Signed Areas and Other Integrals - CALCULUS - MATHEMATICS IN HISTORY - Mathematics for the liberal arts

Mathematics for the liberal arts (2013)

Part I. MATHEMATICS IN HISTORY

Chapter 4. CALCULUS

4.16 Signed Areas and Other Integrals

You might wonder what an integral represents for a function that takes both positive and negative values. For example, we can use the Fundamental Theorem of Calculus to compute

Figure 4.35 illustrates this situation.

Figure 4.35 Signed area under images from 0 to 4.

images

The integral is zero because the area under the curve over the interval [0,2] is counted as positive area, and the area above the function on the interval [2,4] counts as negative area. There are precisely matching amounts of positive and negative area, and they cancel each other out.

We can be sure this is correct if we think about the Riemann definition of integral. Consider dividing the interval [0,4] into small subintervals, and creating a rectangle approximation. The area of the ith rectangle is

images

If images lands in the left half of the interval, then images is positive. In the right half of the interval, images is negative. The width of a rectangle, Δxi, is always positive.

Since we get a sum of “positive rectangles” from the left half of the interval and “negative rectangles” from the right half of the rectangle, the definition verifies that (when we take the limit) the integral will be the sum of the “signed areas.” In this case, we computed this to be zero, and we can see this is zero in the figure. There is an important principal here.

If a quantity is represented by a limit of Riemann sums, then it is computed by taking an integral.

In geometry, we don’t speak of positive and negative areas. Areas are always positive. But for functions that take positive and negative values, the Riemann sum essentially counts positive areas above the x-axis and negative areas below. So the integral does the same.

Although it is intuitive to think of integrals as (signed) areas, anything that can be represented as the limit of a Riemann sum will be an integral, even if you can’t imagine it as an area.

imagesEXAMPLE 4.44

Imagine that you are in a car driving away from town. To keep a record of your speed, you’ve modified your car with a scrolling paper roll (like a seismograph) attached to the speedometer. When you go faster, the “seismograph” pen moves up, and when you go slower, the pen moves down. Perhaps it creates a graph like Figure 4.36.

Figure 4.36 Speed of a car recorded over time.

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On this trip, the car starts from rest. We can see that, because the curve starts at (0,0). The pen makes a mark on the graph for t = 0 indicating that the speed is y = 0 mi/h. Over the first hour of the trip, the car accelerates evenly until it is traveling 55 mi/h, and it maintains that speed for the next hour before slowing to a stop over the next hour.

Consider dividing the interval into subintervals, and creating the rectangles of a Riemann sum, as in Figure 4.37.

Figure 4.37 Speed of a car recorded over time.

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Each rectangle is formed the same way, by taking some reading of the speedometer (measured in mi/h) for the height of the rectangle and a time interval (measured in hours) for the width of the rectangle.

What does the area of a rectangle represent? Looking at units, we can see that the height of the rectangle is mi/h, so the height of the rectangle is a rate. The width of the rectangle is in hours, so it represents an elapsed time. Since distance = rate × time, the area of the rectangle appears to represent a distance.

When the car moves at a constant rate, as it does between t = 1 and t = 2, then the area of a rectangle is exactly how far the car moves in an interval. For example, if the car travels 55 mi/h for 0.5 h, then it covers a distance of 27.5 mi.

When the car is changing speed (either speeding up or slowing down), then we can’t expect things to work out so exactly. But imagine using more and more rectangles, so that the time intervals are very short (and the speedometer doesn’t have time to change much in any interval). If the speedometer shows approximately 40 mi/h for 10 seconds, which is 1/360 of an hour, then a good estimate for the distance traveled would be

images

If we consider cutting the trip into many 10-second time intervals, we can estimate the total distance traveled by the car as the sum of these rectangular areas. The Riemann sum approximates the length of the trip, and the estimate is better when the intervals are shorter. It seems clear that the true length of the trip is found in the limit: the area under the curve measures the distance the car has moved.

In Section 4.2 we saw that the derivative of position is velocity. In this example we saw that the integral of velocity is (change in) position. This is clear if we think about what the Fundamental Theorem of Calculus (FToC) says.

The FToC tells us that the area over an interval and under a function can be computed by finding an antiderivative and subtracting the values at each end of the interval. Since the velocity of a car is the derivative of the position of the same car, the FToC guarantees that the integral of the velocity is the change in position (the distance the car has moved). In symbols:

images

These ideas apply to any rate. If f(t) is the cash flow of a company (in dollars/day), then images is the total change in cash during a 30-day period, i.e., the total net money earned or lost. If E(t) is the energy consumption of the United States (in quadrillion BTU/year), then images is the total energy consumed in a decade (in quadrillion BTU).

Theorem. If f(t) denotes a rate of change at time t in some quantity F, then

images

denotes the total change in F between time a and time b.

images EXAMPLE 4.45

A lead ball is dropped from a tower. How fast is the ball moving after 1.5 seconds?

Solution: To calculate this, we need to know one empirical fact. The acceleration of gravity is (essentially) a constant: –9.8 m/s2. Acceleration is the rate of change in velocity, so integrating the acceleration function tells us the total change in velocity of the ball:

images

Since the ball was dropped, it started at 0 m/s. The velocity changed by –14.7 m/s as it fell, so the final velocity after 1.5 s is –14.7 m/s. The negative answer indicates that the ball is falling (its height is decreasing). ?

images EXAMPLE 4.46

A lead ball is thrown upward at 5 m/s from the top of a tower and then falls. How fast is the ball moving after 1.5 seconds?

Solution: In this example the acceleration of gravity is still – 9.8 m/s2, and the total change in velocity is still the same integral with the same result, –14.7 m/s. However, since the ball started with a velocity of 5 m/s, the final velocity is 5 m/s –14.7 m/s = –9.7 m/s.

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EXERCISES

4.86 A math professor is pushed from the top of a 30 m building.

a) How fast is the professor falling after 0 seconds?

b) How fast is the professor falling after 0.25 seconds?

c) How fast is the professor falling after t seconds?

4.87 A math professor launched from a cannon has a velocity function of v(t) = 9 – 9.8t m/s, where t is measured in seconds from the time the cannon fires.

a) Is the cannon pointed up or down? How do you know?

b) At what time t does the professor have a velocity of 0 m/s? Why does the professor stop moving?

c) How high is the professor after 0.25 seconds?

d) How high is the professor after 1 second?

e) How high is the professor after t seconds?

f) When does the professor hit the ground?

4.88 The rate of fossil fuel consumption in the United States (in quadrillions of British thermal units per year) is approximately images, where t is years since 1900. How much fuel was used between January 1, 1900 and January 1, 1910?

4.89 A yo-yo company has a profit rate of $20,000 per day when a new advertising campaign begins. Profits rise $1,000 per day, each day, for the next six weeks.

a) What is the profit rate in dollars per day after the six weeks have transpired?

b) What was the profit rate t days after the advertisements began running?

c) What was the total amount of money made during the six week campaign?