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How do you take take this integral?

  1. Feb 22, 2009 #1
    how do you take take this integral?

    [tex] \int_{0}^{r} x^2 e^{-2x} dx [/tex]
     
  2. jcsd
  3. Feb 22, 2009 #2

    gabbagabbahey

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    Re: integral

    Use integration by parts (twice)....the derivatives of [itex]x^2[/itex] are easy to find, and likewise for the antiderivative of [itex]e^{-2x}dx[/itex]
     
  4. Feb 22, 2009 #3
    Re: integral

    whats the upper limit? if it's infinity answer is 1/8
     
  5. Feb 22, 2009 #4
    Re: integral

    do you know how this integral turns into

    [tex] \frac {N!}{a^{N + 1}} [/tex] if I take the integral from 0 to infinity? N = 2 and a = 2
     
  6. Feb 22, 2009 #5
    Re: integral

    answer is 1/4 if infinity
     
  7. Feb 22, 2009 #6

    gabbagabbahey

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    Re: integral

    If N=2 and a=2, then [tex]\frac {N!}{a^{N + 1}}=\frac {2!}{2^{2 + 1}}=\frac{1}{4} [/tex] which is what you should be getting using by parts.

    Are you getting something different?
     
  8. Feb 22, 2009 #7
    Re: integral

    I was trying to derive the general expression

    [tex] \int_{0}^{\infty} x^n e^{-ax} dx = \frac{n!}{a^{n+1}} [/tex]

    how is this so?
     
  9. Feb 22, 2009 #8

    gabbagabbahey

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    Re: integral

    Use integration by parts n times and remember the definition of factorial; [itex]n!=n(n-1)(n-2)\ldots (2)(1)[/itex]
     
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