dx so that 9781118496718-eq190503.tif:

9781118496718-eq190504.png

864. 9781118496718-eq190505.png

The given integral is

9781118496718-eq190506.png

Begin by using integration by parts with u = x so that du = dx and let dv = cscx dx so that v = –cot x:

9781118496718-eq190507.png

865. –x cos x + sin x + C

The given integral is

9781118496718-eq190508.png

Use integration by parts with u = x so that du = dx and let dv = sin x dx so that v = –cos x:

9781118496718-eq190509.png

866. 9781118496718-eq190510.png

Here's the given integral:

9781118496718-eq190511.png

Use integration by parts with u = x so that du = dx and let dv = csc x cot x dx so that v = –csc x:

9781118496718-eq190512.png

867. 9781118496718-eq190513.png

The given integral is

9781118496718-eq190514.png

Use integration by parts with u = ln(3x + 1) so that 9781118496718-eq190515.tif and let dv = dx so that v = x. Using the integration by parts formula gives you

9781118496718-eq190516.png

To evaluate the integral 9781118496718-eq190517.tif, you can use long division to simplify, or you can add 1 and subtract 1 from the numerator and split up the fraction:

9781118496718-eq190518.png

868. 9781118496718-eq190519.png

The given integral is

9781118496718-eq190520.png

Use integration by parts with u = tan−1 x so that 9781118496718-eq190521.tif and let dv = dx so that v = x. Using the integration by parts formula gives you

9781118496718-eq190522.png

To evaluate 9781118496718-eq190523.tif, use the substitution u1 = 1 + x2 so that du1 = 2x dx, or 9781118496718-eq190524.tif. With these substitutions, you get the following:

9781118496718-eq190525.png

869. 9781118496718-eq190526.png

The given integral is

9781118496718-eq190527.png

Use integration by parts with u = ln x so that 9781118496718-eq190528.tif and let dv = x−2 dx so that 9781118496718-eq190529.tif. Using the integration by parts formula gives you the following:

9781118496718-eq190530.png

870. 4e3

Here's the given integral:

9781118496718-eq190531.png

Begin by using the substitution 9781118496718-eq190532.tif so that w2 = x and 2w dw = dx. You can find the new limits of integration by noting that if x = 9, then 9781118496718-eq190533.tif, and if x = 1, then 9781118496718-eq190534.tif. These substitutions give you the following:

9781118496718-eq190535.png

Use integration by parts with u = 2w so that du = 2dw and let dv = edw so that v = ew. Using the integration by parts formula gives you the following:

9781118496718-eq190536.png

871. 9781118496718-eq190537.png

The given integral is

9781118496718-eq190538.png

Use integration by parts with u = ln x so that 9781118496718-eq190539.tif and let dv = x3/2 dx so that 9781118496718-eq190540.tif to get the following:

9781118496718-eq190541.png

872. 9781118496718-eq190542.png

Here's the given integral:

9781118496718-eq190543.png

Use integration by parts with u = cos−1 x so that 9781118496718-eq190544.tif and let dv = 1dx so that v = x:

9781118496718-eq190545.png

Next, use the substitution w = 1 – x2 so that dw = –2x dx, or 9781118496718-eq190546.tif:

9781118496718-eq190547.png

873. 9781118496718-eq190548.png

The given integral is

9781118496718-eq190549.png

Use integration by parts with u = sin−1(5x) so that 9781118496718-eq190550.tif and let dv = 1dx so that v = x:

9781118496718-eq190551.png

Next, use the substitution w = 1 – 25x2 so that dw = –50x dx, or 9781118496718-eq190552.tif:

9781118496718-eq190553.png

874. 9781118496718-eq190554.png

Here's the given integral:

9781118496718-eq190555.png

Begin by using integration by parts with u = x2 so that du = 2x dx and let dv = e−2x so that 9781118496718-eq190556.tif:

9781118496718-eq190557.png

Use integration by parts again with u1 = x so that du1 = dx and let dv1 = e−2x so that 9781118496718-eq190558.tif:

9781118496718-eq190559.png

875. –6e−1 + 3

The given integral is

9781118496718-eq190560.png

Begin by using integration by parts with u = x2 + 1 so that du = 2x dx and let dv = edx so that v = –ex:

9781118496718-eq190561.png

Now use integration by parts again with u1 = 2x so that du1 = 2 dx and let dv1 = edx so that v1 = –ex:

9781118496718-eq190562.png

876. 9781118496718-eq190563.png

Here's the given integral:

9781118496718-eq190564.png

Begin by using the substitution w = x2 so that dw = 2x dx, or 9781118496718-eq190565.tif, to get the following:

9781118496718-eq190566.png

Next, use integration by parts with u = w so that du = dw and let dv = cos w dw so that v = sin w:

9781118496718-eq190567.png

877. 9781118496718-eq190568.png

The given integral is

9781118496718-eq190569.png

Use integration by parts with u = x2 so that du = 2x dx and let dv = sin(mx) dx so that 9781118496718-eq190570.tif. Using the integration by parts formula gives you the following:

9781118496718-eq190571.png

To evaluate 9781118496718-eq190572.tif, use integration by parts again with u1 = x so that du1 = dx and let dv1 = cos(mx) so that 9781118496718-eq190573.tif. This gives you the following:

9781118496718-eq190574.png

Therefore,

9781118496718-eq190575.png

878. 9781118496718-eq190576.png

Here's the given integral:

9781118496718-eq190577.png

Begin by using integration by parts with u = (ln x)2 so that 9781118496718-eq190578.tif and let dv = xdx so that 9781118496718-eq190579.tif. This gives you the following:

9781118496718-eq190580.png

To evaluate the new integral, use integration by parts with u1 = ln x so that 9781118496718-eq190581.tif and let dv1 = xdx so that 9781118496718-eq190582.tif. This gives you the following:

9781118496718-eq190583.png

879. 9781118496718-eq190584.png

The given integral is

9781118496718-eq190585.png

Begin by using integration by parts with u = cos(2x) so that du = –2sin(2x)dx and let dv = edx so that v = ex. Using the integration by parts formula gives you the following:

9781118496718-eq190586.png

Use integration by parts again with u1 = sin(2x) so that du1 = 2 cos(2x) and let dv1 = edx so that v1 = ex. Now you have the following:

9781118496718-eq190587.png

Set the expression equal to the original integral:

9781118496718-eq190588.png

Add 9781118496718-eq190589.tif to both sides of the equation (note the addition of C1 on the right):

9781118496718-eq190590.png

Dividing both sides of the equation by 3 yields the solution:

9781118496718-eq190591.png

Note: Don't be thrown off by the constant of integration that switches from C1 to C; they're both arbitrary constants, but simply using 9781118496718-eq190592.tif in both equations would be poor notation.

880. –cos x ln(cos x) + cos x + C

Here's the given integral:

9781118496718-eq190593.png

Begin by using the substitution w = cos x so that dw = –sin xdx, or –dw = sin x dx. Using these substitutions gives you the following:

9781118496718-eq190594.png

Now use integration by parts with u = ln w so that 9781118496718-eq190595.tif and let dv = –1 dw so that v = –w. Using the integration by parts formula now gives you

9781118496718-eq190596.png

Replacing w with the original substitution, cos x, gives you the solution:

9781118496718-eq190597.png

9781118496718-eq190598.png

881. 9781118496718-eq190599.png

The given integral is

9781118496718-eq190600.png

Begin by using the substitution 9781118496718-eq190601.tif so that w2 = x and 2wdw = dx. This gives you the following:

9781118496718-eq190602.png

Now use integration by parts with u = 2w so that du = 2dw and let dv = cos w dw so that v = sin w:

9781118496718-eq190603.png

Using the original substitution, 9781118496718-eq190604.tif, gives you the solution:

9781118496718-eq190605.png

882. 9781118496718-eq190606.png

Here's the given integral:

9781118496718-eq190607.png

Begin by using integration by parts with u = (ln x)2 so that 9781118496718-eq190608.tif and let dv = xdx so that 9781118496718-eq190609.tif:

9781118496718-eq190610.png

To evaluate the new integral, use integration by parts again with u1 = ln x so that 9781118496718-eq190611.tif and let dv = xdx so that 9781118496718-eq190612.tif:

9781118496718-eq190613.png

883. 9781118496718-eq190614.png

Here's the given integral:

9781118496718-eq190615.png

Use integration by parts with u = tan−1 x so that 9781118496718-eq190616.tif and let dv = x dx so that 9781118496718-eq190617.tif:

9781118496718-eq190618.png

Note: The numerator and denominator are almost the same for the expression 9781118496718-eq190619.tif, allowing you to add 1 and subtract 1 (in the second line) and then break it up into two fractions (in the third line); this approach lets you avoid long division. Another way to simplify 9781118496718-eq190620.tif is to use polynomial long division to arrive at 9781118496718-eq190621.tif.

884. 9781118496718-eq190622.png

Here's the given integral:

9781118496718-eq190623.png

Recall that 9781118496718-eq190624.tif so that 9781118496718-eq190625.tif. Using this identity in the integral gives you

9781118496718-eq190626.png

885. 9781118496718-eq190627.png

The given integral is

9781118496718-eq190628.png

Recall that 9781118496718-eq190629.tif. Using this identity gives you the following:

9781118496718-eq190630.png

886. 9781118496718-eq190631.png

Here's the given integral:

9781118496718-eq190632.png

As long as you remember your fundamental trigonometric identities, this problem is an easy one!

9781118496718-eq190633.png

887. 9781118496718-eq190634.png

The given integral is

9781118496718-eq190635.png

Begin by factoring out sect and using a trigonometric identity on one sect to write it in terms of tan t:

9781118496718-eq190636.png

Now use the substitution u = tan t so that du = sect dt. Using these values in the integral gives you

9781118496718-eq190637.png

888. 9781118496718-eq190638.png

The given integral is

9781118496718-eq190639.png

Begin by applying the identity 9781118496718-eq190640.tif to the integrand to produce the following integral:

9781118496718-eq190641.png

The first term has an elementary antiderivative. To integrate the second term, you can use the substitution u = 5x:

9781118496718-eq190642.png

889. 9781118496718-eq190643.png

The given integral is

9781118496718-eq190644.png

Begin by using the identity 9781118496718-eq190645.tif on the integrand to produce the following integral:

9781118496718-eq190646.png

To integrate, use the substitution u = 3x on the first term and the substitution u = 7x on the second term:

9781118496718-eq190647.png

890. 9781118496718-eq190648.png

The given integral is

9781118496718-eq190649.png

Use the substitution u = sin x so that du = cos x dx:

9781118496718-eq190650.png

891. 9781118496718-eq190651.png

The given integral is

9781118496718-eq190652.png

Begin by using the substitution u = tan x so that du = secx dx:

9781118496718-eq190653.png

892. 9781118496718-eq190654.png

Here's the given integral:

9781118496718-eq190655.png

Begin by applying the identity 9781118496718-eq190656.tif to the integrand to produce the integral 9781118496718-eq190657.tif. Now simplify and integrate:

9781118496718-eq190658.png

893. sec x + C

Here's the given integral:

9781118496718-eq190659.png

There's a simple solution for this function:

9781118496718-eq190660.png

894. 9781118496718-eq190661.png

Start by rewriting the radical:

9781118496718-eq190662.png

Use the substitution u = csc x so that du = –csc x cot x dx, or –du = csc x cot x dx:

9781118496718-eq190663.png

895. 9781118496718-eq190664.png

Here's the given integral:

9781118496718-eq190665.png

Begin by factoring out cos x and using a trigonometric identity to write cos2 x in terms of sin x:

9781118496718-eq190666.png

Now use the substitution u = sin x, which gives you du = cos x dx. Putting these values into the integral gives you the following:

9781118496718-eq190667.png

896. 9781118496718-eq190668.png

The given integral is

9781118496718-eq190669.png

Begin by factoring out secx and using a trigonometric identity on one factor of secx to write it in terms of tan x:

9781118496718-eq190670.png

Now use the substitution u = tan x so that du = secx dx. You can also change the limits of integration by noting that if 9781118496718-eq190671.tif, then 9781118496718-eq190672.tif, and if x = 0, then u = tan 0 = 0. With these new values, the integral becomes

9781118496718-eq190673.png

897. 9781118496718-eq190674.png

The given integral is

9781118496718-eq190675.png

Begin by rewriting the integral and factoring out cscx:

9781118496718-eq190676.png

Now use an identity to get the following:

9781118496718-eq190677.png

Next, use the substitution u = cot x to get du = –csc2 x dx so that

9781118496718-eq190678.png

898. 9781118496718-eq190679.png

Here's the given integral:

9781118496718-eq190680.png

Begin by factoring out cos x and using a trigonometric identity to write cosx in terms of sin x:

9781118496718-eq190681.png

Now use the substitution u = sin x, which gives you du = cos x dx. You can also find the new limits of integration. If 9781118496718-eq190682.tif, then 9781118496718-eq190683.tif, and if x = 0, then u = sin 0 = 0. Using these values in the integral gives you

9781118496718-eq190684.png

899. 9781118496718-eq190685.png

The given integral is

9781118496718-eq190686.png

Start by rewriting the integrand as two terms:

9781118496718-eq190687.png

Recall that 9781118496718-eq190688.tif. To evaluate 9781118496718-eq190689.tif, use the substitution u = sin x so that du = cos x dx:

9781118496718-eq190690.png

Combining the antiderivatives of each indefinite integral gives you the solution:

9781118496718-eq190691.png

900. 9781118496718-eq190692.png

Begin by expanding the integrand:

9781118496718-eq190693.png

Next, use the identity 9781118496718-eq190694.tif to produce the following integral:

9781118496718-eq190695.png

901. 9781118496718-eq190696.png

Here's the given integral:

9781118496718-eq190697.png

Begin by applying the identity sin(2x) = 2 sin x cos x to the denominator of the integrand. Then split up the fraction and simplify:

9781118496718-eq190698.png

You can now apply basic antiderivative formulas to arrive at the solution:

9781118496718-eq190699.png

902. 9781118496718-eq190700.png

Here's the given integral:

9781118496718-eq190701.png

Begin by rewriting the integral:

9781118496718-eq190702.png

Then use the substitution u = tan x so that du = sec2 x dx:

9781118496718-eq190703.png

903. 9781118496718-eq190704.png

The given integral is

9781118496718-eq190705.png

To integrate this function, use integration by parts: 9781118496718-eq190706.tif. Let u = x so that du = dx and let dv = sinx. Note that to find v by integrating dv, you have to use a trigonometric identity:

9781118496718-eq190707.png

Therefore, you get the following:

9781118496718-eq190708.png

904. 9781118496718-eq190709.png

Here's the given integral:

9781118496718-eq190710.png

For this problem, you can factor out either sin x or cos x and then use an identity. For example, you can factor out cos x and then use the identity cos2 x = 1 – sinx:

9781118496718-eq190711.png

Then use the substitution u = sin x so that du = cos x dx to get

9781118496718-eq190712.png

905. 9781118496718-eq190713.png

The given integral is

9781118496718-eq190714.png

Begin by factoring out tan x and using an identity:

9781118496718-eq190715.png

Now use the substitution u = sec x so that du = sec x tan x dx. This gives you

9781118496718-eq190716.png

906. 9781118496718-eq190717.png

Begin by factoring out cot x and using an identity:

9781118496718-eq190718.png

Now use the substitution u = csc x so that du = –csc x cot x dx, or –du = csc x cot x dx. This gives you the following:

9781118496718-eq190719.png

907. 9781118496718-eq190720.png

Here's the given integral:

9781118496718-eq190721.png

Begin by factoring out cos x and using a trigonometric identity to write cos2 x in terms of sin x:

9781118496718-eq190722.png

Now use u-substitution: u = sin x so that du = cos x. Using these values in the integral gives you

9781118496718-eq190723.png

908. 9781118496718-eq190724.png

Begin by factoring out 9781118496718-eq190725.tif and using an identity to get the following:

9781118496718-eq190726.png

Now use the substitution u = cot x so that du = –cscx dx, or –du = cscx dx:

9781118496718-eq190727.png

909. 9781118496718-eq190728.png

Here's the given integral:

9781118496718-eq190729.png

Begin by performing the substitution 9781118496718-eq190730.tif so that 9781118496718-eq190731.tif, or 9781118496718-eq190732.tif. Using these values in the integral gives you 9781118496718-eq190733.tif. Now factor out sin u and use some algebra and a trigonometric identity to write sinu in terms of cos u:

9781118496718-eq190734.png

Again use a substitution to integrate: Let w = cos u so that dw = –sin u du, or –dw = sin u du. Using these values yields the integral 9781118496718-eq190735.tif. Now simply expand the integrand and integrate:

9781118496718-eq190736.png

Then undo all the substitutions:

9781118496718-eq190737.png

910. 9781118496718-eq190738.png

The given integral is

9781118496718-eq190739.png

Begin by factoring out sin x and using a trigonometric identity to write sin2 x in terms of cos x:

9781118496718-eq190740.png

Now use the substitution u = cos x, which gives you du = –sin x dx, or –du = sin x dx. Using these values in the integral gives you

9781118496718-eq190741.png

911. 9781118496718-eq190742.png

To evaluate this integral, you can use integration by parts. You may want to first rewrite secx as secx sec x (rewriting isn't necessary, but it may help you see things a bit better!):

9781118496718-eq190743.png

Now use integration by parts with u = sec x so that du = sec x tan x dx and let dv = secx dx so that v = tan x. This gives you the following:

9781118496718-eq190744.png

Now use an identity to get

9781118496718-eq190745.png

Set this expression equal to the given integral:

9781118496718-eq190746.png

After evaluating 9781118496718-eq190747.tif on the right side of the equation, you get

9781118496718-eq190748.png

Add 9781118496718-eq190749.tif to both sides of the equation and then divide by 2 to get the solution:

9781118496718-eq190750.png

912. –sec x – tan x + C

The given integral is

9781118496718-eq190751.png

Begin by multiplying the numerator and denominator of the integrand by the conjugate of the denominator. Recall that the conjugate of (a – b) is (a + b). In this case, you multiply the numerator and denominator by (sin x + 1):

9781118496718-eq190752.png

Now use the identity sinx – 1 = –cosx and split apart the fraction to further simplify the integrand:

9781118496718-eq190753.png

913. 9781118496718-eq190754.png

Here's the given integral:

9781118496718-eq190755.png

Begin by factoring out sec(4x) tan(4x):

9781118496718-eq190756.png

Then use a trigonometric identity to arrive at

9781118496718-eq190757.png

Now use the substitution u = sec(4x) so that du = 4 sec(4x) tan(4x) dx, or 9781118496718-eq190758.tif. Using these substitutions, you now have

9781118496718-eq190759.png

914. 9781118496718-eq190760.png

The given integral is

9781118496718-eq190761.png

Start by using the substitution 9781118496718-eq190762.tif so that 9781118496718-eq190763.tif. Using these substitutions in the integral, you get the following:

9781118496718-eq190764.png

From the substitution 9781118496718-eq190765.tif, you get 9781118496718-eq190766.tif, from which you can deduce that 9781118496718-eq190767.tif. (To deduce that 9781118496718-eq190768.tif, you can label a right triangle using 9781118496718-eq190769.tif and find the missing side using the Pythagorean theorem.) Using these values, you get

9781118496718-eq190770.png

915. 9781118496718-eq190771.png

Here's the given integral:

9781118496718-eq190772.png

Begin by using the substitution 9781118496718-eq190773.tif so that 9781118496718-eq190774.tif. You can find the newlimits of integration by noting that if 9781118496718-eq190775.tif, then 9781118496718-eq190776.tif so that 9781118496718-eq190777.tif, and if x = 0, then 9781118496718-eq190778.tif so that 9781118496718-eq190779.tif:

9781118496718-eq190780.png

916. 9781118496718-eq190781.png

Here's the given integral:

9781118496718-eq190782.png

Use the trigonometric substitution 9781118496718-eq190783.tif so that 9781118496718-eq190784.tif:

9781118496718-eq190785.png

From 9781118496718-eq190786.tif, you can deduce that 9781118496718-eq190787.tif. (To deduce that 9781118496718-eq190788.tif, you can label a right triangle using 9781118496718-eq190789.tif and then use the Pythagorean theoremto find the missing side.) Therefore, the solution becomes

9781118496718-eq190790.png

Note that you can also solve the problem with the substitution u = x2 – 1.

917. 9781118496718-eq190791.png

The given integral is

9781118496718-eq190792.png

Begin with the trigonometric substitution 9781118496718-eq190793.tif to get 9781118496718-eq190794.tif, which gives you the following:

9781118496718-eq190795.png

From 9781118496718-eq190796.tif, you can deduce that 9781118496718-eq190797.tif. (To deduce that 9781118496718-eq190798.tif,you can label a right triangle using 9781118496718-eq190799.tif and then find the missing side using thePythagorean theorem.) Therefore, you get the following solution:

9781118496718-eq190800.png

918. 9781118496718-eq190801.png

The given integral is

9781118496718-eq190802.png

Begin by using the substitution 9781118496718-eq190803.tif to get 9781118496718-eq190804.tif. Substituting these values into the original integral, you have

9781118496718-eq190805.png

From your original substitution, you have 9781118496718-eq190806.tif. From this, you can deduce that 9781118496718-eq190807.tif. (To deduce that 9781118496718-eq190808.tif, you can label a right triangle using 9781118496718-eq190809.tif and then use the Pythagorean theorem to find the missing side.) Therefore, the solution becomes

9781118496718-eq190810.png

919. 9781118496718-eq190811.png

Here's the given integral:

9781118496718-eq190812.png

Begin by using the trigonometric identity 9781118496718-eq190813.tif so that 9781118496718-eq190814.tif:

9781118496718-eq190815.png

Now use an identity to get

9781118496718-eq190816.png

From the substitution 9781118496718-eq190817.tif, you get 9781118496718-eq190818.tif, so you can deduce that 9781118496718-eq190819.tif and that 9781118496718-eq190820.tif. (To deduce that 9781118496718-eq190821.tif, you can label a right triangle using 9781118496718-eq190822.tif and then find the missing side with the Pythagorean theorem.) Therefore, the solution becomes

9781118496718-eq190823.png

920. 9781118496718-eq190824.png

Start by rewriting the given integral:

9781118496718-eq190825.png

Now you can use the substitution 9781118496718-eq190826.tif to get 9781118496718-eq190827.tif and 9781118496718-eq190828.tif. Substituting these values into the integral, you get

9781118496718-eq190829.png

Then you can use the identity 9781118496718-eq190830.tif to simplify the integral:

9781118496718-eq190831.png

From the substitution 9781118496718-eq190832.tif, you can deduce that 9781118496718-eq190833.tif and also that 9781118496718-eq190834.tif. (To deduce that 9781118496718-eq190835.tif, you can label a right triangle using 9781118496718-eq190836.tif and find the missing side of the triangle with the Pythagorean theorem.) Using these values, you get the solution:

9781118496718-eq190837.png

921. 9781118496718-eq190838.png

The given integral is

9781118496718-eq190839.png

Begin by using the substitutions 9781118496718-eq190840.tif and 9781118496718-eq190841.tif:

9781118496718-eq190842.png

From the substitution 9781118496718-eq190843.tif, you get 9781118496718-eq190844.tif, from which you can deduce that 9781118496718-eq190845.tif and also that 9781118496718-eq190846.tif. (To deduce that 9781118496718-eq190847.tif, you can label a right triangle using 9781118496718-eq190848.tif and find the missing side of the triangle with the Pythagorean theorem.) Using these values, you get the following solution:

9781118496718-eq190849.png

922. 9781118496718-eq190850.png

Rewrite the integral with a radical sign:

9781118496718-eq190851.png

Now begin by using the substitution 9781118496718-eq190852.tif, or 9781118496718-eq190853.tif, so that 9781118496718-eq190854.tif:

9781118496718-eq190855.png

From the substitution 9781118496718-eq190856.tif, you can deduce that 9781118496718-eq190857.tif and that 9781118496718-eq190858.tif. (To deduce that 9781118496718-eq190859.tif, you can label a right triangle using 9781118496718-eq190860.tif and find the missing side of the triangle with the Pythagorean theorem.) Therefore, the solution becomes

9781118496718-eq190861.png

923. 9781118496718-eq190862.png

The given integral is

9781118496718-eq190863.png

Begin by using the trigonometric substitution 9781118496718-eq190864.tif so that 9781118496718-eq190865.tif:

9781118496718-eq190866.png

From the substitution 9781118496718-eq190867.tif, you can deduce that 9781118496718-eq190868.tif and that 9781118496718-eq190869.tif. (To deduce that 9781118496718-eq190870.tif, you can label a right triangle using 9781118496718-eq190871.tif and find the missing side of the triangle with the Pythagorean theorem.) Therefore, the solution becomes

9781118496718-eq190872.png

924. 9781118496718-eq190873.png

The given integral is

9781118496718-eq190874.png

Begin with the trigonometric substitution 9781118496718-eq190875.tif so that 9781118496718-eq190876.tif:

9781118496718-eq190877.png

Now use an identity to get

9781118496718-eq190878.png

From the substitution 9781118496718-eq190879.tif, you can deduce that 9781118496718-eq190880.tif9781118496718-eq190881.tif, and 9781118496718-eq190882.tif. (To find the values of 9781118496718-eq190883.tif and 9781118496718-eq190884.tif, you can label a right triangle using 9781118496718-eq190885.tif and find the missing side of the triangle with the Pythagorean theorem.) Therefore, the solution becomes

9781118496718-eq190886.png

925. 9781118496718-eq190887.png

The given integral is

9781118496718-eq190888.png

Begin with the substitution u = x2 so that du = 2x dx, or 9781118496718-eq190889.tif:

9781118496718-eq190890.png

Now use the substitution 9781118496718-eq190891.tif so that 9781118496718-eq190892.tif:

9781118496718-eq190893.png

From the substitution 9781118496718-eq190894.tif, you can deduce that 9781118496718-eq190895.tif and that 9781118496718-eq190896.tif. Therefore, the answer is

9781118496718-eq190897.png

926. 9781118496718-eq190898.png

Here's the given integral:

9781118496718-eq190899.png

First begin with the substitution u = cos t, which gives you du = –sin t dt, or –du = sin t dt. You can also compute the new limits of integration: If 9781118496718-eq190900.tif, then 9781118496718-eq190901.tif, and if t = 0, then u = cos 0 = 1. Substituting these values into the integral gives you

9781118496718-eq190902.png

Next, use the trigonometric substitution 9781118496718-eq190903.tif, which gives you 9781118496718-eq190904.tif. You can again find the new limits of integration: If u = 1, then 9781118496718-eq190905.tif so that 9781118496718-eq190906.tif, and if u = 0, then 9781118496718-eq190907.tif so that 9781118496718-eq190908.tif. Using these values, you produce the integral

9781118496718-eq190909.png

Now use a trigonometric identity and simplify:

9781118496718-eq190910.png

927. 9781118496718-eq190911.png

Here's the given integral:

9781118496718-eq190912.png

Begin with the substitution 9781118496718-eq190913.tif so that 9781118496718-eq190914.tif:

9781118496718-eq190915.png

Next, use an identity and integrate to get

9781118496718-eq190916.png

From the substitution 9781118496718-eq190917.tif, you can deduce that 9781118496718-eq190918.tif and that 9781118496718-eq190919.tif to get the solution:

9781118496718-eq190920.png

928. 9781118496718-eq190921.png

The given integral is

9781118496718-eq190922.png

Use the trigonometric substitution 9781118496718-eq190923.tif so that 9781118496718-eq190924.tif. You can find the new limits of integration by noting that if x = 1, then 9781118496718-eq190925.tif and 9781118496718-eq190926.tif and that if x = 2, then 9781118496718-eq190927.tif and 9781118496718-eq190928.tif. With these new values, you get

9781118496718-eq190929.png

929. 9781118496718-eq190930.png

Start by rewriting the denominator with a radical:

9781118496718-eq190931.png

Use the trigonometric substitution 9781118496718-eq190932.tif to get 9781118496718-eq190933.tif. These substitutions give you the following:

9781118496718-eq190934.png

From the substitution 9781118496718-eq190935.tif, or 9781118496718-eq190936.tif, you can deduce that 9781118496718-eq190937.tif and that 9781118496718-eq190938.tif. (To deduce that 9781118496718-eq190939.tif, you can label a right triangle using 9781118496718-eq190940.tif and then find the missing side with the Pythagorean theorem.) Therefore, the solution becomes

9781118496718-eq190941.png

930. 9781118496718-eq190942.png

Start by rewriting the integrand:

9781118496718-eq190943.png

Use the substitution 9781118496718-eq190944.tif, or 9781118496718-eq190945.tif, so that 9781118496718-eq190946.tif:

9781118496718-eq190947.png

From 9781118496718-eq190948.tif, or 9781118496718-eq190949.tif, you can deduce that 9781118496718-eq190950.tif. (To deduce that 9781118496718-eq190951.tif, you can label a right triangle using 9781118496718-eq190952.tif and then find the missing side of the triangle with the Pythagorean theorem.) Therefore, the solution becomes

9781118496718-eq190953.png

931. 9781118496718-eq190954.png

Here's the given integral:

9781118496718-eq190955.png

Begin with the substitution 9781118496718-eq190956.tif to get 9781118496718-eq190957.tif. Substituting this value into the original integral gives you the following:

9781118496718-eq190958.png

Now factor out 9781118496718-eq190959.tif and use a trigonometric identity along with a u-substitution:

9781118496718-eq190960.png

Let 9781118496718-eq190961.tif to get 9781118496718-eq190962.tif. Using these values in the integral, you get

9781118496718-eq190963.png

From the first substitution, you get 9781118496718-eq190964.tif, from which you can deduce that 9781118496718-eq190965.tif. (You can find the value 9781118496718-eq190966.tif by using a trigonometric identity for tangent and secant.) Inserting this value into the preceding equation and simplifying a bit gives you the answer:

9781118496718-eq190967.png

932. 9781118496718-eq190968.png

Here's the given integral:

9781118496718-eq190969.png

Begin with the substitution 9781118496718-eq190970.tif so that 9781118496718-eq190971.tif:

9781118496718-eq190972.png

Next, use an identity to get

9781118496718-eq190973.png

Now use the substitution 9781118496718-eq190974.tif so that 9781118496718-eq190975.tif:

9781118496718-eq190976.png

From the substitution 9781118496718-eq190977.tif, you can deduce that 9781118496718-eq190978.tif to get the solution:

9781118496718-eq190979.png

Note that you can also do this problem starting with the substitution 9781118496718-eq190980.tif and a bit of algebra.

933. 9781118496718-eq190981.png

The given expression is

9781118496718-eq190982.png

Let 9781118496718-eq190983.tif to get 9781118496718-eq190984.tif. You can also find the new limits of integration by using the original limits along with the substitution 9781118496718-eq190985.tif. If x = 3, you get 9781118496718-eq190986.tif, or 9781118496718-eq190987.tif, so that 9781118496718-eq190988.tif. Likewise, if x = 0, you get 9781118496718-eq190989.tif, or 9781118496718-eq190990.tif, so that 9781118496718-eq190991.tif. Using these substitutions and the new limits of integration gives you the following:

9781118496718-eq190992.png

Now factor out 9781118496718-eq190993.tif and use a trigonometric identity:

9781118496718-eq190994.png

Use the substitution 9781118496718-eq190995.tif so that 9781118496718-eq190996.tif. You can again find the new limits of integration; with 9781118496718-eq190997.tif, you get 9781118496718-eq190998.tif, and with 9781118496718-eq190999.tif, you get u = sec 0 = 1. Using the new limits and the substitution, you can get the answer as follows:

9781118496718-eq191000.png

934. 9781118496718-eq191001.png

Start by rewriting the integrand:

9781118496718-eq191002.png

Use a substitution, letting w = x3, to get dw = 3xdx, or 9781118496718-eq191003.tif. Putting these values into the integral gives you

9781118496718-eq191004.png

Now use the trigonometric substitution 9781118496718-eq191005.tif, which gives you 9781118496718-eq191006.tif. Using these values in the integral gives you the following:

9781118496718-eq191007.png

Now you can use the identity 9781118496718-eq191008.tif to rewrite the integral:

9781118496718-eq191009.png

From 9781118496718-eq191010.tif, you can deduce that 9781118496718-eq191011.tif and that 9781118496718-eq191012.tif. (Note that you can find the value for 9781118496718-eq191013.tif by using an identity for sine and cosine.) Substituting these values into the integral gives you

9781118496718-eq191014.png

Replacing w with x3 gives you the solution:

9781118496718-eq191015.png

935. 9781118496718-eq191016.png

Rewrite the denominator as a radical:

9781118496718-eq191017.png

Now you complete the square on the quadratic expression:

9781118496718-eq191018.png

So the integral becomes

9781118496718-eq191019.png

Now use the substitution 9781118496718-eq191020.tif, which gives you 9781118496718-eq191021.tif. Substituting these values into the integral gives you

9781118496718-eq191022.png

From the substitution 9781118496718-eq191023.tif, or 9781118496718-eq191024.tif, you can deduce that 9781118496718-eq191025.tif. (To deduce that 9781118496718-eq191026.tif, you can label a right triangle using 9781118496718-eq191027.tif and then find the missing side with the Pythagorean theorem.) Therefore, the solution is

9781118496718-eq191028.png

936. 9781118496718-eq191029.png

Here's the given integral:

9781118496718-eq191030.png

This may not look like a trigonometric substitution problem at first glance because so many of those problems involve some type of radical. However, many integrals that contain quadratic expressions can still be solved with this method.

First complete the square on the quadratic expression:

9781118496718-eq191031.png

So the integral becomes

9781118496718-eq191032.png

Use the substitution 9781118496718-eq191033.tif, which gives you 9781118496718-eq191034.tif. Putting this value into your integral yields

9781118496718-eq191035.png

Because 9781118496718-eq191036.tif, you have

9781118496718-eq191037.png

From the substitution 9781118496718-eq191038.tif, or 9781118496718-eq191039.tif, you can deduce that 9781118496718-eq191040.tif and that 9781118496718-eq191041.tif. (To deduce these values for 9781118496718-eq191042.tif and 9781118496718-eq191043.tif, you can label a right triangle using 9781118496718-eq191044.tif and then find the missing side using the Pythagorean theorem.) Putting these values into the antiderivative gives you the solution:

9781118496718-eq191045.png

937. 9781118496718-eq191046.png

You can begin by rewriting the expression underneath of the radical to make the required substitution clearer:

9781118496718-eq191047.png

Use the substitution 9781118496718-eq191048.tif to get 9781118496718-eq191049.tif so that 9781118496718-eq191050.tif. You can find the new limits of integration by using the original limits of integration and the substitution. If 9781118496718-eq191051.tif, then 9781118496718-eq191052.tif, or 9781118496718-eq191053.tif, so that 9781118496718-eq191054.tif. Likewise, if x = 0, then 9781118496718-eq191055.tif so that 9781118496718-eq191056.tif, or 9781118496718-eq191057.tif. Using this information, you can produce the following integral:

9781118496718-eq191058.png

Now use the substitution 9781118496718-eq191059.tif so that 9781118496718-eq191060.tif, or 9781118496718-eq191061.tif. You can again find the new limits of integration: If 9781118496718-eq191062.tif, then 9781118496718-eq191063.tif. Likewise, if 9781118496718-eq191064.tif, then u = cos 0 = 1. Using the substitution along with the new limits of integration, you get the following (recall that when you switch the limits of integration, the sign on the integral changes):

9781118496718-eq191065.png

Now you have the following:

9781118496718-eq191066.png

938. 9781118496718-eq191067.png

Here's the given integral:

9781118496718-eq191068.png

First complete the square on the quadratic expression underneath the square root:

9781118496718-eq191069.png

Therefore, you have the integral

9781118496718-eq191070.png

Now you can use the substitution 9781118496718-eq191071.tif, which gives you 9781118496718-eq191072.tif. Substituting these values into the integral gives you

9781118496718-eq191073.png

Then use the identity 9781118496718-eq191074.tif to simplify the integral:

9781118496718-eq191075.png

From the substitution 9781118496718-eq191076.tif, you get 9781118496718-eq191077.tif and 9781118496718-eq191078.tif. You can deduce that 9781118496718-eq191079.tif by labeling a right triangle using 9781118496718-eq191080.tif and then finding the missing side of the triangle with the Pythagorean theorem. Putting these values into the antiderivative gives you the solution:

9781118496718-eq191081.png

939. 9781118496718-eq191082.png

Here's the given integral:

9781118496718-eq191083.png

Start with the substitution 9781118496718-eq191084.tif so that 9781118496718-eq191085.tif. Substituting these values into the original integral, you get the following:

9781118496718-eq191086.png

Now factor out 9781118496718-eq191087.tif and use a trigonometric identity along with a u-substitution:

9781118496718-eq191088.png

With 9781118496718-eq191089.tif, you get 9781118496718-eq191090.tif, or 9781118496718-eq191091.tif. Substituting these values into the integral gives you

9781118496718-eq191092.png

From the original substitution, you have 9781118496718-eq191093.tif, from which you can deduce that 9781118496718-eq191094.tif. (To deduce that 9781118496718-eq191095.tif, you can label a right triangle using 9781118496718-eq191096.tif and then find the missing side with the Pythagorean theorem.) With these values, you arrive at the following solution:

9781118496718-eq191097.png

940. 9781118496718-eq191098.png

The given expression is

9781118496718-eq191099.png

Notice that you have the linear factor x raised to the third power and the linear factor (x + 1) raised to the second power. Therefore, in the decomposition, all the numerators will be constants:

9781118496718-eq191100.png

941. 9781118496718-eq191101.png

The given expression is

9781118496718-eq191102.png

Begin by factoring the denominator completely:

9781118496718-eq191103.png

In this expression, you have two distinct linear factors and one irreducible quadratic factor. Therefore, the fraction decomposition becomes

9781118496718-eq191104.png

942. 9781118496718-eq191105.png

Here's the given expression:

9781118496718-eq191106.png

In this expression, a linear factor is being squared, and an irreducible quadratic factor is raised to the third power. Therefore, the fraction decomposition becomes

9781118496718-eq191107.png

943. 9781118496718-eq191108.png

The given expression is

9781118496718-eq191109.png

In this example, you have the linear factor x that's being squared, the linear factor (x – 1) that's being cubed, and the irreducible quadratic factor (x2 + 17). Therefore, the fraction decomposition becomes

9781118496718-eq191110.png

944. 9781118496718-eq191111.png

Begin by factoring the denominator of the expression:

9781118496718-eq191112.png

In the expression on the right side of the equation, you have the distinct linear factors (x – 3) and (x + 3) and the irreducible quadratic factor (x2 + 1) that's being squared. Therefore, the fraction decomposition becomes

9781118496718-eq191113.png

945. 9781118496718-eq191114.png

Start by performing the decomposition:

9781118496718-eq191115.png

Multiply both sides of the equation by (x + 2)(x – 1):

9781118496718-eq191116.png

Now let x = 1 to get 1 = A(1 – 1) + B(1 + 2) so that 1 = 3B, or 9781118496718-eq191117.tif. Also let x = –2 to get 1 = A(–2 – 1) + B(–2 + 2) so that 1 = –3A, or 9781118496718-eq191118.tif. Therefore, you arrive at

9781118496718-eq191119.png

946. 9781118496718-eq191120.png

Start by factoring the denominator:

9781118496718-eq191121.png

Then perform the decomposition:

9781118496718-eq191122.png

Multiply both sides of the equation by x(x2 + 1):

9781118496718-eq191123.png

Note that if x = 0, then 9781118496718-eq191124.tif so that 2 = A. Next, expand the right side of the equation and equate the coefficients to find the remaining coefficients:

9781118496718-eq191125.png

After rearranging, you get

9781118496718-eq191126.png

Now equate coefficients to arrive at the equation 0 = A + B; however, you know that 2 = A, so 0 = 2 + B, or –2 = B. Likewise, by equating coefficients, you immediately find that C = 1. Therefore, the partial fraction decomposition becomes

9781118496718-eq191127.png

947. 9781118496718-eq191128.png

Start by factoring the denominator:

9781118496718-eq191129.png

Then perform the fraction decomposition:

9781118496718-eq191130.png

Multiply both sides of the equation by (x – 3)2:

9781118496718-eq191131.png

By letting x = 3 in the last equation, you arrive at 5(3) + 1 = A(3 – 3) + B, so 16 = B. By expanding the right side of the equation, you get

9781118496718-eq191132.png

So by equating coefficients, you immediately see that 5 = A. Therefore, the fraction decomposition becomes

9781118496718-eq191133.png

948. 9781118496718-eq191134.png

Begin by factoring the denominator:

9781118496718-eq191135.png

Then perform the decomposition:

9781118496718-eq191136.png

Multiply both sides of the equation by (x2 + 1)(x2 + 3) to get

9781118496718-eq191137.png

Expand the right side of the equation and collect like terms:

9781118496718-eq191138.png

Equating coefficients gives you the equations A + C = 0, B + D = 1, 3A + C = 0, and 3B + D = 2. Because A = –C, you get 3(–C) + C = 0, so –2C = 0, or C = 0; that means A = 0 as well. Likewise, because B = 1 – D, you get 3(1 – D) + D = 2, so 3 – 2D = 2, or 9781118496718-eq191139.tif; you also get 9781118496718-eq191140.tif. With these values, you get the following solution:

9781118496718-eq191141.png

949. 9781118496718-eq191142.png

Begin by performing the decomposition:

9781118496718-eq191143.png

Multiplying both sides of the equation by (x2 + 5)2 gives you

9781118496718-eq191144.png

Expand the right side of the equation and regroup:

9781118496718-eq191145.png

By equating coefficients, you find A = 0, B = 1, 5A + C = 0, and 5B + D = 0. Using A = 0 and 5A + C = 0, you get 5(0) + C = 0, or C = 0. Likewise, using B = 1 and 5B + D = 1, you find that 5(1) + D = 1 so that D = –4. Therefore, the fraction decomposition becomes

9781118496718-eq191146.png

950. 9781118496718-eq191147.png

The given integral is

9781118496718-eq191148.png

Notice that the degree of the numerator is equal to the degree of the denominator, so you must divide. You could use long division, but for simple expressions of this type, you can simply subtract 5 and add 5 to the numerator and then split the fraction:

9781118496718-eq191149.png

Now apply elementary antiderivative formulas to get the solution:

9781118496718-eq191150.png

951. 9781118496718-eq191151.png

The given integral is

9781118496718-eq191152.png

The degree of the numerator is greater than or equal to the degree of the denominator, so use polynomial long division to get the following:

9781118496718-eq191153.png

Then apply basic antiderivative formulas:

9781118496718-eq191154.png

952. 9781118496718-eq191155.png

The given integral is

9781118496718-eq191156.png

First perform a fraction decomposition on the integrand:

9781118496718-eq191157.png

Multiplying both sides of the equation by (x + 4)(x – 5) and simplifying gives you

9781118496718-eq191158.png

Now you can find the values of the coefficients by picking appropriate values of x and solving the resulting equations. So if you let x = 5, you get 5 – 3 = A(5 – 5) + B(5 + 4), or 2 = 9B, so that 9781118496718-eq191159.tif. Likewise, if x = –4, then you get –4 – 3 = A(–4 – 5) + B(–4 + 4), or –7 = A(–9), so that 9781118496718-eq191160.tif. This gives you

9781118496718-eq191161.png

Applying elementary antiderivative formulas gives you the solution:

9781118496718-eq191162.png

953. 9781118496718-eq191163.png

The given expression is

9781118496718-eq191164.png

First factor the denominator of the integrand:

9781118496718-eq191165.png

Then perform the fraction decomposition:

9781118496718-eq191166.png

Multiplying both sides of the equation by (x + 1)(x – 1) and simplifying gives you

9781118496718-eq191167.png

Now you can solve for the coefficients by picking appropriate values of x. Notice that if x = 1, you get 1 = A(1 – 1) + B(1 + 1) so that 1 = 2B, or 9781118496718-eq191168.tif. Likewise, if x = –1, you get 1 = A(–1 – 1) + B(–1 + 1) so that 1 = –2A, or 9781118496718-eq191169.tif. This gives you

9781118496718-eq191170.png

Applying elementary antiderivatives gives you the following:

9781118496718-eq191171.png

954. 9781118496718-eq191172.png

The given integral is

9781118496718-eq191173.png

Begin by factoring the denominator:

9781118496718-eq191174.png

And then perform a fraction decomposition:

9781118496718-eq191175.png

Multiplying both sides by (x + 1)2 gives you the following:

9781118496718-eq191176.png

Expanding the right side and equating coefficients gives you 3x + 5 = Ax + (A + B) so that A = 3 and A + B = 5; in turn, that gives you 3 + B = 5 so that B = 2. Now you have

9781118496718-eq191177.png

For the first term in the integrand, simply use an elementary antiderivative:

9781118496718-eq191178.png. For the second term in the integrand, use a substitution on 9781118496718-eq191179.tif where u = x + 1 so that du = dx. Using these values gives you

9781118496718-eq191180.png

Therefore, the answer is

9781118496718-eq191181.png

Note: You can also evaluate 9781118496718-eq191182.tif with the substitution u = x + 1 so that du = dx and u – 1 = x.

955. 9781118496718-eq191183.png

The given expression is

9781118496718-eq191184.png

The degree of the numerator is greater than or equal to the degree of the denominator, so use polynomial long division to get the following:

9781118496718-eq191185.png

Next, perform a fraction decomposition:

9781118496718-eq191186.png

Multiply both sides of the equation by (x – 3)(x + 2):

9781118496718-eq191187.png

If you let x = 2, you get 9 = B(–2 – 3) so that 9781118496718-eq191188.tif. And if you let x = 3, you get 14 = A(3 + 2) so that 9781118496718-eq191189.tif. With these values, you produce the following integral:

9781118496718-eq191190.png

To integrate, use basic antiderivative formulas to get

9781118496718-eq191191.png

956. 9781118496718-eq191192.png

The given expression is

9781118496718-eq191193.png

First factor the denominator of the integrand:

9781118496718-eq191194.png

And then perform the fraction decomposition: