Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Integration By Parts Tutorial & Exa... by empoweringminds 44822 views
- INTEGRATION BY PARTS PPT by 03062679929 357 views
- 8.2 integration by parts by dicosmo178 660 views
- Lesson 30: Integration by Parts by Matthew Leingang 5202 views
- 17 formulas from integration by parts by math266 930 views
- Integration by michaelapetro 2694 views

No Downloads

Total views

1,286

On SlideShare

0

From Embeds

0

Number of Embeds

4

Shares

0

Downloads

57

Comments

0

Likes

1

No embeds

No notes for slide

- 1. Integration by Parts
- 2. Integration by Parts There are two techniques for integrations.
- 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. 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. 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. 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. 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. Integration by Parts Integration by parts: Let u = u(x), v = v(x), then ∫ uv'dx = uv – ∫ vu'dx
- 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. 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. 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. 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. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx
- 14. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ xex dx Match the integrand xexdx to udv, specifically
- 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Integration by Parts Steps for Integration by parts:
- 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. 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. 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. 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. 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. 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. Integration by Parts Example: Find ∫udv = uv – ∫vduLn(x) dx∫
- 35. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx Match the integrand Ln(x)dx to udv,
- 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. 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. 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. 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. 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. 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. 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. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
- 44. Integration by Parts Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx Match the integrand x3Ln(x)dx to udv,
- 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. 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. 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. 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. 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. 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. 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. 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. 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. Integration by Parts ∫udv = uv – ∫vdu Sometime we have to use the integration by parts more than once.
- 55. Integration by Parts Example: Find ∫udv = uv – ∫vdu ∫ x2ex dx Sometime we have to use the integration by parts more than once.
- 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. 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. 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. 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. 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. 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. 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. Integration by Parts ∫ 2xexdxIts easier to do separately.
- 64. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx
- 65. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx so du = dx
- 66. Integration by Parts ∫ 2xexdxIts easier to do separately. set u = x dv = exdx so du = dx v = ex
- 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. 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. 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. 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. 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. Integration by Parts ∫udv = uv – ∫vdu Another example where we have to use the integration by parts twice.
- 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. 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. 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. 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. 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. 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. 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. Integration by Parts let u = cos(x) dv = exdx For ∫ cos(x)exdx
- 81. Integration by Parts let u = cos(x) dv = exdx du dx = -sin(x) du = -sin(x)dx For ∫ cos(x)exdx
- 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. 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. 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. 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. 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. 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

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment