410 Chapter 5
Integration
41. Solve the logistic equation
dP
P(k mP)
dt
by answering the following questions.
(a) Find expressions A and B so that
A
B
1
P(k mP) P k mP
(Note: A and B will involve k and m.)
(b) Evaluate
A
B
dP
P k mP
where A and B are the expressions found in part (a).
(c) Separate the variables in the given differential equation and solve, using the
result of part (b). Express P(t) in the form
P(t) C
1 De kt
where C and D are expressions involving k and m.
dQ
kQ(B Q),
dt
dQ
where k and B are positive constants, then the rate of change
is greatest when
dt
B
Q(t) . What does this result tell you about the inflection point of a logistic
2
curve? Explain.
42. Show that if a quantity Q satisfies the differential equation
4
Integration
by Parts
In this section, you will see a technique you can use to integrate certain products
f(x)g(x). The technique is called integration by parts, and as you will see, it is
a restatement of the product rule for differentiation. Here is a statement of the
technique.
Integration by Parts
■
If G is an antiderivative of g, then
f(x)g(x) dx f(x)G(x) f(x)G(x) dx
Chapter 5 ■ Section 4
WHY INTEGRATION
BY PARTS WORKS
Integration by Parts
411
To see how integration by parts is a restatement of what happens when the product
rule is used to differentiate f(x)G(x), where G is an antiderivative of g, note that,
d
[f(x)G(x)] f(x)G(x) f(x)G(x) f(x)G(x) f(x)g(x)
dx
Expressed in terms of integrals, this says
f(x)G(x) or
f(x)G(x) dx f(x)g(x) dx f(x)G(x) f(x)g(x) dx
f(x)G(x) dx
which is precisely the formula for integration by parts.
HOW AND WHEN TO USE
INTEGRATION BY PARTS
Integration by parts is a technique for integrating products f(x)g(x), in which one of
the factors, say g(x), can be easily integrated and the other, f(x), becomes simpler
when differentiated. To evaluate such an integral,
f(x)g(x) dx, using integration by
parts, first integrate g and multiply the result by f to get
f(x)G(x)
where G is an antiderivative of g. Then multiply the antiderivative G by the derivative of f and subtract the integral of this product from the result of the first step to get
f(x)G(x) f(x)G(x) dx
This expression will be equal to the original integral
lucky, the new integral
f(x)g(x) dx, and if you are
f(x)G(x) dx will be easier to find than the original one.
Here is an informal, step-by-step summary of the procedure.
How to Use Integration by Parts to Integrate a Product
Step 1. Select one of the factors of the product as the one to be integrated and
the other as the one to be differentiated. The factor selected for integration
should be easy to integrate, and the factor selected for differentiation should
become simpler when differentiated.
Step 2. Integrate the designated factor and multiply it by the other factor.
Step 3. Differentiate the designated factor, multiply it by the integrated factor
from step 2, and subtract the integral of this product from the result of step 2.
Step 4. Complete the procedure by finding the new integral that was formed in
step 3. Add the constant of integration C only at the very end.
412
Chapter 5
Integration
Here are some examples illustrating the procedure. In each example, g(x) is used
to denote the factor that is to be integrated and f(x) is used to denote the factor that
is to be differentiated. As reminders, the letters I (for integrate) and D (for differentiate) are placed above the appropriate factors in the integrand.
With practice, you will become familiar with the pattern and should find that you
can do integration by parts without the intermediate step of writing down the functions g(x), f(x), G(x), and f(x).
EXAMPLE 4.1
Find
xe2x dx.
Solution
In this case, both factors x and e2x are easy to integrate. Both are also easy to differentiate, but the process of differentiation simplifies x while it leaves e2x essentially
the same. This suggests that you should try integration by parts with
Then,
DI
xe2x
and so
g(x) e2x
and
f(x) x
1
G(x) e2x
2
and
f(x) 1
dx 12 e (x) 12 e (1) dx
2x
1
1
xe2x 2
2
2x
e2x dx
1
1
1
1 2x
xe2x e2x C x e C
2
4
2
2
EXAMPLE 4.2
Find
xx 5 dx.
Chapter 5 ■ Section 4
Integration by Parts
413
Solution
Again, both factors in the product are easy to integrate and differentiate. However,
the factor x is simplified by differentiation, whereas the derivative of x 5 is even
more complicated than x 5 itself. This suggests that you should try integration
by parts with
g(x) x 5
2
G(x) (x 5)3/2
3
Then,
f(x) 1
and
D I
2
xx 5 dx x(x 5)3/2 3
2
x(x 5)3/2 3
and so
Note
f(x) x
and
2
(x 5)3/2 dx
3
4
(x 5)5/2 C
15
Some integrals can be evaluated by either substitution or integration by parts.
For instance, the integral in Example 4.2 can be found by substituting as
follows:
Let u x 5. Then du dx and x u 5, and
xx 5 dx (u 5)u du (u3/2 5u1/2) du
u5/2 5u3/2
2
10
C (x 5)5/2 (x 5)3/2 C
5/2
3/2
5
3
This form of the integral is not the same as that found in Example 4.2. To
show that the two forms are equivalent, note that the antiderivative in Example 4.2 can be expressed as
2x
4
2x
4
(x 5)3/2 (x 5)5/2 (x 5)3/2
(x 5)
3
15
3
15
2x 4
2
10
(x 5)3/2
(x 5)3/2 (x 5) 3
3
5
3
2
10
(x 5)5/2 (x 5)3/2
5
3
which is the form of the antiderivative obtained by substitution. This example shows that it is quite possible for you to do everything right and still not
get the answer given at the back of the book.
414
Chapter 5
Integration
EXAMPLE 4.3
Find
ln x dx.
Solution
The trick is to write ln x as the product 1(ln x), in which the factor 1 is easy to integrate and the factor ln x is simplified by differentiation. This suggests that you use
integration by parts with
g(x) 1
and
G(x) x
Then,
f(x) ln x
f(x) and
1
x
and so
REPEATED APPLICATIONS OF
INTEGRATION BY PARTS
I D
1 (ln x) dx x ln x 1
dx x ln x x
x ln x x C x(ln x 1) C
ln x dx x
1 dx
Sometimes integration by parts leads to a new integral that also must be integrated
by parts. This situation is illustrated in the next example.
EXAMPLE 4.4
Find
x2ex dx.
Solution
Since the factor ex is easy to integrate and the factor x2 is simplified by differentiation, try integration by parts with
Then,
and so
To find
Then,
g(x) ex
and
f(x) x2
G(x) ex
and
f (x) 2x
D I
2 x
x
x2e2dx x e 2 xe dx
xex dx, you have to integrate by parts again, this time with
g(x) ex
G(x) e
x
and
f(x) x
and
f(x) 1
Chapter 5 ■ Section 4
Integration by Parts
and so
415
DI
x2ex dx x2ex 2 xex dx
x2ex 2 xex ex dx
x2ex 2(xex ex) C (x2 2x 2)ex C
SOLVING A DIFFERENTIAL
EQUATION BY PARTS
In the next example, we solve a differential equation by using integration by parts.
EXAMPLE 4.5
Find the particular solution of the differential equation
dy
xexy
dx
that satisfies the initial condition y ln 2 when x 0.
Solution
By first using the fact that exy ex
and then separating the variables, we get
ey
dy
xex
xexy y
dx
e
ey dy xex dx
The integral on the left is just ey (we will add the C later), but the integral on the
right requires integration by parts. Proceeding as in Example 4.1, we choose
g(x) ex
and
f(x) x
G(x) ex
and
f(x) 1
ex(1) dx xex ex
so that
and
xex dx (ex)(x) 416
Chapter 5
Integration
Returning to the given differential equation, we obtain the general solution
ey dy xex dx
ey xex ex C ex(x 1) C
Finally, since y ln 2 when x 0, we have
eln 2 e0(0 1) C
2 (1)(1) C
and
C3
Thus, the required solution of the differential equation is
e y ex(x 1) 3
or, equivalently,
y ln [ex(x 1) 3]
P . R . O . B . L . E . M . S
4.5
In Problems 1 through 25, use integration by parts to find the given integral.
1.
3.
5.
7.
9.
xex dx
2.
(1 x)ex dx
4.
t ln 2tdt
6.
vev/5 dv
8.
xx 6 dx
10.
xe x/2 dx
(3 2x)ex dx
t ln t 2 dt
we0.1w dw
x1 xdx
Chapter 5 ■ Section 4
11.
13.
15.
17.
19.
21.
23.
24.
Integration by Parts
x(x 1)8 dx
12.
x
dx
x 2
14.
x2ex dx
16.
x3ex dx
18.
x2 ln x dx
20.
ln x
dx
x2
22.
2
x3ex dx
417
(x 1)(x 2)6 dx
x
dx
2x 1
x2e3x dx
x3e2x dx
x(ln x)2 dx
ln x
dx
x3
2
[Hint: Rewrite the integrand as x2(xex ).]
x3(x2 1)10 dx
25.
x7(x4 5)8 dx
26. Find the function whose tangent has slope (x 1)ex for each value of x and
whose graph passes through the point (1, 5).
27. Find the function whose tangent has slope x ln x for each value of x 0 and
whose graph passes through the point (2, 3).
DISTANCE
28. After t seconds, an object is moving with velocity tet/2 meters per second. Express
the position of the object as a function of time.
EFFICIENCY
29. After t hours on the job, a factory worker can produce 100te0.5t units per hour. How
many units does the worker produce during the first 3 hours?
FUND-RAISING
30. After t weeks, contributions in response to a local fund-raising campaign were
coming in at the rate of 2,000te0.2t dollars per week. How much money was raised
during the first 5 weeks?
MARGINAL COST
31. A manufacturer has found that marginal cost is (0.1q 1)e0.03q dollars per unit
when q units have been produced. The total cost of producing 10 units is $200. What
is the total cost of producing the first 20 units?
POPULATION GROWTH
32. It is projected that t years from now the population of a certain city will be changing
at the rate of t ln t 1 thousand people per year. If the current population is 2
million, what will the population be 5 years from now?
418
Chapter 5
Integration
33. (a) Use integration by parts to derive the formula
1
n
xneax dx xneax a
a
(b) Use the formula in part (a) to find
xn1eax dx
x3e5x dx.
34. (a) Use integration by parts to derive the formula
(ln x)n dx x(ln x)n n
(b) Use the formula in part (a) to find
(ln x)n1 dx
(ln x)3 dx.
CHAPTER SUMMARY AND REVIEW PROBLEMS
IMPORTANT TERMS, SYMBOLS,
AND FORMULAS
Antiderivative; indefinite integral:
f(x) dx F(x) C if and only if F(x) f(x)
Power rule:
xn dx The integral of
1
:
x
1
xn1 C
n1
Constant multiple rule:
Sum rule:
1
dx ln x C
x
kf(x) dx k
[f(x) g(x)] dx The integral of ekx :
(for n 1)
f(x) dx
f(x) dx g(x) dx
1
ekx dx ekx C
k
Integration by substitution:
g(u)
du
dx G(u) C
dx
Differential equation
General solution; particular solution
Separable differential equation:
where G is an antiderivative of g