a preview of calculus - pennsylvania state university preview of calc… · a preview of calculus...
TRANSCRIPT
A Preview of Calculus
Page 1 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
A Preview of Calculus
Page 2 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 Preview of Calculus
Page 3 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
A Preview of Calculus
Page 4 A First Look at Average and Instantaneous Velocities
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
A Preview of Calculus
Page 5 A First Look at Average and Instantaneous Velocities
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(
A Preview of Calculus
Page 6 A First Look at Average and Instantaneous Velocities
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. ■
A Preview of Calculus
Page 7 A First Look at Average and Instantaneous Velocities
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.
A Preview of Calculus
Page 8 A First Look at Average and Instantaneous Velocities
• 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.
A Preview of Calculus
Page 9 A First Look at Average and Instantaneous Velocities
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?
A Preview of Calculus
Page 10 A First Look at Average and Instantaneous Velocities
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.)
A Preview of Calculus
Page 11 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
A Preview of Calculus
Page 12 The Tangent Line Problem
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?
A Preview of Calculus
Page 13 The Tangent Line Problem
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?
A Preview of Calculus
Page 14 The Tangent Line Problem
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.
A Preview of Calculus
Page 15 The Tangent Line Problem
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
A Preview of Calculus
Page 16 The Tangent Line Problem
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)
Focus on connecting concepts
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 Preview of Calculus
Page 17 The Tangent Line Problem
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.
A Preview of Calculus
Page 18 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
A Preview of Calculus
Page 19 Derivatives of Polynomial Functions
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
A Preview of Calculus
Page 20 Derivatives of Polynomial Functions
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
A Preview of Calculus
Page 21 Derivatives of Polynomial Functions
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
Focus on applying skills
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.
A Preview of Calculus
Page 22 Derivatives of Polynomial Functions
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)?
Focus on connecting concepts
Refer to the material in the section as needed to answer each of questions 27 and
28, but write your answer in your own words. Address your answer to an
imaginary classmate.
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 ?
A Preview of Calculus
Page 23 Derivatives of Polynomial Functions
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).
A Preview of Calculus
Page 24 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
A Preview of Calculus
Page 25 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. ■
A Preview of Calculus
Page 26 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.
A Preview of Calculus
Page 27 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
Learning Focus on developing skills
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
A Preview of Calculus
Page 28 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.
Focus on applying skills
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
A Preview of Calculus
Page 29 Derivative Formulas
39. Let 4533
1)( 23 xxxxP . What is the minimum slope on the graph
of P(x)?
Focus on connecting concepts
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?
A Preview of Calculus
Page 30 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. ■
A Preview of Calculus
Page 31 Interpretations of the Derivative
Example 5.4 The derivative as slope
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. ■
Example 5.5 The derivative as slope
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. ■
Example 5.7 The derivative as instantaneous rate of change
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.
A Preview of Calculus
Page 32 Interpretations of the Derivative
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.
Example 5.8 The derivative as instantaneous rate of change
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
A Preview of Calculus
Page 33 Interpretations of the Derivative
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
A Preview of Calculus
Page 34 Interpretations of the Derivative
19. What is the rate of change in the area of a circle with respect to its radius?
20. What is the rate of change in the volume of a cube with respect to its edge
length?
Focus on connecting concepts
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.
A Preview of Calculus
Page 35 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
.
A Preview of Calculus
Page 36 Antiderivatives of Polynomial Functions
(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.
A Preview of Calculus
Page 37 Antiderivatives of Polynomial Functions
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?
A Preview of Calculus
Page 38 Antiderivatives of Polynomial Functions
• 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
Learning Focus on developing skills
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
Focus on applying skills
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).
A Preview of Calculus
Page 39 Antiderivatives of Polynomial Functions
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.
Focus on connecting concepts
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.
A Preview of Calculus
Page 40 Antiderivatives of Polynomial Functions
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.
A Preview of Calculus
Page 41 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 Preview of Calculus
Page 42 The Area Problem, Sigma Notation, and Summation Formulas
• 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.
A Preview of Calculus
Page 43 The Area Problem, Sigma Notation, and Summation Formulas
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 .
A Preview of Calculus
Page 44 The Area Problem, Sigma Notation, and Summation Formulas
(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
Learning Focus on developing skills
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.
A Preview of Calculus
Page 45 The Area Problem, Sigma Notation, and Summation Formulas
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
Focus on connecting concepts
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()( .
A Preview of Calculus
Page 46 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.
Figure 8.1a Figure 8.1b
Figures 7.2a and 7.2b are reproduced here as Figures 8.1a and 8.1b.
A Preview of Calculus
Page 47 Riemann Sums and the Definition of Area
(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.
A Preview of Calculus
Page 48 Riemann Sums and the Definition of Area
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)
A Preview of Calculus
Page 49 Riemann Sums and the Definition of Area
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.
A Preview of Calculus
Page 50 Riemann Sums and the Definition of Area
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)
A Preview of Calculus
Page 51 Riemann Sums and the Definition of Area
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,
A Preview of Calculus
Page 52 Riemann Sums and the Definition of Area
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
Learning Focus on developing skills
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]
In Exercises 5-8, 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 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]
In Exercises 9-12, let R be the region under the graph of P(x) over the given
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.
A Preview of Calculus
Page 53 Riemann Sums and the Definition of Area
13. P(x) = 3x2, [0, 2] 14. P(x) = x
2 + 4, [0, 1]
Focus on connecting concepts
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].
A Preview of Calculus
Page 54 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.
A Preview of Calculus
Page 55 Definite Integrals and their Properties
Figure 9.1a Figure 9.1b
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.
A Preview of Calculus
Page 56 Definite Integrals and their Properties
(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,
A Preview of Calculus
Page 57 Definite Integrals and their Properties
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 )( .
A Preview of Calculus
Page 58 Definite Integrals and their Properties
• 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 . ■
A Preview of Calculus
Page 59 Definite Integrals and their Properties
Example 9.6 Using areas to evaluate definite integrals
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. ■
Example 9.7 Using areas to evaluate definite integrals
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
A Preview of Calculus
Page 60 Definite Integrals and their Properties
the next section you will learn a method of evaluation that is both more general
and less cumbersome.
Active
Learning Focus on developing skills
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
A Preview of Calculus
Page 61 Definite Integrals and their Properties
Focus on applying skills
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
Focus on connecting concepts
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 Preview of Calculus
Page 62 Definite Integrals and their Properties
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
)( .
A Preview of Calculus
Page 63 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
.
A Preview of Calculus
Page 64 The Fundamental Theorem of Calculus
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 .
A Preview of Calculus
Page 65 The Fundamental Theorem of Calculus
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.
A Preview of Calculus
Page 66 The Fundamental Theorem of Calculus
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
. ■
Example 10.5 Finding areas
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
A Preview of Calculus
Page 67 The Fundamental Theorem of Calculus
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 )()( .
Example 10.6 Finding areas
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
A Preview of Calculus
Page 68 The Fundamental Theorem of Calculus
|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
Learning Focus on developing skills
In Exercises 1-10, evaluate the definite integral.
A Preview of Calculus
Page 69 The Fundamental Theorem of Calculus
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)(
A Preview of Calculus
Page 70 The Fundamental Theorem of Calculus
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
Focus on applying skills
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.
Focus on connecting concepts
Refer to the material in the section as needed to answer each of questions 39 and
40, but write your answer in your own words. Address your answer to an
imaginary classmate.
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].
A Preview of Calculus
Page 71 The Fundamental Theorem of Calculus
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.
A Preview of Calculus
Page 72 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 .
A Preview of Calculus
Page 73 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
A Preview of Calculus
Page 74 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
,
and evaluating at h = 0 gives the conclusion.
(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),
and evaluating at h = 0 gives the conclusion.
(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. ■
A Preview of Calculus
Page 75 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. ■
A Preview of Calculus
Page 76 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
A Preview of Calculus
Page 77 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
A Preview of Calculus
Page 78 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.
A Preview of Calculus
Page 79 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.
A Preview of Calculus
Page 80 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
A Preview of Calculus
Page 81 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. ■
A Preview of Calculus
Page 82 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 .
A Preview of Calculus
Page 83 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 .
A Preview of Calculus
Page 84 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 .
A Preview of Calculus
Page 85 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). ■
A Preview of Calculus
Page 86 Answers to Selected Exercises
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
A Preview of Calculus
Page 87 Answers to Selected Exercises
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
A Preview of Calculus
Page 88 Answers to Selected Exercises
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
A Preview of Calculus
Page 89 Answers to Selected Exercises
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]
A Preview of Calculus
Page 90 Answers to Selected Exercises
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
A Preview of Calculus
Page 91 Answers to Selected Exercises
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