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Solving an integral

  • Thread starter fk378
  • Start date
  • #1
367
0
1. Homework Statement
Find the integral of x(e^-x) evaluated at 1 to inf


2. Homework Equations
int by parts


3. The Attempt at a Solution
I tried setting u=e^-x and dv=x dx
So then du=-(e^-x)(-1)=e^-x
and v= (x^2)/2
When plugging in uv - integral (v du)
I get another product for the integral which (x^2)/2 * (e^-x)

How do I get out of this?
 

Answers and Replies

  • #2
197
4
Try the other substitution. u=x and dv=e^-xdx.
A good rule of thumb for IBP is to try to choose u to be the part with the "simplest" derivative.
In your case, choosing u=e^-x has du/dx = -e^-x which is not really simpler, but has the same "complexity". But choosing u=x has du/dx=1, which is just a constant (can't get much simpler than that).
 
  • #3
367
0
So would the integral (the dv) of e^-x just be e^-x?
 
  • #4
197
4
almost. Whenever you do an integral (without the limits), you can check if it is right by taking the derivative of your answer. If you are right, then you should get back the function you were integrating.
 
  • #5
367
0
Okay, so I just took the derivative of e^-x and I got back e^-x. How come e^-x wouldn't be the antiderivative then?
 
  • #6
197
4
Don't forget to use the chain rule on the exponent (-x).
[tex]{de^{-x}\over dx}=-e^{-x}[/tex].
 
  • #7
197
4
In detail
[tex]{de^{-x}\over dx}={de^u \over du}{du\over dx}[/tex]


where [tex]u=-x[/tex].
 
  • #8
367
0
Isn't it d/dx (e^-x) = -(e^-x)(-1) = e^-x?
 
  • #9
rock.freak667
Homework Helper
6,230
31
Isn't it d/dx (e^-x) = -(e^-x)(-1) = e^-x?
[tex]\frac{d}{dx}(e^{f(x)})=f'(x)e^{f(x)}[/tex]

in this case f(x)=-x and hence f'(x)=-1
 

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