41. Solve the logistic equation dp. by answering the following questions. (a) Find expressions A and B so that. 1 P(k mp) A P B

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1 40 Chapter 5 Integration 4. Solve the logistic equation dp dt by answering the following questions. (a) Find expressions A and B so that (Note: A and B will involve k and m.) (b) Evaluate A P P(k mp) P(k mp) A P B k mp B k mp dp 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) De kt where C and D are expressions involving k and m. 4 Integration by Parts 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. 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

2 Chapter 5 Section 4 Integration by Parts 4 WHY INTEGRATION BY PARTS WORKS HOW AND WHEN TO USE INTEGRATION BY PARTS 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 or f(x)g(x) f(x)g(x) dx f(x)g(x) dx f(x)g(x) dx f(x)g(x) f(x)g(x) dx which is precisely the formula for 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. 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.

3 42 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. Find xe 2x dx. In this case, both factors x and e 2x are easy to integrate. Both are also easy to differentiate, but the process of differentiation simplifies x while it leaves e 2x essentially the same. This suggests that you should try integration by parts with g(x) e 2x and f(x) x Then, G(x) e 2x 2 and f(x) DI and so xe 2x dx 2 e2x (x) 2 e2x () dx 2 xe2x 2 e 2x dx 2 xe2x 4 e2x C 2 x 2 e2x C EXAMPLE 4.2 Find xx 5 dx.

4 Chapter 5 Section 4 Integration by Parts 43 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) 3 D I and so xx 5 dx 2 3 x(x 5)3/2 2 5) 3(x 3/2 dx 2 3 x(x 5)3/2 4 5 (x 5)5/2 C Note 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 (u 3/2 5u /2 ) du u5/2 5u3/2 5/2 3/2 C 2 5 (x 5)5/2 0 3 (x 5)3/2 C 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 3 (x 5)3/2 4 5 (x 5)5/2 (x 5) 3/2 2x (x 5) (x 5) 3/2 2x (x 2 5)3/2 5 (x 5) (x 5)5/2 0 (x 5)3/2 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.

5 44 Chapter 5 Integration EXAMPLE 4.3 Find ln x dx. The trick is to write ln x as the product (ln x), in which the factor is easy to integrate and the factor ln x is simplified by differentiation. This suggests that you use integration by parts with g(x) and f(x) ln x Then, G(x) x and f(x) and so I D ln x dx (ln x) dx x ln x x x ln x x C x(ln x ) C x x dx x ln x dx REPEATED APPLICATIONS OF INTEGRATION BY PARTS Sometimes integration by parts leads to a new integral that also must be integrated by parts. This situation is illustrated in the next example. Find x 2 e x dx. Since the factor e x is easy to integrate and the factor x 2 is simplified by differentiation, try integration by parts with g(x) e x and f(x) x 2 Then, G(x) e x and f(x) 2x DI and so x 2 e 2 dx x 2 e x 2xe x dx To find EXAMPLE 4.4 xe x dx, you have to integrate by parts again, this time with g(x) e x and f(x) x Then, G(x) e x and f(x)

6 Chapter 5 Section 4 Integration by Parts 45 and so DI x 2 e x dx x 2 e x 2xe x dx x 2 e x 2 xe x e x dx x 2 e x 2(xe x e x ) C (x 2 2x 2)e x 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. By first using the fact that e xy e x e y and then separating the variables, we get The integral on the left is just e y (we will add the C later), but the integral on the right requires integration by parts. Proceeding as in Example 4., we choose g(x) e x and f(x) x so that G(x) e x and f(x) and e y dy xe x dx xe x dx (e x )(x) dy dx xexy xex e y e x () dx xe x e x

7 46 Chapter 5 Integration Returning to the given differential equation, we obtain the general solution e y dy xe x dx e y xe x e x C e x (x ) C Finally, since y ln 2 when x 0, we have e ln 2 e 0 (0 ) C 2 ()() C and C 3 Thus, the required solution of the differential equation is e y e x (x ) 3 or, equivalently, y ln [e x (x ) 3] P. R. O. B. L. E. M. S 4.5 P. R. O. B. L. E. M. S 4.5 In Problems through 25, use integration by parts to find the given integral.. xe x dx 2. xe x/2 dx 3. ( x)e x dx 4. (3 2x)e x dx 5. t ln 2tdt 6. t ln t 2 dt 7. ve v/5 dv 8. we 0.w dw 9. xx 6dx 0.x xdx

8 Chapter 5 Section 4 Integration by Parts 47. x(x ) 8 dx 2. (x )(x 2) 6 dx x x x 2 dx 2x dx 5. x 2 e x dx 6. x 2 e 3x dx 7. x 3 e x dx 8. x 3 e 2x dx 9. x 2 ln xdx 20. x(ln x) 2 dx ln x ln x 2. dx 22. x 3 dx x2 23. x 3 e x2 dx [Hint: Rewrite the integrand as x 2 (xe x2 ).] 24. x 3 (x 2 ) 0 dx 25. x 7 (x 4 5) 8 dx DISTANCE EFFICIENCY FUND-RAISING MARGINAL COST POPULATION GROWTH 26. Find the function whose tangent has slope (x )e x for each value of x and whose graph passes through the point (, 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). 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. 29. After t hours on the job, a factory worker can produce 00te 0.5t units per hour. How many units does the worker produce during the first 3 hours? 30. After t weeks, contributions in response to a local fund-raising campaign were coming in at the rate of 2,000te 0.2t dollars per week. How much money was raised during the first 5 weeks? 3. A manufacturer has found that marginal cost is (0.q )e 0.03q dollars per unit when q units have been produced. The total cost of producing 0 units is $200. What is the total cost of producing the first 20 units? 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 thousand people per year. If the current population is 2 million, what will the population be 5 years from now?

9 48 Chapter 5 Integration 33. (a) Use integration by parts to derive the formula (b) Use the formula in part (a) to find 34. (a) Use integration by parts to derive the formula (ln x) n dx x(ln x) n n (ln x) (b) Use the formula in part (a) to find x n e ax dx a xn e ax n x a n e ax dx x 3 e 5x dx. (ln x) 3 dx. n dx CHAPTER SUMMARY AND REVIEW PROBLEMS 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: The integral of (for n ) Constant multiple rule: kf(x) dx k f(x) dx x n dx n xn C x : x dx ln x C Sum rule: [f(x) g(x)] dx f(x) dx g(x) dx The integral of e kx : e kx dx k ekx C 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