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Math Help - Integral

  1. #1
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    Question Integral

    I'm desperately stuck on the following integral:

    Integral: x^5*e^(-1/2x^2)

    Thanks
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  2. #2
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    Parts reduces the x^5 by a power of 2 each time. So using it twice leaves xe^{\frac{-x^2}{2}} which hopefully you can just integrate.
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  3. #3
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    confused

    Thanks for the reply but I'm still a little confused.

    How does by parts reduce x^5 by 2 each time? Doesn't it only reduce it by one?
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  4. #4
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    I=\int x^5 e^{\frac{-x^2}{2}}dx

    Take u=x^5, dv=e^{\frac{-x^2}{2}}

    So du=5x^4, v=\frac{e^{\frac{-x^2}{2}}}{-x}
    We want I = uv - int(v)du

    So I=-x^4 e^{\frac{-x^2}{2}} + \int 5x^3 e^{\frac{-x^2}{2}}

    etc...
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  5. #5
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    Quote Originally Posted by Thomas154321 View Post
    I=\int x^5 e^{\frac{-x^2}{2}}dx

    Take u=x^5, dv=e^{\frac{-x^2}{2}}

    So du=5x^4, v=\frac{e^{\frac{-x^2}{2}}}{-x} (**)
    We want I = uv - int(v)du

    So I=-x^4 e^{\frac{-x^2}{2}} + \int 5x^3 e^{\frac{-x^2}{2}}

    etc...
    The (**) step is simply not true! The best thing to do first is a substitution u = - \frac{x^2}{2} giving

    -4 \int u^2 e^u\,du the integration by parts twice.
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