Difference between revisions of "Math 22 Integration by Parts and Present Value"

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|and <math>dv=e^{3x}dx</math> and <math>v=\frac{1}{3}e^{3x}</math>
 
|and <math>dv=e^{3x}dx</math> and <math>v=\frac{1}{3}e^{3x}</math>
 
|-
 
|-
|Then, by integration by parts: <math>\int xe^{3x}dx=x\frac{1}{3}e^{3x} -\int\frac{1}{3}e^{3x} dx=x\frac{1}{3}e^{3x}-\frac{1}{9}e^{3x} </math>
+
|Then, by integration by parts: <math>\int xe^{3x}dx=x\frac{1}{3}e^{3x} -\int\frac{1}{3}e^{3x} dx=x\frac{1}{3}e^{3x}-\frac{1}{9}e^{3x} +C </math>
 
|}
 
|}
  
 +
'''3)''' <math>\int x^2e^{-x}dx</math>
 +
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Solution: &nbsp;
 +
|-
 +
|Let <math>u=x^2</math>, <math>du=2xdx</math>
 +
|-
 +
|and <math>dv=e^{-x}dx</math> and <math>v=-e^{-x}</math>
 +
|-
 +
|Then, by integration by parts: <math>\int x^2e^{-x}dx=x^2(-e^{-x}) -\int-e^{-x}2x dx=-x^2e^{-x}+\int 2xe^{-x}dx </math>
 +
|-
 +
|Now, we apply integration by parts the second time for <math>\int 2xe^{-x}dx</math>
 +
|-
 +
|Let <math>u=2x</math>, <math>du=2dx</math>
 +
|-
 +
|and <math>dv=e^{-x}dx</math> and <math>v=-e^{-x}</math>
 +
|-
 +
|So <math>\int 2xe^{-x}dx=2x(-e^{-x})-\int -e^{-x} 2dx=-2xe^{-x}-e^{-x}+C</math>
 +
|-
 +
|Therefore, <math>\int x^2e^{-x}dx=-x^2e^{-x}-2xe^{-x}-e^{-x}+C</math>
 +
|}
  
 +
'''4)''' <math>\int xe^{3x}dx</math>
 +
{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Solution: &nbsp;
 +
|-
 +
|Let <math>u=x</math>, <math>du=dx</math>
 +
|-
 +
|and <math>dv=e^{3x}dx</math> and <math>v=\frac{1}{3}e^{3x}</math>
 +
|-
 +
|Then, by integration by parts: <math>\int xe^{3x}dx=x\frac{1}{3}e^{3x} -\int\frac{1}{3}e^{3x} dx=x\frac{1}{3}e^{3x}-\frac{1}{9}e^{3x} +C </math>
 +
|}
  
  

Revision as of 07:10, 18 August 2020

Integration by Parts

 Let  and  be differentiable functions of .
 
 

Exercises Use integration by parts to evaluation:

1)

Solution:  
Let ,
and and
Then, by integration by parts:

2)

Solution:  
Let ,
and and
Then, by integration by parts:

3)

Solution:  
Let ,
and and
Then, by integration by parts:
Now, we apply integration by parts the second time for
Let ,
and and
So
Therefore,

4)

Solution:  
Let ,
and and
Then, by integration by parts:



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