SECTION 6.2
Partial Fractions and Logistic Growth
429
Section 6.2
Partial Fractions
and Logistic
Growth
■ Use partial fractions to find indefinite integrals.
■ Use logistic growth functions to model real-life situations.
Partial Fractions
In Sections 5.2 and 6.1, you studied integration by substitution and by parts. In
this section you will study a third technique called partial fractions. This
technique involves the decomposition of a rational function into the sum of two
or more simple rational functions. For instance, suppose you know that
x⫹7
2
1 .
⫽
⫺
x2 ⫺ x ⫺ 6 x ⫺ 3 x ⫹ 2
Knowing the “partial fractions” on the right side would allow you to integrate the
left side as shown.
冕
x2
x⫹7
dx ⫽
⫺x⫺6
冕冢
冕
冣
2
1
⫺
dx
x⫺3 x⫹2
1
1
⫽2
dx ⫺
dx
x⫺3
x⫹2
⫽ 2 ln x ⫺ 3 ⫺ ln x ⫹ 2 ⫹ C
ⱍ
ⱍ
冕
ⱍ
ⱍ
This method depends on the ability to factor the denominator of the original
rational function and on finding the partial fraction decomposition of the function.
STUDY TIP
Recall that finding the partial
fraction decomposition of a
rational function is a precalculus
topic. Explain how you could
verify that
1
2
⫹
x⫺1 x⫹2
is the partial fraction decomposition of
3x
.
2
x ⫹x⫺2
Partial Fractions
To find the partial fraction decomposition of the proper rational function
p共x兲兾q共 x兲, factor q共x兲 and write an equation that has the form
p共x兲
⫽ 共sum of partial fractions).
q共x兲
For each distinct linear factor ax ⫹ b, the right side should include a term
of the form
A .
ax ⫹ b
For each repeated linear factor 共ax ⫹ b兲n, the right side should include n
terms of the form
A2
A1
An
.
⫹
⫹. . .⫹
ax ⫹ b 共ax ⫹ b兲2
共ax ⫹ b兲n
STUDY TIP
A rational function p共x兲兾q共x兲 is proper if the degree of the numerator is less
than the degree of the denominator.
430
CHAPTER 6
Techniques of Integration
Example 1
Finding a Partial Fraction Decomposition
Write the partial fraction decomposition for
x⫹7 .
x2 ⫺ x ⫺ 6
SOLUTION Begin by factoring the denominator as x2 ⫺ x ⫺ 6 ⫽ 共x ⫺ 3兲共x ⫹ 2兲.
Then, write the partial fraction decomposition as
x2
x⫹7
A
B .
⫽
⫹
⫺x⫺6 x⫺3 x⫹2
To solve this equation for A and B, multiply each side of the equation by the least
common denominator 共x ⫺ 3兲共x ⫹ 2兲. This produces the basic equation as shown.
x ⫹ 7 ⫽ A共x ⫹ 2兲 ⫹ B共x ⫺ 3兲
Algebra Review
You can check the result in
Example 1 by subtracting the
partial fractions to obtain the
original fraction, as shown in
Example 1(a) in the Chapter 6
Algebra Review, on page 470.
Basic equation
Because this equation is true for all x, you can substitute any convenient values
of x into the equation. The x-values that are especially convenient are the ones that
make particular factors equal to 0.
To solve for B, substitute x ⫽ ⫺2:
x⫹7
⫺2 ⫹ 7
5
⫺1
⫽ A共x ⫹ 2兲 ⫹ B共x ⫺ 3兲
⫽ A共⫺2 ⫹ 2兲 ⫹ B共⫺2 ⫺ 3兲
⫽ A共0兲 ⫹ B共⫺5兲
⫽B
Write basic equation.
Substitute ⫺2 for x.
Simplify.
Solve for B.
To solve for A, substitute x ⫽ 3:
x ⫹ 7 ⫽ A共x ⫹ 2兲 ⫹ B共x ⫺ 3兲
3 ⫹ 7 ⫽ A共3 ⫹ 2兲 ⫹ B共3 ⫺ 3兲
10 ⫽ A共5兲 ⫹ B共0兲
2⫽A
Write basic equation.
Substitute 3 for x.
Simplify.
Solve for A.
Now that you have solved the basic equation for A and B, you can write the
partial fraction decomposition as
x⫹7
2
1
⫽
⫺
x2 ⫺ x ⫺ 6 x ⫺ 3 x ⫹ 2
as indicated at the beginning of this section.
✓CHECKPOINT 1
Write the partial fraction decomposition for
x⫹8
.
x2 ⫹ 7x ⫹ 12
■
STUDY TIP
Be sure you see that the substitutions for x in Example 1 are chosen for their
convenience in solving for A and B. The value x ⫽ ⫺2 is selected because
it eliminates the term A共x ⫹ 2兲, and the value x ⫽ 3 is chosen because it
eliminates the term B共x ⫺ 3兲.
SECTION 6.2
TECHNOLOGY
Example 2
冕
Partial Fractions and Logistic Growth
431
Integrating with Repeated Factors
The use of partial
fractions depends on
the ability to factor the
denominator. If this cannot
be easily done, then partial
fractions should not be used.
For instance, consider the
integral
Begin by factoring the denominator as x共x ⫹ 1兲2. Then, write the
partial fraction decomposition as
5x 2 ⫹ 20x ⫹ 6
dx.
⫹ 2 x2 ⫹ x ⫹ 1
To solve this equation for A, B, and C, multiply each side of the equation by the
least common denominator x共x ⫹ 1兲2.
冕
x3
This integral is only slightly
different from that in Example 2,
yet it is immensely more
difficult to solve. A symbolic
integration utility was unable
to solve this integral. Of course,
if the integral is a definite
integral (as is true in many
applied problems), then you can
use an approximation technique
such as the Midpoint Rule.
Algebra Review
You can check the partial
fraction decomposition in
Example 2 by combining the
partial fractions to obtain the
original fraction, as shown in
Example 1(b) in the Chapter 6
Algebra Review, on page 470.
Also, for help with the algebra
used to simplify the answer, see
Example 1(c) on page 470.
Find
5x 2 ⫹ 20x ⫹ 6
dx.
x3 ⫹ 2x 2 ⫹ x
SOLUTION
5x 2 ⫹ 20x ⫹ 6 A
B
C
.
⫽ ⫹
⫹
x共x ⫹ 1兲2
x
x ⫹ 1 共x ⫹ 1兲2
5x 2 ⫹ 20x ⫹ 6 ⫽ A共x ⫹ 1兲2 ⫹ Bx共x ⫹ 1兲 ⫹ Cx
Basic equation
Now, solve for A and C by substituting x ⫽ ⫺1 and x ⫽ 0 into the basic equation.
Substitute x ⫽ ⫺1:
5共⫺1兲2 ⫹ 20共⫺1兲 ⫹ 6 ⫽ A共⫺1 ⫹ 1兲2 ⫹ B共⫺1兲共⫺1 ⫹ 1兲 ⫹ C共⫺1兲
⫺9 ⫽ A共0兲 ⫹ B共0兲 ⫺ C
9⫽C
Solve for C.
Substitute x ⫽ 0:
5共0兲2 ⫹ 20共0兲 ⫹ 6 ⫽ A共0 ⫹ 1兲2 ⫹ B共0兲共0 ⫹ 1兲 ⫹ C共0兲
6 ⫽ A共1兲 ⫹ B共0兲 ⫹ C共0兲
6⫽A
Solve for A.
At this point, you have exhausted the convenient choices for x and have yet to
solve for B. When this happens, you can use any other x-value along with the
known values of A and C.
Substitute x ⫽ 1, A ⫽ 6, and C ⫽ 9:
5共1兲2 ⫹ 20共1兲 ⫹ 6 ⫽ 共6兲共1 ⫹ 1兲2 ⫹ B共1兲共1 ⫹ 1兲 ⫹ 共9兲共1兲
31 ⫽ 6共4兲 ⫹ B共2兲 ⫹ 9共1兲
⫺1 ⫽ B
Solve for B.
Now that you have solved for A, B, and C, you can use the partial fraction decomposition to integrate.
冕
5x 2 ⫹ 20x ⫹ 6
dx ⫽
x3 ⫹ 2x 2 ⫹ x
冕冢
冣
6
1
9
⫺
⫹
dx
x
x ⫹ 1 共x ⫹ 1兲2
共x ⫹ 1兲⫺1
⫽ 6 ln x ⫺ ln x ⫹ 1 ⫹ 9
⫹C
⫺1
x6
9
⫽ ln
⫺
⫹C
x⫹1
x⫹1
ⱍⱍ
ⱍ ⱍ
✓CHECKPOINT 2
Find
冕
3x2 ⫹ 7x ⫹ 4
dx.
x3 ⫹ 4x2 ⫹ 4x
■
ⱍ
ⱍ
432
CHAPTER 6
Techniques of Integration
You can use the partial fraction decomposition technique outlined in
Examples 1 and 2 only with a proper rational function—that is, a rational
function whose numerator is of lower degree than its denominator. If the
numerator is of equal or greater degree, you must divide first. For instance, the
rational function
x3
⫹1
is improper because the degree of the numerator is greater than the degree of the
denominator. Before applying partial fractions to this function, you should divide
the denominator into the numerator to obtain
x2
x2
x3
x .
⫽x⫺ 2
⫹1
x ⫹1
Example 3
Find
Algebra Review
You can check the partial
fraction decomposition in
Example 3 by combining the
partial fractions to obtain the
original fraction, as shown in
Example 2(a) in the Chapter 6
Algebra Review, on page 471.
冕
Integrating an Improper Rational Function
x5 ⫹ x ⫺ 1
dx.
x 4 ⫺ x3
SOLUTION This rational function is improper—its numerator has a degree greater
than that of its denominator. So, you should begin by dividing the denominator
into the numerator to obtain
x5 ⫹ x ⫺ 1
x3 ⫹ x ⫺ 1 .
⫽
x
⫹
1
⫹
x 4 ⫺ x3
x 4 ⫺ x3
Now, applying partial fraction decomposition produces
x3 ⫹ x ⫺ 1 A
B
C
D .
⫽ ⫹ 2⫹ 3⫹
3
x 共x ⫺ 1兲
x
x
x
x⫺1
Multiplying both sides by the least common denominator x3共x ⫺ 1兲 produces the
basic equation.
x 3 ⫹ x ⫺ 1 ⫽ Ax 2共x ⫺ 1兲 ⫹ Bx共x ⫺ 1兲 ⫹ C共x ⫺ 1兲 ⫹ Dx 3
Basic equation
Using techniques similar to those in the first two examples, you can solve for A,
B, C, and D to obtain
A ⫽ 0,
B ⫽ 0,
C ⫽ 1,
and D ⫽ 1.
So, you can integrate as shown.
冕
x5 ⫹ x ⫺ 1
dx ⫽
x 4 ⫺ x3
⫽
⫽
冕冢
冕冢
3
x2
1
⫹ x ⫺ 2 ⫹ ln x ⫺ 1 ⫹ C
2
2x
ⱍ
✓CHECKPOINT 3
Find
冕
x3 ⫹ x ⫺ 1
dx
x 4 ⫺ x3
冣
1
1
x⫹1⫹ ⫹
dx
x
x ⫺ 1冣
x⫹1⫹
x 4 ⫺ x 3 ⫹ 2x 2 ⫹ x ⫹ 1
.
x3 ⫹ x2
■
ⱍ
SECTION 6.2
433
Partial Fractions and Logistic Growth
Logistic Growth Function
y
y=L
Logistic
growth model:
growth is
restricted.
t
FIGURE 6.3
In Section 4.6, you saw that exponential growth occurs in situations for which the
rate of growth is proportional to the quantity present at any given time. That is,
if y is the quantity at time t, then dy兾dt ⫽ ky. The general solution of this
differential equation is y ⫽ Ce kt. Exponential growth is unlimited. As long as
C and k are positive, the value of Ce kt can be made arbitrarily large by choosing
sufficiently large values of t.
In many real-life situations, however, the growth of a quantity is limited and
cannot increase beyond a certain size L, as shown in Figure 6.3. This upper limit
L is called the carrying capacity, which is the maximum population y共t兲 that can
be sustained or supported as time t increases. A model that is often used for this
type of growth is the logistic differential equation
冢
dy
y
⫽ ky 1 ⫺
dt
L
冣
Logistic differential equation
where k and L are positive constants. A population that satisfies this equation does
not grow without bound, but approaches L as t increases. The general solution of
this differential equation is called the logistic growth model and is derived in
Example 4.
STUDY TIP
The graph of
L
y⫽
1 ⫹ be⫺kt
Example 4
Deriving the Logistic Growth Model
Solve the equation
冢
SOLUTION
is called the logistic curve, as
shown in Figure 6.3.
Algebra Review
For help with the algebra used
to solve for y in Example 4, see
Example 2(c) in the Chapter 6
Algebra Review, on page 471.
冣
dy
y
⫽ ky 1 ⫺ .
dt
L
冢
dy
y
⫽ ky 1 ⫺
dt
L
冕
冕冢
冣
1
dy ⫽ k dt
y共1 ⫺ y兾L兲
Write in differential form.
1
dy ⫽
y共1 ⫺ y兾L兲
k dt
Integrate each side.
k dt
Rewrite left side using partial fractions.
冣
1
1
⫹
dy ⫽
y
L⫺y
ⱍⱍ
ⱍ
冕
冕
ⱍ
ln y ⫺ ln L ⫺ y ⫽ kt ⫹ C
✓CHECKPOINT 4
Show that if
y⫽
1
, then
1 ⫹ be⫺kt
dy
⫽ ky共1 ⫺ y兲.
dt
[Hint: First find ky共1 ⫺ y兲 in terms
of t, then find dy兾dt and show that
they are equivalent.] ■
Write differential equation.
ⱍ ⱍ
ⱍ ⱍ
ln
Find antiderivative of each side.
L⫺y
⫽ ⫺kt ⫺ C
y
Multiply each side by ⫺1 and simplify.
L⫺y
⫽ e⫺kt⫺C ⫽ e⫺Ce⫺kt
y
Exponentiate each side.
L⫺y
⫽ be⫺kt
y
Let ± e⫺C ⫽ b.
Solving this equation for y produces the logistic growth model y ⫽
L
.
1 ⫹ be⫺kt
434
CHAPTER 6
Techniques of Integration
Example 5
Comparing Logistic Growth Functions
Use a graphing utility to investigate the effects of the values of L, b, and k on the
graph of
y⫽
L
.
1 ⫹ be⫺kt
Logistic growth function 共L > 0, b > 0, k > 0兲
SOLUTION The value of L determines the horizontal asymptote of the graph to
the right. In other words, as t increases without bound, the graph approaches a
limit of L (see Figure 6.4).
4
4
y=
y = 1 −t
1+e
−3
3
4
2
1 + e−t
y=
−3
3
0
3
1 + e−t
−3
0
3
0
FIGURE 6.4
The value of b determines the point of inflection of the graph. When b ⫽ 1, the
point of inflection occurs when t ⫽ 0. If b > 1, the point of inflection is to the
right of the y-axis. If 0 < b < 1, the point of inflection is to the left of the y-axis
(see Figure 6.5).
4
y=
4
2
1 + 0.2e −t
y=
−3
3
4
2
1 + e −t
y=
−3
3
0
2
1 + 5e − t
−3
0
3
0
FIGURE 6.5
The value of k determines the rate of growth of the graph. For fixed values of b
and L, larger values of k correspond to higher rates of growth (see Figure 6.6).
4
y=
4
2
1 + e −0.2t
y=
−3
3
y=
2
1 + e−t
−3
0
4
3
−3
0
3
0
FIGURE 6.6
✓CHECKPOINT 5
Find the horizontal asymptote of the graph of y ⫽
4
.
1 ⫹ 5e⫺6t
2
1 + e −5t
■
SECTION 6.2
Example 6
Partial Fractions and Logistic Growth
435
Modeling a Population
The state game commission releases 100 deer into a game preserve. During the
first 5 years, the population increases to 432 deer. The commission believes that
the population can be modeled by logistic growth with a limit of 2000 deer. Write
a logistic growth model for this population. Then use the model to create a table
showing the size of the deer population over the next 30 years.
SOLUTION Let y represent the number of deer in year t. Assuming a logistic
growth model means that the rate of change in the population is proportional to
both y and 共1 ⫺ y兾2000兲. That is
冢
冣
dy
y
⫽ ky 1 ⫺
,
dt
2000
100 ≤ y ≤ 2000.
The solution of this equation is
y⫽
Daniel J. Cox/Getty Images
2000 .
1 ⫹ be⫺kt
Using the fact that y ⫽ 100 when t ⫽ 0, you can solve for b.
100 ⫽
2000
1 ⫹ be⫺k共0兲
b ⫽ 19
Then, using the fact that y ⫽ 432 when t ⫽ 5, you can solve for k.
432 ⫽
2000
1 ⫹ 19e⫺k共5兲
k ⬇ 0.33106
So, the logistic growth model for the population is
y⫽
✓CHECKPOINT 6
2000
.
1 ⫹ 19e⫺0.33106t
Logistic growth model
The population, in five-year intervals, is shown in the table.
Write the logistic growth model
for the population of deer in
Example 6 if the game preserve
could contain a limit of 4000 deer.
Time, t
0
5
10
15
20
25
30
Population, y
100
432
1181
1766
1951
1990
1998
■
CONCEPT CHECK
1. Complete the following: The technique of partial fractions involves the
decomposition of a ______ function into the ______ of two or more simple
______ functions.
2. What is a proper rational function?
3. Before applying partial fractions to an improper rational function, what
should you do?
4. Describe what the value of L represents in the logistic growth function
L
.
yⴝ
1 1 beⴚkt
436
CHAPTER 6
Skills Review 6.2
Techniques of Integration
The following warm-up exercises involve skills that were covered in earlier sections. You will use
these skills in the exercise set for this section. For additional help, review Sections 0.4 and 0.5.
In Exercises 1–8, factor the expression.
1. x 2 ⫺ 16
2. x2 ⫺ 25
3. x2 ⫺ x ⫺ 12
4. x 2 ⫹ x ⫺ 6
5. x 3 ⫺ x 2 ⫺ 2x
6. x 3 ⫺ 4x 2 ⫹ 4x
7. x 3 ⫺ 4x 2 ⫹ 5x ⫺ 2
8. x 3 ⫺ 5x 2 ⫹ 7x ⫺ 3
In Exercises 9–14, rewrite the improper rational expression as the sum of a
proper rational expression and a polynomial.
9.
x2 ⫺ 2x ⫹ 1
x⫺2
10.
2x 2 ⫺ 4x ⫹ 1
x⫺1
11.
x 3 ⫺ 3x 2 ⫹ 2
x⫺2
12.
x 3 ⫹ 2x ⫺ 1
x⫹1
13.
x 3 ⫹ 4x 2 ⫹ 5x ⫹ 2
x2 ⫺ 1
14.
x 3 ⫹ 3x 2 ⫺ 4
x2 ⫺ 1
Exercises 6.2
See www.CalcChat.com for worked-out solutions to odd-numbered exercises.
In Exercises 1–12, write the
decomposition for the expression.
partial
fraction
25.
冕
冕
冕
冕
x2 ⫺ 4x ⫺ 4
dx
x3 ⫺ 4x
26.
x⫹2
dx
x 2 ⫺ 4x
28.
2x ⫺ 3
dx
共x ⫺ 1兲2
30.
3x 2 ⫹ 3x ⫹ 1
dx
x 共x 2 ⫹ 2x ⫹ 1兲
32.
冕
冕
冕
冕
x2 ⫹ 12x ⫹ 12
dx
x 3 ⫺ 4x
1.
2共x ⫹ 20兲
x 2 ⫺ 25
2.
3x ⫹ 11
x 2 ⫺ 2x ⫺ 3
27.
3.
8x ⫹ 3
x 2 ⫺ 3x
4.
10x ⫹ 3
x2 ⫹ x
29.
5.
4x ⫺ 13
x 2 ⫺ 3x ⫺ 10
6.
7x ⫹ 5
6 共2x 2 ⫹ 3x ⫹ 1兲
31.
7.
3x 2 ⫺ 2x ⫺ 5
x3 ⫹ x2
8.
3x2 ⫺ x ⫹ 1
x共x ⫹ 1兲2
In Exercises 33 – 40, evaluate the definite integral.
9.
x⫹1
3共x ⫺ 2兲2
8x2 ⫹ 15x ⫹ 9
11.
共x ⫹ 1兲3
10.
3x ⫺ 4
共x ⫺ 5兲2
6x 2 ⫺ 5x
12.
共x ⫹ 2兲3
In Exercises 13–32, use partial fractions to find the
indefinite integral.
13.
15.
17.
19.
21.
23.
冕
冕
冕
冕
冕
冕
1
dx
x2 ⫺ 1
14.
⫺2
dx
x ⫺ 16
16.
2
1
dx
2x2 ⫺ x
18.
10
dx
⫺ 10x
20.
3
dx
x2 ⫹ x ⫺ 2
22.
5⫺x
dx
2x 2 ⫹ x ⫺ 1
24.
x2
冕
冕
冕
冕
冕
冕
冕
冕
冕
冕
5
33.
4
5
35.
1
1
37.
0
2
冕
冕
冕
冕
4x2 ⫹ 2 x ⫺ 1
dx
x 3 ⫹ x2
x4
dx
共x ⫺ 1兲3
3x
dx
x 2 ⫺ 6x ⫹ 9
1
1
dx
9 ⫺ x2
34.
x⫺1
dx
x2 共x ⫹ 1兲
36.
x3
dx
x2 ⫺ 2
38.
0
1
0
1
x 3 ⫺ 4x 2 ⫺ 3x ⫹ 3
dx 40.
x 2 ⫺ 3x
0
4
3
dx
2x 2 ⫹ 5x ⫹ 2
x2 ⫺ x
dx
x ⫹x⫹1
2
x3 ⫺ 1
dx
x2 ⫺ 4
x4 ⫺ 4
dx
x2 ⫺ 1
4
dx
x2 ⫺ 4
39.
⫺4
dx
x ⫺4
In Exercises 41– 44, find the area of the shaded region.
1
2
2
dx
x2 ⫺ 2x
x2
5
dx
⫹x⫺6
1
dx
4 x2 ⫺ 9
x⫹1
dx
x2 ⫹ 4x ⫹ 3
41. y ⫽
14
16 ⫺ x 2
2
42. y ⫽
⫺4
x2 ⫺ x ⫺ 6
y
y
7
6
5
4
3
2
1
x
−3 −2 −1
1 2 3 4
x
−2 −1
1
2
3
SECTION 6.2
x⫹1
x2 ⫺ x
44. y ⫽
x 2 ⫹ 2x ⫺ 1
x2 ⫺ 4
y
y
1
2
x
1
−1
x
1
2
3
4
1
−1
5
In Exercises 45 and 46, find the area of the region
bounded by the graphs of the given equations.
45. y ⫽
12
, y ⫽ 0, x ⫽ 0, x ⫽ 1
x2 ⫹ 5x ⫹ 6
46. y ⫽
⫺24
, y ⫽ 0, x ⫽ 1, x ⫽ 3
x2 ⫺ 16
1
a2 ⫺ x2
48.
1
x共x ⫹ a兲
1
50.
51. Writing
What is the first step when integrating
冕
x⫺5
共x ⫹ 1兲共a ⫺ x兲
dx? Explain. (Do not integrate.)
52. Writing State the method you would use to evaluate
each integral. Explain why you chose that method. (Do not
integrate.)
(a)
冕
2x ⫹ 1
dx
x2 ⫹ x ⫺ 8
(b)
冕
7x ⫹ 4
dx
x2 ⫹ 2x ⫺ 8
53. Biology A conservation organization releases 100
animals of an endangered species into a game preserve.
During the first 2 years, the population increases to 134
animals. The organization believes that the preserve has a
capacity of 1000 animals and that the herd will grow
according to a logistic growth model. That is, the size y of
the herd will follow the equation
冕
1
dy ⫽
y共1 ⫺ y兾1000兲
冕
冕
1
dx
共x ⫹ 1兲共500 ⫺ x兲
where t is the time in hours.
(a) Find the time it takes for 75% of the population to
become infected (when t ⫽ 0, x ⫽ 1).
(b) Find the number of people infected after 100 hours.
2t
dS
⫽
dt
共t ⫹ 4兲2
1
49.
x共a ⫺ x兲
x2
t ⫽ 5010
55. Marketing After test-marketing a new menu item, a
fast-food restaurant predicts that sales of the new item will
grow according to the model
In Exercises 47–50, write the partial fraction
decomposition for the rational expression. Check
your result algebraically. Then assign a value to
the constant a and use a graphing utility to check the
result graphically.
47.
437
54. Health: Epidemic A single infected individual enters a
community of 500 individuals susceptible to the disease.
The disease spreads at a rate proportional to the product of
the total number infected and the number of susceptible
individuals not yet infected. A model for the time it takes
for the disease to spread to x individuals is
k dt
where t is measured in years. Find this logistic curve.
(To solve for the constant of integration C and the
proportionality constant k, assume y ⫽ 100 when t ⫽ 0
and y ⫽ 134 when t ⫽ 2.) Use a graphing utility to graph
your solution.
where t is the time in weeks and S is the sales (in thousands
of dollars). Find the sales of the menu item at 10 weeks.
56. Biology One gram of a bacterial culture is present at
time t ⫽ 0, and 10 grams is the upper limit of the culture’s
weight. The time required for the culture to grow to y
grams is modeled by
kt ⫽
冕
1
dy
y共1 ⫺ y兾10兲
where y is the weight of the culture (in grams) and t is the
time in hours.
(a) Verify that the weight of the culture at time t is
modeled by
10
.
y⫽
1 ⫹ 9e⫺kt
Use the fact that y ⫽ 1 when t ⫽ 0.
(b) Use the graph to determine the constant k.
Bacterial Culture
y
Weight (in grams)
43. y ⫽
Partial Fractions and Logistic Growth
10
9
8
7
6
5
4
3
2
1
(2, 2)
t
2
4
6
8
Time (in hours)
10
12
438
CHAPTER 6
Techniques of Integration
57. Revenue The revenue R (in millions of dollars per year)
for Symantec Corporation from 1997 through 2005 can be
modeled by
R⫽
1340t 2 ⫹ 24,044t ⫹ 22,704
⫺6t 2 ⫹ 94t ⫹ 100
where t ⫽ 7 corresponds to 1997. Find the total revenue
from 1997 through 2005. Then find the average revenue
during this time period. (Source: Symantec Corporation)
58. Environment The predicted cost C (in hundreds of
thousands of dollars) for a company to remove p% of a
chemical from its waste water is shown in the table.
p
0
10
20
30
40
C
0
0.7
1.0
1.3
1.7
p
50
60
70
80
90
C
2.0
2.7
3.6
5.5
11.2
60. Population Growth The population of the United
States was 76 million people in 1900 and reached
300 million people in 2006. From 1900 through 2006,
assume the population of the United States can be modeled
by logistic growth with a limit of 839.1 million people.
(Source: U.S. Census Bureau)
(a) Write a differential equation of the form
冢
dy
y
⫽ ky 1 ⫺
dt
L
冣
where y represents the population of the United States
(in millions of people) and t represents the number of
years since 1900.
L
(b) Find the logistic growth model y ⫽
for this
1 ⫹ be⫺kt
population.
(c) Use a graphing utility to graph the model from part (b).
Then estimate the year in which the population of the
United States will reach 400 million people.
A model for the data is given by
C⫽
124p
, 0 ≤ p < 100.
共10 ⫹ p兲共100 ⫺ p兲
Business Capsule
Use the model to find the average cost for removing
between 75% and 80% of the chemical.
59. Biology: Population Growth The graph shows the
logistic growth curves for two species of the single-celled
Paramecium in a laboratory culture. During which time
intervals is the rate of growth of each species increasing?
During which time intervals is the rate of growth of each
species decreasing? Which species has a higher limiting
population under these conditions? (Source: Adapted
from Levine/Miller, Biology: Discovering Life, Second
Edition)
Paramecium Population
Number
P. aurelia
P. caudatum
2
4
6
8
10
Days
12
14
16
Photo courtesy of Susie Wang and Ric Kostick
usie Wang and Ric Kostick graduated from
the University of California at Berkeley with
degrees in mathematics. In 1999, Wang used
$10,000 to start Aqua Dessa Spa Therapy, a
high-end cosmetics company that uses natural
ingredients in their products. Now, the company
run by Wang and Kostick has annual sales of
over $10 million, operates under several brand
names, including 100% Pure, and has a global
customer base. Wang and Kostick attribute the
success of their business to applying what they
learned from their studies.
S
61. Research Project Use your school’s library, the
Internet, or some other reference source to research
the opportunity cost of attending graduate school
for 2 years to receive a Masters of Business
Administration (MBA) degree rather than working
for 2 years with a bachelor’s degree. Write a short
paper describing these costs.
A104
Answers to Selected Exercises
1. u ⫽ x; dv ⫽ e3x dx
5.
1 3x
3 xe
⫺
1 3x
9e
⫹C
7. ⫺x
13.
17.
⫺xe⫺x⫺e⫺x
21.
1 2
2t
⫺
1 4x
16 e
ln 共t ⫹
⫹C
⫹C
1兲⫺ 12
23. xe x ⫺ 2e x ⫹ C
ⱍ
2e⫺x
⫺
11. 14 e 4x ⫹ C
2
15. 12 e x ⫹
9. x ln 2x ⫺ x ⫹ C
1 4x
4 xe
67. (a) 3.2 ln 2 ⫺ 0.2 ⬇ 2.018
3. u ⫽ ln 2x; dv ⫽ x dx
19.
2x 2e x
ⱍ
1 2
4t
ln t ⫹ 1 ⫺
2xe⫺x
⫺
2e⫺x
(b) 12.8 ln 4 ⫺ 7.2 ln 3 ⫺ 1.8 ⬇ 8.035
⫹C
69. $18,482.03
C
⫺
4e xx
⫹
1
2t
⫹
4e x
⫹C
⫹C
25. ⫺e1兾t ⫹ C
1
共ln x兲3 ⫹ C
3
33.
2
3 x 共x
⫺ 1兲
3兾2
(b) $1,094,142.27
79. (a) $17,378.62
(b) $3681.26
SECTION 6.2
⫺
⫺ 1兲
5兾2
1
2
1
35. x 4 ⫹ x3 ⫹ x 2 ⫹ C
4
3
2
39. e共2e ⫺ 1兲 ⬇ 12.060
43.
5 6
36 e
1
⫹ 36
⬇ 56.060
81. 4.254
(page 436)
1. 共x ⫺ 4兲共x ⫹ 4兲
2. 共x ⫺ 5兲共x ⫹ 5兲
⫹C
3. 共x ⫺ 4兲共x ⫹ 3兲
4. 共x ⫺ 2兲共x ⫹ 3兲
e2x
⫹C
37.
4共2x ⫹ 1兲
5. x共x ⫺ 2兲共x ⫹ 1兲
6. x共x ⫺ 2兲2
7. 共x ⫺ 2兲共x ⫺ 1兲2
8. 共x ⫺ 3兲共x ⫺ 1兲2
41. ⫺12e⫺2 ⫹ 4 ⬇ 2.376
9. x ⫹
45. 2 ln 2 ⫺ 1 ⬇ 0.386
47. Area ⫽ 2e2 ⫹ 6 ⬇ 20.778
1
x⫺2
10. 2x ⫺ 2 ⫺
11. x 2 ⫺ x ⫺ 2 ⫺
2
x⫺2
12. x2 ⫺ x ⫹ 3 ⫺
4
x⫹1
60
0
77. $45,957.78
(page 436)
Skills Review
1
31. ⫺ 共ln x ⫹ 1兲 ⫹ C
x
4
15 共x
73. $4103.07
75. (a) $1,200,000
1
1
1
27. 2 x2共ln x兲2 ⫺ 2 x2 ln x ⫹ 4 x2 ⫹ C
29.
71. $931,265.10
13. x ⫹ 4 ⫹
6
,
x⫺1
x ⫽ ⫺1
14. x ⫹ 3 ⫹
1
,
x⫹1
x⫽1
2
0
1
49. Area ⫽ 9共2e3 ⫹ 1兲 ⬇ 4.575
1
1⫺x
8
0
1.
5
3
⫺
x⫺5 x⫹5
7.
3
5
⫺ 2
x
x
3
0
51. Proof
e5x
共25x 2 ⫺ 10x ⫹ 2兲 ⫹ C
53.
125
1
55. ⫺ 共1 ⫹ ln x兲 ⫹ C
x
57. 1 ⫺ 5e⫺4 ⬇ 0.908
1
59. 4 共e2 ⫹ 1兲 ⬇ 2.097
61.
63.
1,171,875
␲ ⬇ 14,381.070
256
65.
12,000
3
128
⫺8
⫺ 379
⬇ 0.022
128 e
11.
9.
3.
ⱍ ⱍ
ⱍ ⱍ
ⱍ
ⱍ
ⱍ ⱍ
x ⫺ 10
⫹C
19. ln
x
21. ln
23.
3
2
ⱍ
ⱍ
x共x ⫹ 2兲
⫹C
x⫺2
ⱍⱍ
33.
37.
1
6
1
2
ⱍ ⱍ
x⫺1
⫹C
x⫹2
ln 2x ⫺ 1 ⫺ 2 ln x ⫹ 1 ⫹ C
ⱍ
27.
1
共3 ln x ⫺ 4 ⫺ ln x 兲 ⫹ C
2
ⱍ
ⱍ
ⱍⱍ
1
⫹C
x⫺1
ⱍ
ⱍ
31. ln x ⫹ 2 ln x ⫹ 1 ⫹
10
(c) 11,321 units兾yr
ⱍ ⱍ
ⱍ ⱍ
1
3
⫹
x⫺5 x⫹2
1 x⫺1
ln
⫹C
2 x⫹1
2x ⫺ 1
⫹C
17. ln
x
25. ln
(b) 113,212 units
13.
1 x⫹4
⫹C
15. ln
4 x⫺4
ⱍ
(a) Increase
5.
1
1
⫹
3共x ⫺ 2兲 共x ⫺ 2兲2
1
8
2
⫺
⫹
x ⫹ 1 共x ⫹ 1兲2 共x ⫹ 1兲3
29. 2 ln x ⫺ 1 ⫹
0
10,000
9
1
⫺
x⫺3
x
ln 47 ⬇ ⫺0.093
⫺ ln 2 ⬇ ⫺0.193
7
41. 12 ⫺ 2 ln 7 ⬇ 5.189
1
⫹C
x⫹1
4
5
35. ⫺ 5 ⫹ 2 ln 3 ⬇ 0.222
1
39. 4 ln 2 ⫹ 2 ⬇ 3.273
43. 5 ln 2 ⫺ ln 5 ⬇ 1.856
A105
Answers to Selected Exercises
45. 24 ln 3 ⫺ 36 ln 2 ⬇ 1.413
冢
1
1
1
⫹
47.
2a a ⫹ x a ⫺ x
冣
冢
1 1
1
⫹
49.
a x
a⫺x
21.
冣
51. Divide x2 by 共x ⫺ 5兲 because the degree of the numerator
is greater than the degree of the denominator.
53. y ⫽
1000
1 ⫹ 9e⫺0.1656t
23.
冤
冪3
3
1200
33.
30
35.
55. $1.077 thousand
57. $11,408 million; $1426 million
ⱍ
冪3 ⫹ s ⫺ 冪3
冪3 ⫹ s ⫹ 冪3
ⱍ
ⱍ
⫹C
ⱍ
1
⫺1
2
9
⫹
⫺
⫹C
8 2共3 ⫹ 2x兲2 共3 ⫹ 2x兲3 4共3 ⫹ 2x兲4
冤
29. ⫺
−100
ln
ⱍ
25. ⫺ 12 x 共2 ⫺ x兲 ⫹ ln x ⫹ 1 ⫹ C
27.
−5
ⱍ冥 ⫹ C
1
4
2
⫺
⫹ ln 2 ⫹ 3t
27 2 ⫹ 3t 共2 ⫹ 3t兲2
冥
冪1 ⫺ x 2
x
⫹C
1 3
x 共⫺1 ⫹ 3 ln x兲 ⫹ C
9
31.
冢
ⱍ
ⱍ冣 ⫹ C
1
9 共3 ln x ⫺ 4 lnⱍ4 ⫹ 3 ln xⱍ兲 ⫹ C
25
1
3x ⫺
⫹ 10 ln 3x ⫺ 5
27
3x ⫺ 5
37. Area ⫽ 40
3
59. The rate of growth is increasing on 关0, 3兴 for P. aurelia and
on 关0, 2兴 for P. caudatum; the rate of growth is decreasing
on 关3, ⬁兲 for P. aurelia and on 关2, ⬁兲 for P. caudatum;
P. aurelia has a higher limiting population.
4
−1
61. Answers will vary.
9
−1
SECTION 6.3
(page 447)
Skills Review
Area ⫽ 13.3
39. Area ⫽
(page 447)
1. x 2 ⫹ 8x ⫹ 16
冣冥
0.5
2. x2 ⫺ 2x ⫹ 1
3. x 2 ⫹ x ⫹ 14
冢
冤
1
2
4 ⫹ ln
2
1 ⫹ e4
4. x 2 ⫺ 23 x ⫹ 19
5.
2
2
⫺
x
x⫹2
7.
3
2
3
⫺ ⫺
2共x ⫺ 2兲 x 2 2x
6. ⫺
−1
3
3
⫹
4x 4共x ⫺ 4兲
8.
3
2
4
⫺
⫹
x
x⫹1 x⫺2
x3
10. x3 ln x ⫺ ⫹ C
3
9. 2e x共x ⫺ 1兲 ⫹ C
3
−0.5
Area ⬇ 0.3375
41. Area ⫽ 14 关21冪5 ⫺ 8 ln共冪5 ⫹ 3兲 ⫹ 8 ln 2兴
20
冢
ⱍ冣 ⫹ C
1.
1
2
⫹ ln 2 ⫹ 3x
9 2 ⫹ 3x
3.
2共3x ⫺ 4兲
冪2 ⫹ 3x ⫹ C
27
7.
ⱍ
1 2
2
共x ⫺ 1兲e x ⫹ C
2
ⱍ
ⱍ
ⱍ
ⱍ
2 ⫹ 冪4 ⫺ x 2
x
15. 14 x 2共⫺1 ⫹ 2 ln x兲 ⫹ C
19.
1
4
ⱍ ⱍ
9. ln
1 3 ⫹ 冪x2 ⫹ 9
11. ⫺ ln
⫹C
3
x
1
13. ⫺ ln
2
5. ln共x 2 ⫹ 冪x 4 ⫺ 9 兲 ⫹ C
−1
5
−2
x
⫹C
1⫹x
Area ⬇ 9.8145
43.
⫺2冪2 ⫹ 4
⬇ 0.3905
3
冢
ⱍ ⱍ冣
ⱍ ⱍ冣
47. 12 2 ⫹ ln
⫹C
17. 3x 2 ⫺ ln共1 ⫹ e3x 兲 ⫹ C
共x 2冪x 4 ⫺ 4 ⫺ 4 lnⱍx 2 ⫹ 冪x 4 ⫺ 4ⱍ兲 ⫹ C
2
2
1 ⫹ e2
⬇ 6.7946
49. ⫺ 15
4 ⫹ 8 ln 4 ⬇ 7.3404
53. ⫺
冢1x ⫹ ln x ⫹x 1
5
9
45. ⫺ ⫹ ln ⬇ 0.2554
9
4
⫹C
51. 共x 2 ⫺ 2x ⫹ 2兲e x ⫹ C
55. (a) 0.483
(b) 0.283