16 integration by parts
TRANSCRIPT
![Page 1: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/1.jpg)
Integration by Parts
![Page 2: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/2.jpg)
Integration by Parts
There are two techniques for integrations.
![Page 3: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/3.jpg)
Integration by Parts
Integration by substitution finds antiderivatives of derivatives that are obtained by using chain rule.
There are two techniques for integrations.
![Page 4: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/4.jpg)
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.
![Page 5: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/5.jpg)
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
![Page 6: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/6.jpg)
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.
![Page 7: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/7.jpg)
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
![Page 8: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/8.jpg)
Integration by Parts
Integration by parts:Let u = u(x), v = v(x), then
∫ uv'dx = uv – ∫ vu'dx
![Page 9: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/9.jpg)
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:
![Page 10: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/10.jpg)
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
![Page 11: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/11.jpg)
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,
![Page 12: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/12.jpg)
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.
![Page 13: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/13.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
![Page 14: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/14.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
Match the integrand xexdx to udv, specifically
![Page 15: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/15.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
Match the integrand xexdx to udv, specifically
let u = x and dv = exdx
![Page 16: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/16.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo convert, we need du = u'dx and v = ∫ dv
Match the integrand xexdx to udv, specifically
![Page 17: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/17.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo convert, we need du = u'dx and v = ∫ dv
differentiate
Match the integrand xexdx to udv, specifically
![Page 18: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/18.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo convert, we need du = u'dx and v = ∫ dv
differentiate
du dx = 1
Match the integrand xexdx to udv, specifically
![Page 19: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/19.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo convert, we need du = u'dx and v = ∫ dv
differentiate
du dx = 1
du = 1dx
Match the integrand xexdx to udv, specifically
![Page 20: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/20.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo convert, we need du = u'dx and v = ∫ dv
differentiate
du dx = 1
du = 1dx
integrate
Match the integrand xexdx to udv, specifically
![Page 21: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/21.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo 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
![Page 22: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/22.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo 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
![Page 23: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/23.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo 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
![Page 24: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/24.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo 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
![Page 25: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/25.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo 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
![Page 26: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/26.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ xex dx
let u = x and dv = exdxTo 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
![Page 27: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/27.jpg)
Integration by PartsSteps for Integration by parts:
![Page 28: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/28.jpg)
Integration by PartsSteps 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.
![Page 29: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/29.jpg)
Integration by PartsSteps 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.
![Page 30: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/30.jpg)
Integration by PartsSteps 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 – ∫ vdu 3. Rewrite the integral
![Page 31: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/31.jpg)
Integration by PartsSteps 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 – ∫ vdu 3. Rewrite the integral
4. Try to find ∫ vdu
![Page 32: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/32.jpg)
Integration by PartsSteps 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 – ∫ vdu 3. Rewrite the integral
4. Try to find ∫ vdu
Remarks:* Try integration by parts if substitution doesn’t apply
![Page 33: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/33.jpg)
Integration by PartsSteps 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 – ∫ vdu 3. 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
![Page 34: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/34.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu Ln(x) dx∫
![Page 35: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/35.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
![Page 36: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/36.jpg)
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
![Page 37: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/37.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dxdifferentiate
du dx =
1 x
![Page 38: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/38.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dxdifferentiate
du dx =
du =
1 x
x dx
![Page 39: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/39.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dxdifferentiate
du dx =
du =
integrate
v = ∫ dx 1 x
x dx
![Page 40: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/40.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ Ln(x) dx
Match the integrand Ln(x)dx to udv,
let u = Ln(x) dv = dxdifferentiate
du dx =
du =
integrate
v = ∫ dx
v = x
1 x
x dx
![Page 41: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/41.jpg)
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
![Page 42: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/42.jpg)
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
![Page 43: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/43.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ x3Ln(x) dx
![Page 44: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/44.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ x3Ln(x) dx
Match the integrand x3Ln(x)dx to udv,
![Page 45: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/45.jpg)
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,
![Page 46: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/46.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ x3Ln(x) dx
let u = Ln(x) dv = x3dxdifferentiate
du dx =
1 x
Match the integrand x3Ln(x)dx to udv,
![Page 47: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/47.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ x3Ln(x) dx
let u = Ln(x) dv = x3dxdifferentiate
du dx =
du =
1 x
x dx
Match the integrand x3Ln(x)dx to udv,
![Page 48: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/48.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ x3Ln(x) dx
let u = Ln(x) dv = x3dxdifferentiate
du dx =
du =
integrate
v = ∫ x3 dx
1 x
x dx
Match the integrand x3Ln(x)dx to udv,
![Page 49: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/49.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu ∫ x3Ln(x) dx
let u = Ln(x) dv = x3dxdifferentiate
du dx =
du =
integrate
v = ∫ x3 dx v =
1 x
x dx x4
4
Match the integrand x3Ln(x)dx to udv,
![Page 50: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/50.jpg)
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,
![Page 51: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/51.jpg)
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,
![Page 52: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/52.jpg)
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,
![Page 53: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/53.jpg)
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,
![Page 54: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/54.jpg)
Integration by Parts
∫udv = uv – ∫vdu
Sometime we have to use the integration by parts more than once.
![Page 55: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/55.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu
∫ x2ex dx
Sometime we have to use the integration by parts more than once.
![Page 56: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/56.jpg)
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.
![Page 57: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/57.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdxdifferentiate
du dx = 2x
Sometime we have to use the integration by parts more than once.
![Page 58: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/58.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdxdifferentiate
du dx = 2x
du = 2xdx
Sometime we have to use the integration by parts more than once.
![Page 59: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/59.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdxdifferentiate
du dx = 2x
du = 2xdx
integrate
v = ∫ ex dx
Sometime we have to use the integration by parts more than once.
![Page 60: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/60.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu
∫ x2ex dx
Match the integrand x2exdx to udv,
let u = x2 dv = exdxdifferentiate
du dx = 2x
du = 2xdx
integrate
v = ∫ ex dx v = ex
Sometime we have to use the integration by parts more than once.
![Page 61: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/61.jpg)
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.
![Page 62: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/62.jpg)
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
![Page 63: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/63.jpg)
Integration by Parts
∫ 2xexdx Its easier to do separately.
![Page 64: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/64.jpg)
Integration by Parts
∫ 2xexdx Its easier to do separately.
set u = x dv = exdx
![Page 65: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/65.jpg)
Integration by Parts
∫ 2xexdx Its easier to do separately.
set u = x dv = exdx
so du = dx
![Page 66: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/66.jpg)
Integration by Parts
∫ 2xexdx Its easier to do separately.
set u = x dv = exdx
so du = dx v = ex
![Page 67: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/67.jpg)
Integration by Parts
∫ 2xexdx Its easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx)
![Page 68: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/68.jpg)
Integration by Parts
∫ 2xexdx Its easier to do separately.
set u = x dv = exdx
so du = dx v = ex
2 ∫xexdx = 2(xex – ∫exdx) = 2xex – 2ex
![Page 69: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/69.jpg)
Integration by Parts
∫ 2xexdx Its 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
![Page 70: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/70.jpg)
Integration by Parts
∫ 2xexdx Its 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
![Page 71: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/71.jpg)
Integration by Parts
∫ 2xexdx Its 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
![Page 72: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/72.jpg)
Integration by Parts
∫udv = uv – ∫vdu
Another example where we have to use the integration by parts twice.
![Page 73: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/73.jpg)
Integration by Parts
Example: Find ∫udv = uv – ∫vdu
∫ sin(x)ex dx
Another example where we have to use the integration by parts twice.
![Page 74: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/74.jpg)
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.
![Page 75: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/75.jpg)
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,
![Page 76: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/76.jpg)
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,
![Page 77: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/77.jpg)
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,
![Page 78: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/78.jpg)
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,
![Page 79: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/79.jpg)
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,
![Page 80: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/80.jpg)
Integration by Parts
let u = cos(x) dv = exdx
For ∫ cos(x)exdx
![Page 81: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/81.jpg)
Integration by Parts
let u = cos(x) dv = exdx
du dx = -sin(x)
du = -sin(x)dx
For ∫ cos(x)exdx
![Page 82: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/82.jpg)
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
![Page 83: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/83.jpg)
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
![Page 84: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/84.jpg)
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
![Page 85: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/85.jpg)
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
![Page 86: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/86.jpg)
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
![Page 87: 16 integration by parts](https://reader033.vdocuments.site/reader033/viewer/2022052601/558e936a1a28ab57108b4616/html5/thumbnails/87.jpg)
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