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Integration by Parts
Integration by Parts
There are two techniques for integrations.
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rul...
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rul...
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rul...
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rul...
Integration by Parts
Integration by substitution finds antiderivatives
of derivatives that are obtained by using chain
rul...
Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
∫ uv'dx = uv – ∫ vu'dx
Integration by Parts
Integration by parts:
Let u = u(x), v = v(x), then
∫ uv'dx = uv – ∫ vu'dx
An simpler way to describe ...
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'...
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'...
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'...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
Match the integrand xexdx to udv, specifically
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ xex dx
Match the integrand xexdx to udv, specifically
let u = x and d...
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 = ...
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 = ...
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 = ...
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 = ...
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 = ...
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 = ...
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 = ...
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 = ...
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 = ...
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 = ...
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 = ...
Integration by Parts
Steps for Integration by parts:
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the inte...
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the inte...
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the inte...
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the inte...
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the inte...
Integration by Parts
Steps for Integration by parts:
1. Match the integrand to the form "udv",
select a factor of the inte...
Integration by Parts
Example: Find ∫udv = uv – ∫vduLn(x) dx∫
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
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
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
di...
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
di...
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
di...
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
di...
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
He...
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
He...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
Match the integrand x3Ln(x)dx to udv,
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Match the integrand x3Ln(x)dx to u...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
differentiate
du
dx = 1
x
Match th...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
differentiate
du
dx =
du =
1
x
x
d...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
differentiate
du
dx =
du =
integra...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
differentiate
du
dx =
du =
integra...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Hence ∫ Ln(x) dx = Ln(x)
different...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Hence ∫ Ln(x) dx = Ln(x) – ∫
diffe...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Hence ∫ Ln(x) dx = Ln(x) – ∫
diffe...
Integration by Parts
Example: Find ∫udv = uv – ∫vdu∫ x3Ln(x) dx
let u = Ln(x) dv = x3dx
Hence ∫ Ln(x) dx = Ln(x) – ∫
diffe...
Integration by Parts
∫udv = uv – ∫vdu
Sometime we have to use the integration by
parts more than once.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Sometime we have to use the integration by
parts more than o...
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
Some...
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
diff...
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
diff...
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
diff...
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
diff...
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
Henc...
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdx
Henc...
Integration by Parts
∫ 2xexdxIts easier to do separately.
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx)
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx) ...
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx) ...
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx) ...
Integration by Parts
∫ 2xexdxIts easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx) ...
Integration by Parts
∫udv = uv – ∫vdu
Another example where we have to use the
integration by parts twice.
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ sin(x)ex dx
Another example where we have to use the
integration by ...
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...
Integration by Parts
Example: Find
∫udv = uv – ∫vdu
∫ sin(x)ex dx
let u = sin(x) dv = exdx
du
dx = cos(x)
Another example ...
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
An...
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 ...
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 ...
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 ...
Integration by Parts
let u = cos(x) dv = exdx
For ∫ cos(x)exdx
Integration by Parts
let u = cos(x) dv = exdx
du
dx = -sin(x)
du = -sin(x)dx
For ∫ cos(x)exdx
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
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...
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...
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...
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...
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...
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16 integration by parts

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Transcript of "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
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