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Math Help - Simple integration question help

  1. #1
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    Simple integration question help

    Hey guys, after differentiating y=x^{2}e^{-x^{2}}by the product rule to get y=2xe^{-x^{2}}-2x^{3}e^{-x^{2}}<br />
    how do I use this result when integrating \displaystyle\int x^{3}e^{-x^{2}}\,dx

    thanks a lot
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  2. #2
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    \displaystyle \frac{d}{dx}(x^2e^{-x^2}) = 2x\,e^{-x^2} - 2x^3e^{-x^2} is equivalent to

    \displaystyle x^2e^{-x^2} = \int{2x\,e^{-x^2} - 2x^3e^{-x^2}\,dx}

    \displaystyle x^2e^{-x^2} = -e^{-x^2} - 2\int{x^3e^{-x^2}\,dx}.

    Now solve for your integral.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    \displaystyle \frac{d}{dx}(x^2e^{-x^2}) = 2x\,e^{-x^2} - 2x^3e^{-x^2} is equivalent to

    \displaystyle x^2e^{-x^2} = \int{2x\,e^{-x^2} - 2x^3e^{-x^2}\,dx}

    \displaystyle x^2e^{-x^2} = -e^{-x^2} - 2\int{x^3e^{-x^2}\,dx}.

    Now solve for your integral.
    so this is right?
    \displaystyle\int x^3e^{-x^2}\,dx=\frac {-(x^{2}e^{-x^2}+e^{-x^2})}{2}
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  4. #4
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    Quote Originally Posted by aonin View Post
    so this is right?
    \displaystyle\int x^3e^{-x^2}\,dx=\frac {-(x^{2}e^{-x^2}+e^{-x^2})}{2}
    Yes
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