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Homework Help: Integration of exponential functions

  1. Oct 28, 2007 #1
    I have a test tomorrow and I am hoping someone can help me with the following two integrals:

    1. [tex]\int[/tex]ye^[tex]^{-y^{2}}[/tex]/2dy

    For this one I am using substition with u=-y[tex]^{2}[/tex]/2 and du = -y/4

    What I don't understand is how we account for the first y in this integral.

    2. [tex]\int[/tex](1/[tex]{\sqrt{2}[/tex])e^[tex]^{-1/6(y-1)^{2}}[/tex]dy

    If I'm having trouble with the first one, you can assume I'm not getting far on this one either.
  2. jcsd
  3. Oct 28, 2007 #2
    you're problem is unclear.

    can you re-type that plz.
  4. Oct 28, 2007 #3
    sure. one moment...
  5. Oct 28, 2007 #4
    That is not right. Check the derivative of the y2 and then find du/dy.

    This one's a bit unclear. Is it [tex]\exp\left(\frac{-1}{6}(y-1)^2\right)[/tex] or [tex]\exp\left(\frac{-1}{6(y-1)^2}\right)[/tex]? Also should there be another y term in the integral somewhere?
  6. Oct 28, 2007 #5


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    Gold Member

    The first one is probably a lot easier than you think.

    When I see integrals like these, I always solve the derivative of a the exponential part and see what it looks like. What's the derivative of e^(-y^2)?
  7. Oct 28, 2007 #6
    The first one [tex]\exp\left(\frac{-1}{6}(y-1)^2\right)[/tex]
  8. Oct 28, 2007 #7
    -2ye^(-y^2) ?
  9. Oct 28, 2007 #8
    I don't think it's possible to solve the second integral in terms of elementary functions.
  10. Oct 28, 2007 #9

    the derivative of -y^2/2 would be -y?
  11. Oct 28, 2007 #10
    Would it help if it was being integrated from negative infinity to infinity?

    Should have put that up initially :rolleyes:
  12. Oct 28, 2007 #11
    So... for the first one would the answer be -2ye^(-y/2)?
  13. Oct 28, 2007 #12
  14. Oct 28, 2007 #13
    or is it -2ye^(y^2)?
  15. Oct 28, 2007 #14
    Neither. The derivative of [tex]e^{-\frac{y^2}{2}}[/tex], with respect to y, [tex]-ye^{-\frac{y^2}{2}}[/tex].
  16. Oct 28, 2007 #15
    Oh. Now I see why you gave me that great hint.

    Now as far as your hint on the second one... very interesting, but a bit complex for me. The last time I took a calc class was about three years ago.

    Does this look like I'm getting close?

    [tex]=\int_{0}^{\infty}\int_{0}^{\infty} e^{-{(1/6(y-1)^2+1/6(x-1)^2)}} dxdy[/tex]
  17. Oct 28, 2007 #16
    You could use a sub. like u = (1/6)(y-1) to simplify things.
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