a preview of calculus - pennsylvania state university preview of calc… · a preview of calculus...

A Preview of Calculus

Table of Contents

Table of Contents

Section 0: Introduction 2

Section 1: A First Look at Average and Instantaneous Velocities 3

Section 2: The Tangent Line Problem 11

Section 3: Derivatives of Polynomial Functions 18

Section 4: Derivative Formulas 24

Section 5: Interpretations of the Derivative 30

Section 6: Antiderivatives of Polynomial Functions 35

Section 7: The Area Problem, Sigma Notation, and Summation Formulas 41

Section 8: Riemann Sums and the Definition of Area 46

Section 9: Definite Integrals and their Properties 54

Section 10: The Fundamental Theorem of Calculus 63

Appendix: Proofs of Theorems 72

Answers to Selected Exercises 86


Introduction

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 1-4 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 5 and 6 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 7-10 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.


A First Look at Average and Instantaneous Velocities

Section 1 A First Look at Average and Instantaneous Velocities

As indicated in the previous section, differential calculus deals with problems

involving rates of change. As an introduction to the subject, let’s focus on some

questions about velocity, which is a rate of change in the position of a moving

object with respect to time. As specific examples, consider the two sentences, “I

have driven 100 miles in the past 2 hours,” and “My speedometer reading is

currently 50.” Both sentences describe a velocity of 50 miles per hour, but with

different meanings of the word velocity. The first sentence describes an average

velocity over a time interval, while the second describes an instantaneous velocity

at a particular instant. The average velocity of the driver in the first sentence can

be calculated by the formula in the following definition.

Definition An object that moves along a coordinate axis is said to be in rectilinear motion.

The average velocity of such an object over a time interval is the change in its

position divided by the change in time. If its position on the axis at time t is s(t),

its average velocity over the interval [a, b] is ab

asbs

)()(.

If the car in the previous paragraph is traveling on a straight road, the road can be

regarded as a coordinate axis with units measured in miles, the origin at the car’s

initial position, and the car traveling in the positive direction. Its average velocity

over the 2-hour interval is 5002

0100

. However, because neither position nor

time changes in an instant, the algebraic formula for average velocity cannot be

used to calculate instantaneous velocity. Example 1.1 provides a preliminary

exploration of the concepts of average and instantaneous velocity.

Example 1.1 The difference between average and instantaneous velocities

A car accelerates from a stop sign and moves in a straight line, traveling s(t) = 2t2

feet in t seconds, 0 ≤ t ≤ 20.

(a) What is the car’s average velocity during the time interval [3, 5]?

(b) What is its instantaneous velocity (its speedometer reading) 3 seconds

after it leaves the stop sign?

Solution (a) The line in which the car travels can be represented by a coordinate axis,

with units measured in feet, the stop sign at the origin, and the car

traveling in the positive direction. The car’s position is s(3) = 2(3)2 = 18

after 3 seconds and s(5) = 2(5)2 = 50 after 5 seconds. Its average velocity

in the time interval [3, 5] is

162

32

35

1850

35

)3()5(

ssft/sec.

(b) We can approach this question by reasoning that the car’s speedometer

reading changes by only a small amount over a small time interval. For



example, the speedometer reading at t = 3 seconds should be about equal

to the car’s average velocity over the time interval [3, 3.1], which is

2.121.0

1822.19

31.3

)3()1.3(

ssft/sec, or about 8.3 miles per hour.

We can get a more satisfactory result by calculating the car’s average

velocity over a time interval of some small duration h, beginning at t = 3.

The average velocity over such an interval depends on the duration of the

interval, so it is a function of h, given by

h

h

h

shshP

18)3(2

3)3(

)3()3()(

2

h

hh 18692 2

h

hh 2212

= 12 + 2h for h ≠ 0.

It is reasonable to regard an instant as a time interval of duration 0 and

conclude that the car’s speedometer reading at t = 3 is 12 ft/sec (about 8.2

miles per hour). ■

Based on Example 1.1b, we can propose the following 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 1.1b shows that f(h) is a rational function

that is undefined at 0. For h ≠ 0, however, f(h) is equal to a polynomial, and the

polynomial can be evaluated at h = 0. Theorem 1.1 guarantees that this will

always happen when the position of the object is a polynomial function.

Theorem 1.1 Let P be a polynomial function. Then the expression h

xPhxP )()( reduces to

a polynomial P*(x, h) for h ≠ 0.

Theorem 1.1 allows us to give a precise definition of instantaneous velocity for an

object in rectilinear motion if its position is a polynomial function of time. A

more general definition will be given later.

Definition 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



h

tshts )()( 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 )0,(* 0ts .

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

1.2.

Example 1.2 The difference between velocity and speed

A rock thrown upward from ground level is at a height of 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 (a) The rock travels on a vertical coordinate axis where the origin is at ground

level and the positive direction is upward. Its average velocity over a time

interval [2, 2 + h] is

h

hh

h

shs 22)2(16)2(96216)2(96)2()2(

h

hhh 128441696192 2

h

hh 21632

= 32 – 16h for h ≠ 0.

The rock’s velocity after 2 seconds is 32 ft/sec, indicating that the rock is

rising at that rate.

(b) The rock’s average velocity over a time interval [5, 5 + h] is

h

hh

h

shs 22 )5(16)5(96)5(16)5(96)5()5(



h

hhh 8010251696480 2

h

hh 21664

= –64 – 16h for h ≠ 0.

The rock’s velocity after 5 seconds is –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. ■

The solution of Example 1.2 involved more calculation than was necessary. A

more efficient strategy would have been to calculate the velocity of the rock at an

arbitrary time t as a function of t, and then evaluate that function at t = 2 and t = 5.

This labor-saving strategy is illustrated in Example 1.3.

Example 1.3 Velocity as a function of time

An object moving on a coordinate axis has coordinate s(t) = t3 – 6t

2 + 9t after t

seconds. What is its velocity after 0, 2, and 4 seconds?

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

h

hththt

h

tshts )(9)(6)()()( 23

h

ttthththththhtt 96)(92633 23223223

h

hhthhthht 961233 2322

h

hththth 961233 22

= 3t2 + 3th + h

2 – 12t – 6h + 9 for h ≠ 0.

The object’s velocity at time t is 3t2 – 12t + 9. Substituting the values 0, 2, and 4

for t gives the respective velocities of 9, -3, and 9. ■



It is also useful to express an object’s velocity as a function of time when we need

to discover certain characteristics of the object’s motion over a time interval, as in

Example 1.4.

Example 1.4 Velocity at a turning point

How high does the rock in Example 2 go?

Solution 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.

The rock’s average velocity in a time interval [t, t + h] is

h

tththt

h

tshts 22 1696)(16)(96)()(

h

tthththt 222 16962169696

h

hthh 2163296

= 96 – 32t – 16h for h ≠ 0.

The rock’s velocity after t seconds is 96 – 32t ft/sec. This is 0 when t = 3. The

maximum height of the rock during its flight is s(3) = 96(3) – 16(3)2 = 144 ft. ■

A Look

Ahead The concept of instantaneous velocity makes sense for every object in motion, not

just for those whose position on an axis is a polynomial function of time. The

method we have used to define and calculate instantaneous velocity is specific to

polynomial functions, but the reasoning can be generalized as follows.

• The average velocity of every object in rectilinear motion over a time

interval of duration h > 0 is h

tshts )()( , where s(t) is its position at

time t.

• For each value of t, the object’s instantaneous velocity at time t is close to

its average velocity over the interval [t, t + h] when h is small, so it is

reasonable to consider an instant to be a time interval of duration zero.



• For every t, there should be a unique number v(t) that extends the domain

of the expression h

tshts )()( in a reasonable way to include h = 0.

Calculus provides such an extension and defines the number v(t) to be the object’s

instantaneous velocity at time t.

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 each of the following.

a. the object’s average velocity over each of the time intervals [3, 4],

[3, 3.1], [3, 3.01], and [3, 3.001]

b. the object’s instantaneous velocity at t = 3

1. s(t) = 2t + 7 2. s(t) = 5 – 10t

3. s(t) = t2 – 6 4. s(t) = t

2 – 6t

In Exercises 5-12, s(t) is the position at time t ≥ 0 of an object moving on a

coordinate axis. Find each of the following.

a. the object’s average velocity in the time interval [2, 5]

b. the object’s velocity at time t = 2

c. the object’s speed at time t = 2

d. an expression for the object’s average velocity in the time interval

[t, t + h]

e. a polynomial function v(t) to describe the object’s velocity at time t

f. the times, if any, when the object’s velocity is zero

g. the time intervals, if any, during which the object’s velocity is

negative

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

7. s(t) = 2t2 + 3t 8. s(t) = 8t – t

2

9. s(t) = 10 10. s(t) = t3 – 12

11. s(t) = t3 – 12t 12. s(t) = t

3 – 12t

2

Focus on applying skills

13. 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?

14. 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.



In Exercises 15-18, 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

15. Earth, s(t) = 20t – 4.9t2 16. Moon, s(t) = 20t – 0.8t

2

17. Mars, s(t) = 20t – 1.86t2 18. Jupiter, s(t) = 20t – 11.44t

2

19. Refer to the rock in Exercise 13.

a. After how many seconds does the rock strike the ground?

b. With what velocity does the rock strike the ground?

20. A second rock is thrown downward from the roof in Exercise 13. 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?

21. 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)?

22. 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 concepts

Refer to the material in the section as needed to answer each of questions 23-28,

but write your answer in your own words. Address your answer to an imaginary

classmate.

23. What is meant by rectilinear motion?

24. What is the difference between average velocity and instantaneous

velocity?

25. Why can’t average and instantaneous velocities be calculated in the same

way?

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

27. What is the difference between velocity and speed?



28. Why do you think that the words velocity and speed are used

interchangeably in everyday conversation?

29. 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.)

30. In Example 1.4, h

hthh

h

tshts 2163296)()(

, and

hthts 163296),(* .

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 h

tshts )()( be used to

calculate an average velocity for the object in Example 1.4? 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 object in Example 1.4? Can it be used to calculate

an instantaneous velocity? Explain each of your answers.

31. Refer to Example 1.4.

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.)


The Tangent Line Problem

Section 2 The Tangent Line Problem

As you saw in Section 1, calculus

can be used to define and calculate

the velocity of an object in rectilinear

motion. In this section you will see

that it can also be used to define and (3, 18) •

calculate 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 in the window -5 ≤ x ≤ 5, Figure 2.1

-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. A more precise definition of a tangent line is based

on the definition of a secant line to a graph.

Definition Let P(x) be a polynomial function. The secant line to the graph of P(x) over an

interval [a, b] is the line through the points (a, P(a)) and (b, P(b)).

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 to

conclude 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 2.1. Figure 2.2

Example 2.1 The difference between secant lines and tangent lines

(a) What is the slope of the secant line on the graph of P(x) = 2x2 over [3, 5]?

(b) What is the slope of the tangent line to the graph of P(x) at x = 3?

Solution (a) The slope is

162

32

35

1850

35

)3()5(

PP.

(b) The slope of the tangent line at (3, 18) should be close to the slope of the

secant line joining (3, 18) to (3.1, 19.2), the highest point on the graph that

is visible in Figure 2.2. That slope is



2.121.0

22.1

31.3

1822.19

31.3

)3()1.3(

PP.

More generally, the slope of the secant line over an interval [3, 3 + h] is

h

h

h

PhP 18)3(2

3)3(

)3()3( 2

h

hh 18692 2

h

hh 2212

=12 + 2h for h ≠ 0.

Reasoning that the approximation is best when h is small leads us to

conclude that the slope of the tangent line at (3, 18) should be 12. ■

The reasoning in Example 2.1 leads to the following definition of a tangent line to

a polynomial graph. A more general definition of a tangent line to a curve will be

given later.

Definition Let P be a polynomial function, and let P*(x, h) be the polynomial that coincides

with

h

xPhxP )( 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 00 , xPx 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 2.2 Slopes and equations of tangent lines

Let P(x) = 96x – 16x2.

(a) What is the slope of the graph of P(x) at x = 2?

(b) What is the slope of the graph at x = 5?

(c) What are the equations of the tangent lines to the graph at x = 2 and x = 5?



Solution (a) The slope of the secant line over the interval [2, 2 + h] is

h

hh

h

PhP 22 )2(16)2(96)2(16)2(96)2()2(

h

hhh 128441696192 2

h

hh 21632

= 32 – 16h for h ≠ 0.

The slope of the graph at x = 2 is 32.

(b) The slope of the secant line over the interval [5, 5 + h] is

h

hh

h

PhP 22 )5(16)5(96)5(16)5(96)5()5(

h

hhh 8010251696480 2

h

hh 21664

= –64 – 16h for h ≠ 0.

The slope of the graph at x = 5 is -64.

(c) At x = 2, the point on the graph is )128,2()2(,2 P , and the slope of the

graph is 32. The tangent line has the equation y = 128 + 32(x – 2).

At x = 5, the point on the graph is )80,5()5(,5 P , and the slope of the

graph is -64. The tangent line has the equation y = 80 – 64(x – 5). ■

In Example 2.2, as in Example 1.2, a more efficient strategy would have been to

calculate the slope of the curve as a function of the x-coordinate on the curve, and

then evaluate that function at x = 2 and x = 5. This strategy is illustrated in

Example 2.3.

Example 2.3 Slope of a graph as a function

What is the slope of the graph of P(x) = x3 – 6x

2 + 9x at x = 0, 2, and 4?



Solution The slope of the secant line over an interval [x, x + h] is

h

xxxhxhxhx

h

xPhxPhxP

96)(9)(6)()()(),(*

2323

h

xxxhxhxhxhxhhxx 96)(92633 23223223

h

hhxhhxhhx 961233 2322

h

hxhxhxh 961233 22

= 3x2 + 3xh + h

2 – 12x – 6h + 9 for h ≠ 0.

The slope of the graph at a point (x, P(x)) is P*(x, 0) = 3x2 – 12x + 9. Substituting

the values 0, 2, and 4 for x gives the respective slopes of 9, -3, and 9. ■

Example 2.4 Slope at a turning point

Where is the vertex on the graph of P(x) = 96x – 16x2?

Solution Recall that the vertex on the graph of y = ax2 + bx + c occurs at x =

a

b

2 . In this

case, the x-coordinate is 3)16(2

96 , and the y-coordinate is P(3) = 144. Let’s

see if we reach the same conclusion by considering the slope of the graph.

At the vertex the graph appears to have a horizontal tangent line, so the slope of

the graph should be 0 there. The slope of the secant line over an interval [x, x + h]

is

h

xxhxhx

h

xPhxPhxP

22 1696)(16)(96)()(),(*

h

xxhxhxhx 222 16962169696

h

hxhh 2163296

= 96 – 32x – 16h for h ≠ 0.



As a function of x, the slope of the tangent line is P*(x, 0) = 96 – 32x. The slope

is 0 when x = 3. This agrees with our previous conclusion that the vertex on the

graph is at (3, P(3)) = (3, 144). ■

A Look

Ahead If P(x) is a polynomial function, then for every x and for small values of h, the

values of the polynomial P*(x, h) tend toward the unique number P*(x, 0). This

observation makes it possible to define a tangent line to the graph of P(x) in a

reasonable way and guarantees that the graph of every polynomial function has a

tangent line at every point. For a non-polynomial function f that is defined in an

interval around a number x, the expression h

xfhxf )()( that defines the slope

of a secant line may or may not tend toward a unique number for small values of

h. That is, the graph of f(x) may or may not have a tangent line at each point.

Calculus provides a tool for deciding whether the graph of a function has a

tangent line at a given point, and for calculating the slope of the tangent line when

one exists.

Active Learning Focus on developing skills

For the functions P(x) in Exercises 1-4, find each of the following.

a. the slope of the secant line to the graph of P(x) over the intervals

[3, 4], [3, 3.1], [3, 3.01], and [3, 3.001]

b. the slope of the tangent line to the graph of P(x) at x = 3.

1. P(x) = 5x – 4 2. P(x) = x2 + 1

3. P(x) = x2 – 5x + 6 4. P(x) = x

3

For the functions P(x) in Exercises 5-12, find each of the following. (The

functions are the same as those in Exercises 5-12 of Section 1.)

a. the slope of the secant line to the graph of P(x) over the interval

[2, 5]

b. the slope of the tangent line to the graph of P(x) at x = 2

c. the equation of the tangent line to the graph of P(x) at x = 2

d. the expression P*(x, h) for the slope of the secant line to the graph

of P(x) over the interval [x, x + h]

e. the expression P*(x, 0) for the slope of the tangent line to the graph

of P(x) as a function of x

f. the values of x, if any, where the tangent line to the graph of P(x) is

horizontal

g. the values of x, if any, for which the slope of the tangent line is

negative

5. P(x) = x2 – 2x + 5 6. P(x) = 4x+ 1

7. P(x) = 2x2 + 3x 8. P(x) = 8x – x

2

9. P(x) = 10 10. P(x) = x3 – 12



11. P(x) = x3 – 12x 12. P(x) = x

3 – 12x

2

In Exercises 13 and 14, 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.

13. a. P(x) = 4x2 b. P(x) = 4x

2 – 3

c. P(x) = 4x2 + 7.3 d. P(x) = 4x

2 + c (c is any constant)

14. a. P(x) = x3 b. P(x) = x

3 – 3

c. P(x) = x3 + 2 d. P(x) = x

3 + c (c is any constant)


Refer to the material in the section as needed to answer each of questions 15 and

16, but write your answer in your own words. Address your answer to an

imaginary classmate.

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

16. Why can’t slopes of secant lines and slopes of tangent lines be calculated

in the same way?

17. 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 5-12.)

Answer each of Questions 18-20 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.

18. Is it possible for the tangent line to intersect the graph of P(x) at some

point (b, P(b)), b ≠ a?

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

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

21. 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 stops

b. slope of a curve (2) average velocity

c. turning point (3) instantaneous velocity

d. point with a horizontal (4) instant when the object

tangent changes direction

22. 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 13 and 14.) Write a few sentences to say why

this is not surprising.


Derivatives of Polynomial Functions

Section 3 Derivatives of Polynomial Functions

Before continuing, look back at Examples 1.1 and 2.1. Notice that except for the

differences in the letters that represent the variables and the functions, the two

solutions are identical. The same parallels exist between the other examples in

Section 1 and the corresponding examples in Section 2. That is, the function that

defines the slope of a graph is the same as the function that defines the velocity of

an object in rectilinear motion. This observation suggests that it will be useful to

study that function independently of any context. Let’s begin by giving the

function a name.

Definition Let P be a polynomial function, and let P*(x, h) be the polynomial that coincides

with h

xPhxP )()( for h ≠ 0. The derivative of P(x) with respect to x is the

polynomial 0,* xP . The process of calculating a derivative is referred to as

differentiation.

The notation P*(x, 0) is not in common use. The derivative of a polynomial

function P(x) with respect to x is usually written as )(xP (“P prime of x”), dx

dP

(“dee P by dee x”), or )(xPdx

d (“dee by dee x of P(x)”.) Example 2.4 showed

that if P(x) = 96x – 16x2, then xxP 3296)( . This result can also be

expressed by writing xdx

dP3296 or xxx

dx

d32961696 2 .

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 variable

Let y = 2r2s – 3rs

2.

(a) Treating s as a constant, find dr

dy, the derivative of y with respect to r.

(b) Treating r as a constant, find ds

dy, the derivative of y with respect to s.

Solution (a)

h

rssrshrshrhry

2222 32)(3)(2),(*

h

rssrshrshrhr 22222 32)(322



h

rssrhsrsshrhssr 222222 3233242

h

hsshrhs 22 324

=4rs + 2hs – 3s2 for h ≠ 0,

so 234 srsdr

dy .

(b)

h

rssrhsrhsrhsy

2222 32)(3)(2),(*

h

rssrhshsrhsr 22222 3223)(2

h

rssrrhrshrshrsr 222222 3236322

h

rhrshhr 22 362

= 2r2 – 6rs – 3rh for h ≠ 0,

so rsrds

dy62 2 . Note that this is not equal to

dr

dy. ■

Example 3.2 The derivative as velocity and as slope

(a) If P(x) = 2x2, find )3(P and provide two possible interpretations of the

result.

(b) If P(x) = x3 – 6x

2 + 9x, find )(xP and provide two possible interpretations

of the result.

(c) If P(x) = 96x – 16x2, find a value of x for which 0)( xP , provide two

possible interpretations of the result, and say why this value of x is

significant in each case.

Solution (a) The solutions of Examples 1.1b and 2.1b show that 12)3( P . This says

that the graph of P(x) = 2x2 has a slope of 12 when x = 3. It also says that

if the position of an object on a coordinate axis at time x seconds is 2x2,

then the velocity of the object at x = 3 seconds is 12 units/sec.

(b) The solutions of Examples 1.3 and 2.3 show that 9123)( 2 xxxP .

This says that the graph of P(x) has a slope of 3x2 – 12x + 9 at each value



of x. It also says that if the position of an object on a coordinate axis at

time x seconds is x3 – 6x

2 + 9x, then the velocity of the object at time x

seconds is 3x2 – 12x + 9.

(c) The solutions of Examples 1.4 and 2.4 show that 0)( xP when x = 3.

This says that the graph of y = 96x – 16x2 has a slope of 0 when x = 3.

This value of x is significant because it is the location of a turning point on

the graph, specifically the vertex. The result also says that if the position

of an object on a coordinate axis at time x seconds is 96x – 16x2, then its

velocity is 0 at time x = 3 seconds. This value of x is significant because it

is a time at which the object changes direction. ■

A more general interpretation of a derivative is as a rate of change in a function.

Definition Let P be a polynomial function. The average rate of change in P(x) with

respect to x over an interval [x0, x0 + h] is h

xPhxP )()( 00 . The instantaneous

rate of change in P(x) with respect to x at x = x0 is )( 0xP .

By viewing derivatives as rates of change, we can discover other ways of

interpreting them physically. For example, consider an object in rectilinear

motion. The rate at which its velocity changes with respect to time is called its

acceleration. That is, the object’s acceleration is the derivative of its velocity,

which is the derivative of its position. This observation motivates the following

definition.

Definition The second derivative of a polynomial function P(x) with respect to x is the

derivative of )(xP , written )(xP or 2

2

dx

Pd. The third derivative of P(x) with

respect to x is the derivative of )(xP , and derivatives of higher order are defined

similarly.

In the “prime” notation, the kth

derivative of P(x) is written as )()( xP k for k > 3.

For example, the fourth derivative of P(x) is usually written as P(4)

(x), rather than

)(xP .

Definition If P(t) is the position of an object on a coordinate axis at time t, the acceleration

of the object at time t is )(tP .

Example 3.3 Calculation of acceleration

What is the acceleration of the rock in Example 1.2 as a function of t?

Solution The rock’s height after t seconds is P(t) = 96t – 16t2. The solution of Example 1.4

shows that its velocity after t seconds is ttP 3296)( . To calculate the

acceleration )(tP , begin by calculating



h

tht

h

tPhtP 3296)(3296)()(

h

tht 3296323296

h

h32

= –32 for h ≠ 0.

The rock’s acceleration is 32)( tP ft/sec2. This constant acceleration is the

acceleration due to gravity. ■

Active

Learning Focus on developing skills

In Exercises 1-8, find )2( and),2( , )( PPxP .

1. P(x) = -5 2. P(x) = 3x – 1

3 P(x) = 9 – x2 4. P(x) = x

2 – 9

5. 532

)(2

xx

xP 6. 123)( 2 xxxP

7. P(x) = x3 – 12x 8. P(x) = 2x

3 + 3x

2

In Exercises 9-12, find dx

dP.

9. P(x) = mx + b 10. P(x) = ax2 + bx + c

11. P(x) = Ax3 12. P(x) = Ax

4

In Exercises 13-16, find )( and )( xPxP .

13. P(x) = 2x – 97 14. P(x) = 10x2

15. P(x) = 3 + 2x – x2 16. P(x) = 4x

3 + x

In Exercises 17 and 18, find the derivatives of all orders for P(x).

17. P(x) = x3 – x

2 + x – 1 18. 1

2

1

6

1)( 23 xxxxP


In Exercises 19-22, an object moving on a coordinate axis has position s(t) at time

t. Find its velocity and its acceleration as functions of t.



19. s(t) = 3 – 2t 20. s(t) = 15 – 3t2

21. 01.03

2)( 2 ttts 22. s(t) = 8 – t

3

23. On a certain airless planet, a dropped object falls 2.5t2 meters in t seconds.

What is the acceleration due to gravity on this planet?

24. On another airless 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?

25. 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?

26. 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)?





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

28. When calculating a derivative, why is it important to specify the

independent variable?

29a-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.

30. 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?

31. a. Find )(xP if P(x) = x2 – 6x + 5. For what values of x is

0)( xP ?



b. Graph P(x). How does the graph support your answer to part (a)?

c. Graph )(xP . How does the graph support your answer to part (a)?

32. An object in rectilinear motion has position s(t) = t3 – 6t

2 + 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).


Derivative Formulas

Section 4 Derivative Formulas

The previous sections have had a twofold purpose, to develop the concept of the

derivative of a polynomial function and to illustrate the usefulness of derivatives

in several contexts, both mathematical and physical. 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. In this section you will

discover that more efficient methods do exist, and you will learn to apply them.

Theorems 4.1, 4.2, and 4.3 establish a set of formulas that can be used to

differentiate every polynomial function.

Theorem 4.1 (Power Rule, special cases)

(a) For every constant c, 0cdx

d.

(b) 1xdx

d.

Theorem 4.2 (Power Rule, general case) For every integer n ≥ 1, 1 nn nxxdx

d.

Example 4.1 Using the Power Rule

Find: (a) 5dx

d (b) x

dx

d (c) 17x

dx

d

Solution (a) By Theorem 4.1a, 05 dx

d.

(b) By Theorem 4.1b, 1xdx

d.

(c) By Theorem 4.2, 1617 17xxdx

d . ■

Theorem 4.3 Let P and Q be polynomial functions, and let c be a constant. Then:

(a) (Constant Multiple Rule) )()( xPcxcPdx

d

(b) (Sum and Difference Rules) )()()()( xQxPxQxPdx

d

(c) (Product Rule) )()()()()()( xPxQxQxPxQxPdx

d

(d) (Chain Rule) )()()( xQxQPxQPdx

d


Derivative Formulas

Example 4.2 Using the Constant Multiple Rule

Find: (a) 75xdx

d (b)

4

8x

dx

d (c) 33x

dx

d.

Solution (a) 667 35755 xxxdx

d

(b) 778 284

1

4

1xxx

dx

d

(c) 223 33333 xxxdx

d ■

Example 4.3 Using the Sum and Difference Rules

Find the derivatives of all orders of P(x) = 2x3 – 9x

2 + 4.

Solution xxxxxP 186)2(932)( 22 ,

1812)1(18)2(6)( xxxP ,

12)1(12)( xP ,

0)()4( xP ,

and P(n)

(x) = 0 for n ≥ 4. ■

Example 4.4 Using the Product Rule

Find )67( 23 xxxdx

d.

Solution Let P(x) = x3 and Q(x) = x

2 – 7x + 6. Then by the Product Rule,

)()()()()()( xPxQxQxPxQxPdx

d

223 36772 xxxxx

= 2x4 – 7x

3 + 3x

4 – 21x

3 + 18x

2

= 5x4 – 28x

3 + 18x

2.

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

34523 6767( xxxdx

dxxx

dx

d

= 5x4 – 28x

3 + 18x

2. ■


Derivative Formulas

When the Chain Rule is used to differentiate a function of the form P(Q(x)), it is

useful to let u = Q(x) and y = P(Q(x)) = P(u). The Chain Rule can then be written

as dx

du

du

dy

dx

dy . The following example illustrates the use of this idea.

Example 4.5 Using the Chain Rule

Find 43 4 xxdx

d.

Solution Let u = x3 + x – 4 and y = u

4. Then 43 4 xxy , and

dx

du

du

dy

dx

dy

134 23 xu

1344 233 xxx . ■

A careful look at Example 4.5 suggests the following formula for differentiating a

power of a polynomial function. The formula is a special case of the Chain Rule.

Theorem 4.4 (Chain Rule, special case) If Q(x) is a polynomial function, then for every positive

integer n, )()()(1

xQxQnxQdx

d nn

.

Example 4.6 Using the special case of the Chain Rule

Find dx

dP if P(x) = (x

4 – 2x)

7.

Solution Theorem 4.4 implies that

xxdx

dxxxx

dx

d2272 46474

2427 364 xxx . ■

Example 4.7 Using several rules in succession

Find 53)72(5 xx

dx

d.


Derivative Formulas

Solution Let P(x) = (x – 5)3 and Q(x) = (2x + 7)

5. Then by the Product Rule,

)()()()()()( xPxQxQxPxQxPdx

d .

The calculations of )(xP and )(xQ require the Chain Rule. Theorem 4.4 can be

applied to obtain

22 )5(3)1()5(3)( xxxP and

44 )72(10)2()72(5)( xxxQ .

Therefore

254353 )5(3)72()72(10)5()72()5( xxxxxxdx

d

)72(3)5(10)72()5( 42 xxxx

= (x – 5)2(2x + 7)

4(16x – 29). ■

Active


In Exercises 1-16, find )(xP .

1. P(x) = x7 2. P(x) = 10x

7

3. 4

3)( xP 4. P(x) = 0.3x

5. P(x) = 2x4 – 3x

2 + 1 6. P(x) = 4x – x

3

7. 345

)(345 xxx

xP 8. 1052

2)(

2

xx

xP

9. P(x) = (x – 5)(x + 5) 10. P(x) = x3(2x

2 – 7x + 6)

11. P(x) = (x + 2)(x2 – 2x + 4) 12. P(x) = (x

3 + 3x – 6)(x

4 – 2x

2 – 8)

13. P(x) = x(x + 1)(x + 2) 14. P(x) = x2(x + 4)(x

2 – 4x)

15. 2

234 23)(

x

xxxxP

, x ≠ 0

16. 1

1)(

2

x

xxP , x ≠ 1

In Exercises 17 and 18, find the derivatives of all orders for P(x).

17. P(x) = x6

18. P(x) = x6 + 6x

5 + 30x

4 + 120x

3 + 360x

2 + 720x + 720


Derivative Formulas

In Exercises 19-22, use the Chain Rule to find dx

dP.

19. P(u) = 3u – 1, u = Q(x) = 2x + 10

20. P(u) = u2 + 4u, u = Q(x) = x

2 – 9

21. P(u) = u3 – 6u, u = Q(x) = x

2 + x

22. P(u) = u2 + u, u = Q(x) = x

3 – 6x

In Exercises 23-30, find )(xP .

23. P(x) = (2x – 5)3 24. P(x) = (x

3 + 2x)

2

25. P(x) = –4(3 – 7x)5 26. P(x) = 10(x

3 – 5x

2 + 7x – 1)

4

27. 3242 1432

1)( xxxP 28. P(x) = x(x + 2)

3(x – 6)

2

29. P(x) = (x2 + x – 2)

3 – 3(x

2 + x – 2)

2 + 3(x

2 + x – 2) – 1

30. 4239234)( xxxP

31. Let P(x) = (x2 – 3x + 8)(x

2 + 3x – 8). Find )(xP 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.

32. Let 22 3)( xxP . Find )(xP in three different ways.

a. Use the Chain Rule.

b. Write P(x) as (x2 + 3)(x

2 + 3) and use the Product Rule.

c. Write P(x) as x4 + 6x

2 + 9 and use the Power Rule.

Verify that each calculation leads to the same result.


In Exercises 33-36, 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 time

b. the object’s acceleration as a function of time

c. the times, if any, at which the object’s velocity is zero

33. s(t) = t3 + 4t 34. s(t) = t

4 – 4t

3

35. s(t) = t2(t

2 – 4) 36. s(t) = (2x + 10)

3(3x – 6)

4

In Exercises 37 and 38, find the values of a for which the graph of P(x) has:

a. no horizontal tangents

b. exactly one horizontal tangent

c. two horizontal tangents

37. P(x) = x3 + ax 38. P(x) = x

3 + ax

2


Derivative Formulas

39. Let 4533

1)( 23 xxxxP . What is the minimum slope on the graph

of P(x)?


40. What is the maximum number of horizontal tangent lines on the graph of a

polynomial function of degree n? Explain your answer.

41. 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.

42. Explain why Theorem 4.1 can be viewed simply as a restatement in new

language of some facts you learned in your high school algebra course.

43. Refer to the four displayed lines in the proof of Theorem 4.3c (the Product

Rule) in Appendix A. Explain why the expression in each of the first

three displayed lines is equivalent to the expression on the following line.

44. The proof of Theorem 4.3d (the Chain Rule) in Appendix A 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?

45. For nearly all polynomial functions P(x) and Q(x), the derivative of the

product P(x)Q(x) is not the product )()( xQxP . Which polynomial

products are the exceptions to this rule?


Interpretations of the Derivative

Section 5 Interpretations of the Derivative

The examples in this section illustrate several mathematical and physical

interpretations of derivatives.

Example 5.1 The derivative as velocity and as acceleration

A 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 ttP 6.1)( m/sec, and its downward

acceleration is 6.1)( tP m/sec2. This is the acceleration due to gravity on the

moon. ■

Example 5.2 The derivative as velocity and as acceleration

A 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, ttP 1080)( , and its

acceleration, 10)( tP .

(a) The car’s initial velocity is 80)0( P ft/sec.

(b) Because the acceleration is 10)( tP , 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. ■

Example 5.3 The derivative as slope

Let 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?

Solution The slope of the graph is kxxP 23)( . The slope is negative exactly when

33303 22 k

xkk

xkx .

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

3. ■




The graphs of P(x) = x2 and Q(x) = a + bx – x

2 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 xxP 2)( . The slope of the

tangent line to the graph of Q(x) is xbxQ 2)( . 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. ■


For 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 3

00 4xxP and

contains the point 1,,4

0000 xxxPx . The equation of the tangent line is

0

3

0

4

0 41 xxxxy .

The tangent line contains the origin if and only if

0

3

0

4

0 0410 xxx

134

0 x

40

3

1x . ■

Example 5.6 The derivative as instantaneous rate of change

During a flood, the height of a river above its normal level is approximately

P(t) = t4 – 8t

3 + 16t

2 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 ttttP 32244)( 23 feet per day. At

the end of the first day, the river is rising at a rate of 12)1( P feet per day. ■


Suppose that a manufacturer can produce x television sets per week at a cost of

C(x) = 50,000 + 300x + 0.01x2 dollars.

a. Find the weekly cost of increasing production from 100 to 101 sets per

week.



b. Find )100(C .

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 xxC 02.0300)( , so 302)100( C .

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.


The manufacturer in Example 5.7 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 xxR 4.0500)( , so 460)100( R .

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

xxxxCxRxP 42.0200)02.0300()4.0500()()()( ,

so 158)100( P .

d. The profit derived from the sale of the 101st set is the difference between

the revenue derived from the sale (Example 5.8b) and the cost of

production (Example 5.7a), which is $459.80 - $302.01 = $157.79. This

differs from the result in part (c) by only $0.21. ■

Active

Learning 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) = 7.5t

2

3. 12

11)(

3tts 4. s(t) = 2.5t

3

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 stop

b. 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

slope

b. 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) = x

2 + 2x

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

3

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 – 0.1x

2 + 0.001x

3

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?


21. The volume of a cylinder of radius r and height h is V = r2h.

a. Find dr

dV, treating h as a constant, and evaluate

dr

dV when r = 5

and h = 4. What does your result say about a cylinder whose shape

is changing?

b. Find dh

dV, treating r as a constant, and evaluate

dh

dV 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.


Antiderivatives of Polynomial Functions

Section 6 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.

Definition If )()( xPxQ , then Q(x) is an antiderivative of P(x).

Theorems 6.1-6.3 establish some important facts about antiderivatives.

Theorem 6.1 If 0)( xP for all x, then there is a constant C such that P(x) = C for all x.

Theorem 6.2 If )()( xQxP for all x, then there is a constant C such that P(x) = Q(x) + C for

all x.

Theorem 6.3 (a) (Power Rule)

If nxxP )( , then Cxn

xP n

1

1

1)( for some constant C.

(b) (Constant Multiple Rule)

If )()( xQkxP , then P(x) = kQ(x) + C for some constant C.

(c) (Sum and Difference Rules)

If )()()( xRxQxP , then P(x) = Q(x) ± R(x) + C for some constant C.

An equation that specifies a formula for )(xP 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 6.1 illustrates the

solution of three first-order differential equations.

Example 6.1 Finding antiderivatives

(a) Find all possible functions P(x) if 7106)( 2 xxxP .

(b) Find all possible functions P(r) if 2

1)(

4

rrP .

(c) Find all possible functions P(t) if 42)( 2 tttP .

Solution (a) Think of )(xP as 6x2 – 10x

1 + 7x

0. Then

CxxxCxxx

xP

7527

210

36)( 23

23

.



(b) First write 2

1

2

1 as )( 4 rrP . Then

CrrCrr

rP

2

1

10

1

2

1

52

1)( 5

5

.

(c) First write 842 as )( 23 ttttP . Then

CttttCtttt

tP

82

3

2

4

18

24

32

4)( 234

234

. ■

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 6.4 guarantees that the solution

of every initial value problem of the type you will encounter here consists of

exactly one function.

Theorem 6.4 Let Q(x) be a polynomial function. Every initial value problem of the form

1

)1(

210

)( )(,,)(,)(,)(),()(

n

nn yayyayyayyayxQxy

has a unique solution.

Example 6.2 Finding functions from slopes

The 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 14)( xxP and P(3) = 11. First find the

solution of the differential equation.

CxxCxx

xP

2

2

22

4)(

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 6.3 and 6.4.



Example 6.3 Finding position from acceleration

A 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 32)( ts , and the initial velocity and height

provide initial conditions 80)0( s and s(0) = 100. Then )(ts is an

antiderivative of )(ts , so Ctts 32)( . The initial condition on )(ts implies

that 80)0(32 C , so C = -80. Thus the velocity of the rock after t seconds

is 8032)( tts ft/sec.

Continue by noting that s(t) is an antiderivative of )(ts , 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 6.4 Finding position from acceleration

A 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 11)( ts , and the car

accelerates from a stop at time t = 0, giving initial conditions 0)0(,0)0( ss .

The car’s velocity after t seconds is Ctts 11)( ft/sec, and the condition

0)0( s implies that C = 0, so the velocity is tts 11)( 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 ttg 02.16)( . 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.

Active


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

1. 2)( xP 2. xxP 45)(

3. 47

3

5

3

8)( xxxP 4.

20

7

4

3

2)(

49

xx

xP

5. 43)( 23 xxxP 6. 653)( 4 xxxxP

7. 5432)( 2 xxxxP 8. 23 8)( xxP

In Exercises 9-16, solve the initial value problem.

9. xxP 26)( , P(3) = 8

10. 32 26)( xxxP , P(2) = 0

11. 2)1(,24)( 3 PxxP

12. 6

1)1(,1)( 2 PxxxP

13. 4)0(,4)0(,12)( 2 PPxxP

14. 0)1(,3)1(,8512)( 23 PPxxxxP

15. 2)1(,1)1(,0)1(),1(8)( PPPxxP

16. 0)0(,0)0(,0)0(,0)0(,24)()4( PPPPxP


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) = 80

19. v(t) = 144 – 32t, s(0) = 80 20. v(t) = 20 – 9.8t, s(0) = 10

21. a(t) = –32, v(1) = 40, s(1) = 100

22. a(t) = –9.8, v(3) = 0.6, s(3) = 0

23. a(t) = 6t, v(4) = 60, s(4) = 156

24. 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?

27. On a certain airless 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.


31. Use the connection between derivatives and slopes to explain why

Theorem 6.1 must be true.

32. Theorem 6.3 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 )()( xQxP , then P(x) – Q(x) is a linear function.

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

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 0,2)( yayxxy has a unique solution.

36. Two students are asked to find the family of all antiderivatives of 342)( xxxP .

• Anna says, “One antiderivative of P(x) is x2 + x

4, so the family of

all antiderivatives is x2 + x

4 + C.”

• Albert says, “The family of antiderivatives of 2x is x2 + C, and the

family of antiderivatives of 4x3 is x

4 + C, so the family of

antiderivatives of 2x + 4x3 is (x

2 + C) + (x

4 + C) = x

2 + x

4 + 2C.”

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


The Area Problem, Sigma Notation, and Summation Formulas

Section 7 The Area Problem, Sigma Notation, and Summation Formulas

This section begins your study of

integral calculus and the solution of

the area problem, as described in

Section 0. To gain a better

understanding of what the area

problem is, consider Figure 7.1. The

figure shows the graph of P(x) = x2

over the interval [0, 1]. The problem

of finding the area of the region R

under the graph is an instance of the

area problem. Figure 7.1

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 the region R, we must ask how to define the

area of such a region in a meaningful way. One strategy is to compare the portion

of the plane covered by R to that covered by regions whose area is known. Figure

7.2a shows a region S, made up of rectangles, that covers R. Figure 7.2b 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.

Figure 7.2a Figure 7.2b

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 both less than the area of any region that covers R and greater than the area

of any region that is covered by R. We can then define that number to be the area

of R.

To implement this strategy, we need to deal with three obstacles.



• A close approximation of the shape of a region requires a large number of

rectangles. It will be helpful to create specialized notation to describe

unwieldy sums of large numbers of areas and develop formulas to

calculate them.

• We need to ensure that the process of approximating the shape of a region

more closely will lead to a uniquely defined area.

• After we define the area of a region bounded by polynomial graphs, we

need to find a practical way of calculating it.

We will overcome those obstacles one at a time over the next few sections. In

this section you will learn the notation and the formulas that are needed to address

the first obstacle.

Definition Let f(k) be a function (not necessarily a polynomial) that is defined for the

integers k = a, a + 1, a + 2, …, b. The sum f(a) + f(a + 1) + f(a + 2) + … + f(b) is

written in sigma notation as

b

ak

kf )( .

Example 7.1 Using sigma notation

(a) Evaluate

6

3

2 4k

k . (b) Evaluate

6

1

7k

.

Solution (a) Evaluate the function f(k) = k2 – 4 for k = 3, 4, 5, and 6, and add to obtain

46454443 2222 = 5 + 12 + 21 + 32 = 70.

(b) Evaluate the function f(k) = 7 for k = 1, 2, 3, 4, 5, and 6, and add to obtain

f(1) + f(2) + f(3) + f(4) + f(5) + f(6) = 7 + 7 + 7 + 7 + 7 + 7 = 42. ■

With regard to priority of operations, the sigma operation is performed after

multiplication and division, but before addition and subtraction. For example,

20

1

5k

k is the same as

20

1

)5(k

k , and

3

0

3 4j

j means

3

0

3

j

j + 4, not

3

0

3 4j

j .

A sum can be represented in sigma notation in more than one way, as Example

7.2 illustrates.

Example 7.2 Representation of a sum in sigma notation is not unique

Show that

3

0

14k

k and

4

1

34j

j represent the same sum.



Solution The sums are

2813951)134()124()114()104(143

0

k

k

and

2813951)344()334()324()314(344

1

j

j . ■

In order to use the sigma operator in solving the area problem, you will need to

know two additional algebraic properties of the operator. These are established

by Theorem 7.1.

Theorem 7.1 (a) (Constant Multiple Rule)

n

mk

n

mk

kfckcf )()(

(b) (Sum and Difference Rules)

n

mk

n

mk

n

mk

kgkfkgkf )()()()(

Example 7.3 Using the rules for sigma notation

Given that 38510

1

2 k

k and 5510

1

k

k , find

10

1

2 83k

kk .

Solution 715)55(8)385(383838310

1

10

1

210

1

10

1

210

1

2

kkkkk

kkkkkk . ■

The final tool you will need to explore the area problem is a set of formulas for

sums of powers of integers, given in Theorem 7.2.

Theorem 7.2 (a) mm

k

1

1 (b) 2

)1(

1

mmk

m

k

(c) 6

)12)(1(

1

2

mmmk

m

k

(d)

222

1

3

2

)1(

4

1

mmmmk

m

k

Example 7.4 Using the summation formulas

Evaluate each sum.

(a) 20

1

3k (b)

100

1

92k

k

Solution (a) By Theorem 7.2d, 100,442102

2120 2

220

1

3

k

k .



(b) By Theorem 7.1b,

100

1

100

1

100

1

9292kkk

kk .

By Theorem 7.1a, this is

100

1

100

1

192kk

k .

By Theorems 7.2a and 7.2b, this is 9200)100(92

1011002

. ■

Active


Evaluate the sums in Exercises 1-6.

1.

7

4

32k

k 2.

2

2

4

k

k

3.

4

1

12

k k 4.

2

0

124k

k

5.

5

1 1

11

k kk 6.

9

1

1k

kk

In Exercises 7-10, verify that both expressions represent the same sum.

7.

6

3

25j

j and

7

4

35k

k 8.

6

2

2

j

j and

4

0

22

k

k

9.

6

1

2 1j

j and

7

2

2 2k

kk 10.

5

32

1

j jj and

6

42

1

k kk

Suppose that 25)(10

1

k

kf and 32)(10

1

k

kg . Use the Constant Multiple Rule

and the Sum and Difference Rules to evaluate each sum in Exercises 11-14.

11.

10

1

)(3k

kf 12.

10

1

)()(k

kgkf

13.

10

1

)()(2k

kgkf 14.

10

1

)(25)(32k

kgkf

Use the formulas in Theorem 7.2 to evaluate each sum in Exercises 15-20.



15.

200

1k

k 16.

10

1

3

k

k

17.

100

1

2 52k

k 18.

30

1

2

30

1

3

2

2

1

k

kk

19.

100

51k

k 20.

20

11

3

k

k


21. On a certain airless planet, a dropped object falls s(t) = t2 feet in t seconds.

a. Write an expression for the distance that the object drops during

the kth

second.

b. Use sigma notation to write an expression for the sum of the

distances that the object drops during the first 100 seconds.

c. Use the formulas in Theorem 7.2 to evaluate the sum, and verify

that it is equal to s(100).

22. It costs C(x) = 1000 + 50x + 0.2x2 dollars to produce x items of a certain

commodity.

a. Write an expression for the cost of producing the kth

item.

b. Use sigma notation to write an expression for the sum of the costs

of producing the 101st through 200

th items.

c. Use the formulas in Theorem 7.2 to evaluate the sum, and verify

that it is equal to C(200) – C(100).

23. Verify each identity.

a. 2

1 1

1 nn

k

n

j

b. 2

)1(1

1 1

nnn

k

k

j

c. 6

)2)(1(

1 1

nnnj

n

k

k

j

24. Verify that

n

k

n

j

k

j

jnjfjf1 11

)1()( .


Riemann Sums and the Definition of Area

Section 8 Riemann Sums and the Definition of Area

You are now prepared to address the second of the three obstacles listed in

Section 7:

• We need to ensure that the process of approximating the shape of a region

more closely will lead to a uniquely defined area.

It is reasonable to demand that 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.

Let’s begin to find that number for a specific region.

Example 8.1 Approximating the shape of a region

Let R be the region under the graph of P(x) = x2 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.


Figures 7.2a and 7.2b are reproduced here as Figures 8.1a and 8.1b.



(a) Region S is shown in Figure 8.1a. First find the areas of the five

rectangles that make up region S, as in the following table.

Rectangle Height Width Area

1 P(0.2) = 0.04 0.2 (0.04)(0.2) = 0.008

2 P(0.4) = 0.16 0.2 (0.16)(0.2) = 0.032

3 P(0.6) = 0.36 0.2 (0.36)(0.2) = 0.072

4 P(0.8) = 0.64 0.2 (0.64)(0.2) = 0.128

5 P(1.0) = 1.00 0.2 (1.00)(0.2) = 0.200

By Principle A2, the area of region S is the sum of the areas of the five

rectangles, which is 0.44.

(b) Region T is shown in Figure 8.1b. First find the areas of the five

rectangles that make up region T, as in the following table. (Note that the

leftmost “rectangle” has a height of 0.)

Rectangle Height Width Area

1 P(0.0) = 0.00 0.2 (0.00)(0.0) = 0.000

2 P(0.2) = 0.04 0.2 (0.04)(0.2) = 0.008

3 P(0.4) = 0.16 0.2 (0.16)(0.2) = 0.032

4 P(0.6) = 0.36 0.2 (0.36)(0.2) = 0.072

5 P(0.8) = 0.64 0.2 (0.64)(0.2) = 0.128

By Principle A2, the area of region T is the sum of the areas of the five

rectangles, which is 0.24.

(c) By Principle A1, the area of region R must lie between 0.24 and 0.44. ■

Our plan is to narrow the bounds established in Example 8.1c by using more

rectangles to approximate the shape of the region R more closely. It will be

convenient to introduce some terminology to describe the sums involved.

Definition A partition of a closed interval [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].

Definition Let P(x) be a polynomial function, let an interval [a, b] be partitioned into m

subintervals with widths (x)1, (x)2, …, (x)m. For 1 ≤ k ≤ m, let ck belong to the

kth

subinterval. The sum

mm

m

k

kk xcPxcPxcPxcPxcP

332211

1

is a Riemann sum for P(x) on [a, b]. A Riemann sum is called an upper sum if

P(ck) ≥ P(x) for all x in the kth

subinterval and a lower sum if P(ck) ≤ P(x) for all x

in the kth

subinterval.



Riemann sums can be constructed for any polynomial function over any finite

closed interval. For now, in order to define the concept of area under a

polynomial graph, let’s focus only on examples for which P(x) ≥ 0 on [a, b]. In

this case kcP and kx are, respectively, the height and width of a rectangle

constructed over the kth

subinterval of the partition, and the product kk xcP is

the area of that rectangle.

In Example 8.2 we will improve the bounds on the area of the region R of

Example 8.1. Example 8.2 also demonstrates the usefulness of summation

formulas in finding upper and lower sums for the area of a region.

Example 8.2 Improving the bounds for the area of a region

(a) Find an upper sum for the region R of Example 8.1 using 100 rectangles of

equal width.

(b) Find a lower sum for R using 100 rectangles of equal width.

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

R must lie.

Solution The interval [0, 1] is 1 unit wide. If it is partitioned into 100 subintervals of equal

width, then each subinterval has width 100

1 . Therefore if

100

1k

kk xcP is any

Riemann sum using 100 rectangle of equal width, then 100

1 kx for each value

of k.

(a) For the upper sum, kcP must be the largest value of P(x) on the kth

subinterval. Because P(x) is increasing on [0, 1], the largest value in each

subinterval occurs at its right endpoint. The x-coordinates at the right

endpoints of the subintervals are 100100

1003

1002

1001 ,,,, , so the right endpoint

of the kth

subinterval has x-coordinate 100

k . This means that the upper sum

is

100

12

2100

1 100

1

100100

1

100 kk

kkP

100

1

2

3100

1

k

k

6

201101100

100

13

(by Theorem 7.2c)



33835.01006

2011012

.

(b) For the lower sum, kcP must be the smallest value of P(x) on the kth

subinterval. Because P(x) is increasing on [0, 1], the smallest value in

each subinterval occurs at its left endpoint. The x-coordinates at the left

endpoints of the subintervals are 10099

1002

1001

1000 ,,,, , so the left endpoint

of the kth

subinterval has x-coordinate 100

1k . This means that the lower sum

is

100

1 100

1

100

1

k

kP

100

12

2

100

1

100

)1(

k

k

100

1

2

312

100

1

k

kk

100

13

100

13

100

1

2

31

100

1

100

2

100

1

kkk

kk

100100

1

2

101100

100

2

6

201101100

100

1333

(by Theorems 7.2a, 7.2b, and 7.2c)

32835.0100

1

100

101

1006

201101222

.

(c) The area of R must lie between 0.32835 and 0.33835. ■

Notice that when five rectangles are used (Example 8.1), the difference between

the upper and lower bounds for the area of R is 0.44 – 0.24 = 0.2. When 100

rectangles are used (Example 8.2), the difference between the upper and lower

bounds is only 0.33835 – 0.32835 = 0.01. The following two theorems guarantee

that we can continue to improve the bounds to within any desired degree of

closeness and define a unique area, not only for R, but for every region under the

graph of a polynomial function over a finite closed interval.

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 guarantees that for every polynomial function on every finite closed

interval, there is at least one number that is both a lower bound for all upper sums

and an upper bound for all lower sums. In order to show that there is only one

such number, it is enough to show that we can construct upper and lower sums

whose difference is arbitrarily small. Theorem 8.2 guarantees that it is always

possible to do so.

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.

Definition Let P 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 8.3 Finding the area under a polynomial graph

Let R be the region in Examples 8.1 and 8.2, let Sm be the partition of [0, 1] into m

subintervals of equal width, and let Um and Lm be the upper and lower sums for

Sm.

(a) Find an expression in m for Um.

(b) Find an expression in m for Lm.

(c) Use the results of parts (a) and (b) to find the area of R.

Solution Because the width of [0, 1] is 1, the width of each subinterval is m

x 1 .

(a) Because P is increasing on [0, 1], the largest value of P(x) in each

subinterval occurs at its right endpoint. The x-coordinates at the right

endpoints of the subintervals are mm

mmm,,,, 321 , so the right endpoint of

the kth

subinterval has x-coordinate mk . This means that the upper sum is

m

k

m

k mm

k

mm

kP

12

2

1

11

m

k

km 1

2

3

1

6

)12)(1(13

mmm

m

(by Theorem 7.2c)



2

2

6

132

m

mm .

(b) Because P(x) is increasing on [0, 1], the smallest value in each subinterval

occurs at its left endpoint. The x-coordinates at the left endpoints of the

subintervals are m

mmmm

1210 ,,,, , so the left endpoint of the kth

subinterval has x-coordinate m

k 1 . This means that the lower sum is

m

k mm

kP

1

11

m

k mm

k

12

2 1)1(

m

k

kkm 1

2

312

1

m

k

m

k

m

k mk

mk

m 13

13

1

2

31

121

mm

mm

m

mmm

m 33

1

2

)1(2

6

)12)(1(1

(by Theorems 7.2a, 7.2b, and 7.2c)

222

2 11

6

132

mm

m

m

mm

2

2

6

132

m

mm .

(c) The expressions for the lower and upper sums are, respectively, the

rational functions

2

2

2

2

6

132)( and

6

132)(

m

mmmU

m

mmmL

.

If A is the area of the region under the graph of P(x) = x2 over [0, 1], then

L(m) ≤ A ≤ U(m) for every m. The end behaviors of the rational functions

U(m) and L(m) is determined by the terms of largest degree in the

numerator and denominator. Therefore if m is large,



3

1

6

2)( and

3

1

6

2)(

2

2

2

2

m

mmU

m

mmL .

Therefore the area of R is 31 . ■

Active


In Exercises 1-4, let R be the region under the graph of P(x) over the given

interval.

a. Approximate the area of R with an upper sum using 5 rectangles of

equal width.

b. Approximate the area of R with a lower sum using 5 rectangles of

equal width.

c. Use your results from 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]


interval.

a. Approximate the area of R with an upper sum using 3 rectangles of

equal width.

b. Approximate the area of R with a lower sum using 3 rectangles of

equal width.

c. Use your results from parts (a) and (b) to find an interval in which

the area of R must lie.

d. Repeat parts (a) – (c) using 6 rectangles of equal width.

5. P(x) = x2, [0, 3] 6. P(x) = 9 – x

2, [-3, 3]

7. P(x) = x2 – 6x, [-6, 0] 8. P(x) = x

3, [1, 4]


interval. Find both an upper sum and a lower sum for the area of R using 200

rectangles of equal width, and find an interval in which the area of R must lie.

9. P(x) = 2x, [0, 2] 10. P(x) = x2 + 1, [0, 1]

11. P(x) = 2x – x2, [0, 1] 12. P(x) = 5 – x, [0, 5]

(Hint: P(x) is decreasing on [0, 5],

so the largest value of P(x) occurs at

the left endpoint of each subinterval.)

In Exercises 13 and 14, let R be the region under the graph of P(x) over the given

interval. Use the method of Example 8.3 to find the area of R.



13. P(x) = 3x2, [0, 2] 14. P(x) = x

2 + 4, [0, 1]


Refer to the material in the section as needed to answer questions 15-18, but write

your answer in your own words. Address your answer to an imaginary classmate.

15. 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?

16. Describe the process that will be used to define the area of a region under

the graph of a nonnegative polynomial function.

17. What does Theorem 8.1 say? Give an informal argument to explain why it

is true.

18. Prove Theorem 8.2 in the case where P(x) is decreasing on [a, b].

19. Explain why Theorem 8.2 guarantees that the area of every region under

the graph of a nonnegative polynomial function can be uniquely defined.

20. Let R be the region bounded by the graph of xy , the y-axis, and the

line y = 2.

a. Explain why the area of R is equal to the area of the region under

the graph of P(x) = x2 over [0, 2]. (Hint: Graph both regions.)

Then find the area of R.

b. Use your result from part (a) to find the area of the region under

the graph of xy over [0, 4].


Definite Integrals and their Properties

Section 9 Definite Integrals and their Properties

Because our focus in Sections 7 and 8 was on the area problem, we considered

only Riemann sums of a polynomial function P over an interval [a, b] on which

P(x) ≥ 0. However, Riemann sums can also be constructed if P(x) < 0 over all or

part of [a, b]. Examples 9.1 and 9.2 pose questions for which such Riemann sums

provide the answer.

Example 9.1 Approximating areas below the x-axis

(a) Find a lower sum for P(x) = 1 – x2 over the interval [1, 3] using four

subintervals of equal width.

(b) Find an upper sum for P(x) over [1, 3] using four subintervals of equal

width.

(c) Use the results of parts (a) and (b) to find upper and lower bounds for the

area of the region R between the graph of P(x) and the x-axis over [1, 3].

Solution (a) The width of the interval [1, 3] is 2, so the width of each subinterval is

5.0x . Because P is decreasing on [1, 3], the minimum value of P(x)

in each subinterval occurs at its right endpoint. The right endpoints of the

four subintervals are c1 = 1.5, c2 = 2, c3 = 2.5, and c4 = 3. The lower sum

is

4

1k

k xcP = P(1.5)(0.5) + P(2)(0.5) + P(2.5)(0.5) + P(3)(0.5)

= (-1.25)(0.5) + (-3)(0.5) + (-5.25)(0.5) + (-8)(0.5) = -8.75.

(b) The width of each subinterval is again 5.0x , and the maximum value

of P(x) in each subinterval occurs at its left endpoint. The left endpoints

of the four subintervals are c1 = 1, c2 = 1.5, c3 = 2, and c4 = 2.5. The upper

sum is

)5.0)(5.2()5.0)(2()5.0)(5.1()5.0)(1(4

1

PPPPxcPk

kk

= (0)(0.5) + (-1.25)(0.5) + (-3)(0.5) + (-5.25)(0.5) = -4.75.

(c) Figures 9.1a and 9.1b each show regions composed of four rectangles of

equal width that approximate the shape of the region R.




With the numbers ck defined as they were in part (a), the right edge of the

the kth

rectangle in Figure 9.1a has x-coordinate ck. Because

25.1)5.1( P , the height of that rectangle is –P(1.5) = 1.25, and its area

is –P(1.5)(0.5) = –(–1.25)(0.5) = 0.625. A similar statement applies to the

other three rectangles in the figure. Because the rectangles in Figure 9.1a

cover R, an upper sum for the area is

75.84

1

k

k xcP ,

so an upper bound for the area of R is 8.75. (Note that because P(x) ≤ 0 on

[1, 3], the upper bound for the area of R corresponds to the lower sum for

P(x).) By a similar argument, a lower bound for the area is 4.75. ■

Example 9.2 Approximating distance from velocity

A sky diver plans to jump from an airplane and fall for 10 seconds before opening

her parachute. The laws of physics dictate that, taking air resistance into account,

the velocity attained by a person of her weight at two-second intervals after

jumping will be as in the following table.

time (sec) 0 2 4 6 8 10

velocity (ft/sec) 0 -58 -105 -144 -176 -202

(a) Find an upper sum for the velocity function V(t) over the interval [0, 10].

(b) Find a lower sum for V(t) over [0, 10].

(c) Use the results of parts (a) and (b) to find lower and upper bounds for the

distance that the skydiver will fall in the first 10 seconds after jumping.

Solution (a) The table represents a partition of the interval [0, 10] into five subintervals

of length 2. The velocity function is decreasing on [0, 10], so the

maximum value in each subinterval occurs at its left endpoint. The upper

sum is

(0)(2) + (-58)(2) + (-105)(2) + (-144)(2) + (-176)(2) = -966.



(b) The minimum value of the velocity function in each subinterval occurs at

its right endpoint. The lower sum is

(0)(2) + (-58)(2) + (-105)(2) + (-144)(2) + (-176)(2) + (-202)(2) = -1370.

(c) During the first two seconds the skydiver’s downward speed is at least 0

ft/sec, so a lower bound for the distance she falls during those two seconds

is 0·2 = 0 ft. During the next two seconds her downward speed is at least

58 ft/sec, so a lower bound for the distance she falls during those two

seconds is 58·2 = 116 ft. In the same way, a lower bound for the distance

she falls in each two-second interval is the absolute value of the

corresponding term of the upper sum in part (a). Similarly, an upper

bound for the distance she falls in each two-second interval is the absolute

value of the corresponding term of the lower sum in part (b). The lower

and upper bounds for the distance she falls during the entire ten-second

interval are, respectively, 966 ft and 1370 ft. ■

Examples 9.1 and 9.2 illustrate that if P(x) < 0 on [a, b], then the Riemann sums

for P(x) on that interval are approximating a negative number. The absolute value

of that number can represent many things, such as an area (Example 9.1) or a

distance (Example 9.2). This observation motivates the following definition.

Definition Let P be a polynomial function. If a < b, the definite integral of P(x) from a to b

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]. (The existence and uniqueness of I is guaranteed by

Theorem 8.2.) It is customary to write b

adxxP )( to represent the definite integral

of P(x) from a to b. 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 b

adxxP )( is the area under the graph of P(x) over

],[ ba . For example, the result of Example 8.3 shows that 31

1

0

2 dxx .

Theorem 9.1 establishes some fundamental properties of definite integrals.

Theorem 9.1 Let P and Q be polynomial functions. If a < b, then:

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

a

b

adxxPkdxxkP )()( .

(b) (Sum and Difference Rules) b

a

b

a

b

adxxQdxxPdxxQxP )()()()(

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



b

c

c

a

b

adxxPdxxPdxxP )()()( .

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

a

b

adxxQdxxP )()( .

It is customary to define 0)( a

adxxP and

b

a

a

bdxxPdxxP )()( . The latter

definition in particular may appear to be arbitrary and capricious, but in Section

10 you will discover that it is both reasonable and practical.

Example 9.3 Using the properties of definite integrals

Suppose that 6)( and ,4)(,2)(5

0

10

5

5

0 dxxQdxxPdxxP . Find:

a. 5

0)(3)(7 dxxQxP b.

0

5)( dxxQ c.

10

0)( dxxP

Solution a. 5

0

5

0

5

0)(3)(7)(3)(7 dxxQdxxPdxxQxP

(by Theorem 9.1b)

5

0

5

0)(3)(7 dxxQdxxP

(by Theorem 9.1a)

= 7·2 – 3·6 = –4.

b. 6)()(5

0

0

5 dxxQdxxQ .

c. 10

5

5

0

10

0)()()( dxxPdxxPdxxP

(by Theorem 9.1c)

= 2 + (–4) = –2. ■

Let’s return now to the solution of the area problem. 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

b

adxxP )( .

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

the x-axis is b

adxxP )( .



• If P(x) changes sign on [a, b], the area of the region between the graph of

P(x) and the x-axis can be found by partitioning [a, b] into subintervals on

which P(x) does not change sign. The area is the sum of the values of the

appropriate integrals on those subintervals.

Example 9.4 Expressing areas as definite integrals

Let P(x) = x2 – 4. Write one or more integrals to express the area between the

graph of P(x) and the x-axis over each of the following intervals.

(a) [-1, 1] (b) [-1, 3] (c) [-3, 3]

Solution (a) Because P(x) ≤ 0 on [-1, 1], the area is 1

1

2 4 dxx .

(b) Because P(x) ≤ 0 on [-1, 2] and P(x) ≥ 0 on [2, 3], the area is

3

2

22

1

2 44 dxxdxx .

(c) Because P(x) ≥ 0 on both [-3, -2] and [2, 3] and P(x) ≤ 0 on [-2, 2], the

area is

3

2

22

2

22

3

2 444 dxxdxxdxx . ■

Example 9.5 Using areas to evaluate definite integrals

Use area formulas from geometry to evaluate each of the following definite

integrals.

(a) 5

12 dxx (b)

4

1)3( dx (c)

3

0)42( dxx

Solution (a) The integral is the area of the region under the graph of P(x) = x + 2 over

[1, 5]. That region is a trapezoid with width 5 – 1 = 4 and average height

5)73()5()1(21

21 PP . The area is the product of the width and the

average height, which is 20. Thus 2025

1 dxx .

(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, 15)3(4

1 dx .

(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 44221 . The

right-hand triangle has base 1 and altitude 2, so its area is 12121 . The

left-hand triangle is below the x-axis and the right hand triangle is above

it, so 314)42()42()42(3

2

2

0

3

0 dxxdxxdxx . ■




Use area formulas from geometry to show that bdxb

01 for all real numbers b.

Solution If b > 0, the integral represents the area of the rectangle under the graph of

1)( xP over the interval [0, b]. This rectangle has width b and height 1, so

bbdxb

110

.

If b < 0, then |b| = –b is the area of the rectangle under the graph of P(x) = 1 over

the interval [b, 0]. This area is represented by the integral b

bdxdx

0

0

11 , so

bdxb

01 in this case as well.

Finally, if b = 0, then bdxdxb

0110

00. ■


Use area formulas from geometry to show that 2

2

0

bdxx

b

for all real numbers b.

Solution If b > 0, the integral represents the area of the triangle under the graph of P(x) = x

over the interval [0, b]. This triangle has base b and altitude b, so

22

1 2

0

bbbdxx

b

.

If b < 0, then 2

2b is the area of the triangle between the graph of P(x) = x and the

x-axis over the interval [b, 0]. Because this triangle is below the x-axis,

b

bdxxdxx

b

0

02

2.

Finally, if b = 0, then 2

02

0

00

bdxxdxx

b

. ■

As Examples 9.6 and 9.7 illustrate, geometric area formulas can be used to

evaluate any definite integral that involves a constant or linear function. The

evaluation of integrals that involve nonlinear polynomial functions is less

straightforward. You can deal with polynomials of degree up to 3 by forming

Riemann sums and using the summation formulas of Theorem 7.2. However, in



the next section you will learn a method of evaluation that is both more general

and less cumbersome.

Active


In Exercises 1-6, find both a lower sum and an upper sum for P(x) over the given

interval using 4 subintervals of equal width. Then use your results 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.

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

3. P(x) = x2 – 4, [-1, 1] 4. P(x) = x

2 – 4, [-2, 2]

5. P(x) = x3 – 64, [0, 4] 6. P(x) = 3x

2 – 4x

3, [-4, 0]

In Exercises 7-14, 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.

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

9. P(x) = x2 – 2x, [0, 2] 10. P(x) = x

2 – 2x, [-2, 4]

11. P(x) = x3 – 3x, [-3, 3] 12. P(x) = x

3 + x

2 – 6x, [-3, 2]

13. P(x) = x3 – x

4, [-1, 1] 14. P(x) = 4x

3 – x

5, [-4, 4]

Suppose that ,6)(,6)(5

1

5

2 dxxPdxxP and 7)(

5

1 dxxQ . Evaluate each

integral in Exercises 15-20.

15. 5

2)(3 dxxP 16.

5

1)()( dxxQxP

17. 5

1)()(2 dxxPxQ 18.

2

5)( dxxP

19. 1

5

5

1)()( dxxQdxxQ 20.

1

2)( dxxP

In Exercises 21-28, use area formulas from geometry to evaluate the definite

integrals.

21. 3

27 dx 22.

8

4)5( dx

23. 4

33 dxx 24.

4

393 dxx

25. 3

172 dxx 26.

6

472 dxx

27. 8

06 dxx 28.

12

06 dxx




In each of Exercises 29 and 30, the table shows the velocity of an object in

rectilinear motion at several times. Assuming that the acceleration of the object is

always positive, find lower and upper bounds for the distance traveled by the

object over the time interval shown.

29. time (sec) 0 3 6 9 12 15

velocity (cm/sec) 0 2 5 10 18 30

30. time (sec) 0 0.5 1 1.5 2 2.5 3

velocity (ft/sec) 10 16 25 30 34 40 42


Refer to the material in the section as needed to answer each of questions 31-33,

but write your answer in your own words. Address your answer to an imaginary

classmate.

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

32. What does Theorem 9.1d say about areas in the case where P(x) and Q(x)

are nonnegative on [a, b]?

33. Why is it reasonable to make the definition ?0)( a

adxxP

34. Extend the results of Examples 9.6 and 9.7 to prove each of the following

formulas.

a. abdxb

a 1 b.

22

22 abdxx

b

a

35-42. Use the formulas in Exercise 34 to evaluate the integrals in Exercises 21-

28, and verify that the values agree with those you obtained in Exercises

21-28.

43. A dropped object falls with a downward acceleration of 32 ft/sec2.

a. Make a table to show the object’s downward velocity v(t) at times

t = 0, 1, 2, 3, 4, and 5 seconds after it is dropped.

b. Use the table to find lower and upper bounds for the distance s(t)

that the object falls during the first 5 seconds.

c. Solve an initial value problem to find the distance that the object

falls during the first 5 seconds, and verify that your answer falls

within the bounds established in part (b).

44. An object in rectilinear motion moves with constant acceleration k,

starting from rest at time t = 0.



a. Solve an initial value problem to find the object’s velocity function

v(t) and its position function s(t).

b. Use the result of Example 9.7 to show that the distance traveled by

the object in the time interval [0, b] is b

dttv0

)( .


The Fundamental Theorem of Calculus

Section 10 The Fundamental Theorem of Calculus

You are now prepared to address the last of the three obstacles to the solution of

the area problem listed in Section 7.

• After we define the area of a region bounded by polynomial graphs, we

need to find a practical way of calculating it.

More generally, we are seeking a practical method of evaluating definite integrals.

The method you are about to learn relies on what might appear to be a most

unlikely tool, namely the concept of a derivative. How can derivatives, which

were designed to find slopes of curves, help us to evaluate definite integrals,

which were designed to find areas under curves? One clue can be found in

Example 9.2, in which Riemann sums formed from information about a

skydiver’s velocity were used to estimate the distance she falls. (See Exercises 43

and 44 in Section 9 for additional clues.) The example suggests that if an object’s

velocity is known as a polynomial function of time, its position at any time can be

found by evaluating a definite integral. Example 10.1 illustrates that this is in fact

the case.

Example 10.1 Recovering position from velocity

A rock thrown upward from ground level has velocity v(t) = 96 – 32t ft/sec t

seconds after it is released.

(a) Use antiderivatives to find the change in the rock’s height during the time

interval [1, 3].

(b) Use a definite integral to obtain the same information.

Solution (a) The rock’s height s(t) after it is released is an antiderivative of v(t), so s(t)

= 96t – 16t2 + C for some constant C. Because the rock starts at ground

level, s(0) = 0. It follows that C = 0, and s(t) = 96t – 16t2. The change in

its height over [1, 3] is s(3) – s(1) = 144 – 80 = 64 ft.

(b) To approximate the change in the height of the rock over the time interval

[1, 3], begin by partitioning the interval into m subintervals, and let the kth

subinterval have duration (t)k. If (t)k is small, the rock’s velocity in the

kth

subinterval varies only slightly and may be approximated by kcv ,

where ck is an instant within the kth

subinterval. The change in the rock’s

height during that subinterval is approximately (velocity)(time) ≈

kk tcv . The total change in height over the time interval [1, 3] is

approximately

m

k

kk tcv1

.



Every such sum is a Riemann sum for 3

1)( dttv . The change in height is

overestimated by every upper sum for this integral and underestimated by

every lower sum, so the change in height must be exactly equal to

3

1)( dttv . To calculate its value, note that

3

1

3

1)3296()( dttdttv

3

1

3

13296 dttdt (by Theorem 9.1b)

3

1

3

132196 dttdt (by Theorem 9.1a)

1

0

3

0

1

0

3

0321196 dttdttdtdt (by Theorem 9.1c)

2

1

2

332)13(96

22

(by Examples 9.5 and 9.6)

= 96·2 - 32·4 = 64 ft. ■

Example 10.1 suggests the following procedure for evaluating a definite integral

of a polynomial function.

• Find an antiderivative of the integrand.

• Subtract the value of the antiderivative at the lower limit of integration

from the value at the upper limit.

The Fundamental Theorem of Calculus for polynomial functions asserts that this

procedure gives the correct result for every definite integral of every polynomial

function.

Theorem 10.1 (The Fundamental Theorem of Calculus for Polynomial Functions)

Let P(x) be a polynomial function, and let Q(x) be any antiderivative of P(x).

Then:

(a) For all real numbers a and u, u

adxxP )( is a differentiable function of u,

and its derivative with respect to u is P(u).

(b) For all real numbers a and b, )()()( aQbQdxxPb

a .



It is customary to write |)(b

axQ to represent the quantity Q(b) – Q(a).

Example 10.2 Evaluating definite integrals

Evaluate:

a. 2

1

35 486 dxxx b. 4

2)4)(2(3 dxxx

Solution a. One antiderivative of 6x5 – 8x

3 – 4 is x

6 – 2x

4 – 4x. By Theorem 10.1,

|2

1

462

1

35 42486 xxxdxxx

= (64 – 32 – 8) – (1 – 2 – 4) = 29.

b. The integrand is 3(x – 2)(x – 4) = 3x2 – 18x + 24. One antiderivative is

x3 – 9x

2 + 24x, so by Theorem 10.1,

|4

2

234

2249)4)(2(3 xxxdxxx

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

Example 10.3 Finding derivatives of definite integrals

Let x

dtttxQ1

3 544)( . Find )(xQ in two different ways:

a. Use part (a) of Theorem 10.1.

b. Use part (b) of Theorem 10.1.

Solution a. Part (a) of Theorem 10.1 implies that 544)( 3 xxxQ .

b. One antiderivative of the integrand is t4 + 2t

2 – 5t, so part (b) of Theorem

10.1 implies that

|1

24

1

3 52544)(xx

tttdtttxQ

52152 24 xxx

= x4 + 2x

2 – 5x + 2.

Therefore 544)( 3 xxxQ . ■

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.



Example 10.4 Finding areas

Find the area of the region bounded by the graph of P(x) = x3 – 3x

2 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

3

0

323

0

23 33 dxxxdxxx

|3

0

43

4

xx

4

270

4

8127

. ■


Find the area of the region R bounded by the graphs of P(x) = x2 – 1 and

1)( xxQ .

Solution The left and right boundaries of

the region R are the points of

intersection of the graphs of P(x)

and Q(x), as shown in Figure

10.1. The x-coordinates of these

points are the solutions of the

equation x2 – 1 = x + 1. This

equation can be rewritten as

x2 – x – 2 = 0 or, equivalently, as

(x + 1)(x – 2) = 0, and the

solutions are x = -1 and 2.

Figure 10.1

To approximate the area of R, first partition [-1, 2] into m subintervals, and let the

width of the kth

subinterval be (x)k. Choose an x-coordinate ck within the kth

subinterval, and construct a rectangle over that subinterval with height

kk cPcQ . The area of the kth

rectangle is kkk xcPcQ )( . The region

S, consisting of all such rectangles, approximates the shape of R, and the area of S

is

m

k

kkk xcPcQ1

)( .

Each such sum is a Riemann sum for 2

1)()( dxxPxQ , so the area of R is



2

1

22

1

2 21)1( dxxxdxxx

|2

1

32

322

xxx

2

9

3

1

2

12

3

824

. ■

Example 10.5 can be generalized. 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

b

adxxPxQ )()( .


Find 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

10.2. The graphs of P(x)

intersect at points with x-

coordinates that satisfy x3 = 4x.

This equation can be rewritten as

x3 – 4x = 0 or, equivalently, as

x(x + 2)(x – 2) = 0, so the points

of intersection have x-

coordinates -2, 0, and 2.

Figure 10.2

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

2

0

0

2)()()()( dxxPxQdxxQxP .

These two integrals represent the areas of two subregions of R, so we can save

some work by evaluating one of them and doubling the result. The area of R is

2

0

32

042)()(2 dxxxdxxPxQ



|2

0

42

422

xx

8)0()48(2 . ■

Example 10.7 Finding position from velocity

A 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

5

0

5

032160)( dttdttv

|5

0

216160 tt

ft 400)0()400800( . ■

Example 10.8 Finding position from velocity

The 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

11

0

11

0)888()( dttdttv

|11

0

2488 tt

ft 484)0()484968( . ■

Active


In Exercises 1-10, evaluate the definite integral.



1. 2

146 dxx 2.

3

124 dxx

3. 3

0

2 9 dxx 4. 0

2

26 dxxx

5. 1

1

23 124 dxxx 6. 2

0

38 dxxx

7. 3

0

5

27dx

x 8.

1

1

78 dxxx

9. 2

2

23 2112 dxxx 10. 5

1

23 43dx

x

xx

In Exercises 11-18, find the area between the graph of P(x) and the x-axis over the

given interval.

11. P(x) = x2 + 3, [-2, 4] 12. P(x) = 10x

4, [0, 2]

13. P(x) = x3 – 1, [-1, 1] 14. P(x) = 4x – x

2, [1, 3]

15. P(x) = x3 + 2x, [-2, 2] 16. P(x) = x

3 + 6x

2, [-3, 1]

17. P(x) = 4x4 – x

2, [-1, 1] 18. P(x) = x

3 – 3x

2 + 2x, [-1, 3]

In Exercises 19-26, find the area between the graphs of P(x) and Q(x) over the

given interval.

19. P(x) = x + 1, Q(x) = 3x + 1, [0, 6]

20. P(x) = 1 – 2x, Q(x) = 7 – 2x, [-3, 5]

21. P(x) = x2, Q(x) = –x

2, [-1, 5]

22. P(x) = x2 + 1, Q(x) = 2x

3, [0, 1]

23. P(x) = x + 2, Q(x) = 6 – x, [0, 3]

24. P(x) = x2, Q(x) = x

3, [-1, 1]

25. P(x) = 3 – x2, Q(x) = 1 – x, [-2, 4]

26. P(x) = x3 – 2x, Q(x) = 2x, [-3, 3]

In Exercises 27-30, find the total area of all regions bounded by the graphs of P(x)

and Q(x).

27. P(x) = x2 – 4, Q(x) = x – 2

28. P(x) = x2, Q(x) = 8 – x

2

29. P(x) = x3 – 6x, Q(x) = 3x

30. P(x) = x5, Q(x) = x

3

In Exercises 31-34, find )(xQ in two different ways:

a. Use part (a) of Theorem 10.1.

b. Use part (b) of Theorem 10.1.

31. x

dttxQ0

56)( 32. x

dtttxQ2

2 82)(



33. x

dtttxQ1

3 12)( 34. x

dttxQ0

100)(

In Exercises 35 and 36, find )(xQ in two different ways.

a. Use part (b) of Theorem 10.1.

b. Use part (a) of Theorem 10.1 and the Chain Rule.

35.

32

014)(

x

dttxQ 36. 2

0

23)(x

dtttxQ


37. 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 10.1 to

evaluate the integral.

38. 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 30 seconds in two ways.

a. Solve an initial value problem.

b. Express the change in height over the time interval [0, 30] as a

definite integral, and use Theorem 10.1 to evaluate the integral.





39. How are definite and indefinite integrals similar to each other? How are

they different from each other?

40. 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.

41. The displacement of an object in rectilinear motion over a time interval is

its net change in position over the interval.

a. Show that if the object has velocity function v(t), its displacement

over a time interval [a, b] is b

adttv )( .

b. Using the rock in Exercise 38 as an example, explain why the

integral b

adttv )( does not always represent the total distance

traveled by the object over the time interval [a, b].



c. Find the total distance that the rock in Exercise 38 travels over

[0, 30].

d. Give a general description of a method that can be used to find the

total distance traveled by an object in rectilinear motion over a

given time interval.


Proofs of Theorems

Appendix A Proofs of Theorems on Derivatives

Theorem 1.1 Let P(x) be a polynomial function. Then the expression h

xPhxP )()( reduces

to a polynomial P*(x, h) for h ≠ 0.

Proof Recall that (x + h)n = nnnnn hxh

n

nhx

nhx

nx

1221

121 , where

the binomial coefficient

k

n is defined as

k

knnnn

321

)1()2)(1(.

Now let P(x) = anxn + an-1x

n-1 + …+ a1x + a0. Then

xhxaxhxaxhxaxPhxP nn

n

nn

n

)()()()()( 1

11

1 .

Note that

nnnnnnnn xhxhn

nhx

nhx

nxxhx

1221

121)(

nnnn hxhn

nhx

nhx

n

1221

121 .

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

h

xPhxP )()( , resulting in a polynomial in x and h. ■

Theorem 4.1 (Power Rule, special cases)

(a) For every constant c, 0cdx

d.

(b) 1xdx

d.

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

0)()(

),(*

h

cc

h

xPhxPhxP for h ≠ 0.

Thus 0)0,(*)( xPxP .


Proofs of Theorems

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

1)()()(

),(*

h

h

h

xhx

h

xPhxPhxP for h ≠ 0.

Thus 1)0,(*)( xPxP . ■

Theorem 4.2 (Power Rule, general case) For every integer n ≥ 1, 1 nn nxxdx

d.

Proof Recall that for all real numbers a and b,

an – b

n = (a – b)(a

n-1 + a

n-2b + a

n-3b

2 + … + ab

n-2 + b

n-1).

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

P(x + h) – P(x) = (x + h)n - x

n

122321 )()()()()( nnnnn xxhxxhxxhxhxxhx

122321 )()()()( nnnnn xxhxxhxxhxhxh .

Therefore

h

xhxhxP

nn

)(),(*

122321 )()()()( nnnnn xxhxxhxxhxhx for h ≠ 0.

There are n terms in this expression, and each term simplifies to xn-1

when h = 0.

Therefore 1)( nnxxP . ■

Theorem 4.3 Let P and Q be polynomial functions, and let c be a constant. Then:

(a) (Constant Multiple Rule) )()( xPcxcPdx

d

(b) (Sum Rule) )()()()( xQxPxQxPdx

d

(c) (Product Rule) )()()()()()( xPxQxQxPxQxPdx

d

(d) (Chain Rule) )()()( xQxQPxQPdx

d

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


Proofs of Theorems

),(*)()()()(

),(* hxPch

xPhxPc

h

xcPhxcPhxR

,

and evaluating at h = 0 gives the conclusion.

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

h

xQxPhxQhxPhxR

)()()()(),(*

),(*),(*)()()()(

hxQhxPh

xQhxQ

h

xPhxP

,


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

h

xQxPhxQhxPhxR

)()()()(),(*

h

xQxPxQhxPxQhxPhxQhxP )()()()()()()()(

h

xPhxPxQ

h

xQhxQhxP

)()()(

)()()(

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


(d) If Q is a constant function, then both sides are 0, so the conclusion is true.

Otherwise let )()( xQPxR . Note that

h

xQhxQ

j

xQPjxQPhxQjxQP

)()()()(),(*),(*

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

),(*

)()()()(

)()(

)()(hxR

h

xQPhxQP

h

xQhxQ

xQhxQ

xQPhxQP

except, for each x, at the finite number of values of h for which

0)()( xQhxQ . Evaluating P*(Q(x), j) at j = 0 and Q*(x, h) and

R*(x, h) at h = 0 gives the conclusion. ■


Proofs of Theorems

Appendix B Proofs of Theorems on Antiderivatives

FYI The proof of Theorem 6.1 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.

Theorem 6.1 If 0)( xP for all x, then there is a constant C such that P(x) = C for all x.

Proof If P(x) has positive degree n and leading term axn with a ≠ 0, then the leading

term of )(xP is anxn-1

, and the leading coefficient is an, which is not 0. It

follows that if 0)( xP , then the degree of P(x) is 0, so P(x) is a constant

function. ■

Theorem 6.2 If )()( xQxP 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 0)()()( xQxPxR . It follows from Theorem

6.1 that R(x) = C for some constant C, so P(x) = Q(x) + R(x) = Q(x) + C. ■

Theorem 6.3 (a) (Power Rule)

If nxxP )( , then Cxn

xP n

1

1

1)( for some constant C.

(b) (Constant Multiple Rule)

If )()( xQkxP , then P(x) = kQ(x) + C for some constant C.

(c) (Sum and Difference Rules)

If )()()( xRxQxP , then P(x) = Q(x) ± R(x) + C for some constant C.

Proof Note that 1

1

1

nxn

is an antiderivative of xn, kQ(x) is an antiderivative of )(xQk ,

and Q(x) ± R(x) is an antiderivative of )()( xRxQ . The results then follow

from Theorem 6.2. ■

Theorem 6.4 Let Q(x) be a polynomial function. Every initial value problem of the form

1

)1(

210

)( )(,,)(,)(,)(),()(

n

nn yayyayyayyayxQxy

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), …, )(xy , and y(x) are specified as unique

functions. ■


Proofs of Theorems

Appendix C Proofs of Theorems on Sums and Definite Integrals

Theorem 7.1 (a) (Constant Multiple Rule)

n

mk

n

mk

kfckcf )()(

(b) (Sum and Difference Rules)

n

mk

n

mk

n

mk

kgkfkgkf )()()()(

Proof (a) The definition of the sigma operator implies that

)()2()1()()( ncfmcfmcfmcfkcfn

mk

n

mk

kfcnfmfmfmfc )()()2()1()( .

(b) The definition of the sigma operator implies that

n

mk

kgkf )()(

)()()1()1()()( ngnfmgmfmgmf

)()1()()()1()( ngmgmgnfmfmf

n

mk

kgkf )()( .

This verifies the Sum Rule. To verify the Difference Rule, note that

n

mk

n

mk

kgkfkgkf )()()()(

n

mk

n

mk

kgkf )()( (by the Sum Rule)

n

mk

n

mk

kgkf )()( (by the Constant Multiple Rule). ■

The proofs of Theorems 7.2c and 7.2d utilize the principle of mathematical

induction, which may be stated as follows.

Let S(n) be a statement made about an unspecified integer n, and suppose

that


Proofs of Theorems

• S(1) is true, and

• whenever S(n) is true for a particular integer n, S(n + 1) must also

be true.

Then S(n) is true for all integers n ≥ 1.

To make sense of this principle, imagine an infinite row of dominos numbered 1,

2, 3, and so on. Let S(n) be the statement, “Domino n will fall.” If you knock

over Domino 1, then S(1) is true. If you have set up the row so that each domino

will knock over the next one, then the truth of S(n) implies the truth of S(n + 1).

That is, if Domino n falls, then Domino n + 1 must fall. In such a situation every

domino will eventually fall, so S(n) is true for every positive integer n.

Theorem 7.2 For every integer, m ≥ 0,

n

k

mk1

is a polynomial function of n. The functions for

m ≤ 3 are as follows.

(a) nn

k

1

1 (c) 6

)12)(1(

1

2

nnnk

n

k

(b) 2

)1(

1

nnk

n

k

(d)

4

122

1

3

nnk

n

k

Proof (a) By definition, 111111

m

k

(m terms) = m.

(b) The sum is written down twice in the following display.

1 + 2 + 3 + … + (m – 2) + (m – 1) + m

m + (m – 1) + (m – 2) + … + 3 + 2 + 1

There are m pairs of vertically aligned terms, each having a sum of m + 1,

so the sum of all the numbers in the display is m(m + 1). This is twice

m

k

k1

, so 2

)1(

1

mmk

m

k

.

(c) Let S(n) be the assertion that 6

)12)(1(

1

2

nnnk

n

k

. Then S(1) is the

assertion that 6

)112)(11)(1(1

1

2

k

k . Both sides reduce to 1, so S(1) is

true.

If S(n) is true for a particular integer n, then 6

)12)(1(

1

2

nnnk

n

k

, and

S(n + 1) is the assertion that


Proofs of Theorems

6

)32)(2)(1(

6

)1)1(2)(1)1)((1(1

1

2

nnnnnnk

n

k

.

To verify that this is true, note that

22

1

21

1

2 )1(6

)12)(1(1

nnnn

nkkn

k

n

k

6

)1(6)12)(1( 2

nnnn 6

)1(6)12()1(

nnnn

6

672)1( 2

nnn =

6

)32)(2)(1( nnn.

Thus the truth of S(n) implies the truth of S(n + 1). This guarantees that

S(n) is true for every integer n ≥ 1.

(d) Let S(n) be the assertion that

4

122

1

3

nnk

n

k

. Then S(1) is the

assertion that

4

111221

1

3

k

k . Both sides reduce to 1, so S(1) is true.

If S(n) is true for a particular integer n, then

4

122

1

3

nnk

n

k

, and

S(n + 1) is the assertion that

4

)2()1( 221

1

3

nnk

n

k

.

To verify that this is true, note that

3221

1

3 )1(4

)1(

nnn

kn

k

4

)1(4)1( 322

nnn 4

)1(4)1( 22

nnn

4

)2()1(

4

44)1( 2222

nnnnn.


Proofs of Theorems

Thus the truth of S(n) implies the truth of S(n + 1). This guarantees that

S(n) is true for every integer n ≥ 1. ■

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.

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

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 increasing on [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. Because the width of [a, b] is

b – a, the width of each subinterval is n

abx . The largest value of P(x) in the

kth

subinterval occurs at the right endpoint, where x = xk, and the smallest value

occurs at the left endpoint, where x = xk-1. Therefore

xxPxxPxxPxxPxxPU mmm 1321

and

xxPxxPxxPxxPxxPL mmm 12210 .

The difference Um – Lm is

m

abaPbPxaPbPxxPxxP m )()()()(0 .

If we choose

r

abaPbPm

)()()( , then Um – Lm < r.


Proofs of Theorems

A similar argument applies if P is decreasing on [a, b]. 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 nr . The union of these partitions gives a partition of [a, b]

with upper and lower sums whose difference is less than r. ■

Theorem 9.1 Let P and Q be polynomial functions. If a < b, then:

(a) (Constant Multiple Rule) For every constant c, b

a

b

adxxPcdxxcP )()( .

(b) (Sum and Difference Rules) b

a

b

a

b

adxxQdxxPdxxQxP )()()()(

(c) (Interval Additivity Rule) If a < c < b,

b

c

c

a

b

adxxPdxxPdxxP )()()( .

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

a

b

adxxQdxxP )()( .

Proof (a) The sum

m

k

kk xcP1

is a Riemann sum for P(x) on [a, b] if and only if

m

k

kk

m

k

kk xcPcxccP11

is a Riemann sum for cP(x) on [a, b].

The definite integral b

adxxP )( 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

a

b

adxxPccIdxxcP )()( .

(b) To prove the Sum Rule, let S be a partition of [a, b], and let the maximum

value of P(x) + Q(x) in the kth

subinterval be P(ck) + Q(ck). Then in the

same subinterval the maximum values of P(x) and Q(x) are, respectively,

P(ck*) ≥ P(ck) and Q(ck**) ≥ Q(ck). If U1, U2, and U3 are the upper sums

for P(x), Q(x), and P(x) + Q(x), respectively, then

m

k

kkk xcQcPU1

3

m

k

kk

m

k

kk xcQxcP11


Proofs of Theorems

21

11

*** UUxcQxcPm

k

kk

m

k

kk

.

It follows that b

a

b

a

b

adxxQdxxPdxxQxP )()()()( . The reverse

inequality b

a

b

a

b

adxxQdxxPdxxQxP )()()()( is established by a

similar argument using lower sums, and the Sum Rule is proved.

To prove the Difference Rule, note that

b

a

b

adxxQxQxPdxxP )()()()(

b

a

b

adxxQdxxQxP )()()( (by the Sum Rule),

and the result follows.

(c) Let S1 and S2 be partitions of [a, c] and [c, b], respectively. Then

213 SSS 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 b

a

b

c

c

adxxPdxxPdxxP )()()( . The reverse inequality

b

a

b

c

c

adxxPdxxPdxxP )()()( is established by a similar argument

using upper sums.

(d) The given inequality implies that every Riemann sum for P(x) is less than

or equal to the corresponding Riemann sum for Q(x), and the result

follows. ■


Proofs of Theorems

Appendix D Proof of the Fundamental Theorem of Calculus for Polynomial Functions

The Fundamental Theorem of Calculus for polynomial functions can be

established as a consequence of three preliminary theorems, identified here as

Lemmas 1, 2, and 3.

FYI Although Lemma 1 is valid for all polynomial functions (and indeed for a much

wider class of functions), we are presently able to prove it only for polynomial

functions of degree up to 3. You will see a more general proof later in your

course.

Lemma 1 For all real numbers b > 0 and all integers n ≥ 0, 1

1

0

n

bdxx

nb

n .

Proof The proof for n = 0 and n = 1 was given in Examples 9.5 and 9.6, respectively.

The proofs for n = 2 and n = 3 follow.

Let S(m) be a partition of [0, b] into m subintervals of equal width mbx , and let

U(m) be the upper sum for P(x) = xn on [0, b] corresponding to S(m). Because

P(x) is increasing on [0, b], the maximum value of P(x) in the kth

subinterval

occurs at its right endpoint, which has the x-coordinate ck = k(x). Therefore

m

k

n

k

m

k

k xcxcPmU11

)(

m

k

nxxk

1

(

m

k

nnkx

1

1

m

k

n

n

n

km

b

11

1

.

For n = 2,

6

)12)(1()(

3

3

1

2

3

3 mmm

m

bk

m

bmU

m

k

2

3

6

)12)(1(

m

mmb

2

3323

6

32

m

bmbmb .


Proofs of Theorems

Thus U(m) is a rational function of m whose end behavior is determined by the

ratio

36

2 3

2

23 b

m

mb .

That is, the graph of U(m) has a horizontal asymptote at y = b3/3. It follows that

every interval around b3/3 contains an upper sum for P(x) = x

2 on [0, b]. By

Theorem 8.2, every interval around b3/3 also contains a lower sum. Hence

3

3

0

2 bdxx

b

.

For n = 3,

4

1)(

22

4

4

1

3

4

4 mm

m

bk

m

bmU

m

k

2

24

4

)1(

m

mb

2

4424

4

2

m

bmbmb .

Thus U(m) is a rational function of m whose end behavior is determined by the

ratio

44

4

2

24 b

m

mb .

That is, the graph of U(m) has a horizontal asymptote at y = b4/4, and it follows

that

4

4

0

3 bdxx

b

. ■

Lemma 2 For all real numbers b and all integers n ≥ 0, 1

1

0

n

bdxx

nb

n .

Proof The case b > 0 is Lemma 1. If b = 0, then

11

00

110

00

n

b

ndxxdxx

nnn

bn .


Proofs of Theorems

If b < 0 and n is even, then

11

|| 11||

0

n

b

n

bdxx

nnb

n (because n + 1 is odd).

This integral represents the area of the region R under the graph of P(x) = xn over

[0, |b|]. The region under the graph and the x-axis over [b, 0] is congruent to R, so

b

n

b

nb

nn

dxxdxxdxxn

b

0

0||

0

1

1.

If b < 0 and n is odd, then

11

|| 11||

0

n

b

n

bdxx

nnb

n (because n + 1 is even).

This integral represents the area of the region R under the graph of P(x) = xn over

[0, |b|]. The region between the graph and the x-axis over [b, 0] is congruent to R,

but it is below the x-axis. Therefore

b

n

b

nb

nn

dxxdxxdxxn

b

0

0||

0

1

1. ■

Lemma 3 For all real numbers a and b and all integers n ≥ 0, 11

11

n

a

n

bdxx

nnb

a

n .

Proof Note that

a

nb

nb

a

n dxxdxxdxx00

(by Theorem 9.1c)

11

11

n

a

n

b nn

(by Lemma 2). ■

Theorem 10.1 (The Fundamental Theorem of Calculus for Polynomial Functions)

Let P(x) be a polynomial function, and let Q(x) be any antiderivative of P(x).

Then:

(a) For all real numbers a, u

adxxP )( is a differentiable function of u, and its

derivative with respect to u is P(u).

(b) For all real numbers a and b, )()()( aQbQdxxPb

a .


Proofs of Theorems

Proof (a) Let 01

1

1)( cxcxcxcxP n

n

n

n

. Then

u

a

n

n

n

n

u

adxcxcxcxcdxxP 01

1

1)(

u

a

u

a

u

a

n

n

u

a

n

n dxcdxxcdxxcdxxc 01

1

1

(by Theorem 9.1b)

u

a

u

a

u

a

n

n

u

a

n

n dxcdxxcdxxcdxxc 101

1

1

(by Theorem 9.1a)

aucau

cn

a

n

uc

n

a

n

uc

nn

n

nn

n

0

22

11

11

2211

(by Lemma 3).

Differentiating with respect to u (and treating all other symbols as

constants) gives

12

2

11

)1(01

1

1 cu

cn

nuc

n

unc

n

n

n

n

)(01

1

1 uPcucucuc n

n

n

n

.

(b) Let 01

1

1)( cxcxcxcxP n

n

n

n

. Then every antiderivative of

P(x) has the form Cxcx

cn

xc

n

xcxQ

n

n

n

n

0

2

1

1

21)( for some

constant C. By the same argument that was used in part (a),

abcab

cn

a

n

bc

n

a

n

bcdxxP

nn

n

nn

n

b

a

0

22

11

11

2211)(

aca

cn

ac

n

acbc

bc

n

bc

n

bc

n

n

n

n

n

n

n

n 0

2

1

1

0

2

1

1

2121

= Q(b) – Q(a). ■


Answers to Selected Exercises


Section 1 1. a. 2; 2; 2; 2 b. 2

3. a. 7; 6.1; 6.01; 6.001 b. 6

5. a. 5 b. 2 c. 2 d. 2t + h – 2

e. v(t) = 2t – 2 f. t = 1 g. [0, 1)

7. a. 17 b. 11 c. 11 d. 4t + 2h + 3

e. v(t) = 4t + 3 f. none g. none

9. a. 0 b. 0 c. 0 d. 0

e. v(t) = 0 f. all real numbers g. none

11. a. 27 b. 0 c. 0 d. 3t2 + 3th + h

2 – 12

e. v(t) = 3t2 – 12 f. t = 2 g. [0, 2)

13. 400 feet

15. a. 20/9.8 ≈ 2.04 sec b. 100/4.9 ≈ 20.4 m

17. a. 20/3.72 ≈ 5.38 b. 100/1.86 ≈ 53.8 m

19. a. 8 sec b. -160 ft/sec

21. 11 sec

29. a. yes; explanations will vary

b. no; explanations will very

Section 2 1. a. 5; 5; 5; 5 b. 5

3. a. 2; 1.1; 1.01; 1.001 b. 1

5. a. 5 b. 2 c. y = 5 + 2(x – 2)

d. P*(x, h) = 2x + h – 2 e. P*(x, 0) = 2x – 2

f. x = 1 g. (-∞, 1)

7. a. 17 b. 11 c. y = 14 + 11(x – 2)

d. P*(x, h) = 4x + 2h + 3

e. P*(x, 0) = 4x + 3 f. x = -3/4 g. (-∞, -3/4)

9. a. 0 b. 0 c. y = 10

d. P*(x, h) = 0 e. P*(x, 0) = 0

f. all real numbers g. none



11. a. 27 b. 0 c. y = -16

d. P*(x, h) = 3x2 + 3xh + h

2 – 12

e. P*(x, 0) = 3x2 – 12 f. x = ±2 g. (-2, 2)

13. P*(x, 0) = 8x for each part of the exercise.

17. a. yes; explanations will vary

b. no; explanations will vary

21. a. (2) b. (3) c. (4) d. (1)

Section 3 1. 0; 0; 0 3. -2x; -4; 4

5. x – 1/3; 5/3; -7/3 7. 3x2 – 12; 0; 0

9. m 11. 3Ax2

13. 2; 0 15. 2 – 2x; -2

17. 0)(,6)(,26)(,123)( )(2 xPxPxxPxxxP n for n ≥ 4

19. v(t) = -2; a(t) = 0 21. v(t) = 2t – 2/3; a(t) = 2

23. 5 m/sec2 25. -16; -4.9

31. a. 62)( xxP ; x < 3

Section 4 1. 7x6 3. 0

5. 8x3 – 6x 7. x

4 – x

3 + x

2

9. 2x 11. 3x2

13. 3x2 + 6x + 2 15. 2x + 3

17. ,360)(,120)(,30)(,6)( 2)4(345 xxPxxPxxPxxP

P(5)

(x) = 720x, P(6)

(x) = 720, P(n)

(x) = 0 for n ≥ 7.

19. 6 21. 3(x4 + 2x

3 + x

2 – 2)(2x + 1)

23. 6(2x – 5)2 25. 140(3 – 7x)

4

27. 21x3(3x

2 – 4)

3(x

2 + 1)

2 29. 3(x

2 + x – 3)

2(2x + 1)

31. 4x3 – 18x + 48



33. a. 3t2 + 4 b. 6t c. none

35. a. 4t3 – 8t b. 12t

2 – 8 c. 2,0 t

37. a. a > 0 b. a = 0 c. a < 0

39. -4 41. yes, k = n + 1; explanations will vary

Section 5 1. a. 5.5 sec b. 8 ft/sec2

3. a. 4 sec b. 22 ft/sec2

5. a. 5 sec b. 100 ft

7. a. 3 sec b. 336 ft

9. 128 ft/sec

11. a. x < 0 b. x = ±1

13. a. x < –1 b. x = 0

15. a. 15)(,20)(,5)( xPxRxC

15)1000(,20)1000(,5)1000( PRC

b. cost = 5, revenue = 20, profit = 15

17. a. xxPxxRxxC 06.048)(,04.050)(),100(02.0)(

12)1000(,10)1000(,22)1000( PRC

b. cost = 22.01, revenue = 9.98, profit = –12.03

19. 2r

21. a. 2rh; 40; explanations will vary

b. r2; 25; explanations will vary

Section 6 1. P(x) = -2x + C 3. CxxxP 58

3

1)(

5. CxxxxP 43

3

4)( 34

7. CxxxxxP 15113

5

2

1)( 234



9. P(x) = –x2 + 6x – 1 11. 12)( 4 xxxP

13. P(x) = x4 + 4x – 4 15. 464

3

1)( 234 xxxxxP

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

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

3 + 12t + 44

25. P(x) = a(x – 3)2 – 4

27. a. 30 m b. 312 m/sec

29. a. 33 ft b. 11 ft/sec2

c. 198 ft d. 231 ft

35. a. yes b. no

c. Explanations will vary.

Section 7 1. 32 3. 25

5. 5/6 11. 75

13. 114 15. 20,100

17. 676,200 19. 3775

21. a. 2k – 1 b.

100

1

12k

k

c. 10,000

Section 8 1. a. 2.4 b. 1.6 c. [1.6, 2.4] d. 2

3. a. 28 b. 28 c. {28} d. 28

5. a. 14 b. 5 c. [5, 14]

d. 91/8; 55/8; [55/8, 91/8]

7. a. 256 b. 112 c. [112, 256]

d. 217; 145; [145, 217]

9. upper sum = 4.02, lower sum = 3.98, interval = [3.98, 4.02]

11. upper sum ≈ 0.669, lower sum ≈ 0.664, interval = [0.664, 0.669]



13. 8

Section 9 1. lower sum = -15/2, upper sum = -9/2, 9/2 ≤ area ≤ 15/2

3. lower sum = -31/4, upper sum = -27/4, 27/4 ≤ area ≤ 31/4

5. lower sum = -220, upper sum = -156, 156 ≤ area ≤ 220

7. 2

1

21

0

2 11 dxxdxx 9. 2

0

2 2 dxxx

11.

3

3

33

0

30

3

33

3

3 333 dxxxdxxxdxxxdxxx

13.

1

0

430

1

43 dxxxdxxx

15. 18 17. 20

19. 0 21. 35

23. -1/2 25. -6

27. 16

29. lower bound = 105 cm, upper bound = 195 cm

35. 35 37. -1/2

39. -6 41. 16

43. a. v(0) = 0, v(1) = 32, v(2) = 64, v(3) = 96, v(4) = 128, v(5) = 160

b. lower bound = 320 ft, upper bound = 480 ft

c. 400 ft

Section 10 1. 0 3. -18

5. -8 7. 9/2

9. -160 11. 42

13. -2 15. 0

17. 14/15 19. 36



21. 84 23. 5

25. 15 27. 9/2

29. 81/2 31. a-b. 6x – 5

33. a-b. x3 – 2x + 1 35. a-b. 16x + 22

37. a. 792 ft b. 4

011176 dtt ; 792 ft

a preview of calculus - pennsylvania state university preview of calc… · a preview of calculus...

Documents