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16 integration by parts

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16 integration by parts

1. 1. Integration by Parts
2. 2. Integration by Parts There are two techniques for integrations.
3. 3. Integration by Parts Integration by substitution finds antiderivatives of derivatives that are obtained by using chain rule. There are two techniques for integrations.
4. 4. Integration by Parts Integration by substitution finds antiderivatives of derivatives that are obtained by using chain rule. Integration by parts finds antiderivatives of derivatives that are obtained by the product rule, There are two techniques for integrations.
5. 5. Integration by Parts Integration by substitution finds antiderivatives of derivatives that are obtained by using chain rule. Integration by parts finds antiderivatives of derivatives that are obtained by the product rule, i.e. There are two techniques for integrations. ∫uv' + u'v dx = uv + c
6. 6. Integration by Parts Integration by substitution finds antiderivatives of derivatives that are obtained by using chain rule. Integration by parts finds antiderivatives of derivatives that are obtained by the product rule, i.e. There are two techniques for integrations. ∫uv' + u'v dx = uv + c However, this version is not useful in practice because its difficult to match the integrand of an integration problem to uv' + u'v.
7. 7. Integration by Parts Integration by substitution finds antiderivatives of derivatives that are obtained by using chain rule. Integration by parts finds antiderivatives of derivatives that are obtained by the product rule, i.e. There are two techniques for integrations. However, this version is not useful in practice because its difficult to match the integrand of an integration problem to uv' + u'v. The following is the workable version. ∫uv' + u'v dx = uv + c
8. 8. Integration by Parts Integration by parts: Let u = u(x), v = v(x), then ∫ uv'dx = uv – ∫ vu'dx
9. 9. Integration by Parts Integration by parts: Let u = u(x), v = v(x), then ∫ uv'dx = uv – ∫ vu'dx An simpler way to describe it is to use the notation of differentials:
10. 10. Integration by Parts Integration by parts: Let u = u(x), v = v(x), then ∫ uv'dx = uv – ∫ vu'dx where dv = v'dx and du = u'dx. An simpler way to describe it is to use the notation of differentials: ∫ udv = uv – ∫ vdu
11. 11. Integration by Parts Integration by parts: Let u = u(x), v = v(x), then ∫ uv'dx = uv – ∫ vu'dx where dv = v'dx and du = u'dx. An simpler way to describe it is to use the notation of differentials: ∫ udv = uv – ∫ vdu When employed, we matched the integrand to the left hand side of the formula,
12. 12. Integration by Parts Integration by parts: Let u = u(x), v = v(x), then ∫ uv'dx = uv – ∫ vu'dx where dv = v'dx and du = u'dx. An simpler way to describe it is to use the notation of differentials: ∫ udv = uv – ∫ vdu When employed, we matched the integrand to the left hand side of the formula, then convert it to the right hand side and hope that the integral on the right hand side will be easier.
13. 13. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx
14. 14. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx Match the integrand xexdx to udv, specifically
15. 15. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx Match the integrand xexdx to udv, specifically let u = x and dv = exdx
16. 16. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv Match the integrand xexdx to udv, specifically
17. 17. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate Match the integrand xexdx to udv, specifically
18. 18. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate du dx = 1 Match the integrand xexdx to udv, specifically
19. 19. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate du dx = 1 du = 1dx Match the integrand xexdx to udv, specifically
20. 20. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate du dx = 1 du = 1dx integrate Match the integrand xexdx to udv, specifically
21. 21. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate du dx = 1 du = 1dx integrate v = ∫ ex dx Match the integrand xexdx to udv, specifically
22. 22. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate du dx = 1 du = 1dx integrate v = ∫ ex dx v = ex Match the integrand xexdx to udv, specifically
23. 23. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate du dx = 1 du = 1dx integrate v = ∫ ex dx v = ex Thus ∫ xexdx = ∫ u dv = uv – ∫ v du Match the integrand xexdx to udv, specifically
24. 24. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate du dx = 1 du = 1dx integrate v = ∫ ex dx v = ex Thus ∫ xexdx = xex ∫ u dv = uv – ∫ v du Match the integrand xexdx to udv, specifically
25. 25. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate du dx = 1 du = 1dx integrate v = ∫ ex dx v = ex Thus ∫ xexdx = xex – ∫ exdx ∫ u dv = uv – ∫ v du Match the integrand xexdx to udv, specifically
26. 26. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx let u = x and dv = exdx To convert, we need du = u'dx and v = ∫ dv differentiate du dx = 1 du = 1dx integrate v = ∫ ex dx v = ex Thus ∫ xexdx = xex – ∫ exdx = xex – ex + c ∫ u dv = uv – ∫ v du Match the integrand xexdx to udv, specifically
27. 27. Integration by Parts Steps for Integration by parts:
28. 28. Integration by Parts Steps for Integration by parts: 1. Match the integrand to the form "udv", select a factor of the integrand to be u and set the rest as dv. It's possible that dv = dx.
29. 29. Integration by Parts Steps for Integration by parts: 1. Match the integrand to the form "udv", select a factor of the integrand to be u and set the rest as dv. It's possible that dv = dx. 2. Take the derivative of u to find du. Take the antiderivative of dv to find v.
30. 30. Integration by Parts Steps for Integration by parts: 1. Match the integrand to the form "udv", select a factor of the integrand to be u and set the rest as dv. It's possible that dv = dx. 2. Take the derivative of u to find du. Take the antiderivative of dv to find v. ∫ udv as uv – ∫ vdu3. Rewrite the integral
31. 31. Integration by Parts Steps for Integration by parts: 1. Match the integrand to the form "udv", select a factor of the integrand to be u and set the rest as dv. It's possible that dv = dx. 2. Take the derivative of u to find du. Take the antiderivative of dv to find v. ∫ udv as uv – ∫ vdu3. Rewrite the integral 4. Try to find ∫ vdu
32. 32. Integration by Parts Steps for Integration by parts: 1. Match the integrand to the form "udv", select a factor of the integrand to be u and set the rest as dv. It's possible that dv = dx. 2. Take the derivative of u to find du. Take the antiderivative of dv to find v. ∫ udv as uv – ∫ vdu3. Rewrite the integral 4. Try to find ∫ vdu Remarks: * Try integration by parts if substitution doesn’t apply
33. 33. Integration by Parts Steps for Integration by parts: 1. Match the integrand to the form "udv", select a factor of the integrand to be u and set the rest as dv. It's possible that dv = dx. 2. Take the derivative of u to find du. Take the antiderivative of dv to find v. ∫ udv as uv – ∫ vdu3. Rewrite the integral 4. Try to find ∫ vdu Remarks: * Try integration by parts if substitution doesn’t apply * dv must be integrable so we may find v
34. 34. Integration by Parts Example: Find ∫udv = uv – ∫vduLn(x) dx∫
35. 35. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx Match the integrand Ln(x)dx to udv,
36. 36. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx Match the integrand Ln(x)dx to udv, let u = Ln(x) dv = dx
37. 37. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx Match the integrand Ln(x)dx to udv, let u = Ln(x) dv = dx differentiate du dx = 1 x
38. 38. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx Match the integrand Ln(x)dx to udv, let u = Ln(x) dv = dx differentiate du dx = du = 1 x x dx
39. 39. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx Match the integrand Ln(x)dx to udv, let u = Ln(x) dv = dx differentiate du dx = du = integrate v = ∫ dx1 x x dx
40. 40. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx Match the integrand Ln(x)dx to udv, let u = Ln(x) dv = dx differentiate du dx = du = integrate v = ∫ dx v = x 1 x x dx
41. 41. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx Match the integrand Ln(x)dx to udv, let u = Ln(x) dv = dx Hence ∫ Ln(x) dx = xLn(x) – ∫ x differentiate du dx = du = integrate v = ∫ dx v = x 1 x x dx x dx
42. 42. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx Match the integrand Ln(x)dx to udv, let u = Ln(x) dv = dx Hence ∫ Ln(x) dx = xLn(x) – ∫ x differentiate du dx = du = integrate v = ∫ dx v = x 1 x x dx x dx = xLn(x) – x + c
43. 43. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
44. 44. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx Match the integrand x3Ln(x)dx to udv,
45. 45. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx let u = Ln(x) dv = x3dx Match the integrand x3Ln(x)dx to udv,
46. 46. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx let u = Ln(x) dv = x3dx differentiate du dx = 1 x Match the integrand x3Ln(x)dx to udv,
47. 47. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx let u = Ln(x) dv = x3dx differentiate du dx = du = 1 x x dx Match the integrand x3Ln(x)dx to udv,
48. 48. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx let u = Ln(x) dv = x3dx differentiate du dx = du = integrate v = ∫ x3 dx1 x x dx Match the integrand x3Ln(x)dx to udv,
49. 49. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx let u = Ln(x) dv = x3dx differentiate du dx = du = integrate v = ∫ x3 dx v = 1 x x dx x4 4 Match the integrand x3Ln(x)dx to udv,
50. 50. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx let u = Ln(x) dv = x3dx Hence ∫ Ln(x) dx = Ln(x) differentiate du dx = du = integrate v = ∫ x3 dx v = 1 x x dx x4 4 x4 4 Match the integrand x3Ln(x)dx to udv,
51. 51. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx let u = Ln(x) dv = x3dx Hence ∫ Ln(x) dx = Ln(x) – ∫ differentiate du dx = du = integrate v = ∫ x3 dx v = 1 x x dx x dx x4 4 x4 4 x4 4 Match the integrand x3Ln(x)dx to udv,
52. 52. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx let u = Ln(x) dv = x3dx Hence ∫ Ln(x) dx = Ln(x) – ∫ differentiate du dx = du = integrate v = ∫ x3 dx v = 1 x x dx x dx x4 4 x4 4 x4 4 = Ln(x) – ∫ x3dx x4 4 1 4 Match the integrand x3Ln(x)dx to udv,
53. 53. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx let u = Ln(x) dv = x3dx Hence ∫ Ln(x) dx = Ln(x) – ∫ differentiate du dx = du = integrate v = ∫ x3 dx v = 1 x x dx x dx = Ln(x) – + c x4 4 x4 4 x4 4 = Ln(x) – ∫ x3dx x4 4 1 4 x4 4 x4 16 Match the integrand x3Ln(x)dx to udv,
54. 54. Integration by Parts ∫udv = uv – ∫vdu Sometime we have to use the integration by parts more than once.
55. 55. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ x2ex dx Sometime we have to use the integration by parts more than once.
56. 56. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ x2ex dx Match the integrand x2exdx to udv, let u = x2 dv = exdx Sometime we have to use the integration by parts more than once.
57. 57. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ x2ex dx Match the integrand x2exdx to udv, let u = x2 dv = exdx differentiate du dx = 2x Sometime we have to use the integration by parts more than once.
58. 58. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ x2ex dx Match the integrand x2exdx to udv, let u = x2 dv = exdx differentiate du dx = 2x du = 2xdx Sometime we have to use the integration by parts more than once.
59. 59. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ x2ex dx Match the integrand x2exdx to udv, let u = x2 dv = exdx differentiate du dx = 2x du = 2xdx integrate v = ∫ ex dx Sometime we have to use the integration by parts more than once.
60. 60. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ x2ex dx Match the integrand x2exdx to udv, let u = x2 dv = exdx differentiate du dx = 2x du = 2xdx integrate v = ∫ ex dx v = ex Sometime we have to use the integration by parts more than once.
61. 61. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ x2ex dx Match the integrand x2exdx to udv, let u = x2 dv = exdx Hence ∫ x2ex dx = x2ex – ∫ 2xexdx differentiate du dx = 2x du = 2xdx integrate v = ∫ ex dx v = ex Sometime we have to use the integration by parts more than once.
62. 62. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ x2ex dx Match the integrand x2exdx to udv, let u = x2 dv = exdx Hence ∫ x2ex dx = x2ex – ∫ 2xexdx differentiate du dx = 2x du = 2xdx integrate v = ∫ ex dx v = ex Sometime we have to use the integration by parts more than once. Use integration by parts again for ∫ 2xexdx
63. 63. Integration by Parts ∫ 2xexdxIts easier to do separately.
64. 64. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx
65. 65. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx so du = dx
66. 66. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx so du = dx v = ex
67. 67. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx so du = dx v = ex 2 ∫xexdx = 2(xex – ∫exdx)
68. 68. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx so du = dx v = ex 2 ∫xexdx = 2(xex – ∫exdx) = 2xex – 2ex
69. 69. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx so du = dx v = ex 2 ∫xexdx = 2(xex – ∫exdx) = 2xex – 2ex Therefore, ∫ x2ex dx = x2ex – ∫ 2xexdx
70. 70. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx so du = dx v = ex 2 ∫xexdx = 2(xex – ∫exdx) = 2xex – 2ex Therefore, ∫ x2ex dx = x2ex – ∫ 2xexdx = x2ex – (2xex – 2ex) + c
71. 71. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx so du = dx v = ex 2 ∫xexdx = 2(xex – ∫exdx) = 2xex – 2ex Therefore, ∫ x2ex dx = x2ex – ∫ 2xexdx = x2ex – (2xex – 2ex) + c = x2ex – 2xex + 2ex + c
72. 72. Integration by Parts ∫udv = uv – ∫vdu Another example where we have to use the integration by parts twice.
73. 73. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ sin(x)ex dx Another example where we have to use the integration by parts twice.
74. 74. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ sin(x)ex dx Match the integrand sin(x)exdx to udv, let u = sin(x) dv = exdx Another example where we have to use the integration by parts twice.
75. 75. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ sin(x)ex dx let u = sin(x) dv = exdx du dx = cos(x) Another example where we have to use the integration by parts twice. Match the integrand sin(x)exdx to udv,
76. 76. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ sin(x)ex dx let u = sin(x) dv = exdx du dx = cos(x) du = cos(x)dx Another example where we have to use the integration by parts twice. Match the integrand sin(x)exdx to udv,
77. 77. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ sin(x)ex dx let u = sin(x) dv = exdx du dx = cos(x) du = cos(x)dx v = ∫ ex dx v = ex Another example where we have to use the integration by parts twice. Match the integrand sin(x)exdx to udv,
78. 78. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ sin(x)ex dx let u = sin(x) dv = exdx Hence ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx du dx = cos(x) du = cos(x)dx v = ∫ ex dx v = ex Another example where we have to use the integration by parts twice. Match the integrand sin(x)exdx to udv,
79. 79. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ sin(x)ex dx let u = sin(x) dv = exdx Hence ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx du dx = cos(x) du = cos(x)dx v = ∫ ex dx v = ex Another example where we have to use the integration by parts twice. Use integration by parts again for ∫ cos(x)exdx Match the integrand sin(x)exdx to udv,
80. 80. Integration by Parts let u = cos(x) dv = exdx For ∫ cos(x)exdx
81. 81. Integration by Parts let u = cos(x) dv = exdx du dx = -sin(x) du = -sin(x)dx For ∫ cos(x)exdx
82. 82. Integration by Parts let u = cos(x) dv = exdx du dx = -sin(x) du = -sin(x)dx v = ∫ ex dx v = ex For ∫ cos(x)exdx
83. 83. Integration by Parts let u = cos(x) dv = exdx Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx du dx = -sin(x) du = -sin(x)dx v = ∫ ex dx v = ex For ∫ cos(x)exdx
84. 84. Integration by Parts let u = cos(x) dv = exdx Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx du dx = -sin(x) du = -sin(x)dx v = ∫ ex dx v = ex For ∫ cos(x)exdx Since ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx
85. 85. Integration by Parts let u = cos(x) dv = exdx Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx du dx = -sin(x) du = -sin(x)dx v = ∫ ex dx v = ex For ∫ cos(x)exdx Since ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx ∫ sin(x)ex dx = sin(x)ex – cos(x)ex – ∫ sin(x)exdx
86. 86. Integration by Parts let u = cos(x) dv = exdx Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx du dx = -sin(x) du = -sin(x)dx v = ∫ ex dx v = ex For ∫ cos(x)exdx Since ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx ∫ sin(x)ex dx = sin(x)ex – cos(x)ex – ∫ sin(x)exdx 2 ∫ sin(x)ex dx = sin(x)ex – cos(x)ex
87. 87. Integration by Parts let u = cos(x) dv = exdx Hence ∫ cos(x)ex dx = cos(x)ex + ∫ sin(x)exdx du dx = -sin(x) du = -sin(x)dx v = ∫ ex dx v = ex For ∫ cos(x)exdx Since ∫ sin(x)ex dx = sin(x)ex – ∫ cos(x)exdx ∫ sin(x)ex dx = sin(x)ex – cos(x)ex – ∫ sin(x)exdx 2 ∫ sin(x)ex dx = sin(x)ex – cos(x)ex ∫ sin(x)ex dx = ½ (sin(x)ex – cos(x)ex) + c