Integration by Parts, and As You Will See, It Is a Restatement of the Product Rule for Differentiation

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

1 A B 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 C P(t) 1 Dekt where C and D are expressions involving k and m.

dQ 42. Show that if a quantity Q satisﬁes the differential equation 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.

4 In this section, you will see a technique you can use to integrate certain products Integration 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 by Parts 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 Integration by Parts 411

WHY INTEGRATION To see how integration by parts is a restatement of what happens when the product BY PARTS WORKS 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) f(x)G(x) dx f(x)g(x) dx or f(x)g(x) dx f(x)G(x) f(x)G(x) dx

which is precisely the formula for integration by parts. HOW AND WHEN TO USE Integration by parts is a technique for integrating products f(x)g(x), in which one of INTEGRATION BY PARTS 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 f(x)g(x) dx, and if you are

lucky, the new integral 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 func- tions g(x), f(x), G(x), and f(x).

EXAMPLE 4.1

2x Find xe 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 g(x) e2x and f(x) x 1 Then, G(x) e2x and f(x) 1 2 DI and so xe2x 1 1 dx e2x(x) e2x(1) dx 2 2 1 1 xe2x e2x dx 2 2 1 1 1 1 xe2x e2x C x e2x 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 and f(x) x 2 Then, G(x) (x 5)3/2 and f(x) 1 3 DI 2 2 and so xx 5 dx x(x 5)3/2 (x 5)3/2 dx 3 3 2 4 x(x 5)3/2 (x 5)5/2 C 3 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: Note 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 Exam- ple 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 f(x) ln x 1 Then, G(x) x and f(x) x and so I D 1 ln x dx 1 (ln x) dx x ln x x dx x ln x 1 dx x x ln x x C x(ln x 1) C

REPEATED APPLICATIONS OF Sometimes integration by parts leads to a new integral that also must be integrated INTEGRATION BY PARTS 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 g(x) ex and f(x) x2 Then, G(x) ex and f(x) 2x DI and so x2e2dx x2ex 2xexdx

To ﬁnd xex dx, you have to integrate by parts again, this time with g(x) ex and f(x) x Then, G(x) ex and f(x) 1 Chapter 5 Section 4 Integration by Parts 415

and so DI x2ex dx x2ex 2 xex dx

x2ex 2xex ex dx

x2ex 2(xex ex) C (x2 2x 2)ex C

SOLVING A DIFFERENTIAL In the next example, we solve a differential equation by using integration by parts. EQUATION 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 ex By ﬁrst using the fact that exy and then separating the variables, we get e y dy xex xexy dx ey

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 so that G(x) ex and f(x) 1 and

xex dx (ex)(x) ex(1) dx xex ex 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 C 3 Thus, the required solution of the differential equation is ey 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. xex dx 2. xex/2 dx

3. (1 x)ex dx 4. (3 2x)ex dx

5. t ln 2tdt 6. t ln t2 dt

7. vev/5 dv 8. we0.1w dw

9.xx 6dx 10. x1 xdx Chapter 5 Section 4 Integration by Parts 417

11. x(x 1)8 dx 12. (x 1)(x 2)6 dx

x x 13. dx 14. dx x 2 2x 1

15. x2ex dx 16. x2e3x dx

17. x3ex dx 18. x3e2x dx

19. x2 ln xdx 20. x(ln x)2 dx

ln x ln x 21. dx 22. dx x2 x3

2 2 23. x3ex dx [Hint: Rewrite the integrand as x2(xex ) .]

24. x3(x2 1)10 dx 25. x7(x4 5)8dx

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 lnx 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 te t/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 100te 0.5t 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 lnt 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 xn1eax dx a a (b) Use the formula in part (a) to ﬁnd x3e5x dx.

34. (a) Use integration by parts to derive the formula

(ln x)n dx x(ln x)n n (ln x)n1 dx

(b) Use the formula in part (a) to ﬁnd (ln x)3 dx.

CHAPTER SUMMARY AND REVIEW PROBLEMS

IMPORTANT TERMS, SYMBOLS, Antiderivative; indeﬁnite integral: AND FORMULAS f(x) dx F(x) C if and only if F(x) f(x)

1 Power rule: xn dx xn1 C (for n 1) n 1 1 1 The integral of : dx ln x C x x

Constant multiple rule: kf(x) dx kf (x) dx

Sum rule: [f(x) g(x)] dx f(x) dx g(x) dx

1 The integral of ekx : ekx dx ekx C k Integration by substitution: du g(u) dx G(u) C where G is an antiderivative of g dx Differential equation General solution; particular solution Separable differential equation: