Download Step Functions and Integrals: Determining Changes in Values and Finding Areas and more Summaries Calculus in PDF only on Docsity!
CHAPTER 6 (3/20/08)
INTEGRALS AND APPLICATIONS
The derivative, which we studied in Chapters 2 through 5, is used to find rates of change of functions. In this chapter we begin the study of the second main tool of calculus, the integral, which is used
to determine changes in values of quantities from their rates of change, to find areas, volumes, weights, average values, lengths of curves, and in many other applications. Section 6.1 gives a preview of the definition of the integral and of the Fundamental Theorem of Calculus in the context of functions whose rates of change are step functions. The definite integral is defined and some of its properties are discussed in Section 6.2/ In Section 6.3 we derive Part I of the Fundamental Theorem of Calculus, which deals with integrals of derivatives. Part II of the Fundamental Theorem concerning derivatives of
definite integrals with variable endpoints is discussed in Section 6.4.†^ The Fundamental Theorem is used
in Section 6.5 to obtain a formula for definite integrals of power functions y = xn^ with constant n 6 = −1. In this section we also introduce the term indefinite integral for “antiderivative.” Section 6.6 deals with finding approximate values of definite integrals of functions given by graphs and tables. Section 6.7 covers integration formulas derived from differentiation formulas for transcendental functions. The technique of integration by substitution is discussed in Section 6.8.
Section 6.
Step function rates of change
Overview: In this section we first look at applications where changes in values of functions can be determined from their rates of change without calculus because the rates of change are step functions. Then we obtain a preview of the definition of definite integrals (Section 6.2) and of Part I of the Fundamental Theorem of Calculus (Section 6.3) by applying the techniques of this section to approximations of continuous rates of change by step functions.
Topics:
- Step function rates of change
- Approximating continuous rates of change with step functions
Step function rates of change
We begin with two examples that use the formula for distance traveled at constant velocity,
[Distance traveled] = [Velocity] × [Time].
Example 1 The step function v = v(t) of Figure 1 gives a mathematical model of the velocity of a bus. The bus travels 50 miles per hour for two hours, 25 miles per hour for one hour, and 75 miles per hour for three more hours. (a) How far does it travel in the entire six hours? (b) How is the answer to part (a) related to the area of the rectangles in Figure 2?
†Instructors who want to cover Part II of the Fundamental Theorem (derivatives of integrals) before Part I (integrals of
derivatives) should cover Section 6.4 before Section 6.3.
p. 172 (3/20/08) Section 6.1, Step function rates of change
1 2 3 4 5 6 t
v (miles per hour)
(hours)
v = v(t)
2 3 6 t
v (miles per hour)
(hours)
A
B
C
FIGURE 1 FIGURE 2
Solution (a) The bus travels
[
miles hour
]
[2 hours] +
[
miles hour
]
[1 hour] +
[
miles hour
]
[3 hours]
= 100 + 25 + 225 = 350 miles
(b) The distance equals the sum of the areas of rectangles A, B, and C in Figure 2.
Example 2 At 10:00 AM one morning a truck driver is 100 miles east of his home town. He drives 75 miles per hour toward the east for two hours to make a delivery. Next, he drives west at 50 miles per hour for two hours to make another delivery and then drives east at 50 miles per hour for two more hours. According to this mathematical model, his velocity toward the east is the step function of Figure 3 with t = 0 at 10 AM. (a) How far is he from his home town at t = 6? (b) How is the answer to part (a) related to the areas of the rectangles in Figure 4?
1 2 3 4 5 6 t
v (miles per hour)
(hours)
v = v(t)
1 5 6 t
v (miles per hour)
(hours)
A
B
C
FIGURE 3 FIGURE 4
Solution (a) At t = 6, the driver is
[100 miles] +
[
miles hour
]
[2 hours] −
[
miles hour
]
[2 hours] +
[
miles hour
]
[2 hours]
= 100 + 150 − 50 + 50 = 250 miles
east of his home town. (b) His location at t = 6 equals his location at t = 0 plus the the sum of the areas of
rectangles A and C in Figure 2, minus the area of rectangle B.
p. 174 (3/20/08) Section 6.1, Step function rates of change
Example 3 Use Theorem 1 to solve Example 1.
Solution We let s = s(t) be the distance the bus of Example 1 travels in t hours. Then v(t) = s′(t) is the positive step function of Figure 1. By Theorem 1 with x replaced by t and f (x) replaced by s(t), s(6) = s(6) − s(0) equals the area 100 + 25 + 225 = 350 of the three
rectangles above the t-axis in Figure 2, as we saw in Example 1.
Example 4 Find the answer to Example 2 by applying Theorem 1.
Solution Let s = s(t) be the truck’s distance east of the driver’s home town at time t. In this case v(t) = s′(t) is the step function of Figure 3 and by Theorem 1, s(6) − s(0) equals the sum of the areas of rectangles A and C above the t-axis, minus the area of rectangle B below the t-axis in Figure 4. Consequently, s(6) − s(0) = 150 + 100 − 100 = 150, and
this gives s(6) = s(0) + 150 = 50 + 150 = 200 miles, as in Example 2.
Example 5 Suppose a water tank contains 300 gallons of water at time t = 0 (minutes) and that the rate of flow r = r(t) (gallons per minute) into the tank for 0 ≤ t ≤ 70 is the step function in Figure 7. How much water is in the tank at t = 70?
10 20 30 40 50 60 70 t
r (gallons per minute)
(minutes)
r = r(t)
20 30 50 70 t
r (gallons per minute)
(minutes)
FIGURE 7 FIGURE 8
Solution Notice from Figure 7 that the water flows into the tank for 0 < t < 10 and for 60 < t < 70, when r(t) is positive and flows out of the tank for 10 < t < 60, when r(t) is negative. Let V = V (t) (gallons) be the volume of water in the tank at time t. Then the derivative of V is the step function r = V ′(t) of Figure 7. We assume that V is continuous on [0, 70]. Then by Theorem 1, V (70) − V (0) equals the area of the first rectangle above the t-axis in Figure 8, minus the area of the two rectangles below the t-axis, plus the area of the last rectangle above the t-axis. We include the units in calculating the areas and write
V (70) − V (0) =
[
gallons minute
]
[10 minutes] −
[
gallons minute
]
[30 minutes]
[
gallons minute
]
[20 minutes] +
[
gallons minute
]
[10 minutes]
= 100 − 150 − 200 + 50 gallons = −200 gallons.
We rewrite this equation in the form,
V (70) = V (0) − 200.
Since V (0) = 300, this gives V (70) = V (0) − 200 = 300 − 200 = 100. At t = 70 there
are 100 gallons of water in the tank.
Section 6.1, Step function rates of change p. 175 (3/20/08)
Approximating continuous rates of change with step functions
Suppose that the continuous function v = v(t) of Figure 9 is the velocity of a car that is at s = s(t) on an s-axis at time t, so that v(t) = s′(t). Notice that the region between the graph of the velocity and the t axis for a ≤ t ≤ b in Figure 9 is in two parts. The region labeled A for the time period a ≤ t < c, when the car’s velocity is positive, is above the t-axis and below the graph. The region labeled B for c < t ≤ b, when the car’s velocity is negative, is below the t-axis and above the graph.
t
v v = v(t)
a
b c
A
B
t
v v = v(t)
a
b
FIGURE 9 FIGURE 10
We cannot apply Theorem 1 in this case because the velocity is not a step function. Instead we approximate v = v(t) with a step function by approximating regions A and B by rectangles, as in Figure 10, where the sides of the rectangles are determined by a partition of [a, b] and the tops are chosen to intersect the graph of v = v(t) at their midpoints. If the car’s velocity were given by the step function, we could apply Theorem 1 and conclude that the change in its position s(b) − s(a) from t = a to t = b equals the area of the two rectangles above the t-axis, minus the area of the three rectangles below the t-axis. Instead, the step function approximates the actual velocity v, and the difference of the areas of the rectangles gives an approximation of the change in the car’s position:
s(b) − s(a) ≈
[
The area of the rectangles above the t-axis
]
[
The area of the rectangles below the t-axis
]
Section 6.1, Step function rates of change p. 177 (3/20/08)
Exercises 6.
AAnswer provided. OOutline of solution provided. CGraphing calculator or computer required.
CONCEPTS:
- Suppose that y = f (x) is a continuous function on [0, 10] such that its derivative is a step function and f (0) = f (10). What can you say about the graph of r = f ′(x)?
- Based on Theorem 1, how are f (a) and f (b) related if f is continuous on [a, b] and f ′(x) = 0 for a < x < b?
BASICS:
3.O^ What is H(10) if y = H(x) is continuous on [2, 10], H(2) = 50, and r = H′(x) is a step function with H′(x) = −10 for 2 < x < 7, and H′(x) = 20 for 7 < x < 10?
4.A^ At 9:00 AM a runer has completed 8 miles of his morning run. He then runs with the constant velocity of 6 miles per hour between 9:00 AM and 9:30 AM and with the constant velocity of 8 miles per hour between 9:30 AM and 10:15 AM. How far has he run by 10:15 AM?
5.O^ The step function v = v(t) of Figure 12 gives the velocity in the positive direction at time t of an object as it moves on an s-axis. The scale on the s-axis is given in feet, t is measured in minutes, and v is measured in feet per minute. Suppose that the object’s position function is continuous on [0, 60] and that object is s = 300 at t = 10. Where is it at t = 60?
10 30 60 t
v (feet per minute)
(minutes)
v = v(t)
FIGURE 12
6.O^ The function y = F (x) is continuous on [0, 5] and its derivative is the step function
F ′(x) =
4 for 0 ≤ x < 2 7 for 2 < x < 3 − 3 for 3 < x < 5.
What is F (5) if F (0) = 12?
7.O^ What is F (5) if y = F (x) is continuous on [0, 5], F (0) = −5, and r = F ′(x) is a step function with F ′(x) = 3 for 0 < x < 3, and F ′(x) = 7 for 3 < x < 5?
8.A^ What is K(100) − K(50) if y = K(x) is continuous on [50, 100] and r = K′(x) is a step function such that K′(x) = 4 for 50 < x < 70, K′(x) = 0 for 70 < x < 75, and K′(x) = 3 for 75 < x < 100?
- What is F (1) if y = F (x) is continuous on [0, 1], F (0) = 0, and F ′(x) equals 10 for 0 < x < 13 , equals 20 for 13 < x < 23 , and equals 30 for 13 < x < 1?
p. 178 (3/20/08) Section 6.1, Step function rates of change
10.A^ The step function of Figure 13 is a mathematical model of the rate of rainfall r = r(t) (inches per year) in Los Angeles from the beginning of 1881 to the beginning of 1886.(1)^ What was the total rainfall in Los Angeles in the years 1881 through 1885?
t
r (inches per year)
50 r^ =^ r(t)
1881188218831884188518851886
10.7 11.
FIGURE 13
- A few plants, including jack in the pulpit, skunk cabbage, and sacred lotus, can create heat from oxygen and other nutrients. The step function in Figure 14 is a model of the rate of consumption of oxygen by a sacred lotus plant from noon one day to midnight 36 hours later. The flower kept its temperature between 30◦C and 37◦C as the air temperature varied between 10◦C and 35◦C by consuming more oxygen in the colder periods.(2)^ How much oxygen did the flower consume during the 36 hours? (t is measured in minutes.)
t
r (milliliters per minute)
3 r^ =^ r(t)
12 PM6 PM12 AM6 AM12 PM6 PM12 AM
2.5 (^) 2.
FIGURE 14
- In a mathematical model of the numbers of manufacturing workers N = N (t) in New York City at time t (years AD), N (1950) = 740 (thousand workers) and N ′(t) (thousand wokers per year) equals −110 for 1950 < t < 1960, equals −130 for 1960 < t < 1970, equals −140 for 1960 < t < 1970, equals −120 for 1980 < t < 1990, and equals −25 for 1990 < t < 1200 (3)^ Based on this data, how many manufacturing workers where there in New York City at the beginning of year 2000?
(1) (^) Data adapted from Los Angeles Times, January 7, 1993, Source: National Weather Service. (2) (^) Data adapted from “Plants that warm themselves” by R. Seymour, Scientific American, March, 1997. (3) (^) Data adapted from“Bad Things Happen”, Scientific American, June, 2001, p.26. Source: US Census Bureau
