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Solving an integral

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the integral of x(e^-x) evaluated at 1 to inf


    2. Relevant 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?
     
  2. jcsd
  3. Mar 31, 2008 #2
    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).
     
  4. Mar 31, 2008 #3
    So would the integral (the dv) of e^-x just be e^-x?
     
  5. Mar 31, 2008 #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.
     
  6. Mar 31, 2008 #5
    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?
     
  7. Mar 31, 2008 #6
    Don't forget to use the chain rule on the exponent (-x).
    [tex]{de^{-x}\over dx}=-e^{-x}[/tex].
     
  8. Mar 31, 2008 #7
    In detail
    [tex]{de^{-x}\over dx}={de^u \over du}{du\over dx}[/tex]


    where [tex]u=-x[/tex].
     
  9. Mar 31, 2008 #8
    Isn't it d/dx (e^-x) = -(e^-x)(-1) = e^-x?
     
  10. Mar 31, 2008 #9

    rock.freak667

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    Homework Helper

    [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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