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Math Help - integration question

  1. #1
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    integration question

    how to integrate this?

    xe^-2x from 0 to infinity
    i used ibp with dv = e^-2x and u = x
    then v = -0.5e^-2x and du = 1

    I'm not getting the right answer though...
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  2. #2
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    e^(i*pi)'s Avatar
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    Your u and v subs are fine.

    What did you do next?


    \int udv = uv - \int vdu

    -0.5xe^{-2} + 0.5 \int e^{-2x}
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  3. #3
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    = uv - int vdu
    int vdu = 1/4e^-2x

    so i get

    -0.5xe^-2x - 0.25e^-2x | 0 to infinity
    not 100% sure how to evaluate the first term at infinity though, i believe it goes to 0 but i might be wrong...
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  4. #4
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    Quote Originally Posted by Kuma View Post
    = uv - int vdu
    int vdu = 1/4e^-2x

    so i get

    -0.5xe^-2x - 0.25e^-2x | 0 to infinity
    not 100% sure how to evaluate the first term at infinity though, i believe it goes to 0 but i might be wrong...
    That's right (and it does go to 0. For the first term e^(-2x) will shrink faster than x will increase)

    0 - (-0.5 \cdot 0 \cdot 1 - 0.25 \cdot 1) = 0.25


    edit: wolfram agrees - http://www.wolframalpha.com/input/?i...0+and+infinity
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