C H A P T E R 6 Techniques of Integration

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© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. C H A P T E R 6 Techniques of Integration S... ection 6.1 Integration by Parts and Present Value..............................................368 Section 6.2 Integration Tables...............................................................................382 Quiz Yourself .............................................................................................................390 Section 6.3 Numerical Integration.........................................................................393 Section 6.4 Improper Integrals ..............................................................................401 Review Exercises ........................................................................................................405 Test Yourself .............................................................................................................414 368 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. C H A P T E R 6 Techniques of Integration Section 6.1 Integration by Parts and Present Value Skills Warm Up 1. ( ) ( ) ( ) ln 1 1 1 f x x f x x = + ′ = + 2. ( ) ( ) ( ) 2 2 ln 1 2 1 f x x f x x x = − ′ = − 3. ( ) ( ) 3 3 2 3 x x f x e f x xe = ′ = 4. ( ) ( ) 2 2 2 x x f x e fx xe − − = ′ = − 5. ( ) ( ) 2 2 2 x x x f x x e f x xe xe = ′ = + 6. ( ) ( ) 2 2 2 2 x x x f x xe fx xe e − − − = ′ = − + 7. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 3 2 2 23 16 16 64 3 3 3 4 4 2 8 8 16 16 b a A fx gx dx x x dx x dx x x − − − = ⎡⎣ − ⎤⎦ = ⎡ − + − − ⎤ ⎣ ⎦ = − + = ⎡⎣− + ⎤⎦ = − + − − = ∫ ∫ ∫ 8. ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 1 2 1 3 1 1 1 3 1 1 3 3 43 2 1 1 1 1 b a A f x g x dx x dx x dx x x − − − = ⎡⎣ − ⎤⎦ = ⎡ − + − ⎤ ⎣ ⎦ = − + = ⎡⎣− + ⎤⎦ = − + − − = ∫ ∫ ∫ 9. ( ) ( ) ( ) ( ) ( ) ( ) 5 2 1 5 2 1 3 2 5 1 1 3 125 1 3 3 4 5 4 5 2 5 50 25 2 5 36 b a A fx gx dx x x dx x x dx x x x − − − = ⎡⎣ − ⎤⎦ = ⎡ − − ⎤ ⎣ ⎦ = − + + = ⎡⎣− + + ⎤⎦ = − + + − + − = ∫ ∫ ∫ 10. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 3 1 1 1 3 2 3 3 2 1 1 1 3 2 3 3 2 1 1 4 3 2 1 4 3 2 3 1 1 1 1 1 1 4 2 4 2 1 1 1 1 81 4 2 4 2 4 3 2 1 1 3 2 3 3 3 3 3 3 1 3 1 3 27 b a A fx gx dx f x g x dx g x f x dx x x x dx x x x dx x x x dx x x x dx x x x x x x x x − − − − = ⎡⎣ − ⎤⎦ = ⎡⎣ − ⎤⎦ + ⎡⎣ − ⎤⎦ = ⎡ − + − − ⎤ + ⎡ − − − + ⎤ ⎣ ⎦ ⎣ ⎦ = − − + + − + + − = ⎡⎣ − − + ⎤⎦ + ⎡⎣− + + − ⎤⎦ = − − + − + − − + − + ∫ ∫ ∫ ∫ ∫ ∫ ∫ ( 9 ) ( 1 1 ) 2 4 2 + − 9 − − + 1 + − 3 = 8 Section 6.1 Integration by Parts and Present Value 369 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. ∫ xe3x dx u = x, dv = e3x dx 2. ∫ x2e3x dx u = x2, dv = e3x dx 3. ∫ x ln 2x dx u = ln 2x, dv = x dx 4. ∫ ln 4x dx u = ln 4x, dv = dx 5. Let u = x and dv = e3x dx. Then du = dx and 1 3 3 v = e x. ( ) 3 3 3 3 3 3 1 1 3 3 1 1 3 9 19 3 1 x x x x x x xe dx xe e dx xe e C e x C = − = − + = − + ∫ ∫ 6. Let u = x and dv = e−x dx. Then du = dx and v = −e−x . ( ) 1 x x x x x x xe dx xe e dx xe e C e x C − − − − − − = − − − = − − + = − + + ∫ ∫ 7. Let u = ln x and dv = x3 dx. Then du 1 dx x = and 4 . 4 v = x ( ) 4 4 3 4 3 4 4 4 ln ln 1 4 4 ln 4 4 ln 4 16 4 ln 1 16 x x dx x x x dx x x x x dx x x x C x x C = − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − = − + = − + ∫ ∫ ∫ 8. Let u = ln x and dv = x4 dx. Then du 1 dx x = and 5 . 5 v = x ( ) 5 5 4 5 4 5 5 5 ln ln 1 5 5 ln 5 5 ln 5 25 5 ln 1 25 x x dx x x x dx x x x x dx x x x C x x C = − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − = − + = − + ∫ ∫ ∫ 9. Let u = ln 2x and dv = dx. Then du 1 dx x = and v = x. ( ) ln 2 ln 2 1 ln 2 ln 2 ln 2 1 x dx x x x dx x x x dx x x x C x x C = − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − = − + = − + ∫ ∫ ∫ 10. ∫ ln x2 dx = 2∫ ln x dx Let u = ln x and dv = dx. Then du 1 dx x = and v = x. ( ) ln 2 2 ln 1 2 ln 2 ln 2 2 ln 1 x dx x x x dx x x x dx x x x C x x C ⎡ ⎛ ⎞ ⎤ = ⎢ − ⎜ ⎟ ⎥ ⎣ ⎝ ⎠ ⎦ = ⎡ − ⎤ ⎣ ⎦ = − + = − + ∫ ∫ ∫ 11. Let u = x2 and dv = e−x dx. Then du = 2x dx and v = −e−x . ∫ x2e−x dx = −x2e−x + 2∫ xe−x dx Let u = x and dv = e−x dx. Then du = dx and v = −e−x . ( ) 2 2 2 2 2 2 2 2 2 x x x x x x x x x e dx x e xe e dx xe xe e C e x x C − − − − − − − − = − + ⎡− + ⎤ ⎣ ⎦ = − − − + = − + + + ∫ ∫ 370 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 12. Let u = x2 and dv = e2x dx. Then du = 2x dx and 1 2 2 v = e x . 2 2 1 2 2 2 2 ∫ x e x dx = x e x − ∫ xe x dx Let u = x and dv = e2x dx. Then du = dx and 1 2 2 v = e x . ( ) 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 2 2 2 1 1 1 2 2 4 1 4 2 2 1 x x x x x x x x x e dx x e xe e dx xe xe e C e x x C = − ⎡ − ⎤ ⎣ ⎦ = − + + = − + + ∫ ∫ 13. Let u = ln x and dv = x1 2 dx. Then du 1 dx x = and 2 3 2. 3 v = x 1 2 3 2 3 2 3 2 1 2 3 2 3 2 ln 2 ln 2 1 3 3 2 ln 2 3 3 2 ln 4 3 9 x x dx x x x dx x x x x dx x x x C = − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − = − + ∫ ∫ ∫ 14. Let u = x2 and ( )1 2 dv = x − 3 dx. Then 2 du x dx = and ( )3 2 23 v = x − 3 . 2 2 2 ( )3 2 4 ( )3 2 2 2 ( )3 2 4 ( )3 2 3 3 3 3 ∫ x x − 3 dx = x x − 3 − ∫ x x − 3 dx = x x − 3 − ∫ x x − 3 dx Let u = x and ( )3 2 dv = x − 3 dx. Then du = dx and 2( )5 2 5 v = x − 3 . ( ) ( ) ( ) ( ) ( ) ( ) 2 2 3 2 5 2 5 2 2 3 2 5 2 7 2 2 42 2 3 35 5 2 8 16 3 15 105 3 3 3 3 3 3 3 x x dx x x xx x dx x x xx x C − = − − ⎡ − − − ⎤ ⎣ ⎦ = − − − + − + ∫ ∫ 15. Let u = 2x2 and dv = ex dx. Then du = 4x dx and v = ex . ∫ 2x2ex dx = 2x2ex − ∫ 4xex dx Let u = 4x and dv = ex dx. Then du = 4 dx and v = ex . ( ) ( ) 2 2 2 2 2 2 4 4 2 4 4 2 2 2 x x x x x x x x x e dx x e xe e dx x e xe e C e x x C = − ⎡ − ⎤ ⎣ ⎦ = − + + = − + + ∫ ∫ 16. 2 2 x x x dx xe dx e ∫ = ∫ − Let u = 2x and dv = e−x dx. Then du = 2 dx and v = −e−x. ( ) ( ) 2 2 2 2 2 2 1 2 1 x x x x x x x x dx xe e dx e xe e C e x C x C e − − − − − = − − − = − − + = − + + + = − + ∫ ∫ 17. 4 1 4 ( ) 1 4 4 4 ∫ e x dx = ∫ e x 4 dx = e x + C 18. 2 1 2 ( ) 1 2 2 2 ∫ e− x dx = − ∫ e− x −2 dx = − e− x + C 19. Let u = x and dv = e4x dx. Then du = dx and 1 4 . 4 v = e x ( ) 4 4 4 4 4 4 1 1 4 4 1 1 4 16 4 1 16 x x x x x x xe dx xe e dx xe e C e x C = − = − + = − + ∫ ∫ 20. Let u = x and dv = e−2x dx. Then du = dx and 1 2 2 v = − e− x. ( ) 2 2 2 2 2 2 1 1 2 2 1 1 2 4 1 4 2 1 x x x x x x xe dx xe e dx xe e C e x C − − − − − − = − − − = − − + = − + + ∫ ∫ 21. Let u = x and dv = e−x 4 dx. Then du = dx and v = −4e−x 4. 4 4 4 4 4 4 4 4 4 4 4 4 16 x x x x x x x xe dx xe e dx xe e dx xe e C − − − − − − − = − − − = − + = − − + ∫ ∫ ∫ Section 6.1 Integration by Parts and Present Value 371 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 22. Let 1 3 2 u x = and . x dv e dx = Then 2 32 du = x dx and v = ex. 1 3 1 3 3 2 2 2 2 ∫ x ex dx = x ex − ∫ x ex dx Let 2 32 u = x and dv = ex dx. Then du = 3x dx and v = ex. 1 3 1 3 3 2 2 2 2 ∫ x ex dx = x ex − ⎡⎣ x ex − ∫ 3xex dx⎤⎦ Let u = 3x and dv = ex dx. Then du = 3 dx and v = ex . ( ) ( ) 3 3 2 3 2 3 2 1 1 3 2 2 2 1 3 2 2 1 2 3 3 3 3 3 6 6 x x x x x x x x x x x e dx x e x e xe e dx x e x e xe e C e x x x C = − + ⎡ − ⎤ ⎣ ⎦ = − + − + = − + − + ∫ ∫ 23. Let u = ln(t + 1) and dv = t dt. Then ( ) 1 1 du dt t = + and 2 . 2 v = t ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 ln 1 ln 1 1 2 2 1 ln 1 1 1 1 2 2 1 ln 1 1 ln 1 2 22 1 2 ln 1 2 2 ln 1 4 t t dt t t t dt t t t t dt t t t t t t C t t t t t C + = + − + = + − ⎛ − + ⎞ ⎜ + ⎟ ⎝ ⎠ ⎡ ⎤ = + − ⎢ − + + ⎥ + ⎣ ⎦ = ⎡⎣ + + − − + ⎤⎦ + ∫ ∫ ∫ 24. Let u = x − 1and dv = ex. Then du = dx and v = ex. ( ) ( ) ( ) ( ) 1 1 1 2 x x x x x x x e dx x e e dx x e e C e x C − = − − = − − + = − + ∫ ∫ 25. Let u 1. t = Then 2 du 1 dt. t = ⎛− ⎞ ⎜ ⎟ ⎝ ⎠ 1 1 2 t e dt eu du eu C e t C t ∫ = −∫ = − + = − + 26. ( ) ( ) ( ) ( ) 3 3 2 2 1 ln 1 ln ln 2 1 2 ln dx x dx x x x x C C x − − = = + − = − + ∫ ∫ 27. Let ( )2 u = ln x and dv = x dx. Then (2 ln x) du dx x = and 2 . 2 v = x ( ) ( ) 2 2 2 ln ln ln 2 ∫ x x dx = x x − ∫ x x dx Let u = ln x and v = x dx. Then du 1 dx x = and 2 . 2 v = x ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 ln ln ln 2 2 2 ln ln 2 2 4 2 ln 2 ln 1 4 x x dx x x x x x dx x x x x x C x x x C ⎡ ⎤ = − ⎢ − ⎥ ⎣ ⎦ = − + + = ⎡ − + ⎤ + ⎣ ⎦ ∫ ∫ 28. Let u = ln 3x and dv = dx. Then du 1 dx x = and v = x. ( ) ( ) ( ) ( ) ln 3 ln 3 ln 3 ln 3 1 x dx x x dx x x x C x x C = − = − + = ⎡⎣ − ⎤⎦ + ∫ ∫ 29. ( )2 ( ) ( )3 ln ln 2 1 ln 3 x x dx x dx C x x = ⎛ ⎞ = + ⎜ ⎟ ⎝ ⎠ ∫ ∫ 372 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 30. Let u = ln 3x. Then du 1 dx. x = (ln 3 ) 1 1 1 ln ln ln 3 x dx u du x u C x C − ⎛ ⎞ = − ⎜ ⎟ ⎝ ⎠ = + = + ∫ ∫ 31. Let u = ln x and 2 dv 1 dx. x = Then du 1 dx x = and v 1. x = − 2 2 ln ln 1 1 ln 1 ln 1 ln 1 x dx x dx x x x x x dx x x x C x x x C x = − − − ⋅ = − + = − + + = − + ∫ ∫ ∫ 32. Let u = ln 2x and 2 dv 1 dx. x = Then du 1 dx x = and v 1. x = − 2 2 ln 2 ln 2 1 1 ln 2 1 ln 2 1 ln 2 1 x dx x dx x x x x x dx x x x C x x x C x = − − − ⋅ = − + = − − + + = − + ∫ ∫ ∫ 33. Let u = x and dv = x − 1 dx. Then du = dx and ( )3 2 23 v = x − 1 . ( ) ( ) ( ) ( ) ( ) ( ) 3 2 3 2 3 2 5 2 3 2 2 2 3 3 2 4 3 15 2 15 1 1 1 1 1 1 3 2 x x dx xx x dx xx x C x x C − = − − − = − − − + = − + + ∫ ∫ 34. Let u = x and ( ) 1 2 dv x 1 dx. − = − Then du = dx and v = 2 x − 1. ( ) ( ) 3 2 2 1 2 1 1 2 1 4 1 3 1 1 2 3 x dx x x x dx x x x x C x x = − − − − = − − − + = − + ∫ ∫ 35. ( ) ( ) ( ) 2 2 3 2 4 3 2 1 2 1 2 2 4 3 2 xx dx xx x dx x x x dx x x xC + = + + = + + = + + + ∫ ∫ ∫ 36. Let u = x and ( ) 1 2 dv 2 3x dx. − = + Then du = dx and 2(2 3 )1 2. 3 v = + x ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 3 2 1 2 2 2 3 2 2 3 2 3 3 3 2 2 3 4 2 3 3 27 2 2 3 9 22 3 27 2 2 3 3 4 27 x dx x x x dx x x x x C x x x C x x C = + − + + = + − + + = + ⎡⎣ − + ⎤⎦ + = + − + ∫ ∫ 37. Let u = xe2x and ( ) 2 dv 2x 1 dx. − = + Then du = e2x (2x + 1) dx and ( ) 1 . 2 2 1 v x = − + ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 1 2 1 2 2 1 2 1 2 2 1 4 4 2 1 x x x x x x xe dx xe e x x xe e C x e C x = − + + + = − + + + = + + ∫ ∫ 38. Let u = x2ex2 and ( ) dv x x2 1 2 dx. − = + Then du = 2xex2 (x2 + 1) dx and ( 2 ) 1 . 2 1 v x = − + ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 2 1 1 2 1 2 2 1 x x x x x x x x e dx x e x dx x x x e xe dx x x e e C x e C x ⎡ ⎤ = ⎢ ⎥ + ⎢ + ⎥ ⎣ ⎦ = − + + = − + + + = + + ∫ ∫ ∫ Section 6.1 Integration by Parts and Present Value 373 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 39. Let u = ln x and dv = x5 dx. Then du 1 dx x = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ and 6 . 6 v = x ] 6 5 5 1 1 1 6 6 1 6 6 6 ln ln 6 6 6 36 + 1 6 36 36 5 1 36 36 56.060 e e e e x x dx x x x dx e x e e e = − ⎡ ⎤ = − ⎢ ⎥ ⎣ ⎦ = − = + ≈ ∫ ∫ 40. Let u = ln x and dv = 2x. Then du 1 dx x = and v = x2. 2 ] 1 1 1 2 2 1 2 2 ln ln 2 1 1 2 2 4.195 e e e e x x x x xdx e x e = − ⎡ ⎤ = − ⎢ ⎥ ⎣ ⎦ = + ≈ ∫ ∫ 41. Let u = ln(1 + 2x) and dv = dx.Then 2 1 2 du dx x = + and v = x. ( ) ( ) ( ) 1 1 1 0 0 0 1 0 1 0 ln 1 2 ln 1 2 2 1 2 ln 3 1 1 2 1 ln 3 1 ln 2 1 2 ln 3 1 1 ln 3 2 3 ln 3 1 2 0.648 x dx x x x dx x dx x x x + = + ⎤⎦ − + = − ⎡ − ⎤ ⎢⎣ + ⎥⎦ = − ⎡ − + ⎤ ⎢⎣ ⎥⎦ = − + = − ≈ ∫ ∫ ∫ 42. Let u = x and dv = e−x 2 dx. Then du = dx and v = −2e−x 2. 4 2 4 4 2 0 2 0 0 2 2 4 0 2 2 2 8 4 12 4 2.376 x x x x x dx xe e dx e e e e − − − − − = − ⎤⎦ + = − − ⎡⎣ ⎤⎦ = − + ≈ ∫ ∫ 43. Let u = x and ( )1 2 dv = x + 1 dx. Then du dx = and ( )3 2 23 v = x + 1 . ( ) ( ) ( ) ( ) 8 3 2 3 2 0 3 2 5 2 8 0 2 2 3 3 2 4 3 15 1192 15 1 1 1 1 1 x x dx xx x dx x x x + = + − + = ⎡ + − + ⎤ ⎣ ⎦ = ∫ ∫ 44. Let u = x and ( ) 1 2 dv x 4 dx. − = + Then du = dx and ( )1 2 v = 2 x + 4 . ( ) ( ) ( ) ( ) ( ) 12 1 2 1 2 1 2 0 1 2 3 2 12 0 43 64 3 4 2 4 2 4 2 4 4 xx dx xx x dx x x x − + = + − + = ⎡ + − + ⎤ ⎣ ⎦ = ∫ ∫ 45. Let u = x2 and dv = ex dx. Then du = 2x dx and v = ex . 2 2 2 2 2 1 1 1 ∫ x ex dx = x ex ⎤⎦ − 2∫ xex dx Let u = x and dv = ex dx. Then du = dx and v = ex. ( ) ( ) ( ) 2 2 2 2 2 1 1 1 2 2 2 1 2 2 2 2 4 2 4 22 4 4 2 2 2 12.060 x x x x x e dx e e xe e dx e e e e e e e e e e e e e = − − ⎡⎣ ⎤⎦ − = − − − − ⎡⎣ ⎤⎦ = − − + + − = − ≈ ∫ ∫ 374 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 46. Let u = x2 and dv = e−3x dx. Then du = 2x dx and 1 3 . 3 v = − e− x 2 2 3 2 3 3 2 3 3 0 1 2 1 2 3 3 3 3 ∫ x e− x dx = − x e− x − ∫ − xe− x dx = − x e− x + ∫ xe− x dx Let u = x and dv = e−3x dx. Then du = dx and 1 3 . 3 v = − e− x 2 2 3 2 3 3 3 0 2 3 3 3 2 2 3 3 3 0 6 1 2 1 1 3 3 3 3 1 2 2 3 9 9 1 2 2 3 9 27 2 50 27 x x x x x x x x x x xe dx xe xe e dx x e xe e dx x e xe e e − − − − − − − − − − − = − + ⎡− − − ⎤ ⎢⎣ ⎥⎦ = − − + = ⎡− − − ⎤ ⎢⎣ ⎥⎦ − = ∫ ∫ ∫ 47. 2 3 0 Area = ∫ x ex dx Let u = x3 and dv = ex dx. Then du = 3x2 dx and v = ex. ∫ x3ex dx = x3ex − 3∫ x2ex dx Let u = x2 and dv = ex dx. Then du = 2x dx and v = ex . 3 3 2 3 2 3 2 3 6 x x x x x x x x e dx x e x e xe dx xe xe xedx = − ⎡ − ⎤ ⎣ ⎦ = − + ∫ ∫ ∫ Let u = x and dv = ex dx. Then du = dx and v = ex. ( ) 3 3 2 3 2 3 6 3 6 6 x x x x x x x e dx x e x e xe e dx x x x e C = − + ⎡ − ⎤ ⎣ ⎦ = − + − + ∫ ∫ ( ) 2 3 3 2 2 0 0 2 Area 3 6 6 2 6 20.778 x ex dx x x x ex e = = ⎡ − + − ⎤ ⎣ ⎦ = + ≈ ∫ 48. ( ) ( ) 1 2 1 1 2 1 Area 1 1 x x x e dx x e dx − − = − − = − ∫ ∫ Let u = 1 − x2 and dv = ex dx. Then du = −2x dx and v = ex . ( ) ( ) ( ) ( )( ) 2 2 2 1 1 2 1 2 Example 1 x x x x x x x e dx x e xe dx x e xe e − = − + = − + − ∫ ∫ ( ) ( ) ( ) ( ) 1 2 1 2 1 1 1 1 Area 1 1 2 2 0 2 2 2 2 4 1.472 x x x x x e dx x e xe e e e e e e − − − − = − = ⎡ − + − ⎤ ⎣ ⎦ = + − − − − = ≈ ∫ 0 0 2 60 −2 −3 2 3 Section 6.1 Integration by Parts and Present Value 375 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 49. 3 3 3 3 0 0 Area 1 1 9 9 = ∫ xe−x dx = ∫ xe−x dx Let u = x and dv = e−x 3 dx. Then du = dx and v = −3e−x 3. 3 3 3 3 3 1 1 3 3 9 9 1 3 9 9 x x x x x xe dx xe e dx xe e C − − − − − = ⎡− + ⎤ ⎣ ⎦ = ⎡− − ⎤ + ⎢⎣ ⎥⎦ ∫ ∫ ( ) 3 3 3 0 3 3 0 Area 1 3 9 9 1 3 3 2 0.2642 x x x xe e e x e e − − − = ⎡− − ⎤ ⎢⎣ ⎥⎦ = ⎡− + ⎤ ⎢⎣ ⎥⎦ − = ≈ 50. 3 1 Area ln e = ∫ x− x dx Let u = ln x and dv = x−3 dx. Then du 1 dx x = and 1 2. 2 v = − x− 3 2 2 3 2 2 2 ln 1 ln 1 1 2 2 ln 1 2 2 ln 1 2 4 x x dx x x x dx x x x dx x x C x x − − − − = − + ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − + = − − + ∫ ∫ ∫ 2 2 2 1 Area ln 1 1 3 0.1485 2 4 4 e x e x x ⎡ ⎤ − − = ⎢− − ⎥ = ≈ ⎣ ⎦ 51. 2 1 Area ln e = ∫ x x dx Let u = ln x and dv = x2. Then du 1 dx x = and 3 . 3 v = x 3 2 2 3 3 ln ln 3 3 ln 3 9 x x dx x x x dx x x x C = − = − + ∫ ∫ 3 3 2 1 1 3 3 3 Area ln ln 3 9 1 2 1 3 9 9 9 9 4.575 e e x x dx x x x e e e ⎡ ⎤ = = ⎢ − ⎥ ⎣ ⎦ ⎛ ⎞ ⎛ ⎞ = ⎜ − ⎟ − ⎜− ⎟ = + ⎝ ⎠ ⎝ ⎠ = ∫ 52. 1 2 Area ln e x dx x = ∫ Let u = ln x and 2 dv 1 dx. x = Then du 1 dx x = and v 1. x = − 2 2 ln x dx 1 ln x 1 dx 1 ln x 1 x x x x x − ∫ = − − ∫ = − − ( ) 1 2 1 Area ln 1 ln 1 1 1 1 1 2 0.264 e e x dx x x x x e e e = = ⎡− − ⎤ ⎢⎣ ⎥⎦ = ⎛− − ⎞ − − ⎜ ⎟ ⎝ ⎠ = − ≈ ∫ 0 3 0 0.5 −2 4 −0.1 0.2 0 0 3 8 0 0 3 1 376 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 53. Let u = ln x and dv = xn dx. Then du 1 dx x = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ and ( ) 1 . 1 xn v n + = + ( ) ( ) 1 1 1 1 1 1 2 ln ln 1 1 1 ln 1 1 1 ln 1 1 1 1 1 1 ln 1 n n n n n n n n x x dx x x x dx n x n x x xdx n n x x x C n n n x n x C n + + + + + + = − ⋅ + + = − + + = − ⋅ + + + + = ⎡⎣− + + ⎤⎦ + + ∫ ∫ ∫ 54. Let u = xn and dv = eax dx. Then du = nxn −1 dx and v 1eax. a = 1 n ax xneax dx x e n xn eax dx a a ∫ = − ∫ − 55. Using n = 2 and a = 5, 2 5 2 5 2 5 . 5 5 x ∫ x e x dx = x e − ∫ xe x dx Now, using n = 1 and a = 5, ( ) 2 5 5 2 5 5 2 5 5 5 5 2 2 1 5 5 5 5 2 2 5 25 125 25 10 2 . 125 x x x x x x x x x e dx x e xe e dx x e xe e C e x x C ⎡ ⎤ = − ⎢ − ⎥ ⎣ ⎦ = − + + = − + + ∫ ∫ 56. Using n = 1 and a = −3, ( ) 3 3 3 3 3 3 1 3 3 1 1 3 9 1 3 1 . 9 x x x x x x xe dx xe e dx xe e C e x C − − − − − − = + − = − − + = − + + ∫ ∫ 57. Using n = −4, ( ) ( ) ( ) ( ) 4 1 4 2 3 3 ln 1 4 1 ln 4 1 1 3 ln 9 1 1 3 ln 9 x x dx x x C x x C x C x − + − − = ⎡⎣− + − + ⎤⎦ + − + = − − + − = + + ∫ 58. Using 3, 2 n = ( ) ( ) ( ) 3 2 1 3 2 2 5 2 5 2 5 2 32 25 4 ln 1 3 1 ln 1 2 1 5 ln 2 4 1 5 ln 25 2 2 2 5 ln 25 x x dx x x C x x C x x C x x C + ⎡ ⎛ ⎞ ⎤ = ⎢− + ⎜ + ⎟ ⎥ + + ⎣ ⎝ ⎠ ⎦ = ⎛− + ⎞ + ⎜ ⎟ ⎝ ⎠ = ⎛− + ⎞ + ⎜ ⎟ ⎝ ⎠ = − + + ∫ 59. 2 3 4 8 0 3 379 0.022 128 128 ∫ t e− t dt = − e− ≈ 60. ( ) 4 2 1 4 ln 224 ln 2 19 32.755 3 ∫ x + x dx = − ≈ 61. ( ) 5 4 2 3 2 0 25 1,171,875 14,381.070 256 ∫ x − x dx = π ≈ 62. 9 10 1 ln 9 1 1982.392 100 100 e∫ x x dx = e + ≈ 63. (a) (b) ( ) ( ) 10 0.1 10 10 0.1 0 0 0 10 0.1 0 500 20 500 20 500 200 500 t t t te dt dt te dt te dt − − − + = ⎡ + ⎤ ⎢⎣ ⎥⎦ = + ∫ ∫ ∫ ∫ Let u = t and dv = e−0.1t dt. Then du = dt and v = −10e−0.1t . { } ( ) ( ) 0.1 10 10 0.1 0 0 1 0.1 10 0 1 1 1 100,000 500 10 10 100,000 500 100 100 100,000 500 100 100 100 100,000 50,000 1 2 113,212 units t t t te e dt e e e e e − − − − − − − = + ⎡⎣− ⎤⎦ + = + − − ⎡⎣ ⎤⎦ = + − − + = + − ≈ ∫ (c) Average 113,212 11,321 per year 10 = ≈ 0 10,000 10 12,000 The company is forecasting an increase in demand over the next decade. Section 6.1 Integration by Parts and Present Value 377 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 64. I = 2000(375 + 68te−0.2t ), 0 ≤ t ≤ 5 (a) The board of trustees expects the gift income to increase over the five-year period. (b) ( ) ( ) 5 0.2 5 5 0.2 0 0 0 5 0.2 0 2000 375 68 2000 375 68 2000 1875 136,000 t t t te dt dt te dt te dt − − − + = ⎡ + ⎤ ⎢⎣ ⎥⎦ = + ∫ ∫ ∫ ∫ Let u = t and dv = e−0.2t dt. Then du = dt and v = −5e−0.2t . { } ( ) ( ) 0.2 5 5 0.2 0 0 1 0.2 5 0 1 1 1 3,750,000 136,000 5 5 3,750,000 136,000 25 25 3,750,000 136,000 25 25 25 3,750,000 3,400,000 1 2 $4,648,419.80 t t t te e dt e e e e e − − − − − − − = + ⎡⎣− ⎤⎦ + = + − − ⎡⎣ ⎤⎦ = + − − + = + − = ∫ (c) ( ) ( ) 5 0.2 0 Average annual gift income 1 2000 375 68 1 4,684,419.80 $936,883.96 5 0 5 = + te− t dt = = − ∫ 65. (1.6 ln 1) 1.6 ln 1.6 ln t t dt dt t t dt t t tdt + = + = + ∫ ∫ ∫ ∫ Let u = ln t and dv = t dt. Then du 1 dt t = and 2 . 2 v = t 2 2 2 2 2 1.6 ln 2 2 1.6 ln 2 4 0.8 ln 0.4 t t t t dt t t t t t t t t ⎡ ⎤ = + ⎢ − ⎥ ⎣ ⎦ ⎡ ⎤ = + ⎢ − ⎥ ⎣ ⎦ = + − ∫ (a) ( ) 2 1 2 2 2 1 Average value 1 1.6 ln 1 2 1 0.8 ln 0.4 3.2 ln 2 0.2 2.018 t t dt t t t t = + − = ⎡⎣ + − ⎤⎦ = − ≈ ∫ (b) ( ) 4 3 2 2 4 3 Average value 1 1.6 ln 1 4 3 0.8 ln 0.4 12.8 ln 4 7.2 ln 3 1.8 8.035 t t dt t t t t = + − = ⎡⎣ + − ⎤⎦ = − − ≈ ∫ 66. Using Exercise 54 twice with n = 2 and 1 30 a = − and with n = 1 and 1 , 30 a = − ( ) ( ) 2 30 2 30 30 2 30 30 30 2 30 30 30 30 2 410.5 25,000 410.5 30 60 25,000 25,000 410.5 30 1800 1800 25,000 410.5 30 1800 54,000 25,000 12,315 60 1800 t t t t t t t t t t t e dt t e te dt t t te te e dt t te te e C t e t t C − − − − − − − − − − + = ⎡− + ⎤ + ⎣ ⎦ = + ⎡− − + ⎤ ⎣ ⎦ = + ⎡⎣− − − ⎤⎦ + = − + + + ∫ ∫ ∫ . (a) ( ) ( ) 90 2 30 0 30 2 90 0 Average value 1 410.5 25,000 90 0 1 25,000 12,315 60 1800 90 $167,068.28 t t t e dt t e t t − − = + − = ⎡ − + + ⎤ ⎣ ⎦ = ∫ 0 5 750,000 1,050,000 378 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (b) ( ) ( ) 365 2 30 274 30 2 365 274 Average value 1 410.5 25,000 365 274 1 25,000 12,315 60 1800 91 $26,253.48 t t t e dt t e t t − − = + − = ⎡ − + + ⎤ ⎣ ⎦ = ∫ (c) ( ) 365 2 30 0 ∫ 410.5t e−t + 25,000 dt = $31,281,948.97 67. ( ) 1 4 0.04 0.04 4 0 0 0 5000 5000 $18,482.03 0.04 V = t c t e−rt dt = e− t dt = e− t ⎤ = − ⎦ ∫ ∫ 68. ( ) 1 10 0.04 0.04 10 0 0 0 450 450 $3708.90 0.04 V = t c t e−rt dt = e− t dt = e− t ⎤ = − ⎦ ∫ ∫ 69. ( ) ( ) 10 0.05 0 10 0.05 10 0.05 0 0 0.05 10 10 0.05 0 0 0.5 10 0.05 0 Present value 100,000 4000 100,000 4000 2,000,000 4000 2,000,000 1 4000 t t t t t t t e dt e dt te dt e te dt e te dt − − − − − − − = + = + = − ⎤⎦ + = − + ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e−0.05t dt. Then du = dt and v = −20e−0.05t . ( ) { } ( ) { } ( ) ( ) 0.5 0.05 10 10 0.05 0 0 0.5 0.5 0.05 10 0 0.5 0.5 2,000,000 1 4000 20 20 2,000,000 1 4000 200 400 2,000,000 1 4000 600 400 $931,265.0973 t t t e te e dt e e e e e − − − − − − − − = − + ⎡⎣− ⎤⎦ + = − + − + ⎡⎣− ⎤⎦ = − + − + = ∫ 70. ( ) ( ) 6 0.07 0 6 0.07 6 0.07 0 0 0.07 6 6 0.07 0 0 0.42 6 0.07 0 Present value 30,000 500 30,000 500 3,000,000 500 7 3,000,000 1 500 7 t t t t t t t e dt e dt te dt e te dt e te dt − − − − − − − = + = + = − ⎤⎦ + = − − + ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e−0.07t dt. Then du = dt and 100 0.07 . 7 v = − e− t ( ) ( ) ( ) 6 6 0.42 0.07 0.07 0 0 6 0.42 0.42 0.07 0 0.42 0.42 0.42 3,000,000 1 500 100 100 7 7 7 3,000,000 1 500 600 10,000 7 7 49 3,000,000 1 500 600 10,000 10,000 7 7 49 4 t t t e te e dt e e e e e e − − − − − − − − − ⎧⎪⎡ ⎤ ⎪⎫ = − − + ⎨⎢− ⎥ + ⎬ ⎪⎩⎣ ⎦ ⎭⎪ ⎧⎪ ⎡ ⎤ ⎪⎫ = − − + ⎨− + ⎢− ⎥ ⎬ ⎩⎪ ⎣ ⎦ ⎪⎭ = − − + − − + ∫ ( 0.42 ) 0.42 9 3,000,000 1 500 14,200 10,000 7 49 49 $153,816.01 e− e− ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − − + ⎛− + ⎞ ⎜ ⎟ ⎝ ⎠ = Section 6.1 Integration by Parts and Present Value 379 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 71. ( ) ( ) 4 2 0.06 0 4 0.06 0.44 0 4 0.06 0.44 0 0.24 1.76 Present value 1000 50 1000 50 1000 50 0.06 0.44 1000 50 1000 50 0.06 0.44 0.06 0.44 $4103.07 t t t t t t e e dt e e dt e e e e − − − − = + = + = ⎡− + ⎤ ⎢⎣ ⎥⎦ = ⎛− + ⎞ − ⎛− + ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = ∫ ∫ 72. ( ) ( ) 10 10 0.06 0 10 0.06 10 0.04 0 0 0.06 10 10 0.04 0 0 0.6 10 0.04 0 Present value 5000 25 5000 25 5000 25 0.06 5000 1 25 0.06 t t t t t t t te e dt e dt te dt e te dt e te dt − − − − = + = + = − ⎤⎦ + = − − + ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e0.04t dt. Then du = dt and v = 25e0.04t . ( ) { } ( ) { } ( ) ( ) 0.6 0.04 10 0.04 0 0.6 0.4 0.04 10 0 0.6 0.4 5000 1 25 25 25 0.06 5000 1 25 250 625 0.06 5000 1 25 375 625 $39,238.17 0.06 t t t e te e dt e e e e e − − − = − − + ⎡⎣ ⎤⎦ − = − − + − ⎡⎣ ⎤⎦ = − − + − + = ∫ 73. ( ) 15 0.05 0 15 0.05 0 0.05 15 0 Present value 100,000 100,000 0.05 0.05 100,000 0.05 $1,055,266.89 t t t e dt e dt e − − − = = − − = ⎡⎣ ⎤⎦ − = ∫ ∫ 74. ( ) 26 0.03 0 26 0.03 0 0.03 26 0 Present value 2,500,000 2,500,000 0.03 0.03 2,500,000 0.03 $45,132,832.39 t t t e dt e dt e − − − = = − − = ⎡⎣ ⎤⎦ − = ∫ ∫ 75. (a) ( ) 4 2 4 0 0 Actual income = ∫ 150,000 + 75,000t dt = 150,000t + 37,500t ⎤⎦ = $1,200,000 (b) ( ) ( ) 4 0.04 0 4 0.04 4 0.04 0 0 0.04 4 4 0.04 0 0 0.16 4 0.04 0 Present value 150,000 75,000 150,000 75,000 3,750,000 75,000 3,750,000 1 75,000 t t t t t t t e dt e dt te dt e te dt e te dt − − − − − − − = + = + = − ⎤⎦ + = − − + ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e−0.04t dt. Then du = dt and v = −25e−0.04t . ( ) { } ( ) { } ( ) ( ) 0.16 0.04 4 4 0.04 0 0 0.16 0.16 0.04 4 0 0.16 0.16 3,750,000 1 75,000 25 25 3,750,000 1 75,000 100 625 3,750,000 1 75,000 725 625 $1,094,142.27 t t t e te e dt e e e e e − − − − − − − − = − − + ⎡⎣− ⎤⎦ + = − − + − − ⎡⎣ ⎤⎦ = − − + − + = ∫ 380 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 76. (a) ( ) 3 0 2 3 0 Actual value 500,000 125,000 500,000 62,500 $2,062,500 t dt t t = + = ⎡⎣ + ⎤⎦ = ∫ (b) ( ) ( ) 3 0.03 0 3 0.03 3 0.03 0 0 3 0.03 3 0.03 0 0 0.09 3 0.03 0 Present value 500,000 125,000 500,000 125,000 500,000 125,000 0.03 500,000 1 125,000 0.03 t t t t t t t e dt e dt te dt e te dt e te dt − − − − − − − = + = + ⎡− ⎤ = ⎢ ⎥ + ⎣ ⎦ − = − + ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e−0.03t dt. Then du = dt and 1 0.03 . 0.03 v = − e− t ( ) ( ) ( ) 3 0.09 0.03 3 0.03 0 0 0.09 0.09 0.03 3 0 0.09 0.09 0.09 500,000 1 125,000 1 1 0.03 0.03 0.03 500,000 1 125,000 100 1 0.03 0.0009 500,000 1 125,000 100 1 0.03 0.0009 t t t e te e dt e e e e e e − − − − − − − − − − ⎧⎪⎡ ⎤ ⎪⎫ = − + ⎨⎢− ⎥ + ⎬ ⎩⎪⎣ ⎦ ⎭⎪ − ⎧ ⎡ ⎤ ⎫ = − + ⎨− − ⎣ ⎦ ⎬ ⎩ ⎭ − = − + − − ∫ 1 0.0009 $1,964,342.48 ⎧ + ⎫ ⎨ ⎬ ⎩ ⎭ ≈ 77. (a) ( ) 4 0 2 4 0 Actual income 3,000,000 750,000 3,000,000 375,000 $18,000,000 t dt t t = + = ⎡⎣ + ⎤⎦ = ∫ (b) ( ) ( ) 4 0.05 0 4 0.05 4 0.05 0 0 0.05 4 4 0.05 0 0 0.2 4 0.05 0 Present value 3,000,000 750,000 3,000,000 750,000 60,000,000 750,000 60,000,000 1 750,000 t t t t t t t e dt e dt te dt e te dt e te dt − − − − − − − = + = + = ⎡⎣− ⎤⎦ + = − − + ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e−0.05t dt. Then du = dt and v = −20e−0.05t . ( ) { } ( ) { } ( ) { } 0.2 0.05 4 4 0.05 0 0 0.2 0.2 0.05 4 0 0.2 0.2 0.2 60,000,000 1 750,000 20 20 60,000,000 1 750,000 80 400 60,000,000 1 750,000 80 400 400 $16,133,083.71 t t t e te e dt e e e e e e − − − − − − − − − = − − + ⎡⎣− ⎤⎦ + = − − + − − ⎡⎣ ⎤⎦ = − − + − − + ≈ ∫ 78. Graph A shows the expected income and graph B shows the present value of the expected income. The present value of a future amount is the amount to be deposited today to produce the future value (see Section 4.2). Because of inflation, a certain amount today buys, or is worth more than, that same amount years from now. Therefore, for any specific year the present value would be less than that of the actual income. Section 6.1 Integration by Parts and Present Value 381 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 79. ( ) ( ) 1 1 0 0.08 10 10 0.08 0 10 0.8 0.08 0 Future value 3000 3000 0.08 $45,957.78 rt t rt t t e fte dt e e dt e e − − − = = = ⎡ ⎤ ⎢⎣− ⎥⎦ ≈ ∫ ∫ 80. ( ) ( ) ( ) 1 1 0 0.1 5 5 0.05 0.1 0 0.5 5 0.05 0 5 0.5 0.05 0 Future value 3000 3000 3000 0.05 $21,881.75 rt t rt t t t t e fte dt e e e dt e e dt e e − − − − = = = = ⎡ ⎤ ⎢⎣− ⎥⎦ ≈ ∫ ∫ ∫ 81. (a) (0.07)(10) 10 0.07 0 0.7 10 0.07 0 10 0.7 0.07 0 Future value 1200 1200 1200 0.07 $17,378.62 t t t e e dt e e dt e e − − − = = = ⎡ ⎤ ⎢⎣− ⎥⎦ ≈ ∫ ∫ (b) (0.10)(15) 15 0.10 (0.09)(15) 15 0.09 0 0 1.5 15 0.1 1.35 15 0.09 0 0 15 15 1.5 0.1 1.35 0.09 0 0 Difference 1200 1200 1200 1200 1200 1200 0.1 0.09 $41,780.27 $ t t t t t t e e dt e e dt e e dt e e dt e e e e − − − − − − = ⎡ ⎤ − ⎡ ⎤ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ = ⎡ ⎤ − ⎡ ⎤ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ = ⎡ ⎤ − ⎡ ⎤ ⎢⎣ − ⎥⎦ ⎢⎣− ⎥⎦ ≈ − ∫ ∫ ∫ ∫ 38,099.01 = $3681.26 82. (0.07)(18) 18 0.07 0 1.26 18 0.07 0 Future value 400 400 t t e te dt e te dt − − = = ∫ ∫ Let u = t and dv = e−0.07t dt. Then du = dt and 1 0.07 . 0.07 v − e− t = 18 1.26 0.07 18 0.07 0 0 1.26 1.26 0.07 18 0 1.26 1.26 1.26 400 1 1 0.07 0.07 400 18 1 0.07 0.0049 400 18 1 1 0.07 0.0049 0.0049 $103,299.71 t t t e te e dt e e e e e e − − − − − − ⎧⎪⎡ ⎤ ⎪⎫ = ⎨⎢− ⎥ + ⎬ ⎩⎪⎣ ⎦ ⎪⎭ ⎧ − ⎡ ⎤ ⎫ = ⎨ − ⎣ ⎦ ⎬ ⎩ ⎭ ⎧ − ⎫ = ⎨ − + ⎬ ⎩ ⎭ ≈ ∫ (a) Four years at The Pennsylvania State University costs $26,276 × 4 = $105,104. No, the fund would not cover the costs. (b) Four years at The Ohio State University costs $23,604 × 4 = $94,416. Yes, the fund would cover the costs. 83. 4 4 1 3 1 3 4 4 4 4 4 1 23 29 35 41 47 53 10 20 20 20 20 20 20 59 65 71 77 4.254 20 20 20 20 dx dx x x x x f f f f f f f f f f = + + ⎛ − ⎞⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎜ ⎟⎢ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ ⎠⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ + ⎜ ⎟ + ⎜ ⎟+ ⎜ ⎟ + ⎜ ⎟⎥ ≈ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ ∫ 382 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 84. 4 4 1 1 10 10 1 10 4 1 9 11 13 15 17 19 21 12 8 8 8 8 8 8 8 23 25 27 29 31 2.691 8 8 8 8 8 x x dx dx xe xe f f f f f f f f f f f f = ⎛ − ⎞⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎜ ⎟⎢ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎝ ⎠⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ ≈ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ ∫ Section 6.2 Integration Tables 1. Formula 4: u = x, du = dx, a = 2, b = 3 ( )2 1 2 ln 2 3 2 3 9 2 3 x dx x C x x = ⎡ + + ⎤ + + ⎢⎣ + ⎥⎦ ∫ 2. Formula 11: u = x, du = dx, a = 2, b = 3 ( )2 1 1 1 1 ln 2 3 2 2 3 2 2 3 dx x C x x x x ⎡ ⎤ = ⎢ + ⎥ + + ⎣ + + ⎦ ∫ 3. Formula 19: u = x, du = dx, a = 2, b = 3 ( ) ( ) 2 4 3 2 3 2 3 27 2 3 4 2 3 27 x x dx x C x x x C − = − + + + − = + + ∫ 4. Formula 29: u = x, du = dx, a = 3 2 ( ) 4 1 4 ln 3 9 23 3 2 ln 3 3 3 dx x C x x x C x − = + − + − = + + ∫ 5. Formula 25: u = x2 , du = 2x dx, a = 3 2 4 4 2 ln 9 9 x dx x x C x = + − + − ∫ 6. Formula 22: u = x, du = dx, a = 3 2 2 1 ( 2 ) 2 2 8 ∫ x x + 9 dx = ⎡⎢⎣x 2x + 9 x + 9 − 81 ln x + x + 9 ⎤⎥⎦ + C 7. Formula 35: u = x2 , du = 2x dx 3 2 1 2 2 1( 2 ) 2 2 2 ∫ x ex dx = ∫ x ex 2x dx = x − 1 ex + C 8. Formula 37: u = x2 , du = 2x dx ( ) 2 2 2 2 1 2 1 2 1 1 ln 1 2 x x x x dx x dx e e x e C = + + = ⎡ − + ⎤ + ⎢⎣ ⎥⎦ ∫ ∫ Skills Warm Up 1. (x + 4)2 = x2 + 8x + 16 2. (x − 1)2 = x2 − 2x + 1 3. ( )1 2 2 1 2 4 x + = x + x + 4. ( )1 2 2 2 1 3 3 9 x − = x − x + 5. Let u = 2x and dv = ex dx. Then du = 2 dx and v = ex . ∫ 2xex dx = 2xex − ∫ 2ex dx = 2xex − 2ex + C = 2ex (x − 1) + C 6. Let u = ln x and dv = 3x2 dx. Then du 1 dx x = and v = x3. 3 2 ln 3 ln 2 3 ln 1 3 3 ln 1 3 3 x x dx = x x − x dx = x x − x + C = x ⎛⎜ x − ⎞⎟ + C ⎝ ⎠ ∫ ∫ Section 6.2 Integration Tables 383 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 9. Formula 10: u = x, du = dx, a = b = 1 ( ) 1 ln 1 1 dx x C x x x = + + + ∫ 10. Formula 11: u = x, du = dx, a = b = 1 ( )2 1 1 ln 1 1 1 dx x C x x x x = + + + + + ∫ 11. Formula 26: u = x, du = dx, a = 3 2 2 1 1 ln 3 9 9 3 dx x C x x x + + = − + + ∫ 12. Formula 25: u = x, du = dx, a = 1 2 2 1 ln 1 1 dx x x C x = + − + − ∫ 13. Formula 32: u = x, du = dx, a = 2 2 2 1 1 ln 2 4 4 2 dx x C x x x + − = − + − ∫ 14. Formula 24: u = x, du = dx, a = 3 2 2 2 2 x 9 dx x 9 ln x x 9 C x x − − ∫ = − + + − + 15. Formula 42: u = 3x, du = 3 dx ( ) ( )( ) ( ) ( ( )) ( ) 2 2 3 ln 3 1 3 ln 3 3 3 1 3 1 2 ln3 3 4 3 1 2 ln3 4 x x dx x x dx x x C x x C = ⎡ ⎤ = ⎢ − + ⎥ + ⎢⎣ ⎥⎦ = ⎡⎣− + ⎤⎦ + ∫ ∫ 16. Formula 42: u = 5x, du = 5 dx ( ) ( ) ( ) ( ) ( ) 2 2 2 2 1 5 1 5 ln 5 ln 5 5 5 2 2 ln 5 ln 5 2 2 ln 5 ln 5 x dx x dx x x x C x x x = = ⎡ − + ⎤ + ⎣ ⎦ = ⎡ − + ⎤ ⎣ ⎦ ∫ ∫ 17. Formula 37: u = 3x2 , du = 6x dx 2 ( 3 2 ) 3 2 6 3 ln1 1 x x x dx x e C e = − + + + ∫ 18. Formula 37: u = x, du = dx 1 ln(1 ) 1 x x dx x e C e = − + + + ∫ 19. Formula 14: u = x, du = dx, a = 3, b = 1 ( )( ( ) ) ( ) ( ) ( ) ( ) 2 2 3 2 2 3 2 1 2 3 2 3 2 3 3 1 2 2 3 2 3 6 3 7 x x dx x x x x dx x x x x dx + = ⎡ + − + ⎤ + ⎣ ⎦ = ⎡ + − + ⎤ ⎣ ⎦ ∫ ∫ ∫ Formula 14: u = x, du = dx, a = 3, b = 1 ( ) ( )( ( ) )( ( ) ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 2 3 2 2 3 2 3 2 1 2 2 3 2 3 2 3 2 3 2 2 2 3 6 2 3 3 3 7 12 1 3 2 3 12 3 3 6 3 7 5 5 2 3 12 3 24 3 7 5 5 2 3 5 12 24 35 x x x x x dx x x x x x dx x x x x x C x x x C ⎧⎪ ⎡ ⎤⎪⎫ = ⎨ + − ⎢ + − + ⎥⎬ ⎩⎪ ⎢⎣ + ⎥⎦⎪⎭ = ⎧ + − + + + ⎫ ⎨ ⎬ ⎩ ⎭ = ⎛ + − + + + ⎞ + ⎜ ⎟ ⎝ ⎠ = + − + + ∫ ∫ 20. Formula 29: u = x2, du = 2x dx, a = 3 ( ) 2 4 2 2 2 1 2 1 ln 3 9 2 9 12 3 x dx x dx x C x x x − = = + − − + ∫ ∫ 21. Formula 8: u = t, du = dt, a = 2, b = 3 ( ) ( ) ( ) 2 3 2 2 1 4 4 ln 2 3 1 4 2 ln 2 3 2 3 27 2 3 2 2 3 27 2 3 2 3 t dt t C t C t t t t t ⎡ ⎤ ⎡ ⎤ = ⎢ − + + ⎥ + = ⎢ − + + ⎥ + + ⎢⎣ + + ⎥⎦ ⎢⎣ + + ⎥⎦ ∫ 384 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 22. Formula 17: u = t, du = dt, a = 3, b = 4 3 4 2 3 4 3 1 3 4 t dt t dt t t t + = + + + ∫ ∫ Formula 15: u = t, du = dt, a = 3, b = 4 3 4 2 3 4 3 ln 3 4 3 2 3 4 3 ln 3 4 3 3 3 4 3 3 4 3 t dt t t C t t C t t t + + − + − = + + + = + + + + + + + ∫ 23. Formula 15: u = x, du = dx, a = 3, b = 4 1 1 ln 3 4 3 3 4 3 3 4 3 dx x C x x x + − = + + + + ∫ 24. Formula 21: u = x, du = dx, a = 3 2 1 2 2 2 ∫ 3 + x dx = ⎡⎢⎣x 3 + x + 3 ln x + 3 + x ⎤⎥⎦ + C 25. Formula 6: u = x, du = dx, a = 1, b = 1 ( ) 2 2 ln1 1 2 x dx x x x C x = − − + + + + ∫ 26. Formula 38: u = x, du = dx, n = 2 ( 2 ) 2 1 1ln 1 1 2 x x dx x e C e = − + + + ∫ 27. Formula 9: u = x, du = dx, a = 3, b = 2, n = 5 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 5 2 3 4 2 3 4 1 1 6 9 3 2 8 2 3 2 3 3 2 4 3 2 1 1 2 9 8 2 3 2 3 2 4 3 2 x dx C x x x x C x x x ⎡ − ⎤ = ⎢ + − ⎥ + + ⎢⎣ + + + ⎥⎦ ⎡ − ⎤ = ⎢ + − ⎥ + ⎢⎣ + + + ⎥⎦ ∫ 28. Formula 28: u = x, du = dx, a = 2 2 2 2 1 4 4 4 dx x C x x x − = + − ∫ 29. Formula 33: u = x, du = dx, a = 1 2 2 2 1 1 1 dx x C x x x − = − + − ∫ 30. Formula 4: u = x, du = dx, a = 1, b = −3 ( )2 2 2 1 ln 1 3 1 3 9 1 3 x dx x C x x = ⎡ + − ⎤ + − ⎢⎣ − ⎥⎦ ∫ 31. Formula 43: u = 2x, du = 2 dx, n = 2 ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ( )) 2 2 2 1 2 3 3 4 ln 2 1 2 ln 2 2 2 1 2 1 2 1 ln2 2 2 1 1 8 1 3 ln2 2 9 4 1 3 ln2 9 x x dx x x dx x x C x x C x x C + = ⎛ ⎞ = ⎜ ⎣⎡− + + ⎤⎦⎟ + ⎜ + ⎟ ⎝ ⎠ ⎛ ⎞ = ⎜ ⎡⎣− + ⎤⎦⎟ + ⎝ ⎠ = − + + ∫ ∫ 32. Formula 34: u = x2 , du = 2x dx ( ) 2 1 2 1 2 2 2 ∫ xex dx = ∫ ex 2x dx = ex + C Section 6.2 Integration Tables 385 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 33. Formula 7: u = x, du = dx, a = −5, b = 3 ( ) 2 2 1 3 25 10 ln 3 5 3 5 27 3 5 x dx x x C x x = ⎡ − + − ⎤ + − ⎢⎣ − ⎥⎦ ∫ 34. Formula 13: u = x, du = dx, a = −1, b = 2 2 ( )2 ( ) 1 1 1 4 1 4 ln 2 2 1 2 1 2 1 dx x x C x x x x x ⎡ − ⎤ = − ⎢ − ⎥ + − ⎢⎣ − − ⎥⎦ ∫ 35. Formula 3: u = ln x, du = (1 x) dx, a = 4, b = 3 ( ) ln ln 1 1 3 ln 4 ln 4 3 ln 4 3ln 4 3ln 9 x dx x dx x x C x x x x = ⎛ ⎞ = ⎡ − + ⎤ + + + ⎜ ⎟ ⎣ ⎦ ⎝ ⎠ ∫ ∫ 36. Formula 43: u = x, du = dx, n = 3 Formula 42: u = x, du = dx ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 2 3 2 3 2 ln ln 3 ln ln 3 2 2 ln ln ln 3 ln 6 ln 6 x dx x x x dx x x x x x C x x x x C = − = − ⎡ − + ⎤ + ⎣ ⎦ = ⎡ − + − ⎤ + ⎣ ⎦ ∫ ∫ 37. Formula 19: u = x, du = dx, a = 1, b = 1 ( ) 1 1 0 0 2 2 1 1 3 2 2 4 3 3 4 2 2 3 0.391 x x dx x x − ⎤ = − + ⎥ + ⎦ = − − ⎛− ⎞ ⎜ ⎟ ⎝ ⎠ − = ≈ ∫ 38. Formula 19: u = x, du = dx, a = 5, b = 2 ( ) 5 5 0 0 2 10 2 5 2 5 2 12 5 5 3 3.7 x x dx x x − − ⎤ = + ⎥ + ⎦ = ≈ ∫ 39. Formula 4: u = x, du = dx, a = 4, b = 1 ( ) ( ) 5 5 0 2 0 4 ln 4 4 4 4 ln 9 1 ln 4 9 5 ln 9 9 4 0.255 x dx x x x = ⎡ + + ⎤ + ⎢⎣ + ⎥⎦ = ⎛ + ⎞ − + ⎜ ⎟ ⎝ ⎠ = − + ≈ ∫ 40. Formula 6: u = x, du = dx, a = −5, b = 3 ( ) ( ) ( ) 2 4 4 2 2 1 3 10 3 25 ln 5 3 3 5 27 2 1 132 25 ln 7 48 27 1 84 25 ln 7 27 4.913 x dx x x x x = ⎡− − − + − + ⎤ − ⎢⎣ ⎥⎦ = ⎡⎣ + − ⎤⎦ = + ≈ ∫ 41. Formula 38: u = x, du = dx, n = 0.5 ( ) { ( ) ( )} 4 4 0.5 0 0.5 0 2 6 1 6 1 ln 1 1 0.5 6 4 2 ln 1 0 2 ln 2 6.795 x x dx x e e e = ⎡ − + ⎤ + ⎢⎣ ⎥⎦ = ⎡ − + ⎤ − − ≈ ⎣ ⎦ ∫ 42. Formula 21: u = x, du = dx, a = 3 ( ( )) ( ( )) 4 4 2 2 2 2 2 1 2 1 2 3 3 3 ln 3 4 19 3 ln 4 19 2 7 3 ln 2 7 6.953 + x dx = ⎡⎢⎣x + x + x + + x ⎤⎥⎦ = ⎡ + + − + + ⎤ ≈ ⎢⎣ ⎥⎦ ∫ 386 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 43. Formula 43: u = x, du = dx, n = 3 ( ) ( ) ( ) ( ) ( ) ( ) 2 3 2 2 3 1 1 2 3 1 2 1 4 2 1 ln 2 ln 2 1 3 1 ln 3 1 2 4 ln 1 16 2 4 ln 2 1 1 1 16 2 4 ln 2 15 16 8 ln 2 15 3.6702 8 x x dx x x dx x x x x + = ⎡⎛ ⎞⎤ = ⎢⎜ ⎡⎣− + + ⎤⎦⎟⎥ ⎢⎜ + ⎟⎥ ⎣⎝ ⎠⎦ ⎡ ⎤ = ⎢ − ⎥ ⎣ ⎦ ⎡ ⎛ ⎞⎤ = ⎢ − − ⎜ − ⎟⎥ ⎣ ⎝ ⎠⎦ = ⎛ − ⎞ ⎜ ⎟ ⎝ ⎠ = − ≈ ∫ ∫ 44. Formula 5: u = x, du = dx, a = 1, b = 3, n = 4 ( ) ( ) ( )( ) ( )( ) ( ) ( ) 3 3 0 4 2 4 2 4 1 0 3 2 3 0 1 1 1 1 3 3 4 2 1 3 4 1 1 3 1 1 1 9 2 1 3 3 1 3 1 1 1 1 1 9 200 3000 2 3 9 500 x dx x x x x x − − ⎡ − ⎤ = ⎢ + ⎥ + ⎢⎣ − + − + ⎥⎦ ⎡ − ⎤ = ⎢ + ⎥ ⎢⎣ + + ⎥⎦ ⎡⎛ ⎞ ⎛ ⎞⎤ = ⎢⎜− + ⎟ − ⎜− + ⎟⎥ ⎣⎝ ⎠ ⎝ ⎠⎦ = ∫ 45. Formula 35: u = x, du = dx, a = 4 ( ) ( ) ( ) ( ) ( ) 2 2 2 2 3 2 2 2 2 2 2 2 2 1 16 4 4 16 16 2 2 16 12 16 12 1 4 12 1 8 3 dx x x x x x − − − ⎡ ⎤ = ⎢ ⎥ − ⎢ − ⎥ ⎢⎣ ⎥⎦ ⎡ ⎤ = ⎢ ⎥ ⎢⎣ − ⎥⎦ − = − = = ∫ Approximate area: 0.0722 46. Formula 38: u = x, du = dx, n = 4 ( ) ( ) 1 0 4 1 4 0 4 2 1 2 1 ln 1 4 2 1 1 ln 1 1 ln 2 4 4 x x A dx e x e e = + = ⎡ − + ⎤ ⎢⎣ ⎥⎦ = ⎡ − + + ⎤ ⎢⎣ ⎥⎦ ∫ Approximate area: 0.3375 −2 2 0 0.05 2 0 −1 2 Section 6.2 Integration Tables 387 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 47. Formula 12: u = x, du = dx, a = 2, b = 3 ( ) ( ) ( ) 2 2 1 2 1 1 1 1 1 3 ln 9 2 3 9 2 2 2 3 1 1 3 ln 1 1 3 ln 1 18 2 2 4 2 5 1 1 3 ln 1 ln 1 18 2 2 4 5 1 1 4 1 3ln 36 1 5 1 1 3 ln 5 36 4 dx x x x x x ⎡ ⎛ ⎞⎤ = ⎢− ⎜ + ⎟⎥ + ⎣ ⎝ + ⎠⎦ ⎡⎛ ⎛ ⎞⎞ ⎛ ⎛ ⎞⎞⎤ = − ⎢⎜ + ⎜ ⎟⎟ − ⎜ + ⎜ ⎟⎟⎥ ⎣⎝ ⎝ ⎠⎠ ⎝ ⎝ ⎠⎠⎦ ⎡ ⎛ ⎞⎤ = − ⎢− + ⎜ − ⎟⎥ ⎣ ⎝ ⎠⎦ ⎡ ⎛ ⎞⎤ = − ⎢− + ⎜⎜ ⎟⎟⎥ ⎢⎣ ⎝ ⎠⎥⎦ = − ⎡− + ⎤ ⎢⎣ ⎥⎦ ∫ Approximate area: 0.0092 48. Formula 29: u = ex , du = ex dx, a = 1 ( ) ( ) 2 1 2 2 1 2 2 1 2 2 1 1 1 1 ln 1 2 1 1 ln 1 1 ln 1 2 1 2 1 x x x x x x A e dx e e dx e e e e e e e − = − = − ⎡ − ⎤ = ⎢ ⎥ ⎣ + ⎦ − − = − + + ∫ ∫ Approximate area: 0.2498 49. Formula 22: u = x, a = 2, du = dx ( ) ( ) 5 2 2 0 5 2 2 2 0 4 1 2 4 4 16ln 4 8 1 42 5 16 ln 5 3 16 ln 2 8 1 42 5 16 ln 2 8 5 3 1 21 5 8 ln 2 4 5 3 A x x x x x x x = + = ⎡ + + − + + ⎤ ⎢⎣ ⎥⎦ = ⎡ − + + ⎤ ⎣ ⎦ ⎡ ⎛ ⎞⎤ = ⎢ + ⎜ ⎟⎥ ⎣ ⎝ + ⎠⎦ ⎡ ⎛ ⎞⎤ = ⎢ + ⎜ ⎟⎥ ⎣ ⎝ + ⎠⎦ ∫ Approximate area: 9.8145 50. Formula 39: u = x2 , du = 2x dx 4 2 0 A = ∫ x ln x dx ( ) ( ) ( ) ( ) 4 2 1 2 2 4 1 1 2 1 2 1 2 1 2 ln 2 1 ln 16 1 ln 16 1 16 ln 16 15 x x dx x x = = ⎡ − + ⎤ ⎣ ⎦ = ⎡⎣ − + − − ⎤⎦ = − ∫ Approximate area: 14.6807 51. (a) Formula 41: , 1 3 3 u = x du = dx ln 3 ln 1 3 3 3 3 1 ln 3 3 1 ln3 x dx x dx x x C x x C ⎛ ⎞ = ⎛ ⎞⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎡ ⎛ ⎛ ⎞⎞⎤ = ⎢ ⎜− + ⎜ ⎟⎟⎥ + ⎣ ⎝ ⎝ ⎠⎠⎦ ⎛ ⎛ ⎞⎞ = ⎜− + ⎜ ⎟⎟ + ⎝ ⎝ ⎠⎠ ∫ ∫ (b) Let ln and . 3 u = ⎛ x ⎞ dv = dx ⎜ ⎟ ⎝ ⎠ Then 1 1 1 and . 3 3 du dx dx v x x x = ⎛ ⎞ = = ⎜ ⎟ ⎝ ⎠ ln ln 1 3 3 ln 3 ln 3 ln 1 3 x dx x x x dx x x x dx x x x C x x C ⎛ ⎞ = ⎛ ⎞ − ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = ⎛ ⎞ − ⎜ ⎟ ⎝ ⎠ = ⎛ ⎞ − + ⎜ ⎟ ⎝ ⎠ ⎛ ⎛ ⎞ ⎞ = ⎜ ⎜ ⎟ − ⎟ + ⎝ ⎝ ⎠ ⎠ ∫ ∫ ∫ 1 2 0 0.05 1.5 −0.5 0 3 −1 −2 5 20 0 5 −2 16 388 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 52. (a) Formula 37: u = 4x, du = 4 dx 4 1 4 ( ) 1( ) 4 4 4 ∫ 4xe x dx = ∫ 4xe x 4 dx = 4x − 1 e x + C (b) Let u = 4x and dv = e4x dx. Then 1 4 4 du = 4 dx and v = e x. ( )( ) ( )( ) ( ) 4 4 4 4 4 4 4 4 1 1 4 4 1 4 1 4 4 4 4 4 1 x x x x x x x x xe dx x e e dx xe e dx xe e C x e C = − = − = − + = − + ∫ ∫ ∫ 53. (a) Formula 19: u = x, du = dx, a = −3, b = 7 ( ( ) ) ( ) ( ) 2 2 2 3 7 2 3 7 7 6 7 3 3 7 3 7 147 x x dx x C x x C x − − = − − + + = + − + − + ∫ (b) Let ( )1 2 u x and dv 7x 3 dx. − = = − Then and 2(7 3)1 2. 7 du = dx v = x − ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 2 1 2 1 2 3 2 1 2 7 3 2 7 3 2 7 3 7 7 2 7 3 4 7 3 7 147 2 7 3 21 27 3 147 2 7 37 6 147 x x dx x x x dx xx x C x x x C x x C − − = − − − = − − − + = − ⎡⎣ − − ⎤⎦ + = − + + ∫ ∫ 54. (a) Formula 42: u = 7x, du = 7 dx ( ) ( )( ) ( ) ( ( )) ( ( )) ( ( )) 2 2 2 7 ln 7 1 7 ln 7 7 7 1 7 1 2 ln7 7 4 1 49 1 2 ln7 7 4 7 1 2 ln7 4 x x dx x x dx x x C x x C x x C = ⎡ ⎤ = ⎢ − + ⎥ + ⎢⎣ ⎥⎦ ⎡ ⎤ = ⎢ − + ⎥ + ⎣ ⎦ = − + + ∫ ∫ (b) Let u = ln(7x) and dv = 7x dx. Then 1 (7) and 7 2. 7 2 du dx v x x = = ( ) ( ) ( ) ( ) ( ( ) ) 2 2 2 2 2 2 7 ln 7 7 ln 7 7 1 2 2 7 ln 7 7 2 2 7 ln 7 7 2 4 7 2 ln 7 1 4 x x dx x x x dx x x x xdx x x x C x x C = − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − = − + = − + ∫ ∫ ∫ 55. Formula 19: u = x, du = dx, a = 4, b = 5 ( ) ( ) ( ) 75 14 4 5 75 2 8 5 4 5 14 75 1 8 5 4 5 7 b a b a b a P a x b x dx x x x x x ≤ ≤ = ⎛ ⎞ ⎜ + ⎟ ⎝ ⎠ ⎡ − ⎤ = ⎢− + ⎥ ⎣ ⎦ = − ⎡ − + ⎤ ⎣ ⎦ ∫ (a) ( ) ( ) ( ) 0.8 0.4 0.4 0.8 1 8 5 4 5 7 1 4 8 6 6 0.483 7 P ≤ x ≤ = − ⎡⎣ − x + x⎤⎦ = − − ≈ (b) ( ) ( ) ( ) 0.5 0 0 0.5 1 8 5 4 5 7 1 5.5 6.5 16 0.283 7 P ≤ x ≤ = − ⎡⎣ − x + x⎤⎦ = − − ≈ Section 6.2 Integration Tables 389 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 56. Formula 35: u = x2 , du = 2x dx ( ) 2 3 2 2 2 (2 ) ( 2 1) 2 b b b x x x a a a P a ≤ x ≤ b = ∫ x e dx = ∫ x e x dx = ⎡⎢⎣ x − e ⎤⎥⎦ (a) ( ) ( ) ( ) 2 0.25 2 0.0625 0 P 0 ≤ x ≤ 0.25 = ⎡⎢⎣ x − 1 ex ⎤⎥⎦ = −0.9375e − −1 ≈ 0.002 (b) ( ) ( ) ( ) 2 1 2 0.25 0.5 P 0.5 ≤ x ≤ 1 = ⎡⎢⎣ x − 1 ex ⎤⎥⎦ = 0 − −0.75e ≈ 0.963 57. Formula 37: u = 4.8 − 1.9t, du = −1.9 dt Average value ( ) ( ( )) ( ( )) 2 0 4.8 1.9 2 0 4.8 1.9 4.8 1.9 2 0 4.8 1 5000 2 0 1 2500 1.9 1.9 1 2500 4.8 1.9 ln 1 1.9 2500 1 ln 1 4.8 ln 1 1.9 401.402 t t t dt e dt e t e e e − − − = − + − = − + = − ⎡ − − + ⎤ ⎣ ⎦ = − ⎡ − + − − + ⎤ ⎣ ⎦ ≈ ∫ ∫ 58. Formula 37: u = 4.20 − 0.25t, du = −0.25 dt Average ( ) ( ) ( ( )) ( ( )) 28 21 4.20 0.25 28 21 4.20 0.25 4.20 0.25 28 21 2.8 1.05 1 375 28 21 1 375 0.25 7 0.25 1 1500 4.20 0.25 ln 1 7 1500 2.8 ln 1 1.05 ln 1 7 323.352 t t t dt e dt e t e e e − − − − − = − + − = − + = − ⎡ − − + ⎤ ⎣ ⎦ = − ⎡ − − + − − − + ⎤ ⎣ ⎦ ≈ ∫ ∫ 59. ( ) ] ( ) 2 2 2 0 2 1 2 0 0 2 1 2 10,000 1 1 10,000 10,000 1 1 0.1 1 0.1 R dt t dt t t = ⎡ − ⎤ = − ⎢ ⎥ ⎢⎣ + ⎥⎦ + ∫ ∫ Formula 25: u = t 0.1, du = 0.1 dt, a = 1 ( ) 2 2 0 20,000 10,000 ln 0.1 0.1 1 $1138.43 0.1 R = − ⎡⎢⎣ t + t + ⎤⎥⎦ ≈ 60. (a) After 16 weeks, t = 16, the rate of sales, dS , dt is approximately $80 per week. (b) Sales are increasing over the entire interval, 0 ≤ t ≤ 52, since dS dt is positive over the entire interval 0 ≤ t ≤ 52. 61. Formula 21: u = 0.00645t, du = 0.00645 dt, a = 0.1673 Average net profit ( ) 5 2 2 5 2 2 5 2 2 2 1 0.00645 0.1673 5 2 1 0.00645 0.1673 0.00645 3 0.00645 1 1 0.00645 0.00645 0.1673 0.1673 ln 0.00645 0.00645 0.1673 3 0.00645 2 $0.50 billion per year t dt t dt t t t t = − − = + = ⎛ ⎞⎡ + + + + ⎤ ⎜ ⎟⎢⎣ ⎥⎦ ⎝ ⎠ ≈ ∫ ∫ 62. Answers will vary. 0 6000 0 10 0 30 0 400 390 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 6 Quiz Yourself 1. Let u = x and dv = e5x dx. Then du = dx and 1 5 . 5 v = e x ( ) 5 5 5 5 5 5 1 1 5 5 1 1 5 25 1 1 5 5 x x x x x x xe dx xe e dx xe e C e x C = − = − + = − + ∫ ∫ 2. ∫ ln x3 dx = 3∫ ln x dx Let u = ln x and dv = dx. Then du 1 dx x = and v = x. ( ) ln 3 3 ln 3 ln 3 3 ln 1 x dx x x dx x x x C x x C = ⎡ − ⎤ ⎣ ⎦ = − + = − + ∫ ∫ 3. Let u = ln x and dv = (x + 1) dx. Then du 1 dx x = and 1 2 . 2 v = x + x ( ) 2 2 2 2 2 2 2 1 ln ln 1 1 1 2 2 ln 1 1 2 2 ln 1 1 2 4 1 ln ln 1 2 4 x xdx x x x x x dx x x x x x dx x x x x x C x x x x x x C + = ⎛ + ⎞ − ⎛ + ⎞⎛ ⎞ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠ = ⎛ + ⎞ − ⎛ + ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = ⎛ + ⎞ − − + ⎜ ⎟ ⎝ ⎠ = + − − + ∫ ∫ ∫ 4. Let u = x and dv = x + 3 dx. Then du = dx and ( )3 2 23 v = x + 3 . ( ) ( ) ( ) ( ) 3 2 3 2 3 2 5 2 2 2 3 3 2 4 3 15 3 3 3 3 3 x x dx xx x dx xx x C + = + − + = + − + + ∫ ∫ 5. ln ln 1 2 1 ln 2 ∫ x x dx = ∫ x x dx = ∫ x x dx Let u = ln x and dv = x dx. Then du 1 dx x = and 1 2. 2 v = x 2 2 2 ln 1 1 ln 1 2 2 2 1 ln 1 4 8 x x dx x x xdx x x x C = ⎡ − ⎤ ⎢⎣ ⎥⎦ = − + ∫ ∫ 6. Let u = x2 and du = e−2x dx. Then du = 2x dx and 1 2 . 2 v = − e− x 2 2 2 2 2 2 ∫ x e− x dx = − x e− x + ∫ xe− x dx Let u = x and dv = e−2x dx. Then du = dx and 1 2 . 2 v = − e− x ( ) 2 2 2 2 2 2 2 2 2 2 2 2 1 2 2 2 1 2 2 4 1 2 2 1 4 x x x x x x x x x e dx x e x e e dx x e xe e C e x x C − − − − − − − − = − − + = − − − + = − + + + ∫ ∫ Chapter 6 Quiz Yourself 391 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 7. x = 1000(45 + 20te−0.5t ) (a) ( ) [ ] 5 0.5 0 5 5 0.5 0 0 5 5 0.5 0 0 5 5 0.5 0 0 5 0.5 0 Total demand 1000 45 20 1000 45 1000 20 45,000 20,000 45,000 20,000 225,000 20,000 t t t t t te dt dt te dt dt te dt t te dt te dt − − − − − = + = + = + = + = + ∫ ∫ ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e−0.5t dt. Then du = dt and v = −2e−0.5t . { } { } 0.5 5 5 0.5 0 0 2.5 0.5 5 0 2.5 2.5 225,000 20,000 2 2 225,000 20,000 10 4 225,000 200,000 80,000 80,000 282,016 units t t t te e dt e e e e − − − − − − = + ⎡⎣− ⎤⎦ + = + − − ⎡⎣ ⎤⎦ = − − + ≈ ∫ (b) ( ) ( ) ( ) 5 0.5 0 5 0.5 0 Average annual demand 1 1000 45 20 5 0 1 1000 45 20 5 1 282,016 56,403 units 5 t t te dt te dt − − = + − = + ≈ ≈ ∫ ∫ 8. (a) 7 2 7 0 0 Actual income = ∫ 32,000t dt = ⎡⎣16,000t ⎤⎦ = $784,000 (b) 7 0.033 7 0.033 0 0 Present value = ∫ 32,000te− t dt = 32,000∫ te− t dt Let u = t and dv = e−0.033t dt. Then and 1 0.033 . 0.033 du = dt v = − e− t ( ) ( ) ( ) ( ) ( ) 7 0.033 7 0.033 0 0 7 0.033 7 0.033 2 0 0.231 0.231 2 2 32,000 1 1 0.033 0.033 32,000 1 7 1 0.033 0.033 32,000 7 1 1 0.033 0.033 0.033 $673,108.31 t t t te e dt e e e e − − − − − − ⎧⎪⎡ ⎤ ⎪⎫ = ⎨⎢− ⎥ + ⎬ ⎩⎪⎣ ⎦ ⎪⎭ ⎧ ⎡ ⎤ ⎫ = ⎪− − ⎢ ⎥ ⎪ ⎨ ⎬ ⎪ ⎢⎣ ⎥⎦ ⎪ ⎩ ⎭ ⎧⎪ ⎪⎫ = ⎨− − + ⎬ ⎩⎪ ⎪⎭ ≈ ∫ 9. Formula 3: u = x, du = dx, a = 1, b = 2 1(2 ln 1 2 ) 1 2 4 x dx x x C x = − + + + ∫ 10. Formula 10: u = x, du = dx, a = 0.1, b = 0.2 ( ) 1 1ln 0.1 0.2 0.1 0.1 0.2 10 ln 0.1 0.2 dx x C x x x x C x = + + + = + + ∫ 392 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 11. Formula 24: u = x, du = dx, a = 4 2 2 2 2 x 16 dx x 16 ln x x 16 C x x − − ∫ = − + + − + 12. Formula 15: u = x, du = dx, a = 4, b = 9 1 1 ln 4 9 2 4 9 2 4 9 2 dx x C x x x + − = + + + + ∫ 13. Formula 38: u = x2 , du = 2x dx, n = 4 2 ( 4 2 ) 4 2 2 1ln 1 1 4 x x x dx x e C e = − + + + ∫ 14. Formula 35: u = x2 + 1, du = 2x dx ∫ 2x(x2 + 1)ex2 +1 dx = x2ex2 +1 + C 15. R = 144t2 + 400 ∫ 144t2 + 400 dt Formula 23: u = 12t, du = 12 dt, a = 20 ( ) ( ) ( ) 2 2 2 2 2 2 1 12 1 1 12 2 1 50 2 3 144 400 144 400 12 12 144 400 400 ln 12 144 400 144 400 ln 12 144 400 t dt t dt t t t t C t t t t C + = + = ⎡ + + + + ⎤ + ⎢⎣ ⎥⎦ = + + + + + ∫ ∫ (a) Total revenue over first three years: ( ) 3 3 2 2 2 0 0 1 50 2 3 144 400 144 400 ln 12 144 400 $84.28112652 million or $84,281,126.52 t + dt = ⎡⎢⎣ t t + + t + t + ⎤⎥⎦ ≈ ∫ (b) Total revenue over first six years: ( ) 6 6 2 2 2 0 0 1 50 2 3 144 400 144 400 ln 12 144 400 $257.39242972 million or $257,392,429.72 t + dt = ⎡⎢⎣ t t + + t + t + ⎤⎥⎦ ≈ ∫ 16. Let u = x and dv = ex 2 dx. Then du = dx and v = 2ex 2. ( ) 0 2 2 0 0 2 2 2 2 1 2 0 2 1 1 2 2 4 4 4 4 4 8 4 1.057 x x x x xe dx xe e dx e e e e e − − − − − − − = ⎤⎦ − = − ⎡⎣ ⎤⎦ = − − = − ≈ − ∫ ∫ 17. Formula 42: u = x, du = dx ( ) ( ) ( ) 2 2 1 1 2 2 1 5 ln 5 ln 5 1 2 ln 4 5 1 2 ln 2 1 1 4 5 3 2 ln 2 4 10 ln 2 15 3.1815 4 x xdx x xdx x x = ⎡ ⎤ = ⎢ − + ⎥ ⎣ ⎦ = ⎡ − + − − ⎤ ⎢⎣ ⎥⎦ = ⎛− + ⎞ ⎜ ⎟ ⎝ ⎠ = − ≈ ∫ ∫ Section 6.3 Numerical Integration 393 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 18. Formula 19: u = x, du = dx, a = 8, b = 1 ( ) ( ) ( ) ( ) 8 8 0 0 2 16 8 8 3 16 4 32 2 2 3 3 64 64 2 3 3 64 2 1 3 8.8366 x x dx x x ⎡ − ⎤ = ⎢− + ⎥ + ⎣ ⎦ ⎛ − ⎞ ⎛ − ⎞ = ⎜ ⎟ − ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ − = + = − ≈ ∫ 19. Formula 42: u = x, du = dx ( )2 ( )2 1 1 ln 2 2 ln ln 2 0.718 e e x dx x x x e = ⎡ − + ⎤ ⎣ ⎦ = − ≈ ∫ 20. Formula 33: u = x, du = dx, a = 3 2 3 3 2 2 2 2 1 9 5 0.124 9 9 18 dx x x x x − ⎤ = − = ≈ ⎥⎥ − ⎦ ∫ 21. Formula 29: u = x2 , du = 2x dx, a = 2 2 6 6 4 4 2 4 2 1 ln 2 4 4 2 1 ln 17 ln 7 4 19 9 0.035 x dx x x x ⎡ − ⎤ = ⎢ ⎥ − ⎢⎣ + ⎥⎦ = ⎡ − ⎤ ⎢⎣ ⎥⎦ ≈ ∫ Section 6.3 Numerical Integration Skills Warm Up 1. ( ) ( ) ( ) 2 3 1 1 2 f x x f x x f x x = ′ = − ′′ = 2. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 2 3 4 4 ln 2 1 2 2 1 4 2 1 16 2 1 96 2 1 f x x f x x f x x f x x f x x = + ′ = + ′′ = − + ′′′ = + = − + 3. ( ) ( ) ( ) ( ) ( )( ) 2 3 4 4 2 ln 2 2 4 12 f x x f x x f x x f x x f x x = ′ = ′′ = − ′′′ = = − 4. ( ) ( ) ( ) 3 2 2 2 7 12 3 4 7 6 4 f x x x x f x x x f x x = − + − ′ = − + ′′ = − 5. ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 4 2 2 4 8 16 x x x x x f x e f x e f x e f x e f x e = ′ = ′′ = ′′′ = = 6. ( ) ( ) ( ) 2 2 2 2 2 2 4 2 x x x x f x e f x xe f x xe e = ′ = ′′ = + 7. f (x) = −x2 + 6x + 9, [0, 4] f ′(x) = −2x + 6 = 0 when x = 3 f (0) = 9, f (3) = 18, and f (4) = 17, so (3, 18) is the absolute maximum. Interval ∞ < x < 3 3 < x < ∞ Test value x = 0 x = 4 Sign of f ′(x) f ′(x) > 0 f ′(x) < 0 Conclusion Increasing Decreasing 394 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. Exact: 2 2 2 3 0 0 1 8 3 3 ∫ x dx = x ⎤⎦ = ≈ 2.6667 Trapezoidal Rule: ( ) ( ) ( ) ( ) 2 2 2 2 2 2 0 1 1 3 11 4 2 2 4 x dx ≈ ⎡⎢0 + 2 + 2 1 + 2 + 2 ⎤⎥ = = 2.75 ∫ ⎣ ⎦ Simpson’s Rule: ( ) ( ) ( ) ( ) 2 2 2 2 2 2 0 1 1 3 8 6 2 2 3 x dx ≈ ⎡⎢0 + 4 + 2 1 + 4 + 2 ⎤⎥ = ≈ 2.6667 ∫ ⎣ ⎦ 2. Exact: 2 3 1 1 0 0 1 7 1.1667 2 6 6 x dx x x ⎛ ⎞ ⎡ ⎤ ⎜ + ⎟ = ⎢ + ⎥ = ≈ ⎝ ⎠ ⎣ ⎦ ∫ Trapezoidal Rule: 1 2 0 1 1 1 2 1 1 2 1 1 2 9 1 3 75 1.1719 2 8 32 8 32 2 64 x dx ⎛ ⎞ ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ⎜ + ⎟ ≈ ⎢ + ⎜ + ⎟ + ⎜ + ⎟ + ⎜ + ⎟ + ⎥ = ≈ ⎝ ⎠ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ Simpson’s Rule: 1 2 0 1 1 1 4 1 1 2 1 1 4 9 1 3 7 1.1667 2 12 32 8 32 2 6 x dx ⎛ ⎞ ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ⎜ + ⎟ ≈ ⎢ + ⎜ + ⎟ + ⎜ + ⎟ + ⎜ + ⎟ + ⎥ = ≈ ⎝ ⎠ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ Skills Warm Up —continued— 8. ( ) 3 f x 8 , x = [1, 2] ( ) 4 f x 24 x ′ = − Because 4 24 0 x − ≠ for any value of x, there are no critical numbers. f (1) = 8 and f (2) = 1, so (1, 8) is the absolute maximum. 9. 2 2 2 2 2 2 1 0.001 4 1 4 0.001 1000 4 250 250 0 250 5 10 n n n n n n n < < < < − = = = ± So, n < −5 10 or n > 5 10. 10. 4 4 4 4 4 1 0.0001 16 1 16 0.0001 10,000 16 625 625 5 n n n n n n < < < < = = ± So, n < −5 or n > 5. Test interval (−∞, −5) (−5, 5) (5, ∞) n-value −6 1 6 Function value 0.000045 0.0625 0.000045 Conclusion Does satisfy Does not satisfy Does satisfy Test interval (−∞, −5 10) (−5 10, 5 10) (5 10, ∞) n-value −16 1 16 Function value 0.00098 0.25 0.00098 Conclusion Does satisfy Does not satisfy Does satisfy Section 6.3 Numerical Integration 395 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 3. Exact: 2 4 4 2 8 0 0 1 1 1 4 4 4 ∫ e− x dx = ⎡⎣− e− x ⎤⎦ = − e− + ≈ 0.2499 Trapezoidal Rule: 2 4 0 1 2 3 4 5 6 7 8 0 18 ∫ e− x dx ≈ ⎡⎣e + 2e− + 2e− + 2e− + 2e− + 2e− + 2e− + 2e− + e− ⎤⎦ ≈ 0.2704 Simpson’s Rule: 2 4 0 1 2 3 4 5 6 7 8 0 1 12 ∫ e− x dx ≈ ⎡⎣e + 4e− + 2e− + 4e− + 2e− + 4e− + 2e− + 4e− + e− ⎤⎦ ≈ 0.2512 4. Exact: 2 2 1 1 1 dx ln x ln 2 0.6931 x ∫ = ⎤⎦ = ≈ Trapezoidal Rule: 2 1 1 1 1 2 4 2 2 2 4 1 0.6970 8 5 3 7 2 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ Simpson’s Rule: 2 1 1 1 1 4 4 2 2 4 4 1 0.6933 12 5 3 7 2 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ 5. Exact: 2 3 4 2 0 0 1 4 ∫ x dx = x ⎤⎦ = 4 Trapezoidal Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 3 3 3 3 3 3 3 0 1 1 1 3 5 3 7 65 8 4 2 4 4 2 4 16 x dx ≈ ⎡⎢0 + 2 + 2 + 2 + 2 1 + 2 + 2 + 2 + 8⎤⎥ = = 4.0625 ∫ ⎣ ⎦ Simpson’s Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 3 3 3 3 3 3 3 3 0 1 1 1 3 5 3 7 12 4 2 4 4 2 4 x dx ≈ ⎡⎢0 + 4 + 2 + 4 + 2 1 + 4 + 2 + 4 + 8⎤⎥ = 4 ∫ ⎣ ⎦ 6. Exact: ( ) 3 3 3 2 1 1 4 4 2 0.6667 3 3 x dx x x ⎡ ⎤ − = ⎢ − ⎥ = − ≈ − ⎣ ⎦ ∫ Trapezoidal Rule: ( ) ( ) ( ) 3 2 1 4 1 3 2 7 2 0 2 9 5 0.75 4 4 4 x dx ⎡ ⎛ ⎞ ⎛ ⎞ ⎤ − ≈ ⎢ + ⎜ ⎟ + + ⎜− ⎟ + − ⎥ = − ⎣ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ Simpson’s Rule: ( ) ( ) ( ) 3 2 1 4 1 3 4 7 2 0 4 9 5 2 0.6667 6 4 4 3 x dx ⎡ ⎛ ⎞ ⎛ ⎞ ⎤ − ≈ ⎢ + ⎜ ⎟ + + ⎜− ⎟ + − ⎥ = − ≈ − ⎣ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ 7. Exact: 2 2 1 1 1 dx ln x ln 2 0.6931 x ∫ = ⎤⎦ = ≈ Trapezoidal Rule: 2 1 1 1 1 2 8 2 4 2 8 2 2 2 8 2 4 2 8 1 0.6941 16 9 5 11 3 13 7 15 2 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ Simpson’s Rule: 2 1 1 1 1 4 8 2 4 4 8 2 2 4 8 2 4 4 8 1 0.6932 24 9 5 11 3 13 7 15 2 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ 8. Exact: 2 2 1 2 1 1 dx 1 0.5 x x = − ⎤ = ⎥⎦ ∫ Trapezoidal Rule: 2 2 2 2 1 2 1 1 1 2 4 2 2 2 4 1 0.5090 8 5 3 7 4 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ ⎢⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎥⎦ ∫ Simpson’s Rule: 2 2 2 2 1 2 1 1 1 4 4 2 2 4 4 1 0.5004 12 5 3 7 4 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ ⎢⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎥⎦ ∫ 9. Exact: 1 2 2 1 3 3 3 0 0 1 1 1 6 6 6 ∫ xe x dx = ⎡⎢⎣ e x ⎤⎥⎦ = e − ≈ 3.1809 Trapezoidal Rule: ( ) ( ) ( ) 1 3 2 3 16 3 4 27 16 3 0 1 1 1 3 8 4 2 4 ∫ xe x dx ≈ ⎡⎣0 + 2 e + 2 e + 2 e + e ⎤⎦ ≈ 3.8643 Simpson’s Rule: ( ) ( ) ( ) 1 3 2 3 16 3 4 27 16 3 0 1 1 1 3 12 4 2 4 ∫ xe x dx ≈ ⎡⎣0 + 4 e + 2 e + 4 e + e ⎤⎦ ≈ 3.3022 396 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 10. Exact: 2 ( )3 2 2 0 0 23 ∫ 1 + x dx = ⎡⎣ 1 + x ⎤⎦ ≈ 2.7974 Trapezoidal Rule: 2 0 1 3 5 4 2 2 ∫ 1 + x dx ≈ ⎡⎢⎣1 + 2 + 2 2 + 2 + 3⎤⎥⎦ ≈ 2.7931 Simpson’s Rule: 2 0 3 5 2 2 1 1 1 4 2 2 4 3 2.7973 6 ∫ + x dx ≈ ⎡⎢⎣ + + + + ⎤⎥⎦ ≈ 11. Exact: 9 9 3 2 4 4 2 38 3 3 ∫ x dx = x ⎤⎦ = ≈ 12.6667 Trapezoidal Rule: 9 4 5 37 21 47 13 57 31 67 16 8 4 8 2 8 4 8 ∫ x dx ≈ ⎡⎢⎣2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 3⎤⎥⎦ ≈ 12.6640 Simpson’s Rule: 9 4 5 37 21 47 13 57 31 67 24 8 4 8 2 8 4 8 ∫ x dx ≈ ⎡⎢⎣2 + 4 + 2 + 4 + 2 + 4 + 2 + 4 + 3⎤⎥⎦ ≈ 12.6667 12. (a) Trapezoidal Rule: 4 0 2 2 2 2 2 8 1 0 2 8 2 8 2 8 8 4.020 3 2 1 3 2 3 3 3 4 3 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ + ⎣ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ + ⎦ ∫ Simpson’s Rule: 4 0 2 2 2 2 2 8 1 0 4 8 2 8 4 8 8 4.458 3 3 1 3 2 3 3 3 4 3 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ + ⎣ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ + ⎦ ∫ 13. (a) Trapezoidal Rule: ( ) ( ) ( ) ( ) 4 0 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 1 1 0 2 1 2 1 2 1 2 1 1 4 1 2 1 1 1 3 2 1 2 1 2 1 2 1 2 1 1 2.540 5 2 1 3 1 7 2 1 4 1 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎢ ⎜⎜ + ⎟⎟ ⎝ + ⎠ ⎜⎜ + ⎟⎟ ⎝ + ⎠ ⎢⎣ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤+ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ ⎜⎜ ⎟⎟ ⎝ + ⎠ ⎜⎜ ⎟⎟ + ⎥ ⎝ + ⎠ ⎝ + ⎠ ⎥⎦ ∫ (b) Simpson’s Rule: ( ) ( ) ( ) ( ) 4 0 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 3 2 1 1 0 4 1 2 1 4 1 2 1 1 6 1 2 1 1 1 3 2 1 2 1 4 1 2 1 4 1 1 2.541 5 2 1 3 1 7 2 1 4 1 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎢ ⎜⎜ ⎟⎟ ⎝ + ⎠ ⎜⎜ ⎟⎟ ⎝ + ⎠ ⎢⎣ ⎝ + ⎠ ⎝ + ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤+ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ ⎜⎜ + ⎟⎟ ⎝ + ⎠ ⎜⎜ + ⎟⎟ + ⎥⎝ ⎠ ⎝ ⎠ ⎥⎦ ∫ 14. (a) Trapezoidal Rule: 2 0 3 9 35 8 8 1 1 1 2 1 2 1 2 1 1 1.397 1 4 2 3 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ + ⎢ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎥ ⎣ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ (b) Simpson’s Rule: 2 0 3 9 35 8 8 1 1 1 4 1 2 1 4 1 1 1.405 1 6 2 3 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ + ⎢ ⎜ ⎟ ⎝ ⎠ ⎜ ⎟ ⎥ ⎣ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ 15. (a) Trapezoidal Rule: 2 3 0 1 9 35 4 8 8 ∫ 1 + x dx ≈ ⎡⎢⎣1 + 2 + 2 2 + 2 + 3⎤⎥⎦ ≈ 3.283 (b) Simpson’s Rule: 2 3 0 1 9 35 6 8 8 ∫ 1 + x dx ≈ ⎡⎢⎣1 + 4 + 2 2 + 4 + 3⎤⎥⎦ ≈ 3.240 Section 6.3 Numerical Integration 397 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 16. (a) Trapezoidal Rule: 1 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 63 15 55 3 39 7 15 16 64 16 64 4 64 16 64 1 − x dx = ⎢⎡1 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 0⎥⎤ ≈ 0.772 ∫ ⎣ ⎦ (b) Simpson’s Rule: 1 2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 63 15 55 3 39 7 15 24 64 16 64 4 64 16 64 1 − x dx = ⎡⎢1 + 4 + 2 + 4 + 2 + 4 + 2 + 4 + 0⎤⎥ ≈ 0.780 ∫ ⎣ ⎦ 17. (a) Trapezoidal Rule: 1 2 (0)2 (1 8)2 (1 4)2 (3 8)2 (1 2)2 (5 8)2 (3 4)2 (7 8)2 (1)2 0 1 16 ex dx ≈ ⎡⎢e + 2e + 2e + 2e + 2e + 2e + 2e + 2e + e ⎤⎥ ≈ 1.470 ⎣ ⎦ ∫ (b) Simpson’s Rule: 1 2 (0)2 (1 8)2 (1 4)2 (3 8)2 (1 2)2 (5 8)2 (3 4)2 (7 8)2 (1)2 0 1 24 ex dx ≈ ⎡⎢e + 4e + 2e + 4e + 2e + 4e + 2e + 4e + e ⎤⎥ ≈ 1.463 ⎣ ⎦ ∫ 18. (a) Trapezoidal Rule: ( ) 2 2 1 4 1 9 4 4 0 1 4 ∫ e−x dx ≈ 1 + 2e− + 2e− + 2e− + e− ≈ 0.881 (b) Simpson’s Rule: ( ) 2 2 1 4 1 9 4 4 0 1 6 ∫ e−x dx ≈ 1 + 4e− + 2e− + 4e− + e− ≈ 0.882 19. (a) Trapezoidal Rule: ( ) 3 0 2 1 1 1 2 4 2 1 2 4 2 1 2 4 1 1.879 2 2 4 2 5 5 2 13 5 dx x x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ − + ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ (b) Simpson’s Rule: ( ) 3 0 2 1 1 1 4 4 2 1 4 4 2 1 4 4 1 1.888 2 2 6 2 5 5 2 13 5 dx x x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ − + ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ 20. (a) Trapezoidal Rule: 3 0 2 1 0 2 2 2 1 2 6 2 1 2 10 3 0.641 2 4 11 4 23 4 43 14 x dx x x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ + + ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ (b) Simpson’s Rule: 3 0 2 1 0 4 2 2 1 4 6 2 1 4 10 3 0.653 2 6 11 4 23 4 43 14 x dx x x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎥ ≈ + + ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎦ ∫ 21. ( ) ( ) 1 0 4 0.07 0 6000 200 t rt t V cte dt t e dt − − = = + ∫ ∫ Using a program similar to the Simpson’s Rule program, when n = 8, V ≈ $21,831.20. 22. ( ) ( ) 1 0 8 3 0.10 0 200,000 15,000 t rt t V cte dt t e dt − − = = + ∫ ∫ Using a program similar to the Simpson’s Rule program, when n = 8, V ≈ $1,215,971.94. 23. 16 3 14 ΔR = ∫ 5 8000 − x dx Using a program similar to the Simpson’s Rule program, when n = 4, ΔR ≈ $678.36. 24. 16 14 ΔR ≈ ∫ 50 x 20 − x dx Using a program similar to the Simpson’s Rule program, when n = 4, ΔR ≈ $863.44. 25. ( ) 1 2 2 0 0 1 1 2 P x e x dx π ≤ ≤ = ∫ − Using a program similar to the Simpson’s Rule program, when n = 6, P ≈ 0.3413 = 34.13%. 26. ( ) 2 2 2 0 0 2 1 2 P x e x dx π ≤ ≤ = ∫ − Using a program similar to the Simpson’s Rule program, when n = 6, P ≈ 0.4772 = 47.72%. 27. ( ) 4 2 2 0 0 4 1 2 P x e x dx π ≤ ≤ = ∫ − Using a program similar to the Simpson’s Rule program, when n = 6, P ≈ 0.499958 = 49.996%. 28. ( ) 1.5 2 2 0 0 1.5 1 2 P x e x dx π ≤ ≤ = ∫ − Using a program similar to the Simpson’s Rule program, when n = 6, P ≈ 0.4332 = 43.32%. 398 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 29. ( ) 1000 125 4(125) 2(120) 4(112) 2(90) 4(90) 2(95) 4(88) 2(75) 4(35) 0 3 10 89,500 square feet A ≈ ⎡⎣ + + + + + + + + + + ⎤⎦ = 30. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 120 75 4 81 2 84 4 76 2 67 4 68 2 69 4 72 2 68 4 56 2 42 4 23 0 3 12 7463.3333 square feet A ≈ ⎡⎣ + + + + + + + + + + + + ⎤⎦ ≈ 31. ( ) ( ) ( ) ( ) ( )( ) 2 4 2 2 2 2 0 0 f x x x f x x f x f x f x = + ′ = + ′′ = ′′′ = = (a) Trapezoidal Rule: Because f ′′(x) is maximum for all x in [0, 2] and f ′′(x) = 2, you have ( ) ( ) ( ) 3 2 2 0 1 Error 2 0.0833. 12 4 12 − ≤ = ≈ (b) Simpson’s Rule: Because f (4) (x) is maximum for all x in [0, 2] and f (4) (x) = 0, you have ( ) ( ) ( ) 5 4 2 0 Error 0 0. 180 4 − ≤ = 32. ( ) 1 1 f x x = + ( ) ( )2 1 1 f x x ′ = − + ( ) ( )3 2 1 f x x ′′ = + ( ) ( )4 6 1 f x x ′′′ = − + ( )( ) ( ) 4 5 24 1 f x x = + (a) Trapezoidal Rule: Because f ′′(x) is maximum in [0, 1]when x = 0 and f ′′(0) = 2, you have ( ) ( ) ( ) 3 2 1 0 1 Error 2 0.01. 12 4 96 − ≤ = ≈ (b) Simpson’s Rule: Because f (4) (x) is maximum in [0, 1]when x = 0 and f (4) (0) = 24, you have ( ) ( ) ( ) 5 4 1 0 1 Error 24 0.00052. 180 4 1920 − ≤ = ≈ 33. ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 3 2 3 4 3 6 3 3 4 8 5 2 3 3 3 3 2 3 9 18 2 9 9 36 20 x x x x x f x e f x x e f x x xe f x x x e f x x x xe = ′ = ′′ = + ′′′ = + + = + + (a) Trapezoidal Rule: Because f ′′(x) is maximum in [0, 1]when x = 1 and f ′′(1) = 15e, you have ( ) ( ) ( ) 3 2 1 0 5 Error 15 0.212. 12 4 64 e e − ≤ = ≈ (b) Simpson's Rule: Because f (4) (x) is maximum in [0, 1]when x = 1 and f (4) (1) = 585e, you have ( ) ( ) ( ) 5 4 1 0 13 Error 585 0.035. 180 4 1024 e e − ≤ = ≈ 34. ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 2 2 2 2 2 2 2 2 2 2 4 4 2 2 2 4 16 4 16 4 3 256 384 48 x x x x x f x e f x xe f x x e f x x x e f x x x e = ′ = ′′ = + ′′′ = + = + + (a) Trapezoidal Rule: Because f ′′(x) is maximum in [0, 1] when x = 1 and f ′′(1) = 20e2 , you have ( ) ( ) 3 2 2 1 0 Error 20 0.7697. 12 4 e − ≤ ≈ (b) Simpson’s Rule: Because f (4) (x) is maximum in [0, 1] when x = 1 and f (4) (1) = 688e2 , you have ( ) ( ) 5 2 4 1 0 Error 688 0.1103. 180 4 e − ≤ ≈ Section 6.3 Numerical Integration 399 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 35. ( ) ( ) ( ) ( ) ( )( ) 4 3 2 4 4 12 24 24 f x x f x x f x x f x x f x = ′ = ′′ = ′′′ = = (a) Trapezoidal Rule: Because f ′′(x) is maximum in [0, 2]when x = 2 and f ′′(2) = 48, you have ( )3( ) 2 2 2 2 0 48 Error 0.0001 12 32 0.0001 320,000 565.69 . n n n n − ≤ < < < < Let n = 566. (b) Simpson’s Rule: Because f (4) (x) = 24 for all x in [0, 2], you have ( ) ( ) 5 4 4 4 2 0 Error 24 0.0001 180 64 0.0001 15 42,666.67 14.37. n n n n − ≤ < < > > Let n = 16. 36. ( ) ( ) ( ) ( ) ( )( ) 2 3 4 4 5 1 1 2 6 24 f x x f x x f x x f x x f x x = ′ = − ′′ = ′′′ = − = (a) Trapezoidal Rule: Because f ′′(x) is maximum in [1, 3] when x = 1 and f ′′(1) = 2, you have Error ≤ ( ) ( ) 3 2 2 2 3 1 2 0.0001 12 4 0.0001 3 13,333.33 115.47. n n n n − < < > > Let n = 116. (b) Simpson’s Rule: Because f (4) (x) is maximum in [1, 3] when x = 1 and f (4) (1) = 24, you have Error ≤ ( ) ( ) 5 4 4 4 3 1 24 0.0001 180 64 0.0001 15 42,666.67 14.4. n n n n − < < > > Let n = 16. (n must be even.) 37. ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 4 2 2 4 8 16 x x x x x f x e f x e f x e f x e f x e = ′ = ′′ = ′′′ = = (a) Trapezoidal Rule: Because f ′′(x) is maximum (b) Simpson’s Rule: Because f (4) (x) is maximum in [1, 3] when x = 3 and f ′′(3) = 4e6 , you have in [1, 3] when x = 3 and f (4) (3) = 16e6 , you have Error ≤ ( ) ( ) 3 6 2 6 2 2 3 1 4 0.0001 12 8 0.0001 3 10,758,101.16 3279.95. e n e n n n − < < > > Error ≤ ( ) ( ) 5 6 4 6 4 4 3 1 16 0.0001 180 128 0.0001 45 11,475,307.90 58.2. e n e n n n − < < > > Let n = 3280. Let n = 60. (n must be even.) 400 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 38. ( ) ( ) ( ) ( ) ( )( ) 2 3 4 4 ln 1 1 2 6 fx x f x x f x x f x x f x x = ′ = ′′ = − ′′′ = = − (a) Trapezoidal Rule: Because f ′′(x) is maximum (b) Simpson’s Rule: Because f (4) (x) is maximum in [3, 5]when x = 3 and (3) 1, 9 f ′′ = you have in [3, 5]when x = 3 and (4) (3) 2 , 27 f = you have Error ≤ ( )3 2 2 2 5 3 1 0.0001 12 9 2 0.0001 27 740.74 27.2. n n n n − ⎛ ⎞ < ⎜ ⎟ ⎝ ⎠ < > > Error ≤ ( )5 4 4 4 5 3 2 0.0001 180 27 16 0.0001 1215 131.69 3.39. n n n n − ⎛ ⎞ < ⎜ ⎟ ⎝ ⎠ < > > Let n = 28. Let n = 4. 39. Using a program similar to the Simpson’s Rule program, when n = 100, 4 1 ∫ x x + 4 dx ≈ 19.5215. 40. Using a program similar to the Simpson’s Rule program, when n = 100, 4 2 1 ∫ x x + 4 dx ≈ 55.6246. 41. Using a program similar to the Simpson’s Rule program, when n = 100, 5 2 ∫ 10xe−x dx ≈ 3.6558. 42. Using a program similar to the Simpson’s Rule program, when n = 100, 5 2 2 ∫ 10x e−x dx ≈ 11.0405. 43. (a) Average median age ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) } 9 1 1 1 1 35.5 4 35.7 2 35.9 4 36.0 2 36.2 4 36.3 9 1 8 3 2 36.5 4 36.7 36.8 36.2 years = f x dx ≈ ⎧⎨ ⎡⎣ + + + + + − ⎩ + + + ⎤⎦ ≈ ∫ (b) Average median age ( ) 9 2 9 1 1 1 35.4 0.16 0.000004 1 35.4 0.08 0.000004 36.2 years 9 1 8 = + t − et dt = ⎡ t + t − et ⎤ ≈ − ⎣ ⎦ ∫ (c) The results are approximately equal. 44. (a) Average residential price ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) } 9 1 1 1 1 8.58 4 8.44 2 8.72 4 8.95 2 9.45 4 10.40 9 1 8 3 2 10.65 4 11.26 11.55 9.75 cents kilowatt-hour = f x dx ≈ ⎧⎨ ⎡⎣ + + + + + − ⎩ + + + ⎤⎦ ≈ ∫ (b) Average residential price ( ) 9 2 3 1 2 1 9 2 3 4 3 2 1 1 8.4 1.39 0.291 0.0160 1.27097 9 1 1 8.4 0.695 0.097 0.004 2.54194 9.75 cents kilowatt-hour 8 3 t t t t dt t t t t t = − + − + − = ⎡ − + − + ⎤ ≈ ⎢⎣ ⎥⎦ ∫ (c) The results are approximately equal. 45. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 12 2 0 1 2 8 ln 2 4 6.61 4 6.82 2 6.05 4 5.28 2 4.67 4 4.19 2 3.80 4 3.46 3.18 58.915 mg C = ⎡⎣ − t − t + ⎤⎦ dt ≈ ⎡⎣ + + + + + + + + ⎤⎦ ≈ ∫ Section 6.4 Improper Integrals 401 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 46. (a) The Trapezoidal Rule tends to become more accurate as n increases, therefore using n = 16 should yield a more accurate approximation. (b) For a given value of n, Simpson’s Rule tends to be more accurate than the Trapezoidal Rule, therefore using Simpson’s Rule with n = 8, should yield a more accurate approximation. 47. [ ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ] 6 2 0 1 6 1000 0 4 151.63 2 367.88 4 502.04 2 541.34 4 513.03 2 448.08 4 369.92 2 293.05 4 224.96 2 168.45 4 123.62 89.24 S = t e−t dt ≈ + + + + + + + + + + + + ∫ ≈ 1878 subscribers 48. ( ) ( ) ( ) ( ) ( )( ) 3 2 3 2 3 3 3 4 3 3 2 6 2 6 0 P x ax bx cx d P x ax bx c P x ax b P x a P x = + + + ′ = + + ′′ = + ′′′ = = ( ) ( )( ) ( ) ( ) 5 5 4 4 3 4 Error max 0 0 180 180 b a b a P x n n − ⎡ ⎤ − ≤ ⎣ ⎦ = = So, Simpson’s Rule is exact when used to approximate a cubic polynomial. ( ) ( ) 3 1 3 3 3 0 1 0 0 4 1 1 1 0 1 1 1 3 2 2 6 2 4 x dx − ⎡ ⎛ ⎞ ⎤ ⎡ ⎤ = ⎢ + ⎜ ⎟ + ⎥ = ⎢ + + ⎥ = ⎢⎣ ⎝ ⎠ ⎥⎦ ⎣ ⎦ ∫ The exact value of this integral is 4 1 1 3 0 0 1, 4 4 x dx x ⎤ = = ⎥⎦ ∫ which is the same as the Simpson approximation. 49. Answers will vary. Section 6.4 Improper Integrals Skills Warm Up 1. ( ) ( ) 2 lim 2 5 2 2 5 9 x x → + = + = 2. 2 ( 2 ) 1 lim 1 2 1 2 1 3 x 1 x → x ⎛ + ⎞ = + = ⎜ ⎟ ⎝ ⎠ 3. 4 2 4 ( )( ) 4 lim 4 lim 4 16 4 4 lim 1 4 1 4 4 1 8 x x x x x x x x x →− →− →− + + = − + − = − = − − = − 4. ( ) ( ) ( ) 2 0 3 2 0 2 0 2 2 lim lim 3 3 lim 2 3 x x x x x x x x x x x x x x → → → − − = + + − = + ( ) ( ) 0 0 lim 2 3 lim 2 3 x x x x x x x x → − → + − = ∞ + − = −∞ + Limit does not exist. 5. 1 lim 1 x→ x 1 = ∞ − Limit does not exist. 402 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 1. The integral is improper because the function has an infinite discontinuity in [0, 1]. 2. The integral is proper. 3. The integral is proper. 4. The integral is improper because the upper limit of integration is infinite. 5. The integral is proper. 6. The integral is improper because both the upper limit of integration and the lower limit of integration are infinite. 7. This integral converges because 1 2 1 1 lim 1 0 1 1. b b dx x x ∞ →∞ = ⎡− ⎤ = + = ⎢⎣ ⎥⎦ ∫ 8. This integral diverges because 1 1 1 lim 2 2 . b b dx x x ∞ →∞ = ⎡ ⎤ = ∞ − = ∞ ∫ ⎣ ⎦ 9. This integral diverges because 3 3 0 0 lim 3 . x x b b e dx e ∞ →∞ ∫ = ⎡⎣ ⎤⎦ = ∞ 10. This integral converges because ( ) 2 2 0 2 0 0 5 5 5 lim 2 5 0 1 5. 2 2 x x b x b dx e dx e e ∞ ∞ − − → ∞ = = − ⎡⎣ ⎤⎦ = − − = ∫ ∫ 11. This integral diverges because 2 5 2 5 lim 16 3 . 16 b b x dx x x ∞ →∞ = ⎡ − ⎤ = ∞ − = ∞ − ⎣ ⎦ ∫ 12. This integral diverges because 5 5 1 lim 2 1 3 . 2 1 b b dx x x ∞ →∞ = ⎡ − ⎤ = ∞ − = ∞ − ⎣ ⎦ ∫ 13. This integral diverges because 0 0 x lim x 1 . a a e− dx e− −∞ →−∞ ∫ = ⎡⎣− ⎤⎦ = − + ∞ = ∞ 14. This integral converges because 1 1 2 1 lim 1 1 0 1. b b dx x x − − −∞ →−∞ = ⎡− ⎤ = + = ⎢⎣ ⎥⎦ ∫ Skills Warm Up —continued— 6. 2 ( )( ) 3 3 3 2 3 3 1 lim lim 3 3 lim 1 3 1 4 x x x x x x x x x x →− →− →− + − + − = + + = − = − − = − 7. ( )3 43 2x − 1 (a) ( )3 43 2b − 1 (b) 4( )3 4 3 3 2 ⋅ 0 − 1 = − 8. ( )2 1 3 x 5 x 2 + − − (a) ( )2 1 3 b 5 b 2 + − − (b) ( )2 1 3 1 3 11 0 5 0 2 5 4 20 + = − + = − − 9. ln(5 − 3x2 ) − ln(x + 1) (a) ln(5 − 3b2 ) − ln(b + 1) (b) ln(5 3 02 ) ln(0 1) ln 5 ln 1 ln 5 1.609 − ⋅ − + = − = ≈ 10. e3x2 + e−3x2 (a) e3b2 + e−3b2 = e−3b2 (e6b2 + 1) (b) 3(0)2 3(0)2 0 0 1 1 2 e + e− = e + e = + = Section 6.4 Improper Integrals 403 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 15. This integral diverges because 1 1 lim 2 lim 2 2 . x b x b b b e dx e x e e ∞ →∞ →∞ = ⎡ ⎤ ⎣ ⎦ = ⎡ − ⎤ ⎣ ⎦ = ∞ ∫ 16. This integral diverges because ( ) 0 0 2 2 lim 1 ln 1 1 2 0 . b b x dx x −∞ x →−∞ = ⎡ + ⎤ + ⎢⎣ ⎥⎦ = − ∞ = −∞ ∫ 17. This integral converges because ( ) ( ) 3 2 0 3 2 3 2 0 2 0 2 3 3 0 1 1 3 3 1 1 3 3 2 2 2 lim lim 0 0 0. x x x b x x a a b xe dx xe dx xe dx e e ∞ − − ∞ − −∞ −∞ − − →−∞ →∞ = + = ⎡− ⎤ + ⎡− ⎤ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ = − + + + = ∫ ∫ ∫ 18. This integral diverges because ( ) ( ) 2 3 0 2 3 2 3 0 3 0 3 0 1 1 3 3 1 1 3 3 lim lim 0 . x x x c x x b c x e dx x e dx x e dx e e ∞ − − ∞ − − ∞ − ∞ − − →−∞ −∞ →∞ = + = ⎡− ⎤ + ⎡− ⎤ ⎢⎣ ⎥⎦ ⎢⎣ ⎥⎦ = − + ∞ + + = ∞ ∫ ∫ ∫ 19. This integral converges because 4 ( )3 ( )2 ( )2 4 1 lim 1 1 . ln 2ln 2 ln 4 b b dx x x x ∞ →∞ ⎡ ⎤ = ⎢− ⎥ = ⎢⎣ ⎥⎦ ∫ 20. This integral diverges because ( ) 2 3 0 0 3 1 3 1 3 lim . 1 a 2 2 a x dx −∞ x →−∞ ⎡ − ⎤ = ⎢− ⎥ = −∞ = −∞ − ⎢⎣ ⎥⎦ ∫ 21. 0 0 lim 1 x x b b A e dx e ∞ − − →∞ = ∫ = ⎡⎣− ⎤⎦ = 22. 0 x 4 lim 4 x 4 0 4 a a A e dx e −∞ →−∞ = ∫ = ⎡⎣ ⎤⎦ = 23. 1 1 3 2 1 lim 1 1 a 2 a 2 A dx x x − − −∞ →−∞ = ⎛− ⎞ = ⎡ ⎤ = ⎜ ⎟ ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ ∫ 24. 0 5 0 lim 10 4 4 a a A dx x −∞ x →−∞ = = ⎡− − ⎤ = ∞ − ⎣ ⎦ ∫ 25. ( 2 ) 0 2 0 6 lim 3 ln 1 1 b b A x dx x x ∞ →∞ = = ⎡ + ⎤ = ∞ + ⎣ ⎦ ∫ 26. ( 2 ) 0 2 0 16 lim 8 ln 4 4 b b A x dx x x ∞ →∞ = = ⎡ + ⎤ = ∞ + ⎣ ⎦ ∫ 27. μ = 63.8, σ = 2.9 ( ) 1 ( )2 2 2 0.137566 ( 63.8)2 16.82 2 f x e x μ σ e x σ π = − − = − − Using a graphing utility: (a) ( ) 72 60 ∫ f x dx ≈ 0.9026 (b) ( ) 68 f x dx 0.0738 ∞ ∫ ≈ (c) ( ) 72 f x dx 0.00235 ∞ ∫ ≈ 28. (a) The probability of choosing a car at random that gets between 26 and 28 miles per gallon is greater because the area of the region bounded by 26 ≤ x ≤ 28 is greater than that bounded by 22 ≤ x ≤ 24. (b) The probability of choosing a car at random that gets at least 30 miles per gallon is greater because the area of the region for x ≥ 30 is greater than that bounded by 20 ≤ x ≤ 22. 29. μ = 36, σ = 0.2 ( ) 1 ( )2 2 2 1.99471 ( 36)2 0.08 2 f x e x μ σ e x σ π = − − = − − Using a graphing utility: (a) ( ) 35.5 f x dx 0.9938 ∞ ∫ ≈ (b) ( ) 35.9 f x dx 0.6915 ∞ ∫ ≈ 30. μ = 9000, σ = 500 ( ) 1 ( )2 2 2 0.000797 ( 9000)2 500,000 2 f x e x μ σ e x σ π = − − = − − Using a graphing utility: (a) ( ) 10,000 8000 ∫ f x dx ≈ 0.9534 (b) ( ) 11,000 f x dx 0.00003 ∞ ∫ ≈ 31. Present value 5000 $66,666.67 0.075 p r = = ≈ 32. Present value 12,000 $200,000 0.06 p r = = = 33. Present value 18,000 $360,000 0.05 p r = = = The amount you need to start the scholarship fund is $360,000. Yes, you have enough money to start the scholarship fund. 404 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 34. Present value 35,000 $437,500 0.08 p r = = = The amount the foundation needs to fund the donation is $437,500. Yes, the foundation has enough money to start the fund. 35. (a) Present value 20 0.09 0 20 0.09 0 500,000 500,000 0.09 $4,637,228.40 t t e dt e − − = = − ⎤⎥⎦ ≈ ∫ (b) Present value 0.09 0 0.09 0 500,000 lim 500,000 0.09 $5,555,555.56 t b t b e dt e ∞ − − →∞ = = ⎡− ⎤ ⎢⎣ ⎥⎦ = ∫ 36. (a) Present value 20 0.08 0 20 0.08 0 75,000 75,000 0.08 $748,222.01 t t e dt e − − = = ⎡− ⎤ ⎢⎣ ⎥⎦ ≈ ∫ (b) Present value 0.08 0 0.08 0 75,000 lim 75,000 0.08 $937,500.00 t b t b e dt e ∞ − − →∞ = = ⎡− ⎤ ⎢⎣ ⎥⎦ = ∫ 37. 0.10 0 0.10 0 650,000 25,000 650,000 250,000 n t t n C e dt e − − = + = − ⎡⎣ ⎤⎦ ∫ (a) For n = 5, you have 650,000 250,000( 0.50 1) $748,367.34. C = − ⎡⎣ e− − ⎤⎦ ≈ (b) For n = 10, you have C = 650,000 − ⎡⎣250,000(e−1 − 1)⎤⎦ ≈ $808,030.14. (c) For n = ∞, you have ( ) 0.10 0 650,000 lim 250,000 650,000 250,000 0 1 $900,000. t n n C e− →∞ = − ⎡⎣ ⎤⎦ = − − = 38. 0.04 0 0.04 0 800,000 30,000 800,000 750,000 n t t n C e dt e − − = + = − ⎡⎣ ⎤⎦ ∫ (a) For n = 5, you have 800,000 750,000( 0.2 1) $935,951.94. C = − ⎡⎣ e− − ⎤⎦ ≈ (b) For n = 10, you have 800,000 750,000( 0.4 1) $1,047,259.97. C = − ⎡⎣ e− − ⎤⎦ = (c) For n = ∞, you have ( ) 0.04 0 800,000 lim 750,000 800,000 750,000 0 1 $1,550,000. t n n C e− →∞ = − ⎡⎣ ⎤⎦ = − − = 39. 0.06 0 0.06 0 300,000 15,000 300,000 15,000 50 2500 3 9 n t n t C te dt t e − − = + ⎡⎛ ⎞ ⎤ = − ⎢⎜ + ⎟ ⎥ ⎣⎝ ⎠ ⎦ ∫ (a) For n = 5, you have 300,000 15,000 3250 0.3 2500 9 9 $453,901.30. C = − ⎛⎜ e− − ⎞⎟ ⎝ ⎠ ≈ (b) For n = 10, you have 300,000 15,000 4000 0.6 2500 9 9 $807,922.43. C = − ⎛⎜ e− − ⎞⎟ ⎝ ⎠ ≈ (c) For n = ∞, you have 0.06 0 300,000 15,000 lim 50 2500 3 9 300,000 15,000 0 2500 9 $4,466,666.67. n t n C t e− →∞ ⎡⎛ ⎞ ⎤ = − ⎢⎜ + ⎟ ⎥ ⎣⎝ ⎠ ⎦ = − ⎛ − ⎞ ⎜ ⎟ ⎝ ⎠ ≈ 40. ( ) ( ) 0.12 0 0.12 0 19 650,000 25,000 1 0.08 650,000 25,000 125 6 n t t n C te dt t e − − = + + = + ⎡⎣− + ⎤⎦ ∫ (a) For n = 5, you have ( 155 0.6 125) 9 9 650,000 25,000 $760,928.32. C = + − e− + ≈ (b) For n = 10, you have ( 185 1.2 125) 9 9 650,000 25,000 $842,441.86. C = + − e− + ≈ (c) For n = ∞, you have ( ) ( ) 0.12 0 1 9 125 9 650,000 25,000 lim 125 6 650,000 25,000 0 $997,222.22. t n n C te− →∞ = + ⎡⎣− + ⎤⎦ = + + ≈ Review Exercises for Chapter 6 405 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Review Exercises for Chapter 6 1. Let u = ln x and dv = (1 x) dx. Then du = (1 x) dx and v = 2 x. 1 2 ln 2 ln 2 1 2 ln 2 2 ln 4 x dx x x x dx x x x x x dx x x x C − = − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − = − + ∫ ∫ ∫ 2. Let u = ln 4x and dv = x dx. Then 1 (4) 1 4 du dx dx x x = = and 1 2. 2 v = x 2 2 2 2 2 ln 4 1 ln 4 1 1 2 2 1 ln 4 1 2 2 1 ln 4 1 2 4 x x dx x x x dx x x x xdx x x x C = − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = − = − + ∫ ∫ ∫ 3. Let u = x − 1 and dv = ex dx. Then du = dx and v = ex . ( ) ( ) ( ) ( ) 1 1 1 2 x x x x x x x e dx x e e dx x e e C x e C − = − − = − − + = − + ∫ ∫ 4. Let u = x and dv = e−3x dx. Then du = dx and 1 3 3 v = − e− x. 3 3 3 3 3 3 3 1 1 3 3 1 1 3 3 1 1 3 9 x x x x x x x xe dx xe e dx xe e dx xe e C − − − − − − − = − − − = − + = − − + ∫ ∫ ∫ 5. Let u = x and ( )1 2 dv = x − 5 dx. Then du dx = and ( ) 23 v = x − 5 . ( ) ( ) ( ) ( ) 3 2 3 2 3 2 5 2 2 2 3 3 2 4 3 15 5 5 5 5 5 x x dx xx x dx xx x C − = − − − = − − − + ∫ ∫ 6. Let u = x and ( ) 1 2 dv x 8 dx. − = + Then du = dx and v = 2(x + 8). ( ) ( ) 1 2 3 2 2 8 2 8 8 2 8 4 8 3 x dx x x x dx x x x x C = + − + + = + − + + ∫ ∫ 7. Let u = 2x2 and dv = e2x. Then du = 4x dx and 1 2 2 v = e x . 2 2 ( 2 )(1 2 ) 1 2 ( ) 2 2 2 2 2 ∫ 2x e x dx = 2x e x − ∫ e x 4x dx = x e x − ∫ 2xe x dx Let u = 2x and dv = e2x . Then du = 2 dx and 1 2 2 v = e x . 2 2 2 2 ( )( 2 ) 2 ( ) 2 2 2 2 2 2 2 2 1 1 2 2 1 2 2 x x 2 x x 2 x x x x x x xe dx xe x e e dx x e xe e dx x e xe e C = − ⎡ − ⎤ ⎣ ⎦ = − + = − + + ∫ ∫ ∫ 8. Let ( )3 u = ln x and dv = dx.Then du 3(ln x)2 1 dx x = and v = x. ( )3 ( )3 ( )2 ∫ ln x dx = x ln x −∫ 3 ln x dx Let ( )2 u = ln x and dv = 3 dx. Then du 2(ln x)1 x = and v = 3x. ( )3 ( )3 ( )2 ∫ ln x dx = x ln x − ⎡⎣3x ln x − ∫ 6 ln x dx⎤⎦ Let u = ln x and dv = 6 dx. Then du 1 dx x = and v = 6x. ( ) ( ) ( ) ( ) ( ) 3 3 2 3 2 ln ln 3 ln 6 ln 6 ln 3 ln 6 ln 6 x dx x x x x x x dx x x x x x x x C = − + ⎡ − ⎤ ⎣ ⎦ = − + − + ∫ ∫ 406 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 9. Let u = ln x and dv = 6x dx. Then du 1 dx x = and v = 3x2. 2 2 1 2 2 1 2 2 2 6 ln 3 ln 3 1 3 ln 3 2 3 3 0 3 2 2 3 3 2 2 12.584 e e x x dx x x x dx x x x x e e e = − ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = ⎡ − ⎤ ⎢⎣ ⎥⎦ = ⎛ − ⎞ − ⎛ − ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = + ≈ ∫ ∫ 10. Let u = ln(1 + 3x) and dv = dx. Then 1 (3) 1 3 du dx x = + and v = x. ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 0 4 0 ln 1 3 ln 1 3 3 1 3 ln 1 3 1 1 1 3 ln 1 3 1 ln 1 3 3 4 ln 13 4 1 ln 13 0 3 13 ln 13 4 3 7.1148 x dx x x x dx x x x dx x x x x x + = + −+ = + − ⎛ − ⎞ ⎜ + ⎟ ⎝ ⎠ = ⎡ + − + + ⎤ ⎢⎣ ⎥⎦ = ⎛ − + ⎞ − ⎜ ⎟ ⎝ ⎠ = − ≈ ∫ ∫ ∫ 11. Let u = x and dv = e−x 4 dx. Then du = dx and v = −4e−x 4. ( ) ( ) 1 4 4 4 0 4 4 1 0 1 4 1 4 1 4 4 4 4 16 4 16 0 16 16 20 0.4240 x x x x x xe dx xe e dx xe e e e e − − − − − − − − = − + = ⎡⎣− − ⎤⎦ = − − − − = − ≈ ∫ ∫ 12. Let u = x2 and dv = e3x dx. Then du = 2x dx and 1 3 3 v = e x. 2 2 3 2 3 3 0 1 2 3 3 ∫ x e x dx = x e x − ∫ xe x dx Let u = x and dv = e3x dx. Then du = dx and 1 3 3 v = e x . ( ) ( ) 2 3 3 2 3 0 2 3 3 3 2 0 6 6 6 6 1 2 1 1 3 3 3 3 1 2 2 3 9 27 4 4 2 2 3 9 27 27 26 2 27 27 0 0 388.4129 x x x x x x x e xe e dx x e xe e e e e e = − ⎡ − ⎤ ⎢⎣ ⎥⎦ = ⎡⎣ − + ⎤⎦ = − + − − + = − ≈ ∫ 13. Present value 5 0.04 0.04 5 0 0 = ∫ 20,000e− t dt = −500,000e− t ⎤⎦ = $90,634.62 14. Present value ( ) ( ) 10 0.06 0 10 0.06 10 0.06 0 0 10 0.06 10 0.06 0 0 0.6 10 0.06 0 10,000 1500 10,000 1500 10,000 1500 0.06 10,000 1 1500 0.06 t t t t t t t e dt e dt te dt e te dt e te dt − − − − − − − = + = + = − ⎤ + ⎥⎦ = − − + ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e−0.06t dt. Then du = dt and 1 0.06 . 0.06 v = − e− t ( ) ( ) ( ) ( ) 10 0.6 0.06 10 0.06 0 0 0.6 0.6 0.06 10 0 0.6 0.6 0.6 10,000 1 1500 1 0.06 0.06 0.06 10,000 1 1500 10 1 0.06 0.06 0.0036 10,000 1 1500 10 1 1 0.06 0.06 0.0036 $12 t t t e t e e dt e e e e e e − − − − − − − − − ⎧⎪⎡ ⎤ ⎪⎫ = − − + ⎨⎢− ⎥ + ⎬ ⎪⎩⎣ ⎦ ⎭⎪ = − − + ⎧− − ⎡ ⎤ ⎫ ⎨ ⎣ ⎦ ⎬ ⎩ ⎭ = − − + ⎡− − − ⎤ ⎢⎣ ⎥⎦ ≈ ∫ 5,990.30 Review Exercises for Chapter 6 407 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 15. Present value 10 0.05 10 0.05 0 0 = ∫ 24,000te− t dt = 24,000∫ e− t dt Let u = t and dv = e−0.05t dt. Then du = dt and v = −20e−0.05t . { } { } ( ) ( ) 0.05 10 10 0.05 0 0 0.5 0.05 10 0 0.5 0.5 0.5 24,000 20 20 24,000 200 400 24,000 200 400 1 24,000 400 600 $865,958.50 t t t te e dt e e e e e − − − − − − − = ⎡⎣− ⎤⎦ + = − − ⎡⎣ ⎤⎦ = ⎡− − − ⎤ ⎣ ⎦ = − = ∫ 16. ( ) ( ) 5 2 0.05 0 5 0.05 0.45 0 5 0.05 0.45 0 0.25 2.25 Present value 20,000 100 20,000 100 400,000 100 0.45 400,000 100 400,000 100 0.45 0.45 $90,365.85 t t t t t t e e dt e e dt e e e e − − − − = + = + = ⎡− + ⎤ ⎢⎣ ⎥⎦ = ⎛− + ⎞ − ⎛− + ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ ≈ ∫ ∫ 17. (a) ( ) 4 0 2 4 0 Actual income 200,000 50,000 200,000 25,000 $1,200,000 t dt t t = + = ⎡⎣ + ⎤⎦ = ∫ (b) ( ) 4 0.06 0 4 0.06 4 0.06 0 0 4 0.06 4 0.06 0 0 4 0.06 0 Present value 200,000 50,000 200,000 50,000 200,000 50,000 0.06 711,240.46 50,000 t t t t t t t e dt e dt te dt e te dt te dt − − − − − − = + = + ⎡ ⎤ = ⎢ ⎥ + ⎣ − ⎦ = + ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e−0.06t dt. Then du = dt and 1 0.06 . 0.06 v = − e− t 0.06 0.06 4 0.06 0.06 0 711,240.46 50,000 1 1 0.06 0.06 711,240.46 50,000 1 1 0.06 0.0036 $1,052,649.52 t t t t te e dt te e − − − − = + ⎡− + ⎤ ⎢⎣ ⎥⎦ = + ⎡− − ⎤ ⎢⎣ ⎥⎦ ≈ ∫ 408 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 18. (a) ( ) 7 0 2 7 0 Actual income 400,000 175,000 400,000 87,500 $7,087,500 t dt t t = + = ⎡⎣ + ⎤⎦ = ∫ (b) ( ) 7 0.04 0 7 0.04 7 0.04 0 0 0.04 7 7 0.04 0 0 7 0.04 0 Present value 400,000 175,000 400,000 175,000 10,000,000 175,000 2,442,162.59 175,000 t t t t t t t e dt e dt te dt e te dt te dt − − − − − − = + = + = ⎡⎣− ⎤⎦ + = + ∫ ∫ ∫ ∫ ∫ Let u = t and dv = e−0.04t dt. Then du = dt and v = −25e−0.04t . 0.04 7 0.04 0 0.04 0.04 7 0 2,442,162.59 175,000 25 25 2,442,162.59 175,000 25 625 $6,007,438.79 t t t t te e dt te e − − − − = + ⎡− + ⎤ ⎢⎣ ⎥⎦ = + ⎡− − ⎤ ⎢⎣ ⎥⎦ = ∫ 19. Formula 6: u = x, du = dx, a = 2, b = 3 ( ) ( ) ( ) ( ) 2 3 2 2 1 3 4 3 4 ln 2 3 2 3 3 2 1 112 9 4 ln 2 3 27 2 19 12 8 ln2 3 54 x dx x x x C x x x x C x x x C = ⎡− − + + ⎤ + + ⎢⎣ ⎥⎦ = ⎡− − + + ⎤ + ⎢⎣ ⎥⎦ = − + + + ∫ 20. Formula 40: u = x, du = dx, n = 6 ( 6 ) 6 1 1 ln 1 1 6 x x dx x e C e = − + + + ∫ 21. Formula 23: u = x, du = dx, a = 4 2 1( 2 2 ) 2 ∫ x − 16 dx = x x − 16 − 16 ln x + x − 16 + C 22. Formula 43: u = x, du = dx, n = 5 ( ) ( ) ( ) 5 1 5 2 6 ln 1 5 1 ln 5 1 1 6 ln 36 x x dx x x C x x C + = ⎡⎣− + + ⎤⎦ + + = − + + ∫ 23. Formula 4: u = x, du = dx, a = 2, b = 3 ( )2 1 2 ln 2 3 2 3 9 2 3 x dx x C x x = ⎛ + + ⎞ + ⎜ + ⎟ + ⎝ ⎠ ∫ 24. Formula 19: u = x, du = dx, a = 2, b = 3 2(4 3 ) 2 3 2 3 27 x x dx x C x − = − + + + ∫ 25. Formula 23: u = x, du = dx, a = 5 2 2 x 25 dx x2 25 5 ln 5 x 25 C x x + + + ∫ = + − + Review Exercises for Chapter 6 409 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 26. Formula 10: u = x, du = dx, a = 4, b = 3 ( ) 1 1ln 4 3 4 4 3 dx x C x x x = + + + ∫ 27. Formula 29: u = x, du = dx, a = 2 2 1 1 ln 2 4 4 2 dx x C x x − = + − + ∫ 28. Formula 42: u = 3x, du = 3 dx ( ) ( ) ( ) 2 2 2 ln 3 1 ln 3 3 3 2 2ln3 ln3 x dx x dx x x x C = = ⎡ − + ⎤ + ⎣ ⎦ ∫ ∫ 29. Formula 19: u = x, du = dx, a = 1, b = 1 ( ( ) ) ( ) ( ) 2 2 2 1 1 1 3 1 2 2 1 3 x x dx x C x x x C − = − + + + − = − + + ∫ 30. Formula 34: u = x, du = dx, a = 4 2 2 2 1 16 16 16 dx x C x x x − = − + − ∫ 31. Formula 17: u = x, du = dx, a = b = 1 1 2 1 1 1 x dx x dx x x x + = + + + ∫ ∫ Formula 15: u = x, du = dx, a = b = 1 1 2 1 ln 1 1 1 1 x dx x x C x x + + − = + + + + + ∫ 32. Formula 30: u = x, du = dx, a = 3, n = 2 ( )2 2 2 2 1 1 1 9 18 9 9 dx x x x x = − ⎡ + ⎤ ⎢⎣ − − ⎥⎦ − ∫ ∫ Formula 29: u = x, du = dx, a = 3 ( )2 2 2 1 1 1 ln 3 9 18 9 6 3 dx x x C x x x ⎡ − ⎤ = − ⎢ + ⎥ + − ⎣ − + ⎦ ∫ 33. Formula 19: u = x, du = dx, a = 9, b = 16 ( ) ( ) ( ) 96 11 9 16 96 2 18 16 9 16 11 768 1 9 8 9 16 22 b a b a b a P a x b x dx x x x x x ≤ ≤ = ⎛ ⎞ ⎜ − ⎟ ⎝ ⎠ ⎡ − ⎤ = ⎢− + ⎥ ⎣ ⎦ = − ⎡ − + ⎤ ⎣ ⎦ ∫ (a) ( ) ( ) ( ) 0.8 0 1 22 1 22 0 0.8 9 8 9 16 2.6 21.8 27 0.675 P ≤ x ≤ = − ⎡⎣ − x + x⎤⎦ = − − ≈ (b) ( ) ( ) ( ) 0.5 0 1 22 1 22 0 0.5 9 8 9 16 5 17 27 0.290 P ≤ x ≤ = − ⎡⎣ − x + x⎤⎦ = − − ≈ 34. Formula 35: u = x1.5, du = 1.5x0.5 dx ( ) ( ) ( ) 2 1.5 1.5 1.5 0.5 1.5 1.5 1.5 1.5 1 b x a b x a b x a P a x b x e dx xe x dx x e ≤ ≤ = = = ⎡ − ⎤ ⎢⎣ ⎥⎦ ∫ ∫ (a) ( ) ( ) 1.5 0.6 1.5 0.4 P 0.4 ≤ x ≤ 0.6 = ⎡⎢⎣ x − 1 ex ⎤⎥⎦ ≈ 0.110 (b) ( ) ( ) 1.5 0.5 1.5 0 P 0 ≤ x ≤ 0.5 = ⎡⎢⎣ x − 1 ex ⎤⎥⎦ ≈ 0.079 35. Trapezoidal Rule: ( ) ( ) ( ) ( ) ( ) 3 1 2 2 2 2 2 2 1 1 1 2 1 2 1 2 1 1 0.705 41 32 2 52 3 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ = ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ Simpson’s Rule: ( ) ( ) ( ) ( ) ( ) 3 1 2 2 2 2 2 2 1 1 1 4 1 2 1 4 1 1 0.6715 61 32 2 52 3 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ ≈ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ Exact value: 3 3 1 2 1 1 1 2 3 dx x x = ⎡− ⎤ = ⎢⎣ ⎥⎦ ∫ 410 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 36. Trapezoidal Rule: ( ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) 2 2 2 2 2 2 2 0 2 2 2 2 1 1 1 3 8 4 2 4 5 3 7 4 2 4 1 0 1 2 1 2 1 2 1 21 1 2 1 2 1 2 1 2 1 4.6875 x dx ⎡ + ≈ + + + + + + + + + ⎢⎣ ⎤ + + + + + + + +⎥⎦ = ∫ Simpson’s Rule: ( ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) (( ) ) 2 2 2 2 2 2 2 0 2 2 2 2 1 1 1 3 12 4 2 4 5 3 7 4 2 4 1 0 1 4 1 2 1 4 1 21 1 4 1 2 1 4 1 2 1 14 4.6 3 x dx ⎡ + ≈ + + + + + + + + + ⎢⎣ ⎤ + + + + + + + +⎥⎦ = = ∫ Exact value: ( ) 2 2 3 2 0 0 1 8 14 3 3 3 ∫ x + 1 dx = ⎡⎣ x + x⎤⎦ = + 2 = 37. Trapezoidal Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 3 3 3 3 3 3 3 3 3 3 1 1 1 2 1 2 1 2 1 2 1 2 1 161 98 54 118 32 138 2 1 2 1 1 0.3786 7 4 15 8 2 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎥ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ≈ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ Simpson’s Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 3 3 3 3 3 3 3 3 3 3 1 1 1 4 1 2 1 4 1 2 1 4 1 241 98 54 118 32 74 2 1 4 1 1 0.3751 15 8 15 8 2 dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎢ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎤ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎥ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ≈ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ Exact value: 2 2 1 3 2 1 1 1 3 0.375 2 8 dx x x = ⎡− ⎤ = = ⎢⎣ ⎥⎦ ∫ 38. Trapezoidal Rule: ( ) ( ) ( ) ( ) ( ) 2 3 3 3 3 3 3 1 5 3 7 243 18 4 2 4 64 x dx ≈ ⎡⎢ 1 + 2 + 2 + 2 + 2 ⎤⎥ = ≈ 3.7969 ⎣ ⎦ ∫ Simpson’s Rule: ( ) ( ) ( ) ( ) ( ) 2 3 3 3 3 3 3 1 1 5 3 7 15 12 4 2 4 4 x dx ≈ ⎡⎢ 1 + 4 + 2 + 4 + 2 ⎤⎥ = ≈ 3.75 ⎣ ⎦ ∫ Exact value: 2 3 4 2 1 1 1 15 4 4 ∫ x dx = ⎡⎣ x ⎤⎦ = = 3.75 39. Trapezoidal Rule: 4 2 0 1 2 2 2 3 2 4 2 0 1 2 ∫ e−x dx ≈ ⎡⎣e + 2e− + 2e− + 2e− + e− ⎤⎦ ≈ 1.7652 Simpson’s Rule: 4 2 0 1 2 2 2 3 2 4 2 0 1 3 ∫ e−x dx ≈ ⎡⎣e + 4e− + 2e− + 4e− + e− ⎤⎦ ≈ 1.7299 Exact value: 4 2 2 4 2 0 0 ∫ e−x dx = ⎡⎣−2e−x ⎤⎦ = 2 − 2e− ≈ 1.7293 40. Trapezoidal Rule: 8 0 1 2 3 0 3 2 1 3 2 2 3 2 3 3 2 4 3 2 5 3 2 6 3 2 7 3 8 3 20.8464 x + dx ≈ ⎡⎣ + + + + + + + + + + + + + + + + + ⎤ ≈ ⎦ ∫ Simpson’s Rule: 8 0 1 3 3 0 3 4 1 3 2 2 3 4 3 3 2 4 3 4 5 3 2 6 3 4 7 3 8 3 20.8577 x + dx ≈ ⎡⎣ + + + + + + + + + + + + + + + + + ⎤ ≈ ⎦ ∫ Exact value: ( ) ( ) 8 3 2 8 0 0 2 2 3 3 ∫ x + 3 dx = ⎡⎣ x + 3 ⎤⎦ = 11 11 − 3 3 ≈ 20.8578 Review Exercises for Chapter 6 411 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 41. (a) Trapezoidal Rule: ( ) ( ) ( ) ( ) ( ) 2 1 1 1 1 2 1 2 1 2 1 1 0.741 1 ln 8 1 ln 1 1 ln 5 4 1 ln 3 2 1 ln 7 4 1 ln 2 dx x ⎡⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ ≈ ⎢⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟⎥ ≈ + ⎣⎢⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠⎦⎥ ∫ (b) Simpson’s Rule: ( ) ( ) ( ) ( ) ( ) 2 1 1 1 1 4 1 2 1 4 1 1 0.737 1 ln 12 1 ln1 1 ln54 1 ln32 1 ln74 1 ln2 dx x ⎡⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ ≈ ⎢⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟⎥ ≈ + ⎢⎣⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠⎥⎦ ∫ 42. (a) Trapezoidal Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 0 3 3 3 3 3 3 3 3 3 3 1 1 1 2 1 2 1 2 1 2 1 1 8 1 0 1 1 4 1 1 2 1 3 4 1 1 2 1 2 1 2 1 1 1.401 1 54 1 32 1 74 1 2 dx x ⎡⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎢⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟ ⎢⎣⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ ≈ ⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟ ⎜⎜ + ⎟⎟⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎥⎦ ∫ (b) Simpson’s Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 0 3 3 3 3 3 3 3 3 3 3 1 1 1 4 1 2 1 4 1 2 1 1 12 1 0 1 1 4 1 1 2 1 3 4 1 1 4 1 2 1 4 1 1 1.402 1 54 1 32 1 74 1 2 dx x ⎡⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎢⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎢⎣⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ ≈ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟⎥ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠ ⎝ + ⎠⎥⎦ ∫ 43. (a) Trapezoidal Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 3 2 3 2 3 2 3 2 1 0 2 2 2 2 2 1 1 4 1 2 3 4 1 0 2 2 2 0.305 2 8 2 14 2 12 2 34 2 1 x dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ ≈ − ⎢ ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟⎥ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ (b) Simpson’s Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 2 3 2 3 2 3 2 3 2 1 0 2 2 2 2 2 1 1 4 1 2 3 4 1 0 4 2 4 0.289 2 12 2 14 2 12 2 34 2 1 x dx x ⎡ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ ≈ ⎢ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ ≈ − ⎢ ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟ ⎜ − ⎟⎥ ⎣ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ 44. (a) Trapezoidal Rule: 1 2 (0)2 (1 6)2 (1 3)2 (1 2)2 (2 3)2 (5 6)2 (1)2 0 1 12 ex dx ≈ ⎡⎢e + 2e + 2e + 2e + 2e + 2e + e ⎤⎥ ≈ 1.475 ⎣ ⎦ ∫ (b) Simpson’s Rule: 1 2 (0)2 (1 6)2 (1 3)2 (1 2)2 (2 3)2 (5 6)2 (1)2 0 1 18 ex dx ≈ ⎡⎢e + 4e + 2e + 4e + 2e + 4e + e ⎤⎥ ≈ 1.463 ⎣ ⎦ ∫ 45. (a) Trapezoidal Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 8 0 2 2 2 2 2 2 2 2 2 2 3 1 3 2 3 2 3 2 3 2 3 2 2 0 2 1 2 2 2 3 2 4 2 2 3 2 3 2 3 3 2.961 5 2 6 2 7 2 8 2 dx x ⎡⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎢⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟ ⎣⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ ≈ ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ (b) Simpson’s Rule: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 8 0 2 2 2 2 2 2 2 2 2 2 3 1 3 4 3 2 3 4 3 2 3 2 3 0 2 1 2 2 2 3 2 4 2 4 3 2 3 4 3 3 2.936 5 2 6 2 7 2 8 2 dx x ⎡⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ≈ ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎢⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟ ⎣⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎤ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟⎥ ≈ ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟ ⎜ + ⎟⎥ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎦ ∫ 46. (a) Trapezoidal Rule: 1 0 1 1 1 3 8 4 2 4 ∫ 1 − x dx ≈ ⎡⎢⎣ 1 − 0 + 2 1 − + 2 1 − + 2 1 − + 1 − 1⎤⎥⎦ ≈ 0.643 (b) Simpson’s Rule: 1 0 1 1 1 3 12 4 2 4 ∫ 1 − x dx ≈ ⎡⎢⎣ 1 − 0 + 4 1 − + 2 1 − + 4 1 − + 1 − 1⎤⎥⎦ ≈ 0.657 412 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 47. ( ) ( ) ( ) ( ) ( )( ) 2 2 2 2 4 2 2 4 8 16 x x x x x f x e f x e f x e f x e f x e = ′ = ′′ = ′′′ = = (a) Trapezoidal Rule: Because f ′′(x) is maximum in [0, 2]when x = 2 and f ′′(x) = 4e4 , you have ( ) ( ) 3 4 4 2 2 0 Error 4 9.0997. 12 4 6 e e − ≤ = ≈ (b) Simpson’s Rule: Because f (4) (x) is maximum in [0, 2]when x = 2 and f (4) (x) = 16e4 , you have ( ) ( ) 5 4 4 4 2 0 Error 16 0.6066. 180 4 90 e e − ≤ = ≈ 48. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 2 3 4 4 5 1 1 1 1 2 1 6 1 24 1 f x x f x x f x x f x x f x x = − ′ = − − ′′ = − ′′′ = − − = − (a) Trapezoidal Rule: Because f ′′(x) is maximum in [2, 4]when x = 2 and f ′′(x) = 2, you have ( ) ( ) ( ) 3 2 4 2 1 Error 2 0.0833. 12 4 12 − ≤ = ≈ (b) Simpson’s Rule: Because f (4) (x) is maximum in [2, 4]when x = 2 and f (4) (x) = 24, you have ( ) ( ) ( ) 5 4 4 2 1 Error 24 0.0167. 180 4 60 − ≤ = ≈ 49. ( ) ( ) ( ) ( ) ( )( ) 2 4 2 2 0 0 f x x f x x f x f x f x = ′ = ′′ = ′′′ = = (a) Trapezoidal Rule: Because f ′′(x) = 2 for all x, you have ( ) ( ) 3 2 2 2 3 0 Error 2 0.0001 12 9 0.0001 2 45,000 212.13. n n n n − ≤ ≤ ≤ ≤ > Let n = 214. (b) Simpson’s Rule: Because f (4) (x) = 0 for all x, n can be any even value and will result in zero error. Let n = 2. 50. ( ) ( ) ( ) ( ) ( )( ) 5 5 5 5 4 5 1 5 1 25 1 125 1 625 x x x x x f x e f x e f x e f x e f x e = ′ = ′′ = ′′′ = = (a) Trapezoidal Rule: Because f ′′(x) is maximum in [0, 5] when x = 5 and ( ) 1 , 25 f ′′ x = e you have ( )3 2 2 2 5 0 1 Error 0.0001 12 25 5 0.0001 12 12,500 3 106.42. e n e n e n n − ⎛ ⎞ ≤ ⎜ ⎟ ≤ ⎝ ⎠ ≤ ≤ > Let n = 108. (b) Simpson’s Rule: Because f (4) (x) is maximum in [0, 5] when x = 5 and (4) ( ) 1 , 625 f x = e you have ( )5 4 4 4 5 0 1 Error 0.0001 180 625 0.0001 36 2500 9 5.24. e n e n e n n − ⎛ ⎞ ≤ ⎜ ⎟ ≤ ⎝ ⎠ ≤ ≤ > Let n = 6. Review Exercises for Chapter 6 413 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 51. This integral converges because 1 1 5 4 1 lim 1 1 0 1. a 4 a 4 4 dx x x − − −∞ →−∞ = ⎡− ⎤ = − + = − ⎢⎣ ⎥⎦ ∫ 52. This integral diverges because 1 1 1 lim 2 2 . b b dx x x ∞ → ∞ = ⎡ ⎤ = ∞ − = ∞ ∫ ⎣ ⎦ 53. This integral diverges because ( ) 2 3 0 0 3 1 3 8 lim 6 . 8 a 2 a x dx −∞ x →−∞ ⎡ − ⎤ = ⎢− ⎥ = − − ∞ = −∞ − ⎢⎣ ⎥⎦ ∫ 54. This integral converges because 2 2 0 0 lim 1 0 1 1. 2 2 2 b x x b e dx e ∞ − − →∞ = ⎡− ⎤ = − ⎛− ⎞ = ⎢ ⎥ ⎜ ⎟ ⎣ ⎦ ⎝ ⎠ ∫ 55. This integral diverges because ( )2 1 1 ln ln lim 0 . 2 b b x x dx x ∞ →∞ ⎡ ⎤ = ⎢ ⎥ = ∞ − = ∞ ⎢⎣ ⎥⎦ ∫ 56. This integral diverges because ( ) 0 0 lim ln 1 ln 2 . 1 x b x x b e dx e e ∞ →∞ = ⎡ + ⎤ = ∞ − = ∞ + ⎣ ⎦ ∫ 57. 4 4 ( ) 0 0 lim 4 0 4 4 x x b b A e dx e ∞ − − →∞ = ∫ = ⎡⎣− ⎤⎦ = − − = 58. ( 2 ) 0 2 0 2 lim ln 2 ln 2 2 b b A x dx x x ∞ →∞ = = ⎡ + ⎤ = ∞ − = ∞ + ⎣ ⎦ ∫ 59. ( ) 2 2 2 2 2 2 0 0 0 4 lim 4 lim 1 b b x x x b b A xe dx e xdx e ∞ − − → ∞ →∞ = = − − = ⎡− ⎤ = ∫ ∫ ⎢⎣ ⎥⎦ 60. ( ) ( ) ( ) ( )0 0 0 2 3 1 3 2 3 3 lim 1 3 3 lim 3 1 3 3 1 3 b b b b A dx x dx x x − − ∞ → −∞ → −∞ = = − − − = ⎡− − ⎤ = − + ∞ = ∞ − ⎣ ⎦ ∫ ∫ 61. Present value ( ) 0.03 0 0.03 0 8000e lim 8000 0.03 0 266,666.67 $266,666.67 t b t b dt e ∞ − − →∞ = = ⎡− ⎤ ⎢⎣ ⎥⎦ = − − = ∫ 62. Present value ( ) 0.05 0 0.05 0 15,000 lim 300,000 0 300,000 $300,000 t t b b e dt e ∞ − − →∞ = = ⎡⎣− ⎤⎦ = − − = ∫ 63. No, you do not have enough money to start the scholarship fund because Present value ( ) 0.07 0 0.07 0 21,000 lim 300,000 0 300,000 $300,000. t t b b e dt e ∞ − − →∞ = = ⎡⎣− ⎤⎦ = − − = ∫ 64. (a) Present value 15 0.06 0 15 0.06 0 100,000 100,000 0.06 $989,050.57 t t e dt e − − = = ⎡− ⎤ ⎢⎣ ⎥⎦ ≈ ∫ (b) Present value 0.06 0 0.06 0 100,000 lim 100,000 0.06 0 100,000 0.06 $1,666,666.67 t b t b e dt e ∞ − − →∞ = = ⎡− ⎤ ⎢⎣ ⎥⎦ = − ⎛− ⎞ ⎜ ⎟ ⎝ ⎠ ≈ ∫ 414 Chapter 6 Techniques of Integration © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 6 Test Yourself 1. Let u = x and dv = ex+1 dx. Then du = dx and v = ex+1. ( ) 1 1 1 1 1 1 1 x x x x x x xe dx xe e dx xe e C x e C + + + + + + = − = − + = − + ∫ ∫ 2. Let u = ln x and dv = 9x2 dx. Then du 1 dx x = and v = 3x3. ( ) 2 3 2 3 3 3 9 ln 3 ln 3 3 ln 3 ln 1 x x dx x x x dx x x x C x x C = − = − + = − + ∫ ∫ 3. Let u = x2 and dv = e−x 3 dx. Then du = 2x dx and v = −3e−x 3. ∫ x2e−x 3 dx = −3x2e−x 3 + ∫ 6xe−x 3 dx Let u = 6x and dv = e−x 3 dx. Then du = 6 dx and v = −3e−x 3. ( ) 2 3 2 3 3 3 2 3 3 3 3 2 3 18 18 3 18 54 3 6 18 x x x x x x x x xe dx xe xe e dx xe xe e C e x x C − − − − − − − − = − − + =− − − + = − + + + ∫ ∫ 4. R = 295.1 + 147.66 t ln t, 1 ≤ t ≤ 9 and t = 1 → 2001 (a) Total revenue ( ) 9 9 9 1 2 1 1 1 = ∫ 295.1 + 147.66 t ln t dt = ∫ 295.1 dt + 147.66∫ t ln t dt Let u = ln t and dv = t1 2 dt. Then du 1 dt t = and 2 2 3. 3 v = t [ ]9 3 2 1 2 1 9 3 2 3 2 1 295.1 147.66 2 ln 2 3 3 2360.8 147.66 2 ln 4 3 9 $6494.47 million t t t t dt t t t = + ⎧ − ⎫ ⎨ ⎬ ⎩ ⎭ = + ⎡ − ⎤ ⎢⎣ ⎥⎦ ≈ ∫ (b) Average revenue ( ) ( ) 9 1 1 1 6494.47 $811.81 million 9 1 8 = R t dt = ≈ − ∫ 5. Formula 4: u = x, du = dx, a = 7, b = 2 ( )2 1 7 ln 7 2 7 2 4 7 2 x dx x C x x = ⎛ + + ⎞ + ⎜ + ⎟ + ⎝ ⎠ ∫ 6. Formula 37: u = x3, du = 3x2 dx ( ) 2 3 3 3 3 ln 1 1 x x x dx x e C e = − + + + ∫ 7. Formula 19: u = x2 , du = 2x dx, a = 1, b = 5 ( ) ( ) 3 2 2 2 2 2 2 2 1 5 1 5 2 2 5 1 5 75 x dx x x dx x x x x C = + + − = − + + ∫ ∫ 8. Formula 39: u = 3 − 2x, du = −2 dx ( ) 1 0 ∫ ln 3 − 2x dx ( ){ ( )} ( ) ( ) 1 0 1 2 1 2 1 2 3 2 1 ln3 2 1 ln1 3 1 ln3 2 3ln3 0.6479 = ⎡⎣− − x − + − x ⎤⎦ = − ⎡⎣− + − − + ⎤⎦ = − − ≈ 9. Let u = x and ( ) 1 2 dv x 2 dx. − = − Then du = dx and ( )1 2 v = 2 x − 2 . ( ) ( ) ( ) ( ) ( ) ( ( )) ( ) 6 1 2 1 2 1 2 3 1 2 3 2 6 3 43 4 4 3 3 26 3 2 2 2 2 2 2 2 2 24 8 6 xx dx xx x dx x x x − − = − − − = ⎡ − − − ⎤ ⎣ ⎦ = − − − = ∫ ∫ Chapter 6 Test Yourself 415 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 10. Formula 23: u = x, du = dx, a = 4 ( ) ( ) 1 1 2 2 2 3 3 16 16 4 ln 4 16 17 4 ln 4 17 5 4 ln 3 4.8613 x dx x x x x − − − − + ⎡ + + ⎤ = ⎢ + − ⎥ ⎢⎣ ⎥⎦ = − − − − − − ≈ − ∫ 11. Trapezoidal Rule: ( ) (( ) ( )) ( ) ( ) ( ) ( ) ( ) ( ) (( ) ( )) 5 2 2 2 2 2 2 2 3 11 11 7 7 8 4 4 2 2 17 17 4 4 585 32 2 2 2 2 2 2 2 2 2 2 5 25 18.28 x − x dx ≈ ⎡⎢ − + ⎛⎜ − ⎞⎟ + ⎛⎜ − ⎞⎟ ⎣ ⎝ ⎠ ⎝ ⎠ + ⎛⎜ − ⎞⎟ + − ⎝ ⎠ = ≈ ∫ Exact value: ( ) 5 2 3 2 5 2 2 1 3 ∫ x − 2x dx = ⎡⎣ x − x ⎤⎦ = 18 12. Exact: Let u = 9x and dv = e3x dx. Then du = 9 dx and 1 3 3 v = e x. ( ) 1 3 3 1 3 0 0 3 3 1 0 3 3 3 9 3 3 3 3 1 2 1 41.1711 x x x x xe dx xe e dx e e e e e = ⎤⎦ − = − ⎡⎣ ⎤⎦ = − − = + ≈ ∫ ∫ Trapezoidal Rule: ( ) ( ) ( ) 1 3 34 3 2 9 4 3 0 1 9 9 27 12 4 2 4 9 0 4 2 4 9 41.3606 xe x dx = ⎡ + e + e + e + e ⎤ ⎣ ⎦ ≈ ∫ 13. This integral converges because 3 3 0 0 1 1 1 3 3 3 lim 0 . x x b b e dx e ∞ − − →∞ ∫ = ⎡⎣− ⎤⎦ = + = 14. This integral converges because 9 9 0 0 2 lim 4 12. a b dx x x → + = ⎡ ⎤ = ∫ ⎣ ⎦ 15. This integral diverges because ( ) ( ) 0 1 3 0 2 3 1 3 lim 4 1 3 . 4 1 4 a a 4 dx x −∞ x →−∞ = ⎡ − ⎤ = − + ∞ = ∞ − ⎣ ⎦ ∫ 16. (a) Present value 0.04 0.04 0 0 19.95 lim 498.75 $498.75 t t b b e dt e ∞ − − →∞ = ∫ = ⎡⎣− ⎤⎦ = (b) The subscriber should choose plan B because the lifetime subscription for $149 is less than $498.75, the present value of plan A. [Show More]

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