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Integral Problem involving e

  1. May 15, 2012 #1
    W=∫0PdV



    P=e-v2



    Do I simply substitue P in the original equation, differentiate, then integrate?

    W=∫0e-v2dV

    W=∫0(e-v2)/2v

    So far is this correct?
     
  2. jcsd
  3. May 15, 2012 #2

    LCKurtz

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    No, it isn't. That integral is usually worked by a trick of multiplying it by itself and changing it to polar coordinates. See
    http://en.wikipedia.org/wiki/Gaussian_integral
    There they do it from ##-\infty## to ##\infty##, but the same idea applies.
     
  4. May 15, 2012 #3
    I was not familiar with the Gaussian intergral. Well since it's only half the distance I would divide by 2 ∴

    W=∫0e-v2dV

    =(√∏)/2|0

    so does that mean that the interval of ∞ to 0 is negligible???
     
  5. May 15, 2012 #4

    LCKurtz

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    The integrand is an even function so, yes, you can just divide the full integral by 2. The integal from on ##[-\infty,0]## equals the integral over ##[0,\infty]##.
     
  6. May 15, 2012 #5
    thank you for your help. my final answer is W=(√∏)/2
     
  7. May 15, 2012 #6

    LCKurtz

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    That's correct.
     
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