﻿ ﻿Big Idea 3: Integrals and the Fundamental Theorems of Calculus. Areas and Volumes - Review the Knowledge You Need to Score High - 5 Steps to a 5: AP Calculus AB 2017 (2016)

## 5 Steps to a 5: AP Calculus AB 2017 (2016)

### Areas and Volumes

IN THIS CHAPTER

Summary: In this chapter, you will be introduced to several important applications of the definite integral. You will learn how to find the area under a curve and the volume of a solid. Some of the techniques that you will be shown include finding area under a curve by using rectangular and trapezoidal approximations, and finding the volume of a solid using cross sections, discs, and washers. These techniques involve working with algebraic expressions and lengthy computations. It is important that you work carefully through the practice problems provided in the chapter, and check your solutions with the given explanations.

The function

Rectangular Approximations

Trapezoidal Approximations

Area Under a Curve

Area Between Two Curves

Solids with Known Cross Sections

The Disc Method

The Washer Method

13.1 The Function

The Second Fundamental Theorem of Calculus defines

and states that if f is continuous on [a , b ], then F ′ (x ) = f (x ) for every point x in [a , b ].

If f ≥ 0, then F ≥ 0. F (x ) can be interpreted geometrically as the area under the curve of f from t = a to t = x . (See Figure 13.1-1 .)

Figure 13.1-1

If f < 0, F < 0, F (x ) can be treated as the negative value of the area between the curve of f and the t -axis from t = a to t = x . (See Figure 13.1-2 .)

Figure 13.1-2

Example 1

If for 0 ≤ x ≤ 2π , find the value(s) of x where f has a local minimum.

.

Method 2: You can solve this problem geometrically by using area. See Figure 13.1-3 .

Figure 13.1-3

The area “under the curve” is above the t -axis on [0, π /2] and below the x -axis on [π /2, 3π /2]. Thus the local minimum occurs at 3π /2.

Example 2

Let and the graph of f is shown in Figure 13.1-4 .

Figure 13.1-4

(a) Evaluate: p (0), p (1), p (4).

(b) Evaluate: p (5), p (7), p (8).

(c) At what value of t does p have a maximum value?

(d) On what interval(s) is p decreasing?

(e) Draw a sketch of the graph of p .

Solution:

(c) Since f ≥ 0 on the interval [0, 4], p attains a maximum at t = 4.

(d) Since f (t ) is below the x -axis from t = 4 to t = 8, if x > 4,

.

Thus, p is decreasing on the interval (4, 8).

(e) . See Figure 13.1-5 for a sketch.

Figure 13.1-5

• Remember differentiability implies continuity, but the converse is not true, i.e., continuity does not imply differentiability, e.g., as in the case of a cusp or a corner.

Example 3

The position function of a moving particle on a coordinate axis is:

, where t is in seconds and s is in feet.

The function f is a differentiable function and its graph is shown below in Figure 13.1-6 .

Figure 13.1-6

(a) What is the particle’s velocity at t = 4?

(b) What is the particle’s position at t = 3?

(c) When is the acceleration zero?

(d) When is the particle moving to the right?

(e) At t = 8, is the particle on the right side or left side of the origin?

Solution:

(a) Since , then v (t ) = s ′ (t ) = f (t ).

Thus, v (4) = –8 ft/sec.

(b) .

(c) a (t ) = v ′ (t ). Since v ′ (t ) = f ′ (t ), v ′ (t ) = 0 at t = 4. Thus, a (4) = 0 ft/sec2 .

(d) The particle is moving to the right when v (t ) > 0. Thus, the particle is moving to the right on intervals (0, 2) and (7, 8).

(e) The area of f below the x -axis from x = 2 to x = 7 is larger than the area of f above the x -axis from x = 0 to x = 2 and x = 7 to x = 8. Thus, and the particle is on the left side of the origin.

• Do not forget that (fg )′ = fg + gf and not fg ′. However, lim(fg ) = (lim f ) (lim g ).

13.2 Approximating the Area Under a Curve

Main Concepts: Rectangular Approximations, Trapezoidal Approximations

Rectangular Approximations

If f ≥ 0, the area under the curve of f can be approximated using three common types of rectangles: left-endpoint rectangles, right-endpoint rectangles, or midpoint rectangles. (See Figure 13.2-1 .)

Figure 13.2-1

The area under the curve using n rectangles of equal length is approximately:

If f is increasing on [a , b ], then left-endpoint rectangles are inscribed rectangles and the right-endpoint rectangles are circumscribed rectangles. If f is decreasing on [a , b ], then left-endpoint rectangles are circumscribed rectangles and the right-endpoint rectangles are inscribed. Furthermore,

inscribed rectangle ≤ area under the curve circumscribed rectangle.

Example 1

Find the approximate area under the curve of f (x ) = x 2 + 1 from x = 0 to x = 2, using 4 left-endpoint rectangles of equal length. (See Figure 13.2-2 .)

Figure 13.2-2

Let Δxi be the length of i th rectangle. The length .

Area under the curve .

Enter Σ (((.5(x – 1))2 + 1) ∗ .5, x , 1, 4) and obtain 3.75.

Or, find the area of each rectangle:

Area of (RectI + RectII + RectIII + RectIV ) = 3.75.

Thus, the approximate area under the curve of f (x ) is 3.75.

Example 2

Find the approximate area under the curve of from x = 4 to x = 9 using 5 right-endpoint rectangles. (See Figure 13.2-3 .)

Figure 13.2-3

Let Δxi be the length of the i th rectangle. The length .

Or, using Σ notation:

.

Enter and obtain 13.160.

Thus the area under the curve is approximately 13.160.

Example 3

The function f is continuous on [1, 9] and f > 0. Selected values of f are given below:

Using 4 midpoint rectangles, approximate the area under the curve of f for x = 1 to x = 9. (See Figure 13.2-4 .)

Figure 13.2-4

Let Δxi be the length of the i th rectangle. The length .

Area of RectI = f (2)(2) = (1.41)2 = 2.82.

Area of RectII = f (4)(2) = (2)2 = 4.

Area of RectIII = f (6)(2) = (2.45)2 = 4.90.

Area of RectIV = f (8)(2) = (2.83)2 = 5.66.

Area of (RectI + RectII + RectIII + RectIV ) = 2.82 + 4 + 4.90 + 5.66 = 17.38.

Thus the area under the curve is approximately 17.38.

Trapezoidal Approximations

Another method of approximating the area under a curve is to use trapezoids. See Figure 13.2-5 .

Figure 13.2-5

Formula for Trapezoidal Approximation

If f is continuous, the area under the curve of f from x = a to x = b is:

.

Example 1

Find the approximate area under the curve of from x = 0 to x = π , using 4 trapezoids. (See Figure 13.2-6 .)

Figure 13.2-6

Since .

Area under the curve:

• When using a graphing calculator in solving a problem, you are required to write the setup that leads to the answer. For example, if you are finding the volume of a solid, you must write the definite integral and then use the calculator to compute the numerical value, e.g., Volume . Simply indicating the answer without writing the integral would get you only one point for the answer. And you will not get full credit for the problem.

13.3 Area and Definite Integrals

Main Concepts: Area Under a Curve, Area Between Two Curves

Area Under a Curve

If y = f (x ) is continuous and non-negative on [a , b ], then the area under the curve of f from a to b is:

.

If f is continuous and f < 0 on [a , b ], then the area under the curve from a to b is:

. See Figure 13.3-1 .

Figure 13.3-1

If x = g (y ) is continuous and non-negative on [c , d ], then the area under the curve of g from c to d is:

Area . See Figure 13.3-2 .

Figure 13.3-2

Example 1

Find the area under the curve of f (x ) = (x – 1)3 from x = 0 to x = 2.

Step 1: Sketch the graph of f (x ). See Figure 13.3-3 .

Figure 13.3-3

Step 2: Set up integrals.

.

Step 3: Evaluate the integrals.

Another solution is to find the area using a calculator.

Enter and obtain .

Example 2

Find the area of the region bounded by the graph of f (x ) = x 2 – 1, the lines x = –2 and x = 2, and the x -axis.

Step 1: Sketch the graph of f (x ). See Figure 13.3-4 .

Figure 13.3-4

Step 2: Set up integrals.

Step 3: Evaluate the integrals.

Note: Since f (x ) = x 2 – 1 is an even function, you can use the symmetry of the graph and set area = .

An alternate solution is to find the area using a calculator.

Enter (abs (x ^2 – 1), x –2, 2) and obtain 4.

Example 3

Find the area of the region bounded by x = y 2 , y = –1, and y = 3. See Figure 13.3-5 .

Figure 13.3-5

Example 4

Using a calculator, find the area bounded by f (x ) = x 3 + x 2 – 6x and the x -axis. See Figure 13.3-6 .

Figure 13.3-6

Step 1: Enter y 1 = x 3 + x 2 – 6x .

Step 2: Enter and obtain 21.083.

Example 5

The area under the curve y = ex from x = 0 to x = k is 1. Find the value of k .

Area = . Take ln of both sides:

ln(ek ) = ln 2; k = ln 2.

Example 6

The region bounded by the x -axis, and the graph of y = sin x between x = 0 and x = π is divided into 2 regions by the line x = k . If the area of the region for 0 ≤ xk is twice the area of the region kxπ , find k . (See Figure 13.3-7.)

Figure 13.3-7

Area Between Two Curves

Area Bounded by Two Curves: See Figure 13.3-8 .

Figure 13.3-8

Example 1

Find the area of the regions bounded by the graphs of f (x ) = x 3 and g (x ) = x . (See Figure 13.3-9 .)

Figure 13.3-9

Step 1: Sketch the graphs of f (x ) and g (x ).

Step 2: Find the points of intersection.

Step 3: Set up integrals.

Note: You can use the symmetry of the graphs and let area = .

An alternate solution is to find the area using a calculator. Enter and obtain .

Example 2

Find the area of the region bounded by the curve y = ex , the y -axis, and the line y = e 2 .

Step 1: Sketch a graph. (See Figure 13.3-10 .)

Figure 13.3-10

Step 2: Find the point of intersection. Set e 2 = ex x = 2.

Step 3: Set up an integral:

Or using a calculator, enter and obtain (e 2 + 1).

Example 3

Using a calculator, find the area of the region bounded by y = sin x and between 0 ≤ xπ .

Step 1: Sketch a graph. (See Figure 13.3-11 .)

Figure 13.3-11

Step 2: Find the points of intersection.

Using the [Intersection ] function of the calculator, the intersection points are x = 0 and x = 1.89549.

Step 3: Enter nInt(sin(x ) – .5x , x , 0, 1.89549) and obtain 0.420798 ≈ 0.421.

(Note: You could also use the function on your calculator and get the same result.)

Example 4

Find the area of the region bounded by the curve xy = 1 and the lines y = –5, x = e , and x = e 3 .

Step 1: Sketch a graph. (See Figure 13.3-12 .)

Figure 13.3-12

Step 2: Set up an integral.

.

Step 3: Evaluate the integral.

.

• Remember: If f ′ > 0, then f is increasing, and if f ″ > 0, then the graph of f is concave upward.

13.4 Volumes and Definite Integrals

Main Concepts: Solids with Known Cross Sections, The Disc Method, The Washer Method

Solids with Known Cross Sections

If A (x ) is the area of a cross section of a solid and A (x ) is continuous on [a , b ], then the volume of the solid from x = a to x = b is:

(See Figure 13.4-1 .)

Figure 13.4-1

Note: A cross section of a solid is perpendicular to the height of the solid.

Example 1

The base of a solid is the region enclosed by the ellipse . The cross sections are perpendicular to the x -axis and are isosceles right triangles whose hypotenuses are on the ellipse. Find the volume of the solid. (See Figure 13.4-2 .)

Figure 13.4-2

Step 1: Find the area of a cross section A (x ).

Step 2: Set up an integral.

Step 3: Evaluate the integral.

The volume of the solid is .

Verify your result with a graphing calculator.

Example 2

Find the volume of a pyramid whose base is a square with a side of 6 feet long, and a height of 10 feet. (See Figure 13.4-3 .)

Figure 13.4-3

Step 1: Find the area of a cross section A (x ). Note each cross section is a square of side 2s .

Step 2: Set up an integral.

Step 3: Evaluate the integral.

The volume of the pyramid is 120 ft3 .

Example 3

The base of a solid is the region enclosed by a triangle whose vertices are (0, 0), (4, 0), and (0, 2). The cross sections are semicircles perpendicular to the x -axis. Using a calculator, find the volume of the solid. (See Figure 13.4-4 .)

Figure 13.4-4

Step 1: Find the area of a cross section.

Equation of the line passing through (0, 2) and (4, 0):

Step 2: Set up an integral.

Step 3: Evaluate the integral.

Enter and obtain 2.0944.

Thus the volume of the solid is 2.094.

• Remember: If f ′ < 0, then f is decreasing, and if f ″ < 0, then the graph of f is concave downward.

The Disc Method

The volume of a solid of revolution using discs:

.

.

(See Figure 13.4-5 .)

Figure 13.4-5

Revolving about a line y = k :

.

Revolving about a line x = h :

.

(See Figure 13.4-6 .)

Figure 13.4-6

Example 1

Find the volume of the solid generated by revolving about the x -axis the region bounded by the graph of , the x -axis, and the line x = 5.

Step 1: Draw a sketch. (See Figure 13.4-7 .)

Figure 13.4-7

Step 2: Determine the radius of a disc from a cross section.

Step 3: Set up an integral.

Step 4: Evaluate the integral.

Verify your result with a calculator.

Example 2

Find the volume of the solid generated by revolving about the x -axis the region bounded by the graph of where , the x -axis, and the y -axis.

Step 1: Draw a sketch. (See Figure 13.4-8 .)

Figure 13.4-8

Step 2: Determine the radius from a cross section.

Step 3: Set up an integral.

Step 4: Evaluate the integral.

Thus, the volume of the solid is π .

Verify your result with a calculator.

Example 3

Find the volume of the solid generated by revolving about the y -axis the region in the first quadrant bounded by the graph of y = x 2 , the y -axis, and the line y = 6.

Step 1: Draw a sketch. (See Figure 13.4-9 .)

Figure 13.4-9

Step 2: Determine the radius from a cross section.

Step 3: Set up an integral.

Step 4: Evaluate the integral.

The volume of the solid is 18π .

Verify your result with a calculator.

Example 4

Using a calculator, find the volume of the solid generated by revolving about the line y = 8 the region bounded by the graph of y = x 2 + 4, and the line y = 8.

Step 1: Draw a sketch. (See Figure 13.4-10 .)

Figure 13.4-10

Step 2: Determine the radius from a cross section.

r = 8 – y = 8 – (x 2 + 4) = 4 – x 2

Step 3: Set up an integral.

To find the intersection points, set 8 = x 2 + 4 ⇒ x = ±2.

Step 4: Evaluate the integral.

.

Thus, the volume of the solid is .

Verify your result with a calculator.

Example 5

Using a calculator, find the volume of the solid generated by revolving about the line y = –3 the region bounded by the graph of y = ex , the y -axis, and the lines x = ln 2 and y = –3.

Step 1: Draw a sketch. (See Figure 13.4-11 .)

Figure 13.4-11

Step 2: Determine the radius from a cross section.

r = y – (–3) = y + 3 = ex + 3

Step 3: Set up an integral.

Step 4: Evaluate the integral.

The volume of the solid is approximately 13.7383π .

• Remember: If f ′ is increasing, then f ″ > 0 and the graph of f is concave upward.

The Washer Method

The volume of a solid (with a hole in the middle) generated by revolving a region bounded by 2 curves:

; where f (x ) = outer radius and g (x ) = inner radius.

; where p (y ) = outer radius and q (y ) = inner radius.

About a line x = h :

.

About a line y = k :

.

Example 1

Using the Washer Method, find the volume of the solid generated by revolving the region bounded by y = x 3 and y = x in the first quadrant about the x -axis.

Step 1: Draw a sketch. (See Figure 13.4-12 .)

Figure 13.4-12

To find the points of intersection, set x = x 3x 3x = 0 or x (x 2 – 1) = 0, or x = –1, 0, 1. In the first quadrant x = 0, 1.

Step 2: Determine the outer and inner radii of a washer, whose outer radius = x ; and inner radius = x 3 .

Step 3: Set up an integral.

Step 4: Evaluate the integral.

Verify your result with a calculator.

Example 2

Using the Washer Method and a calculator, find the volume of the solid generated by revolving the region in Example 1 about the line y = 2.

Step 1: Draw a sketch. (See Figure 13.4-13 .)

Figure 13.4-13

Step 2: Determine the outer and inner radii of a washer.

The outer radius = (2 – x 3 ) and inner radius = (2 – x ).

Step 3: Set up an integral.

Step 4: Evaluate the integral.

.

The volume of the solid is .

Example 3

Using the Washer Method and a calculator, find the volume of the solid generated by revolving the region bounded by y = x 2 and x = y 2 about the y -axis.

Step 1: Draw a sketch. (See Figure 13.4-14 .)

Figure 13.4-14

Intersection points: .

Step 2: Determine the outer and inner radii of a washer, with outer radius:

and inner radius: x = y 2 .

Step 3: Set up an integral.

Step 4: Evaluate the integral.

.

The volume of the solid is .

13.5 Rapid Review

1. If and the graph of g is shown in Figure 13.5-1 . Find f (3).

Figure 13.5-1

2. The function f is continous on [1, 5] and f > 0 and selected values of f are given below.

Using 2 midpoint rectangles, approximate the area under the curve of f for x = 1 to x = 5.

Answer: Midpoints are x = 2 and x = 4 and the width of each rectangle .

Area ≈ Area of Rect1 + Area of Rect2 ≈ 4(2) + 8(2) ≈ 24.

3. Set up an integral to find the area of the regions bounded by the graphs of y = x 3 and y = x . Do not evaluate the integral.

Answer: Graphs intersect at x = –1 and x = 1. (See Figure 13.5-2 .)

Figure 13.5-2

.

Or, using symmetry, .

4. The base of a solid is the region bounded by the lines y = x , x = 1, and the x -axis. The cross sections are squares perpendicular to the x -axis. Set up an integral to find the volume of the solid. Do not evaluate the integral.

Answer: Area of cross section = x 2 .

.

5. Set up an integral to find the volume of a solid generated by revolving the region bounded by the graph of y = sin x , where 0 ≤ xπ and the x -axis, about the x -axis. Do not evaluate the integral.

.

6. The area under the curve of from x = a to x = 5 is approximately 0.916 where 1 ≤ a < 5. Using your calculator, find a .

13.6 Practice Problems

Part A—The use of a calculator is not allowed.

1 . Let where the graph of f is given in Figure 13.6-1 .

Figure 13.6-1

(a) Evaluate F (0), F (3), and F (5).

(b) On what interval(s) is F decreasing?

(c) At what value of t does F have a maximum value?

(d) On what interval is F concave up?

2 . Find the area of the region(s) enclosed by the curve f (x ) = x 3 , the x -axis, and the lines x = –1 and x = 2.

3 . Find the area of the region(s) enclosed by the curve y = |2x – 6|, the x -axis, and the lines x = 0 and x = 4.

4 . Find the approximate area under the curve from x = 1 to x = 5, using four right-endpoint rectangles of equal lengths.

5 . Find the approximate area under the curve y = x 2 + 1 from x = 0 to x = 3, using the Trapezoidal Rule with n = 3.

6 . Find the area of the region bounded by the graphs , y = –x , and x = 4.

7 . Find the area of the region bounded by the curves x = y 2 and x = 4.

8 . Find the area of the region bounded by the graphs of all four equations: ; x -axis; and the lines, and x = π .

9 . Find the volume of the solid obtained by revolving about the x -axis, the region bounded by the graphs of y = x 2 + 4, the x -axis, the y -axis, and the lines x = 3.

10 . The area under the curve from x = 1 to x = k is 1. Find the value of k .

11 . Find the volume of the solid obtained by revolving about the y -axis the region bounded by x = y 2 + 1, x = 0, y = –1, and y = 1.

12 . Let R be the region enclosed by the graph y = 3x , the x -axis, and the line x = 4. The line x = a divides region R into two regions such that when the regions are revolved about the x -axis, the resulting solids have equal volume. Find a .

Part B—Calculators are allowed.

13 . Find the volume of the solid obtained by revolving about the x -axis the region bounded by the graphs of f (x ) = x 3 and g (x ) = x 2 .

14 . The base of a solid is a region bounded by the circle x 2 + y 2 = 4. The cross sections of the solid perpendicular to the x -axis are equilateral triangles. Find the volume of the solid.

15 . Find the volume of the solid obtained by revolving about the y -axis, the region bounded by the curves x = y 2 and y = x – 2. For Problems 16 through 19, find the volume of the solid obtained by revolving the region as described below. (See Figure 13.6-2 .)

Figure 13.6-2

16 . R 1 about the x -axis.

17 . R 2 about the y -axis.

18 . R 1 about the line .

19 . R 2 about the line .

20 . The function f (x ) is continuous on [0, 12] and the selected values of f (x ) are shown in the table.

Find the approximate area under the curve of f from 0 to 12 using three midpoint rectangles.

13.7 Cumulative Review Problems

(Calculator) indicates that calculators are permitted.

21 . If , find in terms of k .

22 . A man wishes to pull a log over a 9-foot-high garden wall as shown in Figure 13.7-1 . He is pulling at a rate of 2 ft/sec. At what rate is the angle between the rope and the ground changing when there are 15 feet of rope between the top of the wall and the log?

Figure 13.7-1

23 . (Calculator) Find a point on the parabola that is closest to the point (4, 1).

24 . The velocity function of a particle moving along the x -axis is v (t ) = t cos(t 2 + 1) for t ≥ 0.

(a) If at t = 0, the particle is at the origin, find the position of the particle at t = 2.

(b) Is the particle moving to the right or left at t = 2?

(c) Find the acceleration of the particle at t = 2 and determine if the velocity of the particle is increasing or decreasing. Explain why.

25 . (Calculator) given f (x ) = xex and g (x ) = cos x , find:

(a) The area of the region in the first quadrant bounded by the graphs f (x ), g (x ), and x = 0.

(b) The volume obtained by revolving the region in part (a) about the x -axis.

13.8 Solutions to Practice Problems

Part A—The use of a calculator is not allowed.

1 . (a)

(b) Since , F is decreasing on the interval [3, 5].

(c) At t = 3, F has a maximum value.

(d) F ′ (x ) = f (x ), F ′ (x ) is increasing on (4, 5) which implies F ≤ (x ) > 0. Thus F is concave upward on (4, 5).

2 . (See Figure 13.8-1 .)

Figure 13.8-1

3 . (See Figure 13.8-2 .)

Figure 13.8-2

Set 2x – 6 = 0; x = 3 and

4 . (See Figure 13.8-3 .)

Figure 13.8-3

5 . (See Figure 13.8-4 .)

Figure 13.8-4

6 . (See Figure 13.8-5 .)

Figure 13.8-5

7 . (See Figure 13.8-6 .)

Figure 13.8-6

Intersection points: 4 = y 2y = ±2.

You can use the symmetry of the region and obtain the . An alternative method is to find the area by setting up an integral with respect to the x -axis and expressing x = y 2 as and .

8 . (See Figure 13.8-7 .)

Figure 13.8-7

9 . (See Figure 13.8-8 .)

Figure 13.8-8

Using the Disc Method:

10 . Area

Set ln k = 1. Thus e ln k = e 1 or k = e .

11 . (See Figure 13.8-9 .)

Figure 13.8-9

Using the Disc Method:

Note: You can use the symmetry of the region and find the volume by

.

12 . Volume of solid by revolving R :

You can verify your result by evaluating

. The result is 96π .

Part B—Calculators are allowed.

13 . (See Figure 13.8-10 .)

Figure 13.8-10

Step 1: Using the Washer Method: Points of Intersection: Set x 3 = x 2x 3x 2 = 0 ⇒ x 2 (x – 1) = 0 or x = 1. Outer radius = x 2 ; Inner radius = x 3 .

Step 2:

Step 3: Enter and obtain .

14 . (See Figure 13.8-11 .)

Figure 13.8-11

Step 1: x 2 + y 2 = 4 ⇒ y 2 = 4 – x 2 Let s = a side of an equilateral triangle .

Step 2: Area of a cross section:

.

Step 3:

Step 4: Enter and obtain .

15 . (See Figure 13.8-12 .)

Figure 13.8-12

Step 1: Using the Washer Method:

Points of Intersection:

y = x – 2 ⇒ x = y + 2

Set y 2 = y + 2

y 2y – 2 = 0

⇒ (y – 2)(y + 1) = 0

or y = –1 or y = 2.

Outer radius = y + 2;

Inner radius = y 2 .

Step 2:

Step 3: Enter

and obtain .

16 . (See Figure 13.8-13 .)

Figure 13.8-13

Step 1: Using the Washer Method:

y = 8, y = x 3

Inner radius = x 3 .

Step 2: Enter ∫ π (82x 6 , x , 0, 2) and obtain .

17 . (See Figure 13.8-14 .)

Figure 13.8-14

Using the Washer Method:

Inner radius: x = y 1/3 .

.

18 . (See Figure 13.8-15 .)

Figure 13.8-15

Step 1: Using the Disc Method:

Radius = (8 – x 3 ).

Step 2: Enter and obtain .

19 . (See Figure 13.8-16 .)

Figure 13.8-16

Using the Disc Method:

Radius = 2 – x = (2 – y 1/3 ).

.

20 . .

xi = midpoint of the i th interval.

Length of .

Area of RectI = f (2)Δx 1 = (2.24)(4) = 8.96.

Area of RectII = f (6)Δx 2 = (3.61)(4) = 14.44.

Area of RectIII = f (10)Δx 3 = (4.58)(4) = 18.32.

Total Area = 8.96 + 14.44 + 18.32 = 41.72.

The area under the curve is approximately 41.72.

13.9 Solutions to Cumulative Review Problems

21 . (See Figure 13.9-1 .)

Figure 13.9-1

Since e x 2 is an even function, thus

22 . (See Figure 13.9-2 .)

Figure 13.9-2

Differentiate both sides:

.

When x = 15, 92 + y 2 = 152y = 12.

Thus, .

23 . (See Figure 13.9-3 .)

Figure 13.9-3

Step 1: Distance Formula:

where the domain is all real numbers.

Step 2: Enter y 1 =

Enter y 2 = d (y 1 (x ), x ).

Step 3: Use the [Zero ] function and obtain x = 2 for y 2 .

Step 4: Use the First Derivative Test. (See Figures 13.9-4 and 13.9-5 .) At x = 2, L has a relative minimum. Since at x = 2, L has the only relative extremum, it is an absolute minimum.

Figure 13.9-4

Figure 13.9-5

Step 5: At . Thus, the point on closest to the point (4, 1) is the point (2, 2).

24 . (a) s (0) = 0 and

.

Enter

and obtain .

Thus, .

(b) v (2) = 2 cos(22 + 1) = 2 cos(5) = 0.567324

Since v (2) > 0, the particle is moving to the right at t = 2.

(c) a (t ) = v ′(t )

Enter d (x * cos(x ^ 2 + 1), x )|x = 2 and obtain 7.95506.

Thus, the velocity of the particle is increasing at t = 2, since a (2) > 0.

25 . (See Figure 13.9-6 .)

Figure 13.9-6

(a) Point of Intersection: Use the [Intersection ] function of the calculator and obtain (0.517757, 0.868931).

(0, 0.51775) and obtain 0.304261. The area of the region is approximately 0.304.

(b) Step 1: Using the Washer Method:

Outer radius = cos x ;