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A Preview of Calculus

Table of Contents

Section 0 Introduction 2

Section 1 Average Rates of Change 3

Section 2 Instantaneous Rates of Change 10

Section 3 Derivatives of Polynomial Functions 22

Section 4 Interpretations of Derivatives 32

Section 5 Antiderivatives of Polynomial Functions 37

Section 6 The Area Problem 43

Section 7 Upper and Lower Sums 51

Section 8 Definite Integrals and the Fundamental Theorem of Calculus 60

Section 9 Interpretations of the Definite Integral 68

Appendix: Proofs of Theorems 73

Answers to Selected Exercises 81

Table of Contents


Section 0 Introduction

As you begin your study of calculus, it is natural to ask what calculus is. Unfortunately, a concise definition is not easy to provide. What can be provided is an indication of what calculus is about, that is, a description of some types of mathematical and physical problems that calculus was created to solve. The solutions of some of these problems will serve as a framework for your initial exposure to the subject.

Calculus is traditionally divided into two branches, differential calculus and integral calculus. The mathematical problem largely responsible for the early development of differential calculus in the 17th century is that of finding a line with the same slope as a specified curve at a specified point. It is sometimes referred to as the tangent line problem. Calculus solves the problem by providing a meaningful way to define the slope of a curve at a point and to calculate the slopes of a wide variety of curves. Because a slope is a rate of change in one variable with respect to another, it is not surprising that differential calculus can also be used to solve physical problems that deal with rates of change. An important example of such a problem is that of analyzing the position, velocity, and acceleration of a moving object as functions of time.

The development of integral calculus, also in the 17th century, was motivated in part by the area problem. Integral calculus provides a meaningful way to define the area of a region bounded by two or more curves and a way to calculate areas of regions bounded by many types of curves. Integral calculus can also be used to find the length of a trajectory, the center of mass of a solid object, and the force exerted on a dam by the water behind it.

In this booklet you will explore the basic problems of calculus in the context of polynomial functions. In Sections 1-3 you will discover the solution of the tangent line problem in the case where the curve is the graph of a polynomial function. In Sections 4 and 5 you will see a small sample of the mathematical and physical problems that can be solved using the methods developed in the first four sections, and you will explore the problem of recovering a polynomial function from knowledge of its slope at each point. In Sections 6-9 you will discover the solution of the area problem for regions bounded by graphs of polynomial functions, and you will investigate a few related problems that can be solved using the same methods. You will also encounter the Fundamental Theorem of Calculus, which unifies the two branches of calculus through a connection that is both deep and surprising.

The proof of each theorem you encounter is discussed in the Appendices. By reading and comprehending the proofs, you will deepen your understanding of the concepts of calculus and increase your ability to employ those concepts in the solution of both mathematical and physical problems.

Introduction


Section 1 Average Rates of Change

As indicated in the previous section, differential calculus deals with questions involving rates of change. Even if you have never studied calculus, however, you have already encountered mathematical questions about rates of change. In this section you will reinforce your previous experience by answering questions about the average velocity of an object that moves in a straight line and the slope of a line that connects two points on a graph. The questions do not require the use of calculus. You will be able to answer them as long as you remember a bit of elementary algebra.

Average Velocity of an Object in Rectilinear Motion

The velocity of a moving object is a rate of change. Specifically, velocity compares changes in position to changes in time. For example, suppose that you drive a car on a straight road at a constant speed of 50 miles per hour. Your speedometer is telling you that the distance you have driven, in miles, is increasing 50 times as fast as the time you have been driving, in hours.

The following terms will be useful in our discussion of velocity. An object that moves along a coordinate axis is said to be in rectilinear motion. The displacement of such an object over a time interval is its net change in position on the axis. Its average velocity is its net change in its position divided by the change in time. If its position on the axis at time t is s(t), its displacement over a time interval of duration h beginning at time t is s(t + h) – s(t), and its average

velocity is .

Example 1.1 Calculating average velocityA car accelerates from a stop sign and moves in a straight line, traveling s(t) = 2t2

feet in t seconds for 0 ≤ t ≤ 20. Find the car’s displacement and average velocity over the time interval [1, 10].

Solution The car is an example of an object in rectilinear motion. The road can be regarded as a coordinate axis with the origin at the car’s initial position and the car traveling in the positive direction. The car has traveled s(1) = 2(1)2 = 2 feet after 1 second and s(10) = 2(10)2 = 200 feet after 10 seconds. Its displacement over the time interval [1, 10] is s(10) – s(1) = 200 – 2 = 198 feet, and its average velocity is

ft/sec. ■

Note that both the displacement and the average velocity of an object over a time interval [t, t + h] are negative if s(t + h) < s(t). In addition, both are zero if

Average Rates of Change


s(t + h) = s(t), even if the object has moved during the time interval. Example 1.2 serves to clarify these statements.

Example 1.2 Calculating average velocityA rock thrown upward from ground level is at a height of s(t) = 96t – 16t2 feet above the ground after t seconds for 0 ≤ t ≤ 6.(a) Find its displacement and average velocity over the time interval [2, 5].(b) Find its displacement and average velocity over the time interval [1, 5].

Solution The rock is in rectilinear motion. The vertical axis is typically regarded as having its origin at ground level and its positive direction pointing upward.

(a) The height of the rock is s(2) = 128 feet after 2 seconds and s(5) = 80 feet after 5 seconds. Its displacement (net change in height) over the time interval [2, 5] is s(5) – s(2) = 80 – 128 = –48 feet, and its average velocity is

ft/sec.

(b) The height of the rock is s(1) = 80 feet after 1 second and s(5) = 80 feet after 5 seconds. Its displacement over the time interval [1, 5] is s(5) – s(1) = 0 feet, and its average velocity is

ft sec. ■

The result of Example 1.2a does not imply that the rock was falling during the entire time interval [2, 5]. In fact, you can verify that s(3) = 144, so the rock was higher after 3 seconds than it was after 2 seconds. Similarly, the result of Example 1.2b does not imply that the rock was motionless during the entire interval [1, 5].

Slope of a Secant Line on a Curve

The slope of a line is a rate of change. Specifically, it compares changes in the value of the dependent variable, typically y, to changes in the value of the independent variable, typically x. If you trace along a line that has a slope of 2, for example, the y-coordinate increases twice as fast as the x-coordinate.

You will shortly consider the problem of extending the concept of slope to nonlinear curves. As a first step, let’s think about how to interpret the slope of a line between two points on a curve (a secant line) as a rate of change related to the curve.



Example 1.3 Slope of a secant line as a rate of changeFind the slope of the secant line over the x-interval [1, 10] on the curve f(x) = 2x2, and interpret the slope as a rate of change.

Solution The curve and the secant line are shown in Figure 1.1. The slope of the secant line is

Figure 1.1

.

To interpret the slope as a rate of change, imagine walking from left to right along the graph of f(x) = 2x2. Because the graph is not a straight line, your y-coordinate does not always increase 22 times as fast as your x-coordinate. However, it makes sense to say that on the average, your y-coordinate increases 22 times as fast as your x-coordinate. ■


Before continuing, look back at the calculations in Examples 1.1, 1.2, and 1.3. They all involve evaluating an expression of the form

in order to compare changes in two quantities. Such an expression is called the average rate of change in the function P over the interval [x, x + h]. The examples you have seen in this section highlight two contexts in which average rates of change occur. Example 1.4 summarizes what those examples have illustrated.

Example 1.4 Finding and interpreting an average rate of changeLet P(x) = x2 + 10x.(a) Find the average rate of change in P(x) over the interval [5, 8].(b) Provide two interpretations of your result from part (a).

Solution (a) The average rate of change is

.



(b) ● The slope of the secant line between the points (5, 75) and (8, 144) on the graph of P(x) is 23.

● If P(x) represents the position of an object on a coordinate axis at time x, then the average velocity of the object over the time interval [5, 8] is 23. ■

Average Rate of Change as a Function

It is sometimes necessary to calculate average rates of change in the same function over intervals of different lengths that begin at the same point. In order to do this efficiently, it is convenient to calculate the average rate of change over an interval [x, x + h] as an expression in the two variables x and h. The average rate of change over any interval can then be found by substituting the left endpoint of the interval for x and the length of the interval for h. Examples 1.5 and 1.6 illustrate this method.

Example 1.5 Finding average velocities over several intervalsAs in Example 1.2, a rock thrown upward is at a height of s(t) = 96t – 16t2 feet above the ground after t seconds for 0 ≤ t ≤ 6. Find the rock’s average velocity over each of the time intervals [2, 3], [2, 2.1], [2, 2.01], and [2, 2.001].

Solution First find the average velocity in a time interval [t, t + h]. It is

= 96 – 32t – 16h ft/sec for h ≠ 0.

For the interval [2, 3], t = 2 and h = 3 – 2 = 1. The average velocity is 96 – 32(2) – 16(1) = 16 ft/sec.For the interval [2, 2.1], t = 2 and h = 2.1 – 2 = 0.1. The average velocity is 96 – 32(2) – 16(0.1) = 30.4 ft/sec.



For the interval [2, 2.01], t = 2 and h = 2.01 – 2 = 0.01. The average velocity is 96 – 32(2) – 16(0.01) = 31.84 ft/sec.For the interval [2, 2.001], t = 2 and h = 2.001 – 2 = 0.001. The average velocity is 96 – 32(2) – 16(0.001) = 31.984 ft/sec. ■

Example 1.6 Finding slopes of several secant linesFind the slopes of the secant lines on the graph of f(x) = 2x2 over each of the intervals [3, 4], [3, 3.1], [3, 3.01], and [3, 3.001].

Solution First find the slope of the secant line over an interval [x, x + h]. It is

= 4x + 2h for h ≠ 0.

For the interval [3, 4], x = 3 and h = 1. The slope of the secant line is 4(3) + 2(1) = 14.For the interval [3, 3.1], x = 3 and h = 0.1. The slope of the secant line is 4(3) + 2(0.1) = 12.2.For the interval [3, 3.01], x = 3 and h = 0.01. The slope of the secant line is 4(3) + 2(0.01) = 12.02.For the interval [3, 3.001], x = 3 and h = 0.001. The slope of the secant line is 4(3) + 2(0.001) = 12.002. ■

Active Learning Focus on developing skills

In Exercises 1-4, s(t) is the position at time t of an object moving on a coordinate axis. Find the object’s displacement and average velocity over the time interval [4, 6].

1. s(t) = 2t + 7 2. s(t) = 5 – 10t 3. s(t) = t2 – 6 4. s(t) = t2 – 6t



In Exercises 5-12, s(t) is the position at time t of an object moving on a coordinate axis. Find:

a. an expression for the object’s average velocity over a time interval [t, t + h] for h ≠ 0

b. the object’s average velocity over each of the time intervals [3, 4], [3, 3.1], [3, 3.01], and [3, 3.001]

5. s(t) = t2 – 2t + 5 6. s(t) = 4t + 1 7. s(t) = 2t2 + 3t 8. s(t) = 8t – t2

9. s(t) = 10 10. s(t) = t3 – 1211. s(t) = t3 – 12t 12. s(t) = t3 – 12t2

For the functions f(x) in Exercises 13-16, find the slope of the secant line to the graph of f(x) over the interval [1, 5].

13. f(x) = 5x – 4 14. f(x) = x2 + 115. f(x) = x2 – 5x + 6 16. f(x) = x3

For the functions f(x) in Exercises 17-24, find:a. an expression for the slope of the secant line to the graph of f(x)

over an interval [x, x + h] for h ≠ 0b. the slope of the secant line over each of the intervals [2, 3],

[2, 2.1], [2, 2.01], and [2, 2.001]

17. f(x) = 5 18. f(x) = –219. f(x) = 5 – 3x 20. f(x) = 0.4x + 2021. f(x) = 2x2 – x – 15 22. f(x) = 6x – x2

23. f(x) = x3 24. f(x) = x3 + x – 4

Focus on applying skills25. A car decelerating to a stop travels s(t) = 72t – 6t2 feet in t seconds after

the brakes are applied. Find the car’s displacement and average velocity during the first four seconds.

26. A toy rocket fired upward from ground level attains a height of s(t) = 160t – 16t2 feet after t seconds.a. Find the rocket’s displacement and average velocity during the

first five seconds of its flight.b. Find the rocket’s displacement and average velocity during the

next five seconds of its flight.c. What do your answers to parts (a) and (b) tell you about the

maximum height attained by the rocket?



Focus on connecting conceptsIn Exercises 27 and 28, find the average rate of change in P(x) over the interval [1, 4], and provide two interpretations of the result.

27. P(x) = 100 – x2

28. P(x) = x4

Refer to the material in the section as needed to answer each of questions 29-31, but write your answer in your own words. Address your answer to an imaginary classmate.

29. What is meant by rectilinear motion?

30. What is meant by average rate of change? Why is that an appropriate phrase to use in referring to the average velocity of an object in rectilinear motion or the average cost of producing a number of items?

31. Compare the calculations in Examples 1.1 and 1.3. Can the average velocity of an object in rectilinear motion over a time interval always be represented as the slope of a secant line on the graph of its position function? Explain your answer.



Section 2 Instantaneous Rates of Change

Instantaneous Velocity of an Object in Rectilinear Motion

As you saw in Section 1, questions about average rates of change can be answered without the use of calculus. Other questions about rates of change, however, are not so easy to answer. As a specific example, suppose that while driving on a long trip, you remark that you have driven 100 miles in the past 2 hours and that your current speedometer reading is 60 miles per hour. Your first statement describes an average velocity of 50 miles per hour over a time interval of 2 hours, but what does your second statement describe? The speedometer reading of 60 miles per hour is not an average velocity over any time interval, because it is a reading at a specific instant. In fact, the reading cannot be calculated as a change in position divided by a change in time, because neither time nor position change in a single instant. This observation suggests that the concepts of average velocity and instantaneous velocity are related but distinct from each other. It should therefore come as no surprise that the mathematical operations that define the two types of velocity are also related but distinct from each other. Example 2.1 illustrates both the similarities and the differences.

Example 2.1 The difference between average and instantaneous velocitiesA car accelerates from a stop sign and moves in a straight line, traveling s(t) = 2t2

feet in t seconds for 0 ≤ t ≤ 20.(a) Find an expression for the car’s average velocity in a time interval

[t, t + h].(b) Find the car’s average velocity over each of the time intervals [5, 6],

[5, 5.1], [5, 5.01], and [5, 5.001].(c) Find the car’s instantaneous velocity (speedometer reading) 5 seconds

after it leaves the stop sign.

Solution (a) The car’s average velocity over a time interval [t, t + h] is

= 4t + 2h for h ≠ 0.

(b) The result of part (a) implies that the car’s average velocity over a time interval of duration h, beginning at t = 5, is 20 + 2h. The car’s average velocity over several such intervals is shown in the following table.

Instantaneous Rates of Change


interval value of h average velocity[5, 6] 1 22[5, 5.1] 0.1 20.2[5, 5.01] 0.01 20.02[5, 5.001] 0.001 20.002

(c) It is reasonable to regard an instant as a time interval of duration 0 and conclude that the car’s speedometer reading at t = 5 is 20 ft/sec (about 13.6 miles per hour). ■

Example 2.1 suggests a strategy for calculating the instantaneous velocity of an object in rectilinear motion at an instant t = t0. First express the object’s average velocity over a time interval [t0, t0 + h] as a function f(h), then evaluate f(0). A potential obstacle to this strategy is that f(0) may not be defined. In fact, a look back at Example 2.1a shows that before simplifying, the expression for the car’s average velocity is a rational function that is undefined at 0. For h ≠ 0, however, the expression is equal to a polynomial, and the polynomial can be evaluated at h = 0. Theorem 2.1 guarantees that this will always happen when the position of the object is a polynomial function.

Theorem 2.1 Let P be a polynomial function. Then the expression reduces to

a polynomial P*(x, h) for h ≠ 0.

Proof The proofs of all theorems in this book can be found in the Appendix. ■

Notice that the polynomial P*(x, h) is the average rate of change in P(x) over an interval [x, x + h] for h ≠ 0.

Theorem 2.1 allows us to form a precise definition of instantaneous velocity for an object in rectilinear motion if its position is a polynomial function of time. Specifically, suppose that the position at time t of an object in rectilinear motion is a polynomial function s(t), and let s*(t, h) be the polynomial that coincides with

for h ≠ 0. The instantaneous velocity of the object at time t0 is

s*(t0, 0). The instantaneous speed of the object at time t0 is .

From now on the words velocity and speed will always mean instantaneous velocity and speed unless otherwise indicated.

In everyday conversation the words velocity and speed are sometimes used interchangeably, but their precise meanings are different. Velocity indicates both speed and direction. In particular, an object moving in the negative direction on a coordinate axis has positive speed but negative velocity, as illustrated in Example 2.2.



Example 2.2 The difference between velocity and speedA rock thrown upward from ground level is at a height of s(t) = 96t – 16t2 feet above the ground after t seconds.(a) What is its velocity 2 seconds after it is thrown?(b) What is its velocity 5 seconds after it is thrown?(c) What is its speed at each of these times?

Solution In Example 1.5 you discovered that the rock’s average velocity over a time interval [t, t + h] is

s*(t, h) = 96 – 32t – 16h ft/sec for h ≠ 0.

According to the definition just given, the rock’s velocity t seconds after it is thrown is

s*(t, 0) = 96 – 32t ft/sec.

(a) After 2 seconds, its velocity is 96 – 32(2) = 32 ft/sec., indicating that the rock is rising at that rate.

(b) After 5 seconds, its velocity is 96 – 32(5) = –64 ft/sec, indicating that the rock is falling at that rate.

(c) After 2 seconds the rock’s speed is |32| = 32 ft/sec. After 5 seconds its speed is |–64| = 64 ft/sec. ■

Expressing an object’s velocity as a function of time allows you to answer certain questions about the object’s motion that would be more difficult to answer by looking only at its position function. Example 2.3 illustrates.

Example 2.3 Finding a turning point(a) How high does the rock in Example 2.2 go?(b) With what velocity does the rock strike the ground?

Solution (a) To answer this question from the given information, it is necessary to recognize that the rock has a positive velocity as it rises and a negative velocity as it falls. Its velocity at the instant it reaches its highest point must be 0. Therefore we should begin by discovering when the rock has a velocity of 0. Recall from Example 2.2 that the rock’s velocity t seconds after it is thrown is

s*(t, 0) = 96 – 32t ft/sec.

The velocity is equal to 0 at time t = 3 seconds after the rock is thrown. Its height at that time is



s(3) = 96(3) – 16(3)2 = 144 ft. ■

(b) In order to answer this question, you need to know when the rock strikes the ground. That will happen when its height s(t) = 0.

96t – 16t2 = 0

16t(6 – t) = 0

t = 0 or t = 6

The solution t = 0 represents the instant when the rock is released from ground level, so it strikes the ground 6 seconds later. Its velocity at that time is

s*(6, 0) = 96 – 32(6) = –96 ft/sec,

indicating a downward speed of 96 ft/sec. ■

Slope of a Curve: The Tangent Line Problem

As you have just seen, calculus can be used to define and calculate the velocity of an object in rectilinear motion. You are about to see that it can also be used to define and to find (3, 18) •the slope of a line that is tangent to a curve at a point. To get an intuitive idea of what is meant by a tangent line to a curve, look at Figure 2.1, which shows the graph of P(x) = 2x2 Figure 2.1in the window -5 ≤ x ≤ 5, –10 ≤ y ≤ 50. Imagine the graph to be an aerial view of a road, and imagine a car at the point (3, 18). Think of the line along which the car’s headlights shine, indicated by the dashed line, as the tangent line to the graph at that point. Let’s follow a line of thought that will make that intuitive idea more precise.



The graph of P(x) = 2x2 for 2 ≤ x ≤ 4, 16.8 ≤ y ≤ 19.2 is shown in Figure 2.2. The graph looks nearly linear, and the point (3, 18) is in the middle of the window. It is reasonable toconclude that the tangent line to the (3, 18) •graph at that point should have a slope close to that of the secant line joining (3, 18) to a nearby point on the graph. This line of thought is followed in Example 2.4. Figure 2.2

Example 2.4 Finding the slope of a tangent line(a) What is the slope of the secant line to the graph of P(x) = 2x2 over each of

the intervals [3, 4], [3, 3.1], [3, 3.01], and [3, 3.001]?(b) What is the slope of the tangent line to the graph of P(x) at x = 3?

Solution (a) In Example 1.6 you discovered that the secant line to the graph over an interval [x, x + h] is

P*(x, h) = 4x + 2h for h ≠ 0.

The slopes of the secant lines over several intervals beginning at x = 3 are shown in the following table.

interval value of h slope of secant line[3, 4] 1 14[3, 3.1] 0.1 12.2[3, 3.01] 0.01 12.02[3, 3.001] 0.001 12.002

(b) A glance at Figures 2.1 and 2.2 should lead you to conclude that the slope of the dashed line in Figure 2.1 can be approximated by the slope of a secant line over a small interval [3, 3 + h]. The smaller the interval, the better the approximation should be. It is therefore reasonable to regard the tangent line as a secant line over an interval of length 0. Following this reasoning, the tangent line should have a slope of

P*(3, 0) = 4(3) + 2(0) = 12. ■

The reasoning in Example 2.4 can be generalized to provide a definition of a tangent line to a polynomial graph. Specifically, let P(x) be a polynomial

function, and let P*(x, h) be the polynomial that coincides with

for h ≠ 0. For a fixed value x0 of x, the slope of the tangent line to the curve y = P(x) at x = x0 is P*(x0, 0), also referred to as simply the slope of the curve.



The line through the point with that slope is the tangent line to the curve at x = x0.

Recall that the equation of a line through a point (x0, y0) with slope m can be written in point-slope form as y = y0 + m(x – x0). You can use this fact to write the equation of the tangent line to a polynomial graph.

Example 2.5 Slopes and equations of tangent linesLet P(x) = x3.(a) Find an expression for the slope of the graph of P(x) as a function of x.(b) Find the slope of the graph and the equation of the tangent line at x = –1

and x = 2.

Solution (a) For h ≠ 0, the slope of the secant line over an interval [x, x + h] is

= 3x2 + 3xh + h2.

The slope of the graph as a function of x is P*(x, 0) = 3x2.

(b) At x = –1, the point on the graph is (–1, P(–1)) = (–1, –1). The slope of the graph, which is the slope of the tangent line to the graph, is P*(–1, 0) = 3(–1)2 = 3. The equation of the tangent line is

y = –1 + 3(x + 1).

At x = 2, the point on the graph is (2, P(2)) = (2, 8). The slope of the graph is P*(2, 0) = 3(2)2 = 12. The equation of the tangent line is

y = 8 + 12(x – 2). ■

Expressing the slope of a curve y = P(x) as a function of x allows you to answer certain questions about the graph that would be more difficult to answer by looking only at the equation y = P(x). Example 2.6 illustrates.



Example 2.6 Finding a turning pointThe graph of P(x) = x3 – 3x for –3 ≤ x ≤ 3, –10 ≤ y ≤ 10 is shown in Figure 2.3. The graph shows two turning points. Where are they?

Solution The slope of the graph at each turning point should be 0, so begin by expressing the slope of the graph as a function of x. If h ≠ 0, the slope of a secant line over an interval [x, x + h] Figure 2.3is

=3x2 + 3xh + h2 – 3.

The slope of the graph as a function of x is

P*(x, 0) = 3x2 – 3.

The slope is 0 precisely when 3x2 – 3 = 0, that is, when x = ±1. The turning points are located at (–1, P(–1)) = (–1, 2) and (1, P(1)) = (1, –2). ■


Before continuing, look back at Examples 2.2 and 2.5. Each posed questions about a polynomial function P(x). To answer the questions, you first needed to form the polynomial P*(x, h) in x and h, representing the average rate of change in P(x) over an interval [x, x + h]. Then you needed to form the polynomial P*(x, 0). This polynomial, in x alone, is called the instantaneous rate of change in P(x). The examples you have seen in this section highlight two contexts in which instantaneous rates of change occur. Example 2.7 summarizes what those examples have illustrated.

Example 2.7 Finding and interpreting an instantaneous rate of changeLet P(x) = x2 + 10x.(a) Find the instantaneous rate of change in P(x) at x = 5.



(b) Provide two interpretations of your result from part (a).

Solution (a) For h ≠ 0, the average rate of change in P(x) over an interval [x, x + h] is

= 2x + h + 10.

The instantaneous rate of change in P(x) is

P*(x, 0) = 2x + 10,

and the instantaneous rate of change at x = 5 is 20.

(b) ● The slope of the tangent line to the graph of P(x) at the point (5, 75) is 20.

● If P(x) represents the position of an object on a coordinate axis at time x, then the instantaneous velocity of the object at time x = 5 is 20. ■

Active Learning Focus on developing skills

In Exercises 1-4, s(t) is the position at time t of an object moving on a coordinate axis. (The functions are the same as those in Exercises 1-4 of Section 1.) Find each of the following.

a. the object’s average velocity over each of the time intervals [4, 5], [4, 4.1], [4, 4.01], [4, 4.001]

b. the object’s instantaneous velocity at t = 4

1. s(t) = 2t + 7 2. s(t) = 5 – 10t 3. s(t) = t2 – 6 4. s(t) = t2 – 6t

In Exercises 5-12, s(t) is the position at time t of an object moving on a coordinate axis. (The functions are the same as those in Exercises 5-12 of Section 1.) Find each of the following.

a. the expression s*(t, h) for the object’s average velocity over a time interval [t, t + h]

b. the expression s*(t, 0) for the object’s instantaneous velocity as a function of t



c. the instantaneous velocity at t = 3d. the values of t, if any, where the object’s velocity is 0

5. s(t) = t2 – 2t + 5 6. s(t) = 4t+ 1 7. s(t) = 2t2 + 3t 8. s(t) = 8t – t2

9. s(t) = 10 10. s(t) = t3 – 1211. s(t) = t3 – 12t 12. s(t) = t3 – 12t2

The functions in Exercises 13-16 are the same as those in Exercises 13-16 of Section 1. Find each of the following.

a. the slope of the secant line to the graph of P(x) over the intervals [1, 2], [1, 1.1], [1, 1.01], and [1, 1.001]

b. the slope of the graph of P(x) at x = 1

13. P(x) = 5x – 4 14. P(x) = x2 + 115. P(x) = x2 – 5x + 6 16. P(x) = x3

The functions f(x) in Exercises 17-24 are the same as those in Exercises 17-24 of Section 1.

a. Find the expression f*(x, h) for the slope of the secant line to the graph of f(x) over an interval [x, x + h].

b. Find the expression f*(x, 0) for the slope of the tangent line to the graph of f(x) as a function of x.

c. Find the equation of the tangent line at x = 2.d. Support your result from part (c) by graphing f(x) and its tangent

line at x = 2 in a small viewing window around the point (2, f(2)).e. Find the values of x, if any, where the graph of f(x) has a horizontal

tangent line.

17. f(x) = 5 18. f(x) = –219. f(x) = 5 – 3x 20. f(x) = 0.4x + 2021. f(x) = 2x2 – x – 15 22. f(x) = 6x – x2

23. f(x) = x3 24. f(x) = x3 + x – 4

Focus on applying skills25. A rock is thrown upward from a roof 256 feet above ground level. Its

height t seconds later is s(t) = 256 + 96t – 16t2 feet. What is the maximum height attained by the rock?

26. You want to throw a baseball to your friend whose dormitory window is 30 feet directly above you. If you throw the ball straight upward at 30 miles per hour (44 ft/sec), its height t seconds later will be s(t) = 44t – 16t2

feet. Will the ball reach your friend? Show a calculation to justify your answer.

27. Refer to the rock in Exercise 25.



a. After how many seconds does the rock strike the ground?b. With what velocity does the rock strike the ground?

28. A second rock is thrown downward from the roof in Exercise 25. Its height t seconds later is s(t) = 256 – 96t – 16t2 feet.a. After how many seconds does the rock strike the ground?b. With what velocity does the rock strike the ground?

In Exercises 29-32, an object is thrown upward from ground level on the indicated planet or satellite. (Ignore the fact that there is no “ground” on the gaseous planet Jupiter.) The object is thrown with an initial velocity of 20 m/sec and reaches a height of s(t) meters after t seconds. In each case, find

a. the length of time required for the object to reach its maximum height

b. the maximum height attained by the object

29. Earth, s(t) = 20t – 4.9t2 30. Moon, s(t) = 20t – 0.8t2

31. Mars, s(t) = 20t – 1.86t2 32. Jupiter, s(t) = 20t – 11.44t2

33. A car accelerating from a stop sign and moving in a straight line travels s(t) = 4t2 feet in t seconds. How long does it take the car to reach a speed of 60 miles per hour (88 ft/sec)?

34. A car approaching a stop sign travels 88t – 4t2 feet in t seconds after the brakes are applied. How far does the car travel before it stops?

Focus on connecting conceptsRefer to the material in the section as needed to answer each of questions 35-41, but write your answer in your own words. Address your answer to an imaginary classmate.

35. What is the difference between average velocity and instantaneous velocity?

36. Why can’t average and instantaneous velocities be calculated in the same way?

37. Why is it reasonable to regard an instant as a time interval of duration 0?

38. What is the difference between velocity and speed?

39. Why do you think that the words velocity and speed are used interchangeably in everyday conversation?

40. What is the difference between a secant line and a tangent line?



41. Why can’t slopes of secant lines and slopes of tangent lines be calculated in the same way?

Answer each of Questions 42-44 about the tangent line to the graph of P(x) at a specific point (a, P(a)). If your answer is yes, give an example of a polynomial function and a tangent line that exhibit the described behavior. If your answer is no, explain why.

42. Is it possible for the tangent line to intersect the graph of P(x) at some point (b, P(b)), b ≠ a?

43. Is it possible for the tangent line to cross the graph of P(x) at x = a?

44. Is it possible for the tangent line to be vertical?

45. Suppose that the position of an object in rectilinear motion is described by a polynomial function s(t).a. Must its velocity be zero at each instant when the object reverses

its direction? Write a sentence or two to explain your answer.b. Must the object reverse its direction at each instant when its

velocity is zero? If not, write a few sentences to describe the possible motions of the object in a time interval during which its velocity is zero at some instant. (Hint: Examples can be found in your answers to Exercises 5-12.)

46. Let P(x) be a polynomial function.a. Must the graph of P(x) have a horizontal tangent at each turning

point? Write a sentence or two to explain your answer.b. Must the graph of P(x) have a turning point whenever it has a

horizontal tangent? If not, give an example of a polynomial function and a point on its graph where there is a horizontal tangent but no turning point. (Hint: Examples can be found among the functions in Exercises 17-24.)

47. In Example 2.2, , and

.a. How do the values of these two expressions compare when

(t, h) = (2, 1)? (2, 0.1)? (2, 0)?b. Write a few sentences to describe the ways in which these two

expressions are alike and the ways in which they are different.

c. Can the unsimplified expression for be used to

calculate an average velocity for the rock in Example 2.2? Can it be used to calculate an instantaneous velocity? Explain each of your answers.



d. Can the polynomial s*(t, h) be used to calculate an average velocity for the rock in Example 2.2? Can it be used to calculate an instantaneous velocity? Explain each of your answers.

48. Refer to Example 2.2.a. Verify that over the time interval [0, 6], the rock has an average

velocity of zero.b. Does it make sense to say that the rock has an average speed of

zero over this time interval? If not, explain how to calculate its average speed, then do it. (Hint: You will need to notice when its velocity is positive and when it is negative.)

49. The terms in the left column refer to the graph of a polynomial function. Match each one with an appropriate term in the right column related to an object in rectilinear motion.a. slope of a secant line (1) instant when the object stopsb. slope of a curve (2) average velocityc. turning point (3) instantaneous velocityd. point with a horizontal (4) instant when the object

tangent changes direction

In Exercises 50 and 51, find the expression P*(x, 0) for the slope of the tangent line to the graph of y = P(x) as a function of x.

50. a. P(x) = 4x2 b. P(x) = 4x2 – 3c. P(x) = 4x2 + 7.3 d. P(x) = 4x2 + c (c is any constant)

51. a. P(x) = x3 b. P(x) = x3 – 3c. P(x) = x3 + 2 d. P(x) = x3 + c (c is any constant)

52. What do you think must be true about two polynomial functions P(x) and Q(x) if their graphs have the same slope for all values of x? (Hint: Refer to the results of Exercises 50 and 51.) Write a few sentences to say why this is not surprising.



Section 3 Derivatives of Polynomial Functions

In Section 2 you discovered the mathematical operation that is used to define and calculate the instantaneous velocity of an object in rectilinear motion. Then you learned that the same operation can be used to define and calculate the slope of a curve. An operation so powerful as to be useful in two such different environments deserves a name. For every polynomial function P(x), the instantaneous rate of change P*(x, 0) is called the derivative of P(x) with respect to x. The process of calculating a derivative is referred to as differentiation.

FYIYou should be aware that the notation P*(x, 0) is not in common use. The derivative of P(x) with respect to x is usually written as:

● (pronounced “P prime of x”),

● (“dee P by dee x”), or

● (“dee by dee x of P(x)”.)

To illustrate the use of these notations, let P(x) = x3. Example 2.5 showed that the instantaneous rate of change in P(x) is P*(x, 0) = 3x2. The conventional ways of expressing this result are:

● ,

● , or

● .

If it is obvious that the independent variable is x, the phrase “with respect to x” is usually omitted when discussing derivatives. However, it is important to identify the independent variable before beginning to calculate a derivative, as Example 3.1 illustrates.

Example 3.1 The importance of specifying the independent variableLet y = 2x + 3t.

(a) Treating t as a constant, find , the derivative of y with respect to x.

(b) Treating x as a constant, find , the derivative of y with respect to t.

Derivatives of Polynomial Functions


Solution (a) For h ≠ 0,

= 2,

so .

(a) For h ≠ 0,

= 3,

so .

Derivative Formulas

At this point it is worthwhile to ask whether derivatives of polynomial functions can be calculated in a more efficient manner than that which has been used so far. Theorems 3.1 through 3.4 assert that more efficient methods do exist. These theorems establish a set of formulas that can be used to differentiate every polynomial function.

Theorem 3.1 (Power Rule, special cases)

(a) For every constant c, .

(b) .

Theorem 3.2 (Power Rule, general case) For every integer n ≥ 1, .

Example 3.2 Using the Power Rule



Find: (a) (b) (c)

Solution (a) By Theorem 3.1a, .

(b) By Theorem 3.1b, .

(c) By Theorem 3.2, . ■

Theorem 3.3 Let P and Q be polynomial functions, and let c be a constant. Then:

(a) (Constant Multiple Rule)

(b) (Sum and Difference Rules)

(c) (Product Rule)

Example 3.3 Using the Constant Multiple Rule

Find: (a) (b) (c) .

Solution (a)

(b)

(c) ■

Example 3.4 Using the Sum and Difference RulesFind if P(x) = 2x3 – 9x2 + 4.

Solution . ■

Example 3.5 Using the Product Rule

Find .



Solution Let P(x) = x3 and Q(x) = x2 – 7x + 6. Then by the Product Rule,

= 2x4 – 7x3 + 3x4 – 21x3 + 18x2

= 5x4 – 28x3 + 18x2.

The same result can be obtained without the use of the product rule as follows.

= 5x4 – 28x3 + 18x2. ■

Theorem 3.4 (Chain Rule) For polynomial functions P and Q, .

A lengthier, but simpler, statement of the Chain Rule is the following.

Theorem 3.4a (Chain Rule, alternate form) If y is a polynomial function of u and u is a

polynomial function of x, then .

Example 3.6 Using the Chain Rule

Find .

Solution Let u = x3 + x – 4 and y = u4. Then , and

. ■

A careful look at Example 3.6 suggests the following formula for differentiating a power of a polynomial function. The formula is a special case of the Chain Rule.



Theorem 3.5 (Chain Rule, special case) If Q(x) is a polynomial function, then for every positive

integer n, .

Example 3.7 Using the special case of the Chain Rule

Find if P(x) = (x4 – 2x)7.

Solution Theorem 3.5 implies that

. ■

Example 3.8 Using several rules in succession

Find .

Solution Let P(x) = (x – 5)3 and Q(x) = (2x + 7)5. Then by the Product Rule,

.

The calculations of and require the Chain Rule. You can apply Theorem 3.5 to obtain

and

.

Therefore

= (x – 5)2(2x + 7)4(16x – 29). ■



Higher-Order Derivatives and Acceleration

The Power Rule, Constant Multiple Rule, and Sum and Difference Rules guarantee that the derivative of every polynomial function is another polynomial function. It therefore makes sense to define the second derivative of a polynomial function P(x) with respect to x as the derivative of , written

or . The third derivative of P(x) with respect to x is the derivative of

, and derivatives of higher order are defined similarly.

In the “prime” notation, the kth derivative of P(x) is written as for k > 3. For example, the fourth derivative of P(x) is usually written as P(4)(x), rather than

.

Example 3.9 Finding higher-order derivativesFind the derivatives of all orders for P(x) = x4 – 2x3 + 5x + 7.

Solution

P(4)(x) = 24P(k)(x) = 0 for k ≥ 5. ■

You can see a physical interpretation of a higher-order derivative in the study of rectilinear motion. Specifically, let s(t) be the position of an object on a coordinate axis at time t, and let be its velocity. The function

, the instantaneous rate of change in the object’s velocity with respect to time, is defined to be the object’s acceleration.

Example 3.10 Finding accelerationA projectile shot upward from ground level reaches a height of s(t) = 160t – 16t2

feet, or equivalently r(t) = 49t – 4.9t2 meters, after t seconds. Find the projectile’s acceleration:(a) in ft/sec2

(b) in m/sec2

Solution (a) The projectile’s velocity is ft/sec. Its acceleration is ft/sec2.

(b) The projectile’s velocity is m/sec. Its acceleration is m/sec2. ■



FYIYou should be aware that the equations in Example 3.10 would not quite describe the motion of an actual projectile. In particular, the acceleration found in each part of the example is the acceleration due only to gravity. Although gravity may be the dominant force acting on the projectile, air resistance and other minor forces would affect the its acceleration as well as its velocity and height.

ActiveLearning Focus on developing skills

In Exercises 1-8, find .

1. P(x) = -5 2. P(x) = 3x – 1 3 P(x) = 9 – x2 4. P(x) = x2 – 9

5. 6.

7. P(x) = x3 – 12x 8. P(x) = 2x3 + 3x2


9. P(x) = x7 10. P(x) = 10x7

11. 12. P(x) = 0.3x

13. P(x) = 2x4 – 3x2 + 1 14. P(x) = 4x – x3

15. 16.

17. P(x) = (x – 5)(x + 5) 18. P(x) = x3(2x2 – 7x + 6)19. P(x) = (x + 2)(x2 – 2x + 4) 20. P(x) = (x3 + 3x – 6)(x4 – 2x2 – 8)21. P(x) = x(x + 1)(x + 2) 22. P(x) = x2(x + 4)(x2 – 4x)

23. , x ≠ 024. , x ≠ 1

25. P(x) = mx + b 26. P(x) = ax2 + bx + c27. P(x) = Ax3 28. P(x) = Ax4


29. P(x) = 2x – 97 30. P(x) = 10x2

31. P(x) = 3 + 2x – x2 32. P(x) = 4x3 + x

In Exercises 33-36, find the derivatives of all orders for P(x).

33. P(x) = x3 – x2 + x – 1 34.

35. P(x) = x4

36. P(x) = x6 + 6x5 + 30x4 + 120x3 + 360x2 + 720x + 720



In Exercises 37-40, use the Chain Rule to find .

37. P(u) = 3u – 1, u = Q(x) = 2x + 1038. P(u) = u2 + 4u, u = Q(x) = x2 – 939. P(u) = u3 – 6u, u = Q(x) = x2 + x40. P(u) = u2 + u, u = Q(x) = x3 – 6x

In Exercises 41-48, find .41. P(x) = (2x – 5)3 42. P(x) = (x3 + 2x)2

43. P(x) = –4(3 – 7x)5 44. P(x) = 10(x3 – 5x2 + 7x – 1)4

45. 46. P(x) = x(x + 2)3(x – 6)2

47. P(x) = (x2 + x – 2)3 – 3(x2 + x – 2)2 + 3(x2 + x – 2) – 148.

49. Let P(x) = (x2 – 3x + 8)(x2 + 3x – 8). Find in two different ways.a. Use the Product Rule.b. Multiply P(x) out and use the Power Rule.Verify that both calculations lead to the same result.

50. Let . Find in three different ways.a. Use the Chain Rule.b. Write P(x) as (x2 + 3)(x2 + 3) and use the Product Rule.c. Write P(x) as x4 + 6x2 + 9 and use the Power Rule.Verify that each calculation leads to the same result.

In Exercises 51-54, the position of an object moving on a coordinate axis is a function s(t) of time. Find:a. the object’s velocity as a function of timeb. the object’s acceleration as a function of timec. the times, if any, at which the object’s velocity is zero

51. s(t) = t3 + 4t 52. s(t) = t4 – 4t3

53. s(t) = t2(t2 – 4) 54. s(t) = (2x + 10)3(3x – 6)4

Focus on applying skills55. On a certain planet a dropped object falls 2.5t2 meters in t seconds. What

is the acceleration due to gravity on this planet?

56. On another planet a dropped object falls at2 meters in t seconds. It is known that the acceleration due to gravity on this planet is 7.2 m/sec2. What is a?



57. On Earth, the height of a thrown or dropped object can be modeled by a quadratic function s(t) = at2 + bt + c if all forces except gravity are ignored. The acceleration due to gravity is about -32 ft/sec2 or -9.8 m/sec2. What is the value of a if t is measured in seconds and s(t) is measured in feet? What is it if s(t) is measured in meters?

58. An object moving on a coordinate axis has position u(t) = 9t2 + 1 after t seconds. A second object on the same axis has position w(t) = t3 + 24t after t seconds. Are their velocities ever equal? Are their accelerations ever equal? If so, at what time(s)?

Focus on connecting conceptsRefer to the material in the section as needed to answer each of questions 59 and 60, but write your answer in your own words. Address your answer to an imaginary classmate.

59. Why is it useful to define the concept of a derivative?

60. When calculating a derivative, why is it important to specify the independent variable?

In Exercises 61 and 62, find the values of a for which the graph of P(x) has:a. no horizontal tangentsb. exactly one horizontal tangentc. two horizontal tangents

61. P(x) = x3 + ax 62. P(x) = x3 + ax2

63. Let . What is the minimum slope on the graph

of P(x)?

64a-h. Write a few sentences to interpret your results in each of Exercises 1-8 in terms of slopes of curves. Then write a few sentences to interpret the same results in terms of the velocity of an object in rectilinear motion.

65. a. What can you say about the velocity of an object in rectilinear motion whose position is a linear function of time?

b. What can you say about the acceleration of an object in rectilinear motion whose position is a linear function of time?

c. What can you say about the acceleration of an object in rectilinear motion whose velocity is a linear function of time?

66. a. Find if P(x) = x2 – 6x + 5. For what values of x is ?b. Graph P(x). How does the graph support your answer to part (a)?c. Graph . How does the graph support your answer to part (a)?



67. An object in rectilinear motion has position s(t) = t3 – 6t2 + 20 after t seconds.a. Graph s(t). For what values of t is the object moving in a negative

direction? Explain how the graph supports your answer.b. Find the object’s velocity v(t), and show how to use the equation of

v(t) to answer the question in part (a).c. Graph v(t), and explain how the graph supports your answer in part

(a).

68. What is the maximum number of horizontal tangent lines on the graph of a polynomial function of degree n? Explain your answer.

69. Let P(x) be a polynomial function of degree n. Does it always happen that for some positive integer k, P(k)(x) = 0 for all x? If so, what is the smallest value of k for which that happens? Explain your answers.

70. Explain why Theorem 3.1 can be viewed simply as a restatement in new language of some facts you learned in your high school algebra course.

71. Refer to the four displayed lines in the proof of Theorem 3.3c (the Product Rule) in the Appendix. Explain why the expression in each of the first three displayed lines is equivalent to the expression on the following line.

72. The proof of Theorem 3.4 (the Chain Rule) in the Appendix depends on the claim that P*(Q(x), j)·Q*(x, h) = R*(x, h) “except, for each x, at the finite number of values of h for which Q(x + h) – Q(x) = 0.” Why must there be only finitely many such values of h?

73. For nearly all polynomial functions P(x) and Q(x), the derivative of the product P(x)Q(x) is not the product . Which polynomial products are the exceptions to this rule?



Section 4 Interpretations of Derivatives

The examples in this section illustrate several mathematical and physical interpretations of derivatives.

Derivatives as Velocity and Acceleration

Example 4.1 A derivative as velocity and as accelerationA dropped object on the moon falls P(t) = 0.8t2 meters in t seconds. What is the acceleration due to gravity on the moon?

Solution The object’s downward velocity is m/sec, and its downward acceleration is m/sec2. This is the acceleration due to gravity on the moon. ■

Example 4.2 A derivative as velocity and as accelerationA car approaching a stop sign travels P(t) = 80t – 5t2 feet in t seconds after the brakes are applied.(a) What is the car’s initial velocity?(b) What is the car’s rate of deceleration?(c) How long does it take for the car to stop?(d) How far does it travel in that time?

Solution To answer the questions, you will need the car’s velocity, , and its acceleration, .

(a) The car’s initial velocity is ft/sec.

(b) Because the acceleration is , the car’s rate of deceleration is 10 ft/sec2.

(c) The car stops when its velocity is 0. Solving 80 – 10t = 0 gives a stopping time of t = 8 sec.

(d) In 8 seconds, the car travels P(8) = 80(8) – 5(8)2 = 320 feet. ■

Derivatives as Slopes

Example 4.3 A derivative as slopeLet P(x) = x3 – kx. For what value of k does the graph of y = P(x) have a negative slope for -1 < x < 1 and a nonnegative slope otherwise?

Interpretations of Derivatives


Solution The slope of the graph is . The slope is negative exactly when

.

For the slope to be negative when -1 < x < 1 and nonnegative otherwise, k must be 3. ■

Example 4.4 A derivative as slopeThe graphs of P(x) = x2 and Q(x) = a + bx – x2 are tangent to each other (that is, they have the same tangent line) at x = 3. Find a and b.

Solution The slope of the tangent line to the graph of P(x) is . The slope of the tangent line to the graph of Q(x) is . These are equal when x = 3, so 2(3) = b – 2(3), and b = 12. Furthermore, the graph of P(x) contains the point (3, P(3)) = (3, 9), and the graph of Q(x) contains the same point, so Q(3) = 9. Therefore a + b(3) – (3)2 = 9, and because b = 12, it follows that a = -18. ■

Example 4.5 A derivative as slopeFor what values of x does the tangent line to the graph of P(x) = x4 + 1 contain the origin?

Solution The tangent line to the graph of P(x) at x = x0 has slope and contains

the point . The equation of the tangent line is

.

The tangent line contains the origin if and only if

. ■

Derivatives as Instantaneous Rates of Change

Example 4.6 A derivative as an instantaneous rate of changeDuring a flood, the height of a river above its normal level is approximately P(t) = t4 – 8t3 + 16t2 feet after t days for 0 ≤ t ≤ 4. About how fast is the river rising at the end of the first day?



Solution The question concerns the instantaneous rate of change in the river’s height with respect to time. That rate after t days is feet per day. At the end of the first day, the river is rising at a rate of feet per day. ■

Example 4.7 A derivative as an instantaneous rate of changeSuppose that a manufacturer can produce x television sets per week at a cost of C(x) = 50,000 + 300x + 0.01x2 dollars.b(a) Find the weekly cost of increasing production from 100 to 101 sets per

week.(b) Find .(c) Explain why the results in parts (a) and (b) should be approximately equal.

Solution (a) The cost is C(101) – C(100) = $80,402.01 – $80,100 = $302.01.(b) The derivative of C(x) is , so .(c) The result in part (a) is the average rate of change in C(x) over the interval

[100, 101]. The result in part (b) is the instantaneous rate of change in C(x) at x = 100. Because an interval of length 1 can be considered small in this context, it is reasonable to expect the two results to be close. ■

In economics, the derivative of a cost function is referred to as marginal cost. The concepts of marginal revenue and marginal profit are defined in a similar manner.

Example 4.8 A derivative as an instantaneous rate of changeThe manufacturer in Example 4.7 obtains a weekly revenue of R(x) = 500x – 0.2x2

dollars from the sale of x television sets per week.(a) Find the marginal revenue function and evaluate it for x = 100.(b) Find the revenue derived from the sale of the 101st television set, and

compare this result with that in part (a).(c) Find the marginal profit function and evaluate it for x = 100.(d) Find the profit derived from the sale of the 101st television set, and

compare this result with that in part (c).

Solution (a) The marginal revenue function is , so .(b) The revenue derived from the sale of the 101st set is R(101) – R(100)

= 48,459.80 – 48,000 = $459.80. This differs from the result in part (a) by only $0.20.

(c) Because profit is the difference between revenue and cost, the manufacturer’s profit function is P(x) = R(x) – C(x). The marginal profit function is

,

so .



(d) The profit derived from the sale of the 101st set is the difference between the revenue derived from the sale (Example 4.8b) and the cost of production (Example 4.7a), which is $459.80 - $302.01 = $157.79. This differs from the result in part (c) by only $0.21. ■

ActiveLearning Focus on applying skills

In Exercises 1-4, a car accelerating from a stop travels s(t) feet in t seconds. Find:a. the length of time required for the car to reach a speed of 30 mph

(44 ft/sec)b. the car’s acceleration when it reaches a speed of 30 mph

1. s(t) = 4t2 2. s(t) = 7.5t2

3. 4. s(t) = 2.5t3

In Exercises 5-8, a car decelerating to a stop travels s(t) feet in t seconds. Find:a. the length of time required for the car to stopb. the distance traveled by the car before it stops

5. s(t) = 4t(10 – t) 6. s(t) = 4t(20 – t) 7. s(t) = 6t(9 – t2) 8. s(t) = 90 – 10(t – 3)2

9. An object thrown upward from ground level with initial velocity v0 ft/sec reaches a height of s(t) = v0t – 16t2 feet in t seconds. What initial velocity will allow the object to attain a maximum height of 256 feet?

10. An object dropped from above the surface of the Moon falls s(t) = 0.8t2

meters in t seconds. Two objects are dropped from the same height one second apart.a. Would you expect the objects to remain the same distance apart

after the second object is dropped? Write a sentence to explain your answer in everyday words.

b. Find a formula D(u) for the distance, in meters, between the objects u seconds after the second object is dropped. Is the formula consistent with your answer to part (a)?

In Exercises 11-14, find:a. the values of x, if any, for which the graph of P(x) has a negative

slopeb. the values of x, if any, for which the tangent line to the graph of

P(x) contains the point (0, 1)

11. P(x) = x2 + 2 12. P(x) = x2 + 2x 13. P(x) = x2 + 2x + 1 14. P(x) = x3



In Exercises 15-18, the cost of producing x items per day is C(x) dollars, and the revenue derived from the sale of x items is R(x) dollars.

a. Find the marginal cost, marginal revenue, and marginal profit functions, and evaluate each at x = 1000.

b. Find the cost, revenue, and profit associated with the 1001st item, and compare each number with the corresponding number in part (a).

15. C(x) = 1000 + 5x 16. C(x) = 1000 + 5x + 0.001x2

R(x) = 20x R(x) = 20x – 0.004x2

17. C(x) = 0.01(x + 100)2 18. C(x) = 200x – 0.1x2 + 0.001x3

R(x) = x(50 – 0.02x) R(x) = 500x

19. What is the rate of change in the area of a circle with respect to its radius?

20. What is the rate of change in the volume of a cube with respect to its edge length?

Focus on connecting concepts21. The volume of a cylinder of radius r and height h is V = r2h.

a. Find , treating h as a constant, and evaluate when r = 5 and

h = 4. What does your result say about a cylinder whose shape is changing?

b. Find , treating r as a constant, and evaluate when r = 5 and

h = 4. What does your result say about a cylinder whose shape is changing?

22. Explain in your own words why the derivative of a cost function is approximately “the cost of producing the next item.”

23. Show that the rate of change in the volume of a sphere with respect to its radius is numerically equal to its surface area.



Section 5 Antiderivatives of Polynomial Functions

In previous sections you have seen how to find the velocity and acceleration of an object in rectilinear motion when its position at all times is known. In practice the forces acting on the object determine its acceleration, which is then used to calculate its velocity and its position through the process of recovering a function from its derivative. In this section you will explore that process for polynomial functions.

Antiderivatives and Differential Equations

If , then Q(x) is an antiderivative of P(x). Theorems 5.1-5.3 establish some important facts about antiderivatives.

Theorem 5.1 If for all x, then there is a constant C such that P(x) = C for all x.

Theorem 5.2 If for all x, then there is a constant C such that P(x) = Q(x) + C for all x.

Theorem 5.3 (a) (Power Rule)

If , then for some constant C.

(b) (Constant Multiple Rule)If , then P(x) = kQ(x) + C for some constant C.

(c) (Sum and Difference Rules)If , then P(x) = Q(x) ± R(x) + C for some constant C.

An equation that specifies a formula for is an example of a differential equation, that is, an equation that involves one or more derivatives of an unknown function. The order of the differential equation is the order of the highest derivative involved. The solution of the differential equation is the family of functions for which the equation is true. Example 5.1 illustrates the solution of three first-order differential equations.

Example 5.1 Finding antiderivatives(a) Find all possible functions P(x) if .

(b) Find all possible functions P(r) if .

(c) Find all possible functions P(t) if .

Solution (a) Think of as 6x2 – 10x1 + 7x0. By the Power Rule, the antiderivatives

of x2, x1, and x0 are and x, respectively. Apply the Constant

Multiple Rule and the Sum and Difference Rules to obtain

Antiderivatives of Polynomial Functions


.

(b) First write . Then

.

(c) Notice that there is no product rule for antiderivatives, so first write . Then

. ■

Initial Value Problems

An initial value problem is a differential equation of order n together with the values of the unknown function and its first n – 1 derivatives at a particular value of the independent variable. These values are called initial conditions. The solution of an initial value problem is a function that satisfies both the differential equation and all the initial conditions. Theorem 5.4 guarantees that the solution of every initial value problem of the type you will encounter here consists of exactly one function.

Theorem 5.4 Let Q(x) be a polynomial function. Every initial value problem of the form

has a unique solution.

Example 5.2 Finding functions from slopesThe graph of P(x) contains the point (3, 11) and has slope 4x – 1 for each value of x. Find an equation for P(x).



Solution The given conditions imply that and P(3) = 11. First find the solution of the differential equation.

Then use the initial condition to place P(x) within the family of solutions.

P(3) = 11 2(3)2 – 3 + C = 11 C = -4

Therefore P(x) = 2x2 – x – 4. ■

The recovery of the position of an object in rectilinear motion from its acceleration involves the solution of a second-order differential equation. To determine the position function uniquely, it is necessary to have two initial conditions, as illustrated in Examples 5.3 and 5.4.

Example 5.3 Finding position from accelerationA rock is thrown from a roof 100 feet above ground level with an initial downward velocity of 80 ft/sec. The acceleration due to gravity is approximately –32 ft/sec2, and the effects of other forces are small enough to be ignored. Find the velocity and the height of the rock as functions of time.

Solution Let the rock have height s(t) feet after t seconds. The condition on its acceleration gives the differential equation , and the initial velocity and height provide initial conditions and s(0) = 100. Then is an antiderivative of , so . The initial condition on implies that , so C = -80. Thus the velocity of the rock after t seconds is ft/sec.

Continue by noting that s(t) is an antiderivative of , so s(t) = -16t2 – 80t + C. The initial condition on s(t) implies that -16(0)2 – 80(0) + C = 100, so C = 100. Thus the height of the rock after t seconds is s(t) = -16t2 – 80t + 100 feet. ■

Example 5.4 Finding position from accelerationA car accelerates at a constant rate from 0 to 60 mph (88 ft/sec) in 8 seconds. How far does it travel in that time?

Solution Let the distance that the car travels in t seconds be s(t) feet. Because the car accelerates at a constant rate and reaches a velocity of 88 ft/sec in 8 seconds, its rate of acceleration is 88/8 = 11 ft/sec2. Therefore , and the car accelerates from a stop at time t = 0, giving initial conditions .

The car’s velocity after t seconds is ft/sec, and the condition implies that C = 0, so the velocity is ft/sec.



The car travels s(t) = 5.5t2 + C feet in t seconds, and the condition s(0) = 0 implies that C = 0, so it travels s(t) = 5.5t2 feet. In 8 seconds it travels 5.5(8)2 = 352 feet. ■

A Look Ahead

You have now seen how derivatives and antiderivatives can be used to answer questions about instantaneous rates of change in polynomial functions. However, questions can also be asked about instantaneous rates of change in functions belonging to other families. For example:

• A weight suspended on a spring bobs up and down so that its height, in centimeters, above a table t seconds after it is released is the trigonometric function f(t) = 10 + 5 sin t. What is its instantaneous velocity as a function of time?

• The estimated population of the world, in billions, t years from now can be modeled approximately by the exponential function . What is the instantaneous rate of growth in the population today?

You will soon learn how to answer questions like these by extending the ideas of differential calculus to trigonometric, exponential, and other families of functions.


In Exercises 1-8, find all possible functions P(x).

1. 2.

3. 4.

5. 6. 7. 8.

In Exercises 9-16, solve the initial value problem. 9. , P(3) = 810. , P(2) = 011.

12.

13.14.15.16.



In Exercises 17-24, an object in rectilinear motion has position s(t), velocity v(t), and acceleration a(t) at time t. Find an equation for s(t).17. v(t) = 100, s(0) = 50 18. v(t) = 8t3, s(0) = 8019. v(t) = 144 – 32t, s(0) = 80 20. v(t) = 20 – 9.8t, s(0) = 1021. a(t) = –32, v(1) = 40, s(1) = 10022. a(t) = –9.8, v(3) = 0.6, s(3) = 023. a(t) = 6t, v(4) = 60, s(4) = 15624. a(t) = 3t2 – 4t, v(0) = 7, s(0) = –5

25. Find equations for all quadratic functions that have a horizontal tangent line at (3, –4).

26. The graph of a quadratic function contains the point (1, 7) and has a slope of –4 there. The same graph contains the point (5, 1). What is its slope there?

Focus on applying skills27. On a certain planet, the acceleration due to gravity is –2.4 m/sec2.

a. A rock is thrown upward from the surface of the planet with initial velocity 12 m/sec. How high does it go?

b. With what initial velocity must the rock be thrown in order to reach a maximum height of 90 m?

28. A car accelerates from a stop at a constant rate of 10 ft/sec2.a. How long does it take for the car to reach a speed of 80 ft/sec?b. How far does the car travel in that time?

A traffic light is to be placed at an intersection of two roads. The speed limit on each road is 45 mph (66 ft/sec). The duration of the yellow signal must be long enough to allow all approaching drivers either to stop safely or to go through the intersection before the light turns red. In Exercises 29 and 30, you can determine the minimum duration for the yellow signal.

29. An approaching driver who is traveling at the speed limit when the light turns yellow decides to stop. Assume that he will take 0.5 seconds to react and apply the brakes. Assume also that after the brakes are applied, the car will decelerate at a constant rate and come to a stop in 6 seconds.a. How far will the car travel before the brakes are applied?b. What is the constant rate of deceleration?c. After the brakes are applied, how far will the car travel before

coming to a stop?d. How far will the car travel altogether after the light turns yellow?

30. Let d be the distance you obtained in Exercise 29d. A second approaching driver is also traveling at the speed limit when the light turns yellow and is



d feet from the intersection. This driver decides to continue driving at the same speed and go through the light. a. How long does it take for this driver to pass through the

intersection?b. Explain why your answer to part (a) is the minimum duration for

the yellow signal.

Focus on connecting concepts31. Use the connection between derivatives and slopes to explain why

Theorem 5.1 must be true.

32. Theorem 5.3 implies that if the derivative of a polynomial function P(x) is constant, then P(x) is a linear function. Use the connection between derivatives and slopes to explain why this must be true.

33. Prove that if , then P(x) – Q(x) is a linear function.

34. What can you say about the graphs of P(x) and Q(x) if ?

35. Consider the family of graphs of all functions P(x) = x2 + C as C ranges over all real values.a. Does this family of graphs cover the entire coordinate plane?b. Do any two graphs in the family intersect?c. Explain why your answers to parts (a) and (b) imply that every

initial value problem has a unique solution.

36. Two students are asked to find the family of all antiderivatives of .

• Anna says, “One antiderivative of P(x) is x2 + x4, so the family of all antiderivatives is x2 + x4 + C.”

• Albert says, “The family of antiderivatives of 2x is x2 + C, and the family of antiderivatives of 4x3 is x4 + C, so the family of antiderivatives of 2x + 4x3 is (x2 + C) + (x4 + C) = x2 + x4 + 2C.”

Is Anna correct? Is Albert correct? Explain your answers.



Section 6 The Area Problem

In this section you will begin your study of integral calculus and the solution of the area problem as described in Section 0. To gain a better understanding of the nature of the area problem, consider Figure 6.1. The figure shows the graph of P(x) = x2 + 1 over the interval [0, 1]. The problem of finding the area of the region R under the graph is aninstance of the area problem. Figure 6.1

A Strategy for Defining Areas

A moment’s reflection reveals the true nature of the problem. Areas of circles, as well as those of triangles, rectangles, and other polygons, are both defined and calculated by formulas that you learned in plane geometry. However, precalculus mathematics offers no definition of area for most other planar regions. Therefore before we ask how to find the area of a region, we must ask how to define the area of a region in a meaningful way. Surely any definition of area should be consistent with the following two principles.

A1 If a region S1 covers a region S2, then the area of S1 is at least equal to the area of S2.

A2 If two regions S1 and S2 have disjoint interiors, then the area of the union S1 S2 is the sum of the areas of S1 and S2.

Principles A1 and A2 have far-reaching consequences. Assuming nothing more than these two simple principles, you will soon see that the area of every region bounded by graphs of polynomial functions can be defined as a unique number.

A first step in a strategy for defining the area of the region R in Figure 6.1 is to compare the portion of the plane covered by R to that covered by regions whose area is known. Figure 6.2a shows a region S, made up of rectangles, that covers R. Figure 6.2b shows a region T, also made up of rectangles, that is covered by R. The areas of both S and T can be calculated, and it is reasonable to demand that the area of R should lie between them. Example 6.1 illustrates this step of the strategy.

The Area Problem


Figure 6.2a Figure 6.2b

Example 6.1 Approximating the shape of a regionLet R be the region under the graph of P(x) = x2 + 1 over the interval [0, 1]. (a) Use five rectangles with equal width and disjoint interiors to create a

region S that approximates the shape of R and covers R, and find the area of S.

(b) Use five rectangles with equal width and disjoint interiors to create a region T that approximates the shape of R and is covered by R, and find the area of T.

(c) Use the results of parts (a) and (b) to find an interval in which the area of R must lie.

Solution Because the interval [0, 1] has width 1, each of the five rectangles has width . The rectangles partition [0, 1] into the five subintervals [0, 0.2],

[0.2, 0.4], [0.4, 0.6], [0.6, 0.8], and [0.8, 1].

(a) Region S is shown in Figure 6.2a. Because P(x) is increasing on [0, 1], the height of the rectangle over each subinterval is equal to the value of P(x) at the right endpoint of the subinterval. The following table shows the areas of the five rectangles that make up region S.

Rectangle Height Width Area 1 P(0.2) = 1.04 0.2 (1.04)(0.2) = 0.208 2 P(0.4) = 1.16 0.2 (1.16)(0.2) = 0.232 3 P(0.6) = 1.36 0.2 (1.36)(0.2) = 0.272 4 P(0.8) = 1.64 0.2 (1.64)(0.2) = 0.328 5 P(1.0) = 2.00 0.2 (2.00)(0.2) = 0.400

By Principle A2, the area of region S is the sum of the areas of the five rectangles, which is 1.44.

(b) Region T is shown in Figure 6.2b. Because P(x) is increasing on [0, 1], the height of the rectangle over each subinterval is equal to the value of P(x) at the left endpoint of the subinterval. The following table shows the areas of the five rectangles that make up region T.

The Area Problem


Rectangle Height Width Area 1 P(0.0) = 1.00 0.2 (1.00)(0.0) = 0.200 2 P(0.2) = 1.04 0.2 (1.04)(0.2) = 0.208 3 P(0.4) = 1.16 0.2 (1.16)(0.2) = 0.232 4 P(0.6) = 1.36 0.2 (1.36)(0.2) = 0.272 5 P(0.8) = 1.64 0.2 (1.64)(0.2) = 0.328

By Principle A2, the area of region T is the sum of the areas of the five rectangles, which is 1.24.

(c) By Principle A1, the area of region R must lie between 1.24 and 1.44. ■

By increasing the number of rectangles, as in Figures 6.3a and 6.3b, we can approximate the shape of R more closely and establish narrower bounds on the area of R.


Example 6.2 Improving the bounds on the area of a regionLet R be the region in Example 6.1. (a) Use ten rectangles with equal width and disjoint interiors to create a region

S that approximates the shape of R and covers R, and find the area of S.(b) Use ten rectangles with equal width and disjoint interiors to create a region

T that approximates the shape of R and is covered by R, and find the area of T.

(c) Use the results of parts (a) and (b) to find an interval in which the area of R must lie.

Solution Each of the ten rectangles has width . The rectangles partition [0, 1] the ten subintervals [0, 0.1], [0.1, 0.2], [0.2, 0.3], …, [0.9, 1].

(a) Region S is shown in Figure 6.3a. The height of the rectangle in each subinterval is equal to the value of P(x) at the right endpoint of the subinterval. The following table shows the areas of the ten rectangles that make up region S.

The Area Problem


Rectangle Height Width Area 1 P(0.1) = 1.01 0.1 (1.01)(0.1) = 0.101 2 P(0.2) = 1.04 0.1 (1.04)(0.1) = 0.104 3 P(0.3) = 1.09 0.1 (1.09)(0.1) = 0.109 4 P(0.4) = 1.16 0.1 (1.16)(0.1) = 0.116 5 P(0.5) = 1.25 0.1 (1.25)(0.1) = 0.125 6 P(0.6) = 1.36 0.1 (1.36)(0.1) = 0.136 7 P(0.7) = 1.49 0.1 (1.49)(0.1) = 0.149 8 P(0.8) = 1.64 0.1 (1.64)(0.1) = 0.164 9 P(0.9) = 1.81 0.1 (1.81)(0.1) = 0.181 10 P(1.0) = 2.00 0.1 (2.00)(0.1) = 0.200

By Principle A2, the area of region S is the sum of the areas of the ten rectangles, which is 1.385.

(b) Region T is shown in Figure 6.3b. The height of the rectangle in each subinterval is equal to the value of P(x) at the left endpoint of the subinterval. The following table shows the areas of the ten rectangles that make up region T.

Rectangle Height Width Area 1 P(0.0) = 1.00 0.1 (1.00)(0.1) = 0.100 2 P(0.1) = 1.01 0.1 (1.01)(0.1) = 0.101 3 P(0.2) = 1.04 0.1 (1.04)(0.1) = 0.104 4 P(0.3) = 1.09 0.1 (1.09)(0.1) = 0.109 5 P(0.4) = 1.16 0.1 (1.16)(0.1) = 0.116 6 P(0.5) = 1.25 0.1 (1.25)(0.1) = 0.125 7 P(0.6) = 1.36 0.1 (1.36)(0.1) = 0.136 8 P(0.7) = 1.49 0.1 (1.49)(0.1) = 0.149 9 P(0.8) = 1.64 0.1 (1.64)(0.1) = 0.164 10 P(0.9) = 1.81 0.1 (1.81)(0.1) = 0.181

By Principle A2, the area of region T is the sum of the areas of the ten rectangles, which is 1.285.

(c) By Principle A1, the area of region R must lie between 1.285 and 1.385. ■

By using a larger number of thinner rectangles, we can approximate the shape of R as closely as we like. It is reasonable to expect that there is a unique number that is less than or equal to the area of any region that covers R and greater than or equal to the area of any region that is covered by R. We can then define that number to be the area of R.

The Area Problem


A Physical Interpretation of Areas

Before carrying out our strategy, let’s try to gain some insight into the full significance of the area problem. Just as the solution of the tangent line problem provides information about more than tangent lines, the solution of the area problem will provide information about more than areas. Example 6.3 illustrates a physical interpretation of areas.

Example 6.3 Estimating distance from velocityA ball thrown upward with initial velocity 64 ft/sec has velocity v(t) = 64 – 32t ft/sec after t seconds. Let R be the region under the graph of v(t) over the interval [0, 2].(a) Use four rectangles with equal width and disjoint interiors to create a

region S that approximates the shape of R and covers R, and find the area of S.

(b) Explain why your result in part (a) is an upper bound for the displacement of the ball during the first 2 seconds of its flight.

(c) Use four rectangles with equal width and disjoint interiors to create a region T that approximates the shape of R and is covered by R, and find the area of T.

(d) Explain why your result in part (c) is a lower bound for the displacement of the ball during the first 2 seconds of its flight.

(e) Use a geometric formula to find the area under the graph of v(t) over the interval [0, 2], and verify that it lies between the bound established in parts (a) and (c).

(f) Use antiderivatives to find the displacement of the ball during the first 2 seconds of its flight, and verify that it is equal to the area under the graph of v(t) over the interval [0, 2].

Solution The interval [0, 2] has width 2, so each of the four rectangles has width . The rectangles partition [0, 2] into the four subintervals [0, 0.5], [0.5, 1], [1, 1.5], and [1.5, 2].


(a) Region S is shown in Figure 6.4a. Because v(t) is decreasing on [0, 2], the height of the rectangle over each subinterval is equal to the value of v(t) at

The Area Problem


the left endpoint of the subinterval. The following table shows the areas of the four rectangles that make up region S.

Rectangle Height Width Area 1 v(0.0) = 64 0.5 (64)(0.5) = 32 2 v(0.5) = 48 0.5 (48)(0.5) = 24 3 v(1.0) = 32 0.5 (32)(0.5) = 16 4 v(1.5) = 16 0.5 (16)(0.5) = 8

By Principle A2, the area of region S is the sum of the areas of the four rectangles, which is 80.

(b) Because v(t) is decreasing in the interval [0, 0.5], the ball’s velocity in this interval never exceeds v(0) = 64 ft/sec. That is, the height of the first rectangle in part (a) is the ball’s maximum velocity in this interval. The width of the rectangle is the duration of the interval, which is 0.5 sec. Therefore the product (64)(0.5) = 32 is an upper bound for both the area under the graph of v(t) over [0, 0.5] and the distance that the ball travels in the first 0.5 seconds of its flight. A similar argument can be made for each of the remaining three time subintervals.

(c) Region T is shown in Figure 6.4b. Because v(t) is decreasing on [0, 2], the height of the rectangle over each subinterval is equal to the value of v(t) at the right endpoint of the subinterval. The following table shows the areas of the four rectangles that make up region T. (Notice that the fourth “rectangle” has a height of 0.)

Rectangle Height Width Area 1 v(0.5) = 48 0.5 (48)(0.5) = 24 2 v(1.0) = 32 0.5 (32)(0.5) = 16 3 v(1.5) = 16 0.5 (16)(0.5) = 8 4 v(2.0) = 0 0.5 (0)(0.5) = 0

By Principle A2, the area of region T is the sum of the areas of the four rectangles, which is 48.

(d) Because v(t) is decreasing in the interval [0, 0.5], the ball’s velocity in this interval is always at least v(0.5) = 48 ft/sec. That is, the height of the first rectangle in part (c) is the ball’s minimum velocity in this interval. The width of the rectangle is the duration of the interval, which is 0.5 sec. Therefore the product (48)(0.5) = 24 is a lower bound for both the area under the graph of v(t) over [0, 0.5] and the distance that the ball travels in the first 0.5 seconds of its flight. A similar argument can be made for each of the remaining three time subintervals.

The Area Problem


(e) The region R is a triangle with altitude 64 and base 2. Its area is . This is between the previously established upper and lower

bounds of 80 and 48.

(f) The height of the ball after t seconds is s(t) = C + 64t – 16t2 for some constant C. The displacement of the ball during the first 2 seconds of its flight is

s(2) – s(0) = (C + 64(2) – 16(2)2) – (C) = 64 ft.

This is equal to the area found in part (e). ■

In its use of antiderivatives, Example 6.3 hints at a connection between the area problem and the tangent line problem. This connection will be established more firmly in Section 8.


In Exercises 1-6, let R be the region under the graph of P(x) over the given interval.

a. Use five rectangles with equal width and disjoint interiors to create a region S that approximates the shape of R and covers R, and find the area of S.

b. Use five rectangles with equal width and disjoint interiors to create a region T that approximates the shape of R and is covered by R, and find the area of T.

c. Use the results of parts (a) and (b) to find an interval in which the area of R must lie.

d. Use a geometric formula to find the area of R, and verify that the area lies within the interval you found in part (c).

1. P(x) = x – 2, [2, 4] 2. P(x) = 6 – 2x, [0, 3] 3. P(x) = 7, [1, 5] 4. P(x) = 3x + 4, [-1, 1] 5. P(x) = 10 – x, [-2, 3] 6. P(x) = 9 – 4x, [1, 2]

In Exercises 7-12, let R be the region under the graph of P(x) over the given interval.

a. Use three rectangles with equal width and disjoint interiors to create a region S that approximates the shape of R and covers R, and find the area of S.

b. Use three rectangles with equal width and disjoint interiors to create a region T that approximates the shape of R and is covered by R, and find the area of T.

c. Use the results of parts (a) and (b) to find an interval in which the area of R must lie.

d. Repeat parts (a) – (c) using six rectangles of equal width.

The Area Problem


7. P(x) = x2, [0, 3] 8. P(x) = 9 – x2, [-3, 3] 9. P(x) = x2 – 6x, [-6, 0] 10. P(x) = x2 + x, [1, 2]11. P(x) = x3, [1, 4] 12. P(x) = x3 – x, [-1, 0]

Focus on applying skillsIn Exercises 13-18, an object moves on a coordinate axis with velocity v(t).

a. Use the method of Example 6.3 to find an upper bound for the displacement of the object during the given time interval. Partition the time interval into four subintervals of equal duration.

b. Use the method of Example 6.3 to find a lower bound for the displacement of the object during the given time interval. Partition the time interval into four subintervals of equal duration.

c. Use antiderivatives to find the displacement of the object during the given time interval, and verify that it lies between the bounds established in parts (a) and (b).

d. Use a geometric formula to find the area under the graph of v(t) over the given interval, and verify that it is equal to the displacement found in part (c).

13. v(t) = 4t, [0, 6] 14. v(t) = 10t – 30, [3, 5]15. v(t) = 25, [-2, 2] 16. v(t) = 128 – 32t, [0, 4]17. v(t) = 12t + 48, [2, 10] 18. v(t) = 85 – 5t, [4, 12]

In each of Exercises 19 and 20, the table shows the velocity of a car at several times. Assuming that the car’s acceleration is always positive, find lower and upper bounds for the distance traveled by the car over the time interval shown.

19. time (sec) 0 3 6 9 12 15velocity (m/sec) 0 2 5 10 18 30

20. time (sec) 0 0.5 1 1.5 2 2.5 3velocity (ft/sec) 15 19 25 30 36 40 42

Focus on connecting conceptsRefer to the material in the section as needed to answer questions 21 and 22, but write your answer in your own words. Address your answer to an imaginary classmate.

21. What do the area principles A1 and A2 say? Do you agree that it is reasonable to require any definition of area to be consistent with these principles?

22. Describe the process that will be used to define the area of a region under the graph of a nonnegative polynomial function.

The Area Problem


Section 7 Upper and Lower Sums

The examples in Section 6 illustrate a strategy that can be used to establish increasingly sharp upper and lower bounds for the area of a region under the graph of a polynomial function. Before we explore this strategy any further, it will be helpful to introduce some language we can use to describe the process we are using to establish our bounds.

Finding Upper and Lower Sums

Let P(x) be a polynomial function defined on a closed interval [a, b]. A partition of [a, b] is a set of points {x0, x1, x2, …, xm} with a = x0 < x1 < x2 < … < xm = b. The interval is said to be partitioned into the subintervals [x0, x1], [x1, x2], …, [xm-1, xm]. An upper sum for P(x) on [a, b] is constructed as follows.

• Choose a partition of [a, b] into m subintervals. It is usually convenient, although not necessary, to make all subintervals of equal width. Denote the width of the kth subinterval by (x)k.

• Identify the maximum value of P(x) within each subinterval. Let x = ck be the value of x that maximizes P(x) within the kth subinterval.

• Calculate the sum P(c1)(x)1 + P(c2)(x)2 + P(c3)(x)3 + … + P(cm)(x)m.

Using the minimum value of P(x) within each subinterval instead produces a lower sum. Examples 6.1a and 6.2a showed upper sums for P(x) = x2 + 1 over [0, 1] for m = 5 and m = 10, respectively. Examples 6.1b and 6.2b showed the corresponding lower sums.

The same process can be used to find upper and lower bounds for the displacement of an object in rectilinear motion when its velocity is known. The upper sum in Example 6.3a and the lower sum in Example 6.3b provided upper and lower bounds for both the area under the graph of v(t) = 64 – 32t over the interval [0, 2] and the displacement of a rock whose velocity is v(t) for 0 ≤ t ≤ 2. Example 7.1 provides an additional illustration of upper and lower sums.

Example 7.1 Finding upper and lower sumsLet P(x) = 2x2 – x.(a) Find an upper sum for P(x)

over [1, 4] using a partition into m = 6 subintervals of equal width.

(b) Find the corresponding lower sum.

(c) Provide two interpretations of your results.

Figure 7.1

Upper and Lower Sums


Solution The interval [1, 4] has a width of 3, so the 6 subintervals each have a width of . The partition is {1, 1.5, 2, 2.5, 3, 3.5, 4}. Notice that P(x) is increasing

on [1, 4], as shown in Figure 7.1.

(a) The maximum value of P(x) in each subinterval occurs at its right endpoint. The upper sum is calculated in the following table.

k ck P ( c k) ( x ) k P ( c k)( x ) k

1 1.5 3 0.5 1.52 2.0 6 0.5 3.03 2.5 10 0.5 5.04 3.0 15 0.5 7.55 3.5 21 0.5 10.56 4.0 28 0.5 14.0

The upper sum is 41.5.

(b) The minimum value of P(x) in each subinterval occurs at its left endpoint. The lower sum is calculated in the following table.

k ck P ( c k) ( x ) k P ( c k)( x ) k

1 1.0 1 0.5 0.52 1.5 3 0.5 1.53 2.0 6 0.5 3.04 2.5 10 0.5 5.05 3.0 15 0.5 7.56 3.5 21 0.5 10.5

The lower sum is 28.0.

(c) The area under the graph of P(x) over [0, 2] must be defined as a number between 28.0 and 41.5. Alternately, if an object moving on a coordinate axis has velocity P(x) at time x, then its displacement over the time interval [0, 2] must be between 28.0 and 41.5. ■

Differences of Upper and Lower Sums

You may have noticed that in each of Examples 6.1, 6.2, 6.3, and 7.1, many of the terms in the upper sum also appear in the lower sum. Specifically, the terms in the upper sum with the exception of the largest coincide with the terms in the lower sum with the exception of the smallest. This potentially labor-saving observation deserves further attention. However, you should be aware that this fortuitous duplication of terms does not occur in every pair of upper and lower sums, but rather depends on our choice of both an interval [a, b] and a partition of that interval. Theorem 7.1 establishes precise conditions under which the duplication of terms occurs.



Theorem 7.1 Suppose that the polynomial function P(x) is either increasing on all of [a, b] or decreasing on all of [a, b]. Let U and L be, respectively, upper and lower sums for P(x) for a partition of [a, b] into m subintervals of equal width . Then

.

In this section you will focus on situations in which Theorem 7.1 applies. That is, you will find upper and lower sums for a function only over intervals where the function has no turning points, and all partitions of intervals will be into subintervals of equal width. In Exercises 29-32 at the end of this section you will discover how to form upper and lower sums over more general intervals.

Example 7.2 Finding the difference between an upper and lower sumLet P(x) = 10 – x3.(a) Find an upper sum for P(x) over [0, 2] using 4 subintervals of equal width.(b) Find the corresponding lower sum.

Solution The 4 subintervals each have a width of . The partition is {0, 0.5, 1, 1.5, 2}. A graph shows that P(x) is decreasing on [0, 2].

(a) The maximum value of P(x) in each subinterval occurs at its left endpoint. The upper sum is calculated in the following table.

k ck P ( c k) x P ( c k)( x ) 1 0.0 10.000 0.5 5.00002 0.5 9.875 0.5 4.93753 1.0 9.000 0.5 4.50004 1.5 6.625 0.5 3.3125

The upper sum is 17.7500.

(b) The difference between the upper and lower sums is

.

The lower sum is 17.75 – 4 = 13.75. ■

Upper and Lower Sums for Negative Functions

You can use the process described at the beginning of this section to construct upper and lower sums for polynomial functions over intervals where the function assumes negative values. Example 7.3 explores the interpretation of such sums in terms of both areas and displacements of objects in rectilinear motion.



Example 7.3 Upper and lower sums for negative functionsA ball thrown down from a roof has velocity v(t) = –32 – 32t ft/sec after t seconds.(a) Find an upper sum for v(t) over the time interval [0, 3] using 5

subintervals of equal width.(b) Find the corresponding lower sum.(c) Use your results to establish upper and lower bounds for the displacement

of the ball during the first 3 seconds of its flight.(d) Use antiderivatives to find the ball’s displacement during the first 3

seconds of its flight, and verify that it lies between the bounds in part (c).(e) Use your results from parts (a) and (b) to establish upper and lower

bounds for the area of the region R between the graph of v(t) and the t-axis over [0, 3].

(f) Use a geometric formula to find the area of the region R, and verify that it lies between the bounds established in part (e).

Solution Each of the 5 subintervals has a width of . The partition is {0, 0.6, 1.2, 1.8, 2.4, 3}. The function v(t) is decreasing on [0, 3].

(a) The maximum value in each subinterval occurs at its left endpoint. The upper sum is calculated in the following table.

k ck v ( c k) t v ( c k)( t ) 1 0.0 v(0.0) = –32.0 0.6 –19.202 0.6 v(0.6) = –51.2 0.6 –30.723 1.2 v(1.2) = –70.4 0.6 –42.244 1.8 v(1.8) = –89.6 0.6 –53.765 2.4 v(2.4) = –108.8 0.6 –65.28

The upper sum is –211.20

(b) The difference between the upper and lower sums is

| v(3) – v(0) | (0.6) = | –128 – (–32) | (0.6) = 57.6.

The lower sum is –211.20 – 57.60 = –268.80.




(c) An upper bound for the displacement of the ball during the first 3 seconds of its flight is –211.20 feet, and a lower bound is –268.80 feet.

(d) The position function s(t) of the ball is an antiderivative of the velocity function v(t) = –32 – 32t, so s(t) = –32t – 16t2 + C for some constant C. The displacement of the ball over the time interval [0, 3] is

s(3) – s(0) = (–32·3 – 16·32 + C) – (C) = –240 feet.

This is between the bounds established in part (c).

(e) The y-values used to construct the upper sum in part (a) are the negative numbers v(0) = –32, v(0.6) = –51.2, v(1.2) = –70.4, v(1.8) = –89.6, and v(2.4) = –108.8. The heights of the rectangles in Figure 7.2a are the absolute values of these five numbers. Each rectangle has a width of 0.6, so the areas of the five rectangles are the absolute values of the terms in the upper sum. The calculation in part (a) shows that the area of the region consisting of the five rectangles is 211.20. Figure 7.2a shows that this is a lower bound for the area of the region R. Similarly, the area of the region consisting of the five rectangles in Figure 7.2b is the absolute value of the lower sum, or 268.80, and Figure 7.2b shows that this is an upper bound for the area of R.

(f) The region R is a trapezoid. Its area is , where w is its width and h1 and h2 are the heights at the left and right edges. The area of R is

.

This is between the bounds established in part (e). ■ Active Learning Focus on developing skills

In Exercises 1-12:a. Find an upper sum for P(x) using a partition of the given interval

into 5 subintervals of equal width.b. Find the corresponding lower sum.c. Use your results from parts (a) and (b) to establish upper and lower

bounds for the area of the region under the graph of P(x) over the given interval.

1. P(x) = x – 3, [3, 6] 2. P(x) = 2x + 5, [-2, 1] 3. P(x) = 7 – 3x, [1, 2] 4. P(x) = –4x, [-5, -1] 5. P(x) = x2 – 2, [2, 6] 6. P(x) = x2 – 2x, [2, 4] 7. P(x) = x2 – 2, [-6, -2] 8. P(x) = x2 – 2x, [-2, 0] 9. P(x) = x3 + 1, [-1, 1] 10. P(x) = x4, [0, 1]



11. P(x) = 1 – x3, [-1, 1] 12. P(x) = 1 – x4, [0, 1]

In Exercises 13-20:a. Find an upper sum for P(x) using a partition of the given interval

into 5 subintervals of equal width.b. Find the corresponding lower sum.c. Use your results from parts (a) and (b) to establish upper and lower

bounds for the area of the region between the graph of P(x) and the x-axis over the given interval.

13. P(x) = x – 3, [0, 3] 14. P(x) = 2x + 5, [-6, -3]15. P(x) = 6 – 2x, [3, 4] 16. P(x) = –3x, [2, 7]17. P(x) = 1 – x2, [1, 3] 18. P(x) = x2 – 4, [-2, 0]19. P(x) = 1 – x2, [-3, -1] 20. P(x) = x2 – 4, [0, 2]

Focus on applying skillsIn Exercises 21-24, an object on a coordinate axis has velocity v(t) at time t.

a. Use an upper sum with 4 subintervals of equal duration to establish an upper bound for the displacement of the object over the given time interval.

b. Use the corresponding lower sum to establish a lower bound for the displacement of the object over the given time interval.

c. Use antiderivatives to find the displacement of the object over the given time interval, and verify that it lies between the bounds established in parts (a) and (b).

21. v(t) = 5t + 25, [2, 6] 22. v(t) = 50 – 10t, [5, 7]23. v(t) = t2 – 2t – 15, [2, 4] 24. v(t) = 20 + 2t – t2, [1, 5]

25. A car accelerates from rest with constant acceleration 10 ft/sec2.a. Solve an initial value problem to find the car’s velocity v(t) after t

seconds.b. Establish upper and lower bounds for the car’s displacement over

the first 3 seconds by partitioning the time interval [0, 3] into 6 subintervals of equal duration and finding the upper and lower sums for v(t).

c. Solve an initial value problem to find the car’s displacement over the first 3 seconds, and verify that it lies between the bounds established in part (b).

26. A ball is thrown upward from ground level with initial velocity 29.4 m/sec.a. Solve an initial value problem to find the ball’s velocity v(t) after t

seconds. The acceleration due to gravity is –9.8 m/sec2.b. Find the time T at which the ball reaches its maximum height.



c. Establish upper and lower bounds for the ball’s maximum height by partitioning the time interval [0, T] into 6 subintervals of equal duration and finding the upper and lower sums for v(t).

d. Solve an initial value problem to find the ball’s height s(t) after t seconds.

e. Find the ball’s maximum height, and verify that it lies between the bounds established in part (c).

Focus on connecting conceptsRefer to the material in the section as needed to answer questions 27 and 28, but write your answer in your own words. Address your answer to an imaginary classmate.

27. Suppose that a polynomial function P(x) is either increasing over all of an interval [a, b] or decreasing over the entire interval.a. Figures 6.2, 6.3, 6.4, and 7.2 suggest that almost every term in a

lower sum for P(x) over [a, b] is equal to a term in the corresponding upper sum “shifted over” by one subinterval. Explain why this is the case.

b. Use your conclusion from part (a) to provide an informal argument that Theorem 7.1 is true.

28. Suppose that an object in rectilinear motion has velocity v(t) ≥ 0 at time t. Explain why an upper sum for v(t) over a time interval [a, b] is an upper bound for the displacement of the object over that interval.

29. Let P(x) = 8 + 2x – x2.a. Find upper and lower sums for P(x) over [-2, 1] using a partition

into 3 subintervals of equal width.b. Find upper and lower sums for P(x) over [1, 4] using a partition

into 3 subintervals of equal width.c. Use your results from parts (a) and (b) to establish upper and lower

bounds for the area of the region under the graph of P(x) over [-2, 4].

In Exercises 30 and 31, round all results to two decimal places.a. Find a value x = c in (a, b) where the graph of P(x) has a turning

point.b. Find upper and lower sums for P(x) over both [a, c] and [c, b]. In

all cases use a partition into 4 subintervals of equal width.c. Use your results from part (b) to establish upper and lower bounds

for the area of the region under the graph of P(x) over [a, b].

30. P(x) = 2x2, [a, b] = [-2, 2]31. P(x) = x3 – 3x2 + 5, [1, 3] (Hint: Turning points occur where .)



32. Generalize Exercises 29-31 as follows. Suppose that P(x) ≥ 0 on [a, b] and that the graph of P(x) has exactly one turning point in (a, b), at x = c. Describe a procedure for obtaining upper and lower bounds for the area of the region under the graph of P(x) over [a, b].

33. Let P(x) = 2x – 6.a. Find upper and lower sums for P(x) over [0, 3] using a partition

into 3 subintervals of equal width.b. Find upper and lower sums for P(x) over [3, 9] using a partition

into 3 subintervals of equal width.c. Use your results from parts (a) and (b) to establish upper and lower

bounds for the area of the region between the graph of P(x) and the x-axis over [0, 9]. (Hint: Graph P(x) over the interval [0, 9].)

d. Use geometry to find the area of the region between the graph of P(x) and the x-axis over [0, 9], and verify that it lies between the bounds established in part (c).

In Exercises 34 and 35, round all results to two decimal places.a. Find an x-intercept on the graph of P(x) at x = c in (a, b).b. Find upper and lower sums for P(x) over both [a, c] and [c, b]. In

each case use a partition into 4 subintervals of equal width.c. Use your results from part (b) to establish upper and lower bounds

for the area of the region between the graph of P(x) and the x-axis over [a, b].

34. P(x) = x2 – 4x, [2, 6] 35. P(x) = 8 – x3, [1, 3]

36. Generalize Exercises 33-35 as follows. Suppose that the graph of P(x) has exactly one x-intercept in (a, b), at x = c. Describe a procedure for obtaining upper and lower bounds for the area of the region between the graph of P(x) and the x-axis over [a, b].

37. Suppose that P(x) ≥ 0 on [a, b]. Let UP and LP be upper and lower sums for P(x) over [a, b], and let U–P and L–P be upper and lower sums for –P(x), with all sums using the same partition of [a, b].a. Explain why U–P = –LP.b. Write a corresponding formula for L–P.

In Exercises 38-41, let U and L be upper and lower sums for P(x) over the given interval using m subintervals of equal width. Use Theorem 7.1 to complete the exercises.

a. Find U – L if m = 10. You do not need to find the sums.b. Find U – L if m = 100. You do not need to find the sums.c. Find U – L if m = 1000. You do not need to find the sums.d. Find the smallest value of m for which U – L ≤ 0.001.



38. P(x) = 4x + 1, [-2, 2] 39. P(x) = x2 + 2x – 3, [0, 5]40. P(x) = 3x2 – x3, [2, 4] 41. P(x) = x4, [-1, 0]

42. Your results in Exercises 38-41 suggest that for a given polynomial function on a given interval, it is always possible to find upper and lower sums whose difference is as small as you please. Explain why this should be true.



Section 8 Definite Integrals and The Fundamental Theorem of Calculus

The Definition of Area

You are now ready to learn how upper and lower sums lead to a solution of the area problem for polynomial functions. The solution is based on two theorems, each of which is intuitively plausible.

Theorem 8.1 Let P be a polynomial function, and let L and U be lower and upper sums for P on a closed interval [a, b]. Then L ≤ U.

Theorem 8.1 states that for a given polynomial function P(x) on a given closed interval [a, b], no lower sum can be greater than any upper sum. As a consequence, there is at least one number I that is both a lower bound for all upper sums for P(x) on [a, b] and an upper bound for all lower sums. The number I will be unique if it is possible to find upper and lower sums whose difference is arbitrarily small. Exercises 38-42 in Section 7 suggest that this is the case. Theorem 8.2 guarantees it.

Theorem 8.2 Let P be a polynomial function, let [a, b] be a closed interval, and let r be any positive real number. Then there is an upper sum U and a lower sum L for P on [a, b] such that U – L < r.

Theorem 8.2 implies that if P(x) ≥ 0 on [a, b], there is exactly one way to define the area under the graph of P(x) consistent with Principles A1 and A2. Specifically, let P(x) be a polynomial function such that P(x) ≥ 0 on [a, b], and let R be the region under the graph of P(x) over [a, b]. The area of R is the unique number A such that L ≤ A ≤ U for every lower sum L and every upper sum U for P on [a, b].

Example 8.1 illustrates that for a region whose area is given by a geometric formula, the definition of area just given leads to the same result as the formula.

Example 8.1 Comparing areas defined by calculus with areas defined by geometryLet R be the region under the graph of P(x) = 3x – 1 over [1, 7].(a) Use a geometric formula to find the area of R.(b) Show that the definition of area given in this section leads to the same

result.

Solution (a) The region R is a trapezoid. Its height at its left edge is P(1) = 2, its height at its right edge is P(7) = 20, and its width is 6. Its geometrically defined area is

.

Definite Integrals and The Fundamental Theorem of Calculus


(b) If U is an upper sum for P(x) over [1, 7], then U is the area of a region that covers R. By Principle A1 and the result of part (a), U ≥ 66, so 66 is a lower bound for all upper sums for P(x) on [1, 7]. A similar argument shows that 66 is an upper bound for all lower sums for P(x) on [1, 7]. If the area of R is defined as the unique number with both of these properties, then the area of R is 66. ■

Definite Integrals

Theorems 8.1 and 8.2 apply to every polynomial function P(x) on every closed interval [a, b], regardless of whether P(x) ≥ 0 on [a, b]. It follows that there is a unique number I such that L ≤ I ≤ U for every lower sum L and every upper sum U for P on [a, b]. This number is called the definite integral of P(x) from a to b, written

.

The function P(x) is the integrand, the symbol is the integral sign, and the numbers a and b are the limits of integration. If P(x) ≥ 0 on [a, b], then

is the area of the region under the graph of P(x) over [a, b]. For

example, the result of Example 8.1 shows that .

Theorem 8.3 establishes some fundamental properties of definite integrals.

Theorem 8.3 Let P(x) and Q(x) be polynomial functions. If a < b, then:

(a) (Constant Multiple Rule) For every constant k, .


(c) (Interval Additivity Rule) For a < c < b, .

(d) (Dominance Rule) If P(x) ≤ Q(x) on [a, b], then .

It is customary to define and . The latter definition in particular may appear to be arbitrary and capricious, but you will shortly discover that it is both reasonable and practical.



Example 8.2 Using the properties of definite integralsSuppose that . Find:

(a) (b) (c)

Solution (a)(by Theorem 8.3b)

(by Theorem 8.3a)

= 7·2 – 3·6 = –4.

(b) .

(c)(by Theorem 8.3c)

= 2 + (–4) = –2. ■

As a special case of Theorem 8.3a, . You can use this observation to express areas of regions below the x-axis as definite integrals, as in Example 8.3.

Example 8.3 Expressing areas below the x -axis as definite integrals Let R be the region between the graph of P(x) = –x2 and the x-axis over [0, 1]. Express the area of R as a definite integral.

Solution Let S be the region under the graph of P(x) = x2 over [0, 1]. The area of R is equal to the area of S, which is . ■

Example 8.3 illustrates that the relationship of definite integrals to areas can be summarized as follows.

• If P(x) ≥ 0 on [a, b], the area of the region under the graph of P(x) is .

• If P(x) ≤ 0 on [a, b], the area of the region between the graph of P(x) and the x-axis is .



Example 8.4 Using areas to evaluate definite integralsUse area formulas from geometry to evaluate each of the following definite integrals.

(a) (b) (c)

Solution (a) Because P(x) = x + 2 ≥ 0 on [1, 5], the integral is the area of the region under the graph of P(x) over [1, 5]. That region is a trapezoid with width 5 – 1 = 4 and average height . The area is the product of the width and the average height, which is 20. Thus

.

(b) The region between the graph of P(x) = –3 and the x-axis over [-1, 4] is a rectangle with height 3 and width 4 – (-1) = 5. The area of the region is 3·5 = 15. Because the region is below the x-axis, .

(c) The graph of P(x) = 2x – 4 crosses the x-axis at x = 2. The region between the graph and the x-axis over [0, 3] is the union of two triangles. The left-hand triangle has base 2 and altitude 4, so its area is . The right-

hand triangle has base 1 and altitude 2, so its area is . The left-hand triangle is below the x-axis and the right hand triangle is above it, so

. ■

The Fundamental Theorem of Calculus

You now know that the definite integral of any polynomial function P(x) over any closed interval [a, b] is a uniquely defined number, and that the area of the region between the graph of P(x) and the x-axis over [a, b] can be expressed in terms of definite integrals. To complete the solution of the area problem, we need to find a practical method of evaluating definite integrals. Such a method is illustrated in Example 8.5.

Example 8.5 Evaluating a definite integralEvaluate .

Solution Let v(t) = t2 be the velocity of an object moving on a coordinate axis for 0 ≤ t ≤ 1, and let be its position. Every upper sum for v(t) on [0, 1] is an upper bound for the object’s displacement over the time interval [0, 1], and every



lower sum is a lower bound for the displacement. It follows that is equal to the object’s displacement over [0, 1]. That is,

. ■

By demonstrating that antiderivatives can be used to evaluate definite integrals, Example 8.4 illustrates the surprisingly close relationship between the area problem and the tangent line problem. The generalization of the method used in Example 8.4 is known as the Fundamental Theorem of Calculus.

Theorem 8.4 (The Fundamental Theorem of Calculus for Polynomial Functions)Let P(t) be a polynomial function, and let Q(t) be any antiderivative of P(t).Then:(a) For all real numbers a and b, .

(b) For all real numbers a and x, is a differentiable function of x, and its derivative with respect to x is P(x).

It is customary to write to represent the quantity Q(b) – Q(a).

Example 8.6 Evaluating definite integralsEvaluate:(a) (b)

Solution (a) One antiderivative of 6x5 – 8x3 – 4 is x6 – 2x4 – 4x. By Theorem 8.4,

= (64 – 32 – 8) – (1 – 2 – 4) = 29.

(b) The integrand is 3(x – 2)(x – 4) = 3x2 – 18x + 24. One antiderivative is x3 – 9x2 + 24x, so by Theorem 8.4,

= (64 – 144 + 96) – (8 – 36 + 48) = –4. ■

8.7 Finding derivatives of definite integralsLet . Find in two different ways:(a) Use Theorem 8.4a.(b) Use Theorem 8.4b.



Solution (a) One antiderivative of the integrand is t4 + 2t2 – 5t, so Theorem 8.4a implies that

= x4 + 2x2 – 5x + 2.

Therefore .(b) Theorem 8.4b implies that

. ■


Suppose that and . Evaluate each integral in Exercises 1-6.

1. 2.

3. 4.

5. 6.In Exercises 7-14, use area formulas from geometry to evaluate the definite integrals.

7. 8.

9. 10.

11. 12.

13. 14.

In Exercises 15-22, write one or more definite integrals to represent the area of the region between the graph of P(x) and the x-axis over the given interval.

15. P(x) = 1 – x2, [0, 2] 16. P(x) = 2x + 4, [-3, 0]17. P(x) = x2 – 2x, [0, 2] 18. P(x) = x2 – 2x, [-2, 4]19. P(x) = x3 – 3x, [-3, 3] 20. P(x) = x3 + x2 – 6x, [-3, 2]21. P(x) = x3 – x4, [-1, 1] 22. P(x) = 4x3 – x5, [-4, 4]



In Exercises 23-32, evaluate the definite integral.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

In Exercises 33-36, find in two different ways:a. Use Theorem 8.4a.b. Use Theorem 8.4b.

33. 34.

35. 36.

In Exercises 37 and 38, find in two different ways.a. Use Theorem 8.4a.b. Use Theorem 8.4b and the Chain Rule.

37. 38.

Focus on connecting conceptsRefer to the material in the section as needed to answer each of questions 39-42, but write your answer in your own words. Address your answer to an imaginary classmate.

39. Why is it useful to define the concept of a definite integral?

40. What does Theorem 8.3d say about areas in the case where P(x) and Q(x) are nonnegative on [a, b]?

41. Suppose that –2 ≤ P(x) ≤ 3 on the interval [7, 9]. Use Theorem 8.3d to establish upper and lower bounds for .

42. Why is it reasonable to make the definition

43. Show that the definition is necessary if we want Theorem 8.3c to apply when a = b.



44. The Fundamental Theorem of Calculus shows that the area problem is, in a certain sense, the inverse of the tangent line problem. Explain this statement.

45. When you use Theorem 8.4a to evaluate a definite integral , why doesn’t it matter which antiderivative of P(x) you use?



Section 9 Interpretations of the Definite Integral

As a result of the Fundamental Theorem of Calculus, the definite integral becomes a powerful tool in the solution of a variety of mathematical and physical problems. The following examples provide only a small sample.

Areas Between a Graph and the x-axis

Example 9.1 Finding areasFind the area of the region bounded by the graph of P(x) = x3 – 3x2 and the x-axis.

Solution The graph of P(x) intersects the x-axis at the points (0, 0) and (3, 0). For 0 ≤ x ≤ 3 the graph is below the x-axis, so the area of the region is

. ■

Areas Between Two Graphs

Example 9.2 Finding areasFind the area of the region R bounded by the graphs of P(x) = x2 – 1 and

.

Solution


The region R is shown in Figure 9.1a. Figure 9.1b shows the congruent region S that is bounded by the graphs of P(x) + 1 and Q(x) + 1. The x-coordinates at the left and right boundaries of both regions are the solutions of the equation P(x) = Q(x), that is, x2 – 1 = x + 1. Rewriting the equation as x2 – x – 2 = 0, and then as (x + 1)(x – 2) = 0, shows that the solutions are x = –1 and x = 2.

Interpretations of the Definite Integral


Region S is the region under the graph of Q(x) + 1 over [–1, 2], minus the region under the graph of P(x) + 1 over [–1, 2]. It follows that the area of S is

. ■

Example 9.2 can be generalized.

Theorem 9.1 If Q(x) ≥ P(x) on [a, b], then the area of the region bounded by the graphs of P(x) and Q(x) over [a, b] is

.

Example 9.3 Finding areasFind the area of the region R bounded by the graphs of P(x) = x3

and Q(x) = 4x.

Solution The region R is shown in Figure 9.2. The graphs of P(x) intersect at points with x-coordinates that satisfy x3 = 4x. This equation can be written as x3 – 4x = 0 or asx(x + 2)(x – 2) = 0, so the points Figure 9.2of intersection occur at x = -2, 0, and 2.



Because P(x) ≥ Q(x) on [-2, 0] and Q(x) ≥ P(x) on [0, 2], the area of R is

.

These two integrals represent the areas of two congruent subregions of R, so we can save some work by evaluating one of them and doubling the result. The area of R is

. ■

Displacement in Rectilinear Motion

Example 9.4 Finding displacement from velocityA toy rocket shot upward has velocity v(t) = 160 – 32t ft/sec during its ascent. How far above its initial height does it rise?

Solution You could write and solve an initial value problem to answer this question, but you can also answer it by evaluating a definite integral. First note that the rocket’s velocity is 0 when t = 5, so the rocket is climbing for 5 seconds. If its height, in feet, after t seconds is s(t), then its maximum height above its initial position is s(5) – s(0). According to the Fundamental Theorem of Calculus, this is the value of

. ■

Example 9.5 Finding displacement from velocityThe driver of a car traveling at 60 mph (88 ft/sec) applies the brakes, and the car decelerates at a constant rate of 8 ft/sec2. How far does the car travel before it stops?

Solution The car’s velocity t seconds after the brakes are applied is v(t) = 88 – 8t ft/sec., which is 0 after 11 seconds. If the car travels s(t) feet in t seconds after the brakes are applied, then it travels s(11) – s(0) feet before stopping. This is



. ■ ActiveLearning Focus on developing skills

In Exercises 1-8, find the area between the graph of P(x) and the x-axis over the given interval.

1. P(x) = x2 + 3, [-2, 4] 2. P(x) = 10x4, [0, 2] 3. P(x) = x3 – 1, [-1, 1] 4. P(x) = 4x – x2, [1, 3] 5. P(x) = x3 + 2x, [-2, 2] 6. P(x) = x3 + 6x2, [-3, 1] 7. P(x) = 4x4 – x2, [-1, 1] 8. P(x) = x3 – 3x2 + 2x, [-1, 3]

In Exercises 9-16, find the area between the graphs of P(x) and Q(x) over the given interval.

9. P(x) = x + 1, Q(x) = 3x + 1, [0, 6]10. P(x) = 1 – 2x, Q(x) = 7 – 2x, [-3, 5]11. P(x) = x2, Q(x) = –x2, [-1, 5]12. P(x) = x2 + 1, Q(x) = 2x3, [0, 1]13. P(x) = x + 2, Q(x) = 6 – x, [0, 3]14. P(x) = x2, Q(x) = x3, [-1, 1]15. P(x) = 3 – x2, Q(x) = 1 – x, [-2, 4]16. P(x) = x3 – 2x, Q(x) = 2x, [-3, 3]

In Exercises 17-20, find the total area of all regions bounded by the graphs of P(x) and Q(x).

17. P(x) = x2 – 4, Q(x) = x – 2 18. P(x) = x2, Q(x) = 8 – x2

19. P(x) = x3 – 6x, Q(x) = 3x20. P(x) = x5, Q(x) = x3

In Exercises 21-30, an object moves on a coordinate axis with velocity v(t) at time t. Find the object’s displacement over the given time interval by evaluating a definite integral.

21. v(t) = 50, [0, 4] 22. v(t) = –15, [2, 5]23. v(t) = 10t – 60, [4, 8] 24. v(t) = 49 – 9.8t, [0, 10]25. v(t) = t2 – 5t + 4, [1, 4] 26. v(t) = 5t – t2, [0, 5]27. v(t) = 4t3 + 2t, [1, 2] 28. v(t) = 16t – t3, [2, 4]29. v(t) = t4, [1, 2] 30. v(t) = 6t5, [0, 1]



Focus on applying skills31. A race car accelerates at a constant rate from a speed of 120 mph (176

ft/sec) to 150 mph (220 ft/sec) in 4 seconds. Find the distance the car travels during that time in two ways.a. Solve an initial value problem.b. Express the distance as a definite integral, and use Theorem 8.4 to

evaluate the integral.

32. A volcanic eruption throws a rock upward from the top of the volcano’s crater at a speed of 384 ft/sec. Find the height of the rock above the crater after 12 seconds in two ways.a. Solve an initial value problem.b. Express the change in height over the time interval [0, 12] as a

definite integral, and use Theorem 8.4 to evaluate the integral.



Appendix Proofs of Theorems

Theorem 2.1 Let P be a polynomial function. Then the expression reduces to

a polynomial P*(x, h) for h ≠ 0.

Proof Recall that (x + h)n = , where the binomial coefficient is

defined as .

Now let P(x) = anxn + an-1xn-1 + …+ a1x + a0. Then

.

Note that

.

Thus after simplifying, every term of P(x + h) – P(x) has at least one factor of h, which may be cancelled with the factor of h in the denominator of

, resulting in a polynomial in x and h. ■

Theorem 3.1 (Power Rule, special cases)

(a) For every constant c, .

(b) .

Proof (a) If P(x) = c, then

for h ≠ 0.

Thus .

Proofs of Theorems


(b) If P(x) = x, then

for h ≠ 0.

Thus . ■

Theorem 3.2 (Power Rule, general case) For every integer n ≥ 1, .

Proof Recall that for all real numbers a and b,

an – bn = (a – b)(an-1 + an-2b + an-3b2 + … + abn-2 + bn-1).

In particular, if P(x) = xn, then

P(x + h) – P(x) = (x + h)n - xn

.

Therefore

for h ≠ 0.

There are n terms in this expression, and each term simplifies to xn-1 when h = 0. Therefore . ■

Theorem 3.3 Let P and Q be polynomial functions, and let c be a constant. Then:

(a) (Constant Multiple Rule)


(c) (Product Rule)

Proof (a) Let R(x) = cP(x). Then

,

and evaluating at h = 0 gives the conclusion.

(b) Let R(x) = (P + Q)(x). Then

Proofs of Theorems


,

and evaluating at h = 0 gives the conclusion.

(c) Let R(x) = P(x)Q(x). Then

=P(x + h)Q*(x, h) + Q(x)P*(x, h),

and evaluating at h = 0 gives the conclusion. ■

Theorem 3.4 (Chain Rule) For polynomial functions P and Q, .

Proof If Q is a constant function, then both sides are 0, so the conclusion is true. Otherwise let . Note that

for h ≠ 0 and j ≠ 0. Letting j = Q(x + h) – Q(x), this expression is equal to

except, for each x, at the finite number of values of h for which . Evaluating P*(Q(x), j) at j = 0 and Q*(x, h) and R*(x, h) at h

= 0 gives the conclusion. ■

Theorem 3.5 (Chain Rule, special case) If Q(x) is a polynomial function, then for every positive

integer n, .

Proof Let P(x) = xn. Then , , and the result follows from Theorem 3.4. ■

Proofs of Theorems


Theorem 5.1 If for all x, then there is a constant C such that P(x) = C for all x.

FYIThe proof of Theorem 5.1 requires an assumption that only polynomial functions have polynomial derivatives. This assumption can be validated after the concept of a derivative has been extended to a wider class of functions.

Proof If P(x) has positive degree n and leading term axn with a ≠ 0, then the leading term of is anxn-1, and the leading coefficient is an, which is not 0. It follows that if , then the degree of P(x) is 0, so P(x) is a constant function. ■

Theorem 5.2 If for all x, then there is a constant C such that P(x) = Q(x) + C for all x.

Proof Let R(x) = P(x) – Q(x). Then . It follows from Theorem 5.1 that R(x) = C for some constant C, so P(x) = Q(x) + R(x) = Q(x) + C. ■

Theorem 5.3 (a) (Power Rule)

If , then for some constant C.

(b) (Constant Multiple Rule)If , then P(x) = kQ(x) + C for some constant C.

(c) (Sum and Difference Rules)If , then P(x) = Q(x) ± R(x) + C for some constant C.

Proof Note that is an antiderivative of xn, kQ(x) is an antiderivative of ,

and Q(x) ± R(x) is an antiderivative of . The results then follow from Theorem 5.2. ■

Theorem 5.4 Let Q(x) be a polynomial function. Every initial value problem of the form

has a unique solution.

Proof Let Pn(x) = Q(x). The function y(n-1)(x) must be an antiderivative of Q(x), so it has the form Pn-1(x) + C, where Pn-1(x) is a particular antiderivative of Q(x). The initial condition y(n-1)(a) = yn-1 implies that Pn-1(a) + C = yn-1. This is a linear equation in C, and its unique solution specifies y(n-1)(x) as a unique function. In the same way the functions y(n-2)(x), y(n-3)(x), …, , and y(x) are specified as unique functions. ■

Proofs of Theorems


Theorem 7.1 Suppose that the polynomial function P(x) is either increasing on all of [a, b] or decreasing on all of [a, b]. Let U and L be, respectively, upper and lower sums for P(x) for a partition of [a, b] into m subintervals of equal width . Then

.

Proof Let the partition of [a, b] be {a = x0, x1, x2, …, xm = b}. First suppose that P(x) is increasing on [a, b]. Then

U = P(x1)x + P(x2)x + P(x3)x + … + P(xm)x, and

L = P(x0)x + P(x1)x + P(x2)x + … + P(xm-1)x, so

U – L = P(xm)x – P(x0)x

.

Similarly, if P(x) is decreasing on [a, b], then

L = P(x1)x + P(x2)x + P(x3)x + … + P(xm)x, and

U = P(x0)x + P(x1)x + P(x2)x + … + P(xm-1)x, so

U – L = P(x0)x – P(xm)x

. ■

Theorem 8.1 Let P be a polynomial function, and let L and U be lower and upper sums for P on a closed interval [a, b]. Then L ≤ U.

Proof The conclusion is true if L and U correspond to the same partition. To establish the general result, first consider the special case where S1 = {x0, x1, x2, …, xn} and S2 = {x0, x1, x2, …, xk-1, x*, xk, …, xn} are partitions with xk-1 < x* < xk. Let L1 and U1 be the lower and upper sums for P(x) using S1, and let L2 and U2 be the lower and upper sums for P(x) using S2. The lower sum using S1 contains a term P(ck)(x)k = P(ck)(xk – xk-1). In the lower sum using S2 this term is replaced with two terms P(c*)(x* – xk-1) + P(c**)(xk – x*). Because P(ck) ≤ P(x) for all x in the subinterval (xk-1, xk), it follows that

Proofs of Theorems


P(ck)(xk – xk-1) = P(ck)(x* – xk-1) + P(ck)(xk – x*) ≤ P(c*)(x* – xk-1) + P(c**)(xk – x*).

All other terms in the two lower sums are the same, so L1 ≤ L2. A similar argument establishes that U1 ≥ U2, so L1 ≤ L2 ≤ U2 ≤ U1.

In the general case, let L and U correspond to partitions S1 and S2, respectively, and let S3 = S1 S2. Repeated application of the argument used in the special case above leads to the conclusion that L ≤ L3 ≤ U3 ≤ U. ■

Theorem 8.2 Let P be a polynomial function, let [a, b] be a closed interval, and let r be any positive real number. Then there is an upper sum U and a lower sum L for P on [a, b] such that U – L < r.

Proof First suppose P is either increasing on [a, b] or decreasing on all of [a, b]. Let Sm = {a = x0, x1, x2, …, xm-1, xm = b} be a partition of [a, b] into m subintervals of equal width , and let Um and Lm, respectively, be the upper and lower

sums for Sm. By Theorem 7.1, If we choose

,

then Um – Lm < r.

In the general case, [a, b] is a union of n subintervals on which P is either increasing or decreasing. On each subinterval we can find a partition with upper and lower sums whose difference is less than . The union of these partitions is a partition of [a, b] with upper and lower sums whose difference is less than r. ■

Theorem 8.3 Let P(x) and Q(x) be polynomial functions. If a < b, then:

(a) (Constant Multiple Rule) For every constant k, .


(c) (Interval Additivity Rule) For a < c < b, .

(d) (Dominance Rule) If P(x) ≤ Q(x) on [a, b], then .

Proofs of Theorems


Proof (a) Note that U is an upper sum for P(x) on [a, b] if and only if cU is an upper sum for cP(x) on [a, b]. Similarly, L is a lower sum for P(x) if and only if cL is a lower sum for cP(x). The definite integral is the unique number I such that L ≤ I ≤ U for every lower sum L and every upper sum U for P(x) on [a, b]. Because cL ≤ cI ≤ cU as well, it follows that

.

(b) To prove the Sum Rule, let S be a partition of [a, b], let the maximum values of P(x), Q(x), and P(x) + Q(x) in the kth subinterval occur at x = ck*, x = ck**, and x = ck, respectively, and let U1, U2, and U3 be the upper sums for P(x), Q(x), and P(x) + Q(x), respectively. Because P(ck) ≤ P(ck*) and Q(ck) ≤ Q(ck**) for all k,

P(ck) + Q(ck) ≤ P(ck*) + Q(ck**).

That is, the kth term of U3 is less than or equal to the sum of the kth terms of U1 and U2 for all k, so U3 ≤ U1 + U2. It follows that

.

The reverse inequality is established by a similar argument using lower sums, and the Sum Rule is proved.

To prove the Difference Rule, note that

(by the Sum Rule),

and the result follows.

(c) Let S1 and S2 be partitions of [a, c] and [c, b], respectively. Then is a partition of [a, b]. If L1 and L2 are lower sums for P(x)

using S1 and S2, then L3 = L1 + L2 is a lower sum for P(x) using S3. It follows that . The reverse inequality is established by a similar argument using upper sums.

(d) The given inequality implies that every upper or lower sum for P(x) is less than or equal to the corresponding upper or lower sum for Q(x), and the result follows. ■

Proofs of Theorems


Theorem 8.4 (The Fundamental Theorem of Calculus for Polynomial Functions)Let P(t) be a polynomial function, and let Q(t) be any antiderivative of P(t).Then:(a) For all real numbers a and b, .

(b) For all real numbers a and x, is a differentiable function of x, and its derivative with respect to x is P(x).

Proof (a) Let P(t) be the velocity of an object in rectilinear motion. Every upper (respectively, lower) sum for P(t) over [a, b] is an upper (respectively, lower) bound for the object’s displacement over [a, b]. Therefore the object’s displacement on [a, b] is . If the object’s position function is s(t), then s(t) = Q(t) + C for some constant C, so the object’s displacement is also (Q(b) + C) – (Q(a) + C) = Q(b) – Q(a), and the result follows.

(b) Let Q(t) be an antiderivative of P(t). By Theorem 8.4a,

,so

. ■

Theorem 9.1 If Q(x) ≥ P(x) on [a, b], then the area of the region bounded by the graphs of P(x) and Q(x) over [a, b] is

.

Proof It is possible to choose a constant K such that both Q(x) + K ≥ 0 and P(x) + K ≥ 0 on [a, b]. The region bounded by the graphs of P(x) and Q(x) over [a, b] is congruent to the region bounded by the graphs of P(x) + K and Q(x) + K over the same interval. The area of the latter region is the area under the graph of Q(x) + K minus the area under the graph of P(x) + K over [a, b], which is

. ■

Proofs of Theorems


Section 1 1. 4; 2 3. 20; 10

5. a. 2t + h – 2 7. a. 4t + 2h + 3b. 5; 4.1; 4.01; 4.001 b. 17; 15.2; 15.02; 15.002

9. a. 0 11. a. 3t2 + 3th + h2 – 12b. 0; 0; 0; 0 b. 25; 15.61; 15.0601; 15.006001

13. 5 15. 1

17. a. 0 19. a. –3b. 0; 0; 0; 0 b. –3; –3; –3; –3

21. a. 4x + 2h – 1 23. a. 3x2 + 3xh + h2

b. 9; 7.2; 7.02; 7.002 b. 19; 12.61; 12.0601; 12.006001

25. 192 ft; 48 ft/sec 27. –5; interpretations will vary

Section 2 1. a. 2; 2; 2; 2 3. a. 9; 8.1; 8.01; 8.001

b. 2 b. 8

5. a. 2t + h – 2 7. a. 4t + 2h + 3b. 2t – 2 b. 4t + 3c. 4 c. 15d. t = 1 d.

9. a. 0 11. a. 3t2 + 3th + h2 – 12b. 0 b. 3t2 – 12c. 0 c. 15d. all real values of t d. t = ±2

13. a. 5; 5; 5; 5 15. a. –2; –2.9; –2.99; –2.999b. 5 b. –3

17. a. 0 19. a. –3b. 0 b. –3c. y = 5 c. y = –1 – 3(x – 2)e. all real values of x e. none

21. a. 4x + 2h – 1 23. a. 3x2 + 3xh + h2

b. 4x – 1 b. 3x2

c. y = –9 + 7(x – 2) c. y = 8 + 12(x – 2)e. e. x = 0

Answers to Selected Exercises


25. 400 ft 27. a. 8 secb. –160 ft/sec

29. a. sec 31. a. sec

b. m b. m

33. 11 sec 43. yes (examples will vary)

45. a. yes (explanations will vary) 49. a. (2)b. no (explanations will vary) b. (3)

c. (4)d. (1)

51. a-d. 3x2

Section 3 1. 0; 0; 0 3. –2x; –4; 4

5. 7. 3x2 – 12; 0; 0

9. 7x6 11. 0

13. 8x3 – 6x 15. x4 – x3 + x2

17. 2x 19. 3x2

21. 3x2 +6x + 2 23. 2x + 3, x ≠ 0

25. m 27. 3Ax2

29. 2; 0 31. 2 – 2x; –2

33. for k ≥ 4

35. for k ≥ 5

37. 6 39. 3(x4 + 2x3 + x2 – 2)(2x + 1)

41. 6(2x – 5)2 43. 140(3 – 7x)4

45. 21x3(3x2 – 4)3(x2 + 1)2 47. 3(x4 + 2x3 – 5x2 – 4x + 9)(2x + 1)

49. a-b. 4x3 – 18x + 48



51. a. 3t2 + 4 53. a. 4t3 – 8tb. 6t b. 12t2 – 8c. none c.

55. 5 m/sec2 57. –16; –4.9

61. a. a > 0 63. –4b. a = 0c. a < 0

65. a. The velocity is constant. 69. Yes; k = n + 1 (explanations will vary)b. The acceleration is zero.c. The acceleration is constant.

Section 4 1. a. 5.5 sec 3. a. 4 sec

b. 8 ft/sec2 b. 22 ft/sec2

5. a. 5 sec 7. a. secb. 100 ft b. ft

9. 128 ft/sec 11. a. x < 0b. x = ±1

13. a. x < –1b. x = 0

15. a.b. $5.00; $20.00; $15.00; all are equal

17. a.

b. $22.01; $9.98; –$12.03; the differences are $0.01, –$0.02, and –$0.03

19. 2r 21. a. 2rh; 40 (explanations will vary)b. r2; 25 (explanations will vary)

Section 5 1. P(x) = –2x + C 3.

5. 7.

9. P(x) = 6x – x2 – 1 11.



13. P(x) = x4 + 4x – 4 15.

17. s(t) = 100t + 50 19. s(t) = 80 + 144t – 16t2

21. s(t) = 44 + 72t – 16t2 23. s(t) = t3 + 12t + 44

25. P(x) = a(x – 3)2 – 4 27. a. 30 mb. m/sec

29. a. 33 ft 35. a. yesb. 11 ft/sec b. noc. 198 ft c. explanations will varyd. 231 ft

Section 6 1. a. 2.4 3. a. 28

b. 1.6 b. 28c. 1.6 ≤ area ≤ 2.4 c. area = 28d. 2 d. 28

5. a. 50 7. a. 14b. 45 b. 5c. 45 ≤ area ≤ 50 c. 5 ≤ area ≤ 14d. 47.5 d. 11.375; 6.875;

6.875 ≤ area ≤ 11.375

9. a. 256 11. a. 99b. 112 b. 36c. 112 ≤ area ≤ 256 c. 36 ≤ area ≤ 99d. 217; 145; 145 ≤ area ≤ 217 d. 80.4375; 48.9375;

48.9375 ≤ area ≤ 80.4375

13. a. 90 15. a. 100b. 54 b. 100c. 72 c. 100d. 72 d. 100

17. a. 1056 19. 105 m; 195 mb. 864c. 960d. 960

Section 7 1. a. 5.4 3. a. 2.8

b. 3.6 b. 2.2c. 3.6 ≤ area ≤ 5.4 c. 2.2 ≤ area ≤ 2.8



5. a. 75.2 7. a. 75.2b. 49.6 b. 49.6c. 49.6 ≤ area ≤ 75.2 c. 49.6 ≤ area ≤ 75.2

9. a. 2.4 11. a. 2.4b. 1.6 b. 1.6c. 1.6 ≤ area ≤ 2.4 c. 1.6 ≤ area ≤ 2.4

13. a. –3.6 15. a. –0.8b. –5.4 b. –1.2c. 3.6 ≤ area ≤ 5.4 c. 0.8 ≤ area ≤ 1.2

17. a. –5.12 19. a. –5.12b. –8.32 b. –8.32c. 5.12 ≤ area ≤ 8.32 c. 5.12 ≤ area ≤ 8.32

21. a. 190 23. a. –21.25b. 170 b. –25.25c. 180 c.

25. a. v(t) = 10t ft/sec 29. a. 22; 13b. 52.5 ft; 37.5 ft b. 22; 13c. 45 ft c. 26 ≤ area ≤ 44

31. a. x = 2b. upper sum over [1, 2] ≈ 2.02; lower sum over [1, 2] ≈ 1.52

upper sum over [2, 3] = 2.80; lower sum over [2, 3] ≈ 1.80c. 3.32 ≤ area ≤ 4.82

33. a. –6; –12b. 48; 24c. 60; 30d. 45

35. a. x = 2b. upper sum over [1, 2] ≈ 5.08; lower sum over [1, 2] ≈ 3.33

upper sum over [2, 3] ≈ –5.95; lower sum over [2, 3] ≈ –10.70c. 9.28 ≤ area ≤ 15.78

39. a. 17.5 41. a. 0.1b. 1.75 b. 0.01c. 0.175 c. 0.001d. m = 175,000 d. m = 1000



Section 8 1. 18 3. 20

5. 0 7. 35

9. 11. –6

13. 16 15.

17. 19.

21. 23. 0

25. –18 27. –8

29. 31. –160

33. a-b. 6x + 5 35. a-b. x3 – 2x + 1

37. a-b. 16x + 22

Section 9 1. 42 3. 2

5. 16 7. 1

9. 36 11. 84

13. 5 15. 15

17. 19.

21. 200 23. 0

25. 27. 18

29. 31. a-b. 792 ft


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