Review of Basic Integration Formulas 6.1 INTEGRATION by SUBSTITUTION

388 CHAPTER 6 Techniques of Integration

6.1 INTEGRATION BY SUBSTITUTION

■ Use the basic integration formulas to find indefinite integrals. ■ Use substitution to find indefinite integrals. ■ Use substitution to evaluate definite integrals. ■ Use integration to solve real-life problems.

Review of Basic Integration Formulas Each of the basic integration rules you studied in Chapter 5 was derived from a corresponding differentiation rule. It may surprise you to learn that, although you now have all the necessary tools for differentiating algebraic, exponential, and logarithmic functions, your set of tools for integrating these functions is by no means complete. The primary objective of this chapter is to develop several techniques that greatly expand the set of integrals to which the basic integration formulas can be applied.

Basic Integration Formulas

1. Constant Rule: k dx kx C

xn1 2. Simple Power Rule n 1: xn dx C n 1 du 3. General Power Rule n 1: un dx un du dx un1 C n 1

4. Simple Exponential Rule: ex dx ex C

du 5. General Exponential Rule: eu dx eu du dx eu C 1 6. Simple Log Rule: dx lnx C x dudx 1 7. General Log Rule: dx du u u lnu C

As you will see once you work a few integration problems, integration is not nearly as straightforward as differentiation. A major part of any integration problem is determining which basic integration formula (or formulas) to use to solve the problem. This requires remembering the basic formulas, familiarity with various procedures for rewriting integrands in the basic forms, and lots of practice. SECTION 6.1 Integration by Substitution 389

Integration by Substitution STUDY TIP There are several techniques for rewriting an integral so that it fits one or more of When you use integration by the basic formulas. One of the most powerful techniques is integration by substitution, you need to realize substitution. With this technique, you choose part of the integrand to be u and that your integral should contain then rewrite the entire integral in terms of u. just one variable. For instance, the integrals x EXAMPLE 1 Integration by Substitution dx x 12 Use the substitution u x 1 to find the indefinite integral. and x dx x 12 u 1 du u2 SOLUTION From the substitution u x 1, are in the correct form, but the du x u 1, 1, and dx du. integral dx x By replacing all instances of x and dx with the appropriate u-variable forms, you dx u2 obtain is not. x u 1 dx du Substitute for x and dx. x 12 u2 u 1 du Write as separate fractions. u2 u2 TRY IT 1 1 1 du Simplify. u u2 Use the substitution u x 2 to find the indefinite integral. 1 lnu C Find antiderivative. u x dx 2 1 x 2 lnx 1 C. Substitute for u. x 1

The basic steps for integration by substitution are outlined in the guidelines below.

Guidelines for Integration by Substitution 1. Let u be a function of x (usually part of the integrand). 2. Solve for x and dx in terms of u and du. 3. Convert the entire integral to u-variable form and try to fit it to one or more of the basic integration formulas. If none fits, try a different substitution. 4. After integrating, rewrite the antiderivative as a function of x. 390 CHAPTER 6 Techniques of Integration

DISCOVERY EXAMPLE 2 Integration by Substitution Suppose you were asked to evaluate the integrals below. Which Find xx2 1 dx. one would you choose? Explain your reasoning. SOLUTION Consider the substitution u x2 1, which produces du 2x dx. To create 2xdxas part of the integral, multiply and divide by 2. x2 1 dx or u12 du

1 2 xx2 1 dx x2 11 2 2x dx Multiply and divide by 2. x x 1 dx 2

1 u1 2 du Substitute for x and dx. 2 1 u32 C Power Rule 2 32

1 u3 2 C Simplify. 3

1 x2 13 2 C Substitute for u. 3 You can check this result by differentiating.

TRY IT 2

Find xx2 4 dx.

EXAMPLE 3 Integration by Substitution

e3x Find dx. 1 e3x

SOLUTION Consider the substitution u 1 e3x, which produces du 3e3x dx. To create 3e3x dx as part of the integral, multiply and divide by 3.

1u du

e3x 1 1 dx 3e3x dx Multiply and divide by 3. 1 e3x 3 1 e3x 1 1 du Substitute for x and dx. 3 u 1 lnu C Log Rule 3 TRY IT 3 1 ln1 e3x C Substitute for u. e2x 3 Find dx. 2x 1 e Note that the absolute value is not necessary in the final answer because the quan- tity 1 e3x is positive for all values of x. SECTION 6.1 Integration by Substitution 391

EXAMPLE 4 Integration by Substitution Find the indefinite integral.

xx 1 dx

SOLUTION Consider the substitution u x 1, which produces du dx and x u 1.

xx 1 dx u 1u1 2 du Substitute for x and dx.

u3 2 u1 2 du Multiply.

u52 u32 C Power Rule 52 32

2 2 x 15 2 x 13 2 C Substitute for u. 5 3 This form of the antiderivative can be further simplified. 2 2 6 10 x 152 x 132 C x 152 x 132 C 5 3 15 15 2 x 1323x 1 5 C 15 2 x 1323x 2 C 15 You can check this answer by differentiating.

TRY IT 4 Find the indefinite integral.

xx 2 dx

Example 4 demonstrates one of the characteristics of integration by substitution. That is, you can often simplify the form of the antiderivative as it exists immediately after resubstitution into x-variable form. So, when working the exercises in this section, don’t assume that your answer is incorrect just because it doesn’t look exactly like the answer given in the back of the text. You may be able to reconcile the two answers by algebraic simplification.

TECHNOLOGY

If you have access to a symbolic integration utility, try using it to find an antiderivative of fx x2x 1 and check your answer analytically using the substitution u x 1. You can also use the utility to solve several of the exercises in this section. 392 CHAPTER 6 Techniques of Integration

Substitution and Definite Integrals The fourth step outlined in the guidelines for integration by substitution on page 389 suggests that you convert back to the variable x. To evaluate definite integrals, however, it is often more convenient to determine the limits of integration for the variable u. This is often easier than converting back to the variable x and evaluat- ing the antiderivative at the original limits.

EXAMPLE 5 Using Substitution with a Definite Integral

5 x Evaluate dx. 1 2x 1

SOLUTION Use the substitution u 2x 1, which implies that u2 2x 1, 1 2 x 2 u 1 , and dx u du. Before substituting, determine the new upper and lower limits of integration. Lower limit: When x 1, u 21 1 1. Upper limit: When x 5, u 25 1 3. y x y Now, substitute and integrate, as shown. 2x 1 5 Integration limits for x 4 Integration limits for u

3 5 x 3 1 u2 1 dx u du Substitute for x, dx, and 2 limits of integration. 5 2x 1 u 2 5, 1 1 (1, 1) ( 3 ) 3 1 1 u2 1 du Simplify. 2 1 x −1 1 2 3 4 5 1 u3 3 Find antiderivative. − u 1 2 3 1 1 1 9 3 1 Apply Fundamental Theorem. FIGURE 6.1 Region Before 2 3 Substitution 16 Simplify. 3 u2 1 f u) ) f(u) 2 5 (3, 5) TRY IT 5

4 2 Evaluate x4x 1 dx. 3 0

2 (1, 1) 1 STUDY TIP u −1 1 2 3 4 5 In Example 5, you can interpret the equation − 1 5 x 3 1 u2 1 dx u du 2x 1 u 2 FIGURE 6.2 Region After 1 1 Substitution graphically to mean that the two different regions shown in Figures 6.1 and 6.2 have the same area. SECTION 6.1 Integration by Substitution 393

Application Integration can be used to find the probability that an event will occur. In such an Researchers such as psychologists application, the real-life situation is modeled by a probability density function f, use definite integrals to represent and the probability that x will lie between a and b is represented by the probability that an event will occur. For instance, a probability of b Pa ≤ x ≤ b fx dx. 0.5 means that an event will occur a about 50% of the time. The probability Pa ≤ x ≤ b must be a number between 0 and 1.

EXAMPLE 6 Finding a Probability

A psychologist finds that the probability that a participant in a memory experi- ALGEBRA REVIEW ment will recall between a and b percent (in decimal form) of the material is For help on the algebra in b 28 Pa ≤ x ≤ b x3 1 x dx, 0 ≤ a ≤ b ≤ 1. Example 6, see Example 1(a) in a 9 the Chapter 6 Algebra Review on page 446. Find the probability that a randomly chosen participant will recall between 0% and 87.5% of the material.

SOLUTION Let u 3 1 x. Then u3 1 x, x 1 u3, and dx 3u2 du. Lower limit: When x 0, u 3 1 0 1. y 3 Upper limit: When x 0.875, u 1 0.875 0.5. 28 3 y 9 x 1 x To find the probability, substitute and integrate, as shown. 1.5 0.875 28 1 2 28 x3 1 x dx 1 u3u3u2 du 1.0 0 9 1 9 28 1 2 u6 u3 du 0.5 3 1 28 u7 u4 12 x 3 7 4 1 0.51.0 1.5 0.865 0.875 So, the probability is about 86.5%, as indicated in Figure 6.3. FIGURE 6.3

TRY IT 6 Use Example 6 to find the probability that a participant will recall between 0% and 62.5% of the material.

T AKE ANOTHER LOOK

Probability

In Example 6, explain how you could find a value of b such that P0 ≤ x ≤ b 0.5. 394 CHAPTER 6 Techniques of Integration

PREREQUISITE The following warm-up exercises involve skills that were covered in earlier sections. You will REVIEW 6.1 use these skills in the exercise set for this section.

In Exercises 1–8, evaluate the indefinite integral. 1 1. 5 dx 2. dx 3

3. x32 dx 4. x23 dx

5. 2xx2 13 dx 6. 3x2x3 12 dx

2 7.6e6x dx 8. dx 2x 1

In Exercises 9–12, simplify the expression. 9.2xx 12 xx 1 10. 6xx 43 3x2 x 42 11.3x 712 2xx 712 12. x 513 5x 523

EXERCISES 6.1

In Exercises 1–38, find the indefinite integral. ex ex 25. dx 26. dx ex 2 1 ex 1. x 24 dx 2. x 532 dx x x2 27. dx 28. dx 2 4 x 14 x 13 3. dt 4. dt t 92 1 t3 x 5x 29. dx 30. dx 2t 1 2y3 3x 12 x 43 5. dt 6. dy t 2 t 2 y4 1 1 1 31. dt 32. dx t 1 x 1 7. 1 x dx 8. 3 x52 dx 2t 1 6x x 33. dt 34. dx 12x 2 6x2 2 t x 9. dx 10. dx 3x2 x x3 x x x2 35. dx 36. dx 1 1 2x 1 x 1 11. dx 12. dx 5x 13 3x 12 37. t21 t dt 38. y23y 1 dy 1 1 13. dx 14. dx x 1 5x 1 In Exercises 39–46, evaluate the definite integral. e3x 4e2x 4 4 15. 3x dx 16. 2x dx 1 e 1 e 39. 2x 1 dx 40. 4x 1 dx 2x ex1 0 2 1 2 17. 2 dx 18. dx 3x 2 e x 1 41. 3xex dx 42. e2x dx x2 2x 0 0 19. dx 20. dx 4 x 1 x 1 x 4 43. dx 44. xx 54 dx x 42 t 0 0 21. xx2 4 dx 22. dt 0.5 0.5 1 t2 45. x1 x3 dx 46. x21 x3 dx 0 0 23. e5x dx 24. tet 21 dt SECTION 6.1 Integration by Substitution 395

In Exercises 47–54, find the area of the region bounded by the 63. Probability The probability of recall in an experiment is graphs of the equations.Then use a graphing utility to graph the modeled by region and verify your answer. b ≤ ≤ 15 47. y xx 3, y 0, x 7 P a x b x 1 x dx a 4 48. y x2x 1, y 0, x 4 where x is the percent of recall (see figure). 49. y x21 x, y 0, x 3 (a) What is the probability of recalling between 40% and 50. y x2x 2, y 0, x 7 80%? x2 1 (b) What is the median percent recall? That is, for what 51. y , y 0, x 1, x 5 ≤ ≤ 2x 1 value of b is P 0 x b 0.5? y 2x 1 1 y 15 52. y , y 0, x , x 6 y x 1 x x 3 2 4 y 1155 x3(1 x)3 2 3 32 53. y x3 x 1, y 0, x 0, x 7 2 Pa( x b) P(a x b) 54. y x3 x 2, y 0, x 2, x 10 2 1 In Exercises 55–58, find the area of the region bounded by the 1 graphs of the equations. x a b 0.5 1 x 55.y xx 2, y 0 56. y x3 1 x, y 0 0.5 a b 1 y y Figure for 63 Figure for 64 64. Probability The probability of finding between a and b 2 0.5 percent iron in ore samples is modeled by 1 b 1155 Pa ≤ x ≤ b x31 x32 dx x x a 32 2 1 0.5 1.0 (see figure). Find the probabilities that a sample will contain between (a) 0% and 25% and (b) 50% and 100% iron. 57.y2 x2 1 x2 58. y 11 x , 65. Meteorology During a two-week period in March in a (Hint: Find the area of y 0, x 0, x 4 small town near Lake Erie, the measurable snowfall S (in the region bounded by y inches) on the ground can be modeled by y x1 x2 and y 0. Then multiply by 4.) 3 St t14 t, 0 ≤ t ≤ 14 y 2 where t represents the day. 1 1 (a) Use a graphing utility to graph the function. x x (b) Find the average amount of snow on the ground during 1234 −2 2 the two-week period. −1 (c) Find the total snowfall over the two-week period. 66. Revenue A company sells a seasonal product that gen- In Exercises 59 and 60, find the volume of the solid generated by erates a daily revenue R (in dollars per day) modeled by revolving the region bounded by the graph(s) of the equation(s) 2 12 ≤ ≤ about the x-axis. R 0.06t 365 t 1250, 0 t 365 59. y x1 x2 where t represents the day. 60. y x1 x2, y 0 (a) Find the average daily revenue over a period of 1 year. (b) Describe a product whose seasonal sales pattern resem- In Exercises 61 and 62, find the average amount by which the bles the model. Explain your reasoning. function f exceeds the function g on the interval. 1 x In Exercises 67 and 68,use a program similar to the Midpoint Rule 61. fx , gx , 0, 1 x 1 x 12 program on page 366 with n 10 to approximate the area of the region bounded by the graph(s) of the equation(s). 62. fx x4x 1, gx 2x3, 0, 2 67. y 3 x 4 x, y 0 68. y2 x2 1 x2