386
Chapter 5
Integration
(a) Find a function P(x) that satisfies these conditions. Use the graphing utility of
your calculator to graph this function.
(b) Use trace and zoom to determine the level of population 9 months from now.
When will the population be 7,590?
(c) Suppose the current population were 2,000 (instead of 5,000). Sketch the graph
of P(x) with this assumption. Then sketch the graph of P(x) assuming current
populations of 4,000 and 6,000. What is the difference between the graphs?
49. A car traveling at 67 ft/sec decelerates at the constant rate of 23 ft/sec2 when the
brakes are applied.
(a) Find the velocity v(t) of the car t seconds after the brakes are applied. Then
find its distance s(t) from the point where the brakes are applied.
(b) Use the graphing utility of your calculator to sketch the graphs of v(t) and s(t)
on the same screen (use [0, 5]1 by [0, 200]10).
(c) Use trace and zoom to determine when the car comes to a complete stop and
how far it travels in that time. How fast is the car traveling when it has traveled 45 feet?
2
Integration by
Substitution
Recall (from Section 2.5) that according to the chain rule, the derivative of the function (x2 3x 5)9 is
d
[(x2 3x 5)9] 9(x2 3x 5)8(2x 3)
dx
Notice that the derivative is a product and that one of its factors, 2x 3, is the derivative of an expression, x2 3x 5, which occurs in the other factor. More precisely,
the derivative is a product of the form
g(u)
du
dx
where, in this case, g(u) 9u8 and u x2 3x 5.
du
by applying the chain rule
dx
in reverse. Specifically, if G is an antiderivative of g, then
You can integrate many products of the form g(u)
g(u)
du
dx G(u) C
dx
since, by the chain rule,
d
du
du
[G(u)] G(u) g(u)
dx
dx
dx
To summarize:
Chapter 5 ■ Section 2
Integration by Substitution
387
Integral Form of the Chain Rule ■ If g is a continuous function
of u and u(x) is a differentiable function of x, then
g(u)
du
dx dx
g(u) du
That is, to integrate a product of the form g(u)
du
in which one of the factors
dx
du
is the derivative of an expression u that appears in the other factor:
dx
1. Find the antiderivative
g(u) du of the factor g(u) with respect to u.
2. Replace u in the answer by its expression in terms of x.
Here are two examples.
EXAMPLE 2.1
Find
9(x2 3x 5)8(2x 3) dx.
Solution
The integrand 9(x2 3x 5)8(2x 3) is a product in which one of the factors
2x 3 is the derivative of an expression x2 3x 5 that appears in the other factor. In particular,
9(x2 3x 5)8(2x 3) g(u)
g(u) 9u8
where
u x2 3x 5
and
Hence, by the integral form of the chain rule,
9(x2 3x 5)8(2x 3) dx du
dx
g(u) du 9u8 du u9 C
(x2 3x 5)9 C
The product to be integrated in the next example is not exactly of the form
du
g(u) . However, it is a constant multiple of such a product, and you can integrate
dx
it by combining the constant multiple rule with the integral form of the chain rule.
388
Chapter 5
Integration
EXAMPLE 2.2
Refer to Example 2.2. Store
1 x2 4
e
3 as (1 4)e^(X^2 4
1
4) 3 in Y1 and eu 3 as
4
(1 4)e^U 3 in Y2 in your
graphing utility. Store .42 as
X and (.42)2 4 as U. Evaluate Y1 and Y2 at X and U,
respectively. Explain why the
results are the same.
Find
4
x3ex
2
dx.
Solution
First use the constant multiple rule to rewrite the integral as
4
x3ex
2
dx 1 3 x4 2
1
) dx (4x e
4
4
4
4x3ex
2
dx
so that the new integrand 4x3ex 2 is a product in which one of the factors 4x3 is the
derivative of an expression x4 2 that appears in the other factor. In particular,
4
4
4x3ex
2
g(u) eu
where
g(u)
du
dx
u x4 2
and
Hence, by the integral form of the chain rule,
4
x3ex
2
dx 1
4
4
4x3ex
2
dx 1
4
g(u) du
1
1
eu du eu C
4
4
1 4
ex 2 C
4
CHANGE OF VARIABLES
The integral form of the chain rule may be thought of as a technique for simplifying
an integral by changing the variable of integration. In particular, you start with an
du
integral g(u)
dx in which the variable of integration is x and transform it into the
dx
simpler integral
g(u) du in which the variable of integration is u. In this transfor-
du
dx in the original integral is replaced in the simplified
dx
du
integral by the symbol du. You can remember this relationship between
dx and du
dx
du
by pretending that
is a quotient and writing
dx
mation, the expression
du
dx = du
dx
Chapter 5 ■ Section 2
Integration by Substitution
389
These observations suggest the following general integration technique, called
integration by substitution, in which the variable u is formally substituted for an
appropriate expression in x and the original integral is transformed into a simpler one
in which the variable of integration is u.
Integration by Substitution
Note
Step 1. Introduce the letter u to stand for some expression in x that is chosen
with the simplification of the integral as the goal.
du
Step 2. Rewrite the integral in terms of u. To rewrite dx, compute
and solve
dx
du
algebraically as if the symbol
were a quotient.
dx
Step 3. Evaluate the resulting integral and then replace u by its expression in
terms of x in the answer.
If the integrand is a product or quotient of two terms and one term is a constant multiple of the derivative of an expression that appears in the other,
then this expression is probably a good choice for u.
The method of integration by substitution is illustrated in the next example using
the integral from Example 2.1.
EXAMPLE 2.3
Find
9(x2 3x 5)8(2x 3) dx .
Solution
The integrand is a product in which one of the factors 2x 3 is the derivative of an
expression x2 3x 5 that appears in the other factor. This suggests that you let
u x2 3x 5. Then
du
2x 3
dx
du (2x 3) dx
and so
Substituting u x2 3x 5 and du (2x 3) dx, you get
9(x2 3x 5)8(2x 3) dx 9u8 du u9 C
(x2 3x 5)9 C
390 Chapter 5
Integration
Here are some additional examples illustrating the method of integration by
substitution.
EXAMPLE 2.4
Find
3x
dx.
x 1
2
Solution
Observe that
d 2
2
(x 1) 2x (3x)
dx
3
Thus the integrand is a quotient in which one term 3x is a constant multiple of the
derivative of an expression x2 1 that appears in the other factor. This suggests that
you let u x2 1. Then,
du
2x
dx
du 2x dx
Substituting u x2 1 and
or
3
du 3x dx
2
3
du 3x dx, you get
2
3x
dx x 1
2
1 3
3 1
du du
u 2
2 u
3
ln |u| C
2
3
ln |x2 1| C
2
EXAMPLE 2.5
Find
3x 6
dx.
2x2 8x 3
Solution
Observe that
4
d
(2x2 8x 3) 4x 8 4(x 2) (3x 6)
dx
3
Chapter 5 ■ Section 2
Integration by Substitution
391
Thus, the integrand is a quotient in which one term 3x 6 is a constant multiple of
the derivative of an expression 2x2 8x 3 that appears in the other factor. This
suggests that you let u 2x2 8x 3. Then,
du
4x 8
dx
4
du (4x 8) dx 4(x 2) dx (3x 6) dx
3
3
du (3x 6) dx
4
or
Substituting u 2x2 8x 3 and
3x 6
dx 2x2 8x 3
3
du (3x 6) dx, you get
4
3 1/2
3 1
du u
du
4 u
4
3
3
(2u1/2) C 2x2 8x 3 C
4
2
EXAMPLE 2.6
Find
(ln x)2
dx.
x
Solution
Observe that
d
1
(ln x) dx
x
Thus, the integrand
(ln x)2
1
(ln x)2
x
x
1
is the derivative of an expression ln x that appears
x
in the other factor. This suggests that you let u = ln x. Then,
is a product in which one factor
du 1
dx x
or
du 1
dx
x
392
Chapter 5
Integration
Substituting u ln x and du (ln x)2
dx x
1
dx, you get
x
1
1
u2 du u3 C (ln x)3 C
3
3
The next example is designed to show you the versatility of the formal method
of substitution. It deals with an integral that does not seem to be of the form
du
g(u)
dx but that nevertheless can be simplified significantly by a clever change
dx
of variables.
EXAMPLE 2.7
Find
x
dx.
x1
Solution
There is no easy way to integrate this quotient as it stands. But watch what happens
if you make the substitution u x 1. Then du dx and x u 1 and so
A COMPARISON OF THE
TWO TECHNIQUES
x
u1
1
dx du 1 du du
x1
u
u
u ln |u| C x 1 ln |x 1| C
The integral form of the chain rule, which was used in Examples 2.1 and 2.2, is attractive because it involves nothing more than a familiar rule for differentiation applied
in reverse. With practice in using this method you should be able to find integrals like
those in Examples 2.1 through 2.6 by inspection, without writing down any intermediate steps.
Nevertheless, many people prefer to use the method of formal substitution. They
like the fact that it involves straightforward manipulation of symbols, and they appreciate the convenience of the notation. This method of substitution is also somewhat
more versatile, as you saw in Example 2.7.
For many of the integrals you will encounter, both methods work well and you
should feel free to use the one with which you are more comfortable.
WHEN SUBSTITUTION FAILS
The method of substitution does not always succeed. In the next example, we consider an integral very like the integral in Example 2.2 but just enough different so no
substitution will work.
Chapter 5 ■ Section 2
Integration by Substitution
393
EXAMPLE 2.8
Evaluate
4
x4ex
2
dx .
Solution
The natural substitution is u x4 2, as in Example 2.2. As before, you find du 1
4x3 dx, so x3 dx du. However, the integrand involves x4, not x3, and the “extra”
4
4
factor of x satisfies x u
2, so when the substitution is made, you have
4
x4ex
2
dx 4
xex
2
4
(x3 dx) u
2 eu du
which is hardly an improvement on the original integral! Try a few other possible
substitutions (say, u x2 or u x3) to convince yourself that nothing works.
AN APPLICATION
INVOLVING SUBSTITUTION
EXAMPLE 2.9
The price p (dollars) of each unit of a particular commodity is estimated to be changing at the rate
dp
135x
dx 9 x2
where x (hundred) units is the consumer demand (the number of units purchased at
that price). Suppose 400 units (x 4) are demanded when the price is $30 per unit.
(a) Find the demand function p(x).
(b) At what price will 300 units be demanded? At what price will no units be
demanded?
(c) How many units are demanded at a price of $20 per unit?
Solution
(a) The price per unit demanded p(x) is found by the integrating p(x) with respect
to x. To perform this integration, use the substitution
u 9 x2,
to get
du 2x dx,
x dx 1
du
2
394
Chapter 5
Integration
135x
135
dx 2
u1/2
9 x
135
135
u 1/2 du 2
2
2
1359 x C
p(x) 12 du
u1/2 C
1/2
Since p 30 when x 4, you find that
30 1359 42 C
C 30 13525 705
so
p(x) 1359 x2 705
(b) When 300 units are demanded, x 3 and the corresponding price is
p(3) 1359 32 705 $132.24 per unit
No units are demanded when x 0 and the corresponding price is
p(0) 1359 0 705 $300 per unit
(c) To determine the number of units demanded at a unit price of $20 per unit, you
need to solve the equation
1359 x2 705 20
1359 x2 685
9 x2 685
5.07
135
By squaring both sides of this equation and simplifying, you find that x 4.09.
That is, roughly 409 units will be demanded when the price is $20 per unit.
P . R . O . B . L . E . M . S
5.2
In Problems 1 through 26, find the indicated integral and check your answer by
differentiation.
1.
3.
(2x 6)5 dx
2.
4x 1 dx
4.
e5x dx
1
dx
3x 5
Chapter 5 ■ Section 2
5.
7.
9.
11.
13.
15.
17.
19.
21.
23.
25.
Integration by Substitution
e1x dx
2
xex dx
6.
8.
t(t2 1)5 dt
10.
x2(x3 1)3/4 dx
12.
2y4
dy
y5 1
14.
(x 1)(x2 2x 5)1/2 dx
16.
3x4 12x3 6
dx
x 5x4 10x 12
18.
3u 3
du
(u 2u 6)2
20.
ln 5x
dx
x
22.
1
dx
x(ln x)2
24.
2x ln (x2 1)
dx
x2 1
26.
5
2
395
[(x 1)5 3(x 1)2 5] dx
2xex 1 dx
2
3tt2 8 dt
6
x5e1x dx
y2
dy
(y3 5)2
(3x 2 1)e x x dx
3
10x3 5x
dx
x4 x2 6
6u 3
du
4u 4u 1
2
1
dx
x ln x
ln x2
dx
x
ex
dx
x
27. Find the function whose tangent has slope xx2 5 for each value of x and
whose graph passes through the point (2, 10).
28. Find the function whose tangent has slope
2x
for each value of x and whose
1 3x2
graph passes through the point (0, 5).
In Problems 29 through 32, the velocity v(t) x(t) at time t of an object moving
along the x axis is given, along with the initial position x(0) of the object. In each
case, find:
(a) The position x(t) at time t.
(b) The position of the object at time t 4.
(c) The time when the object is at x 3.
29. x(t) 2(3t 1)1/2; x(0) 4
396 Chapter 5
Integration
30. x(t) 3(5 2t)3; x(0) 5
31. x(t) t
; x(0) 0
(t 1)2
32. x(t) 2t
; x(0) 4
(1 t2)3/2
1
(x 1)2
meters per year. After 2 years it has reached a height of 5 meters. How tall was it
when it was transplanted?
TREE GROWTH
33. A tree has been transplanted and after x years is growing at the rate of 1 DEPRECIATION
34. The resale value of a certain industrial machine decreases at a rate that changes with
time. When the machine is t years old, the rate at which its value is changing is
960et/5 dollars per year. If the machine was bought new for $5,000, how much
will it be worth 10 years later?
POPULATION GROWTH
35. It is projected that t years from now the population of a certain country will be
changing at the rate of e0.02t million per year. If the current population is 50 million,
what will be the population 10 years from now?
LAND VALUE
36. It is estimated that x years from now, the value V(x) of an acre of farmland will be
increasing at the rate of
V(x) 0.4x3
0.2x4 8,000
dollars per year. The land is currently worth $500 per acre.
(a) Find V(x). Use the graphing utility of your calculator to sketch the graph of
V(x).
(b) Use trace and zoom to determine how much the land will be worth in 10 years.
When will the land be worth $1,000 per acre?
AIR POLLUTION
37. In a certain suburb of Los Angeles, the level of ozone L(t) at 7:00 A.M. is 0.25
parts per million (ppm). A 12-hour weather forecast predicts that the ozone level t
hours later will be changing at the rate of
L(t) 0.24 0.03t
36 16t t 2
parts per million per hour (ppm/hr).
(a) Express the ozone level L(t) as a function of t. When does the peak ozone
level occur? What is the peak level?
(b) Use the graphing utility of your calculator to sketch the graph of L(t) and use
trace and zoom to answer the questions in part (a). Then determine at what other
time the ozone level will be the same as it is at 11:00 A.M.
Chapter 5 ■ Section 2
PRODUCTION
Integration by Substitution
397
38. Bejax Corporation has set up a production line to manufacture a new type of cellular
telephone. The rate of production of the telephones is
t
dP
1,500 2 dt
2t 5
units/month
How many telephones are produced during the third month? [That is, find P(3) P(2).]
DEPRECIATION
39. The resale value of a certain industrial machine decreases at a rate that depends on
its age. When the machine is t years old, the rate at which its value is changing is
960et/5 dollars per year.
(a) Express the value of the machine in terms of its age and initial value.
(b) If the machine was originally worth $5,200, how much will it be worth when
it is 10 years old?
MARGINAL COST
40. At a certain factory, the marginal cost is 3(q 4)2 dollars per unit when the level of
output is q units.
(a) Express the total production cost in terms of the overhead (the cost of producing 0 units) and the number of units produced.
(b) What is the cost of producing 14 units if the overhead is $436?
RETAIL PRICES
41. In a certain section of the country, the price of chicken is currently $3 per kilogram.
It is estimated that x weeks from now the price will be increasing at the rate of
3x 1 cents per week. How much will chicken cost 8 weeks from now?
DEMAND
42. The price p (dollars) of each pair of Yike sports sneakers is estimated to be changing
at the rate
p(x) 150x
(144 x2)3/2
where x (hundred) units is the consumer demand. Suppose 500 pairs of sneakers
(x 5) are demanded when the price is $75 per pair.
(a) Find the demand function p(x).
(b) At what price will 400 pairs of sneakers be demanded? At what price will no
pairs be demanded?
(c) How many pairs will be demanded at a price of $90 per pair?
(d) Find the revenue function R(x) xp(x) and the marginal revenue R(x). For
what quantity demanded x is revenue maximized?
SUPPLY
43. The owner of the Dog Gone hot dog restaurant chain estimates that the dollar price
of her newest product, Weenie Babies, is changing at the rate
p(x) 30x
(3 x)2
398
Chapter 5
Integration
when x (thousand) Weenies are supplied for purchase. The current price is $2.25
per Weenie.
(a) Find the supply function p(x).
(b) At what price will 4,000 additional Weenies (x 4) be supplied?
(c) How many more Weenies will be supplied at a price of $3 per Weenie?
MARGINAL PROFIT
MARGINAL PROFIT
3
Introduction
to Differential
Equations
44. A company determines that the marginal revenue from the production of x units is
R(x) 7 3x 4x2 hundred dollars per unit, and the corresponding marginal cost
is C(x) 5 2x hundred dollars per unit. By how much does the profit change
when the level of production is raised from 5 to 9 units?
11 x
45. Repeat Problem 44 for marginal revenue R(x) and for the
x
14
2
marginal cost C(x) 2 x x .
A differential equation is an equation that involves a derivative or differential. For
example,
dy
3x2 5
dx
or
dP
kP
dt
and
dy
dx
2
3
dy
2y ex
dx
are all differential equations. Differential equations are among the most useful tools
for modeling continuous phenomena, including important situations that occur in business and economics and the social and life sciences. In this section, we introduce techniques for solving basic differential equations and examine a few practical applications.
The simplest type of differential equation has the form
dy
g(x)
dx
in which the derivative of the quantity y is given explicitly as a function of the independent variable x. Such an equation can be solved by simply finding the indefinite
integral of g(x). A complete characterization of all possible solutions of the equation
is called a general solution, and a solution that satisfies specified side conditions is
called a particular solution. This terminology is illustrated in the following examples.
EXAMPLE 3.1
Find the general solution of the differential equation
dy
x2 3x
dx
and the particular solution that satisfies y 2 when x 1.