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Math Help - Improper Integral xe^-(x^2)..

  1. #1
    Member zangestu888's Avatar
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    Improper Integral xe^-(x^2)..

    I cant seem to get this in any way??

    inttegral of xe^(-x^2)dx a=-infinite,b=infinite

    The Answer in my textbook is 0, i dont know how to approach the problem? I know this i have to break it up into to several integrals, but i dont know how to integrate the function. Thank You for Any Help
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  2. #2
    o_O
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    Hello Kurosaki.

    You just have to use this little fact: If f(x) is an odd function, then \int_{-a}^a f(x) \ dx = 0
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    Quote Originally Posted by zangestu888 View Post
    I cant seem to get this in any way??

    inttegral of xe^(-x^2)dx a=-infinite,b=infinite

    The Answer in my textbook is 0, i dont know how to approach the problem? I know this i have to break it up into to several integrals, but i dont know how to integrate the function. Thank You for Any Help
    What is the derivative of e^{-x^2} ?

    Then use the fundamental theorem of calculus.

    .
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    Member zangestu888's Avatar
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    I dont quite understand? So thier is no actual way of doing it only just by the fact that its an odd function? am confused lol
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  5. #5
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    Quote Originally Posted by zangestu888 View Post
    I cant seem to get this in any way??

    inttegral of xe^(-x^2)dx a=-infinite,b=infinite

    The Answer in my textbook is 0, i dont know how to approach the problem? I know this i have to break it up into to several integrals, but i dont know how to integrate the function. Thank You for Any Help
    Because of the symmetry in the origin you should get zero.. To show this split the integral up into two pieces

    \int_{-\infty}^0 x e^{-x^2}\,dx + \int_0^{\infty} x e^{-x^2}\,dx

    and do each separately.
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    Quote Originally Posted by o_O View Post
    Hello Kurosaki.

    You just have to use this little fact: If f(x) is an odd function, then \int_{-a}^a f(x) \ dx = 0
    That does not prove that the double limit exists, it only tells you what its value is if it does.

    .

    .
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    Member zangestu888's Avatar
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    Wel i know how to split it up i got that part, my problem is integrating the functioon after the split?
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    Member zangestu888's Avatar
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    i tried subs. and parts i didnt get anything after the split
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    Quote Originally Posted by zangestu888 View Post
    Wel i know how to split it up i got that part, my problem is integrating the functioon after the split?
    Let  u = -x^2,\;\;\;du = -2x \,dx and switch the limits.
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    Member zangestu888's Avatar
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    okay idid that but then i would have to integrate e^-(u)^2how can i do that
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  11. #11
    Member zangestu888's Avatar
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    i have this integral of -1/2(e^-(u^2))
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    Quote Originally Posted by zangestu888 View Post
    okay idid that but then i would have to integrate e^-(u)^2how can i do that
    Actually no. Under u = -x^2, du = -2x\,dx

    \int x\, e^{-x^2}\, dx = - \frac{1}{2} \int e^u \,du
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  13. #13
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    omg lol am so stupied i didnt see that thanks!!
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    Quote Originally Posted by zangestu888 View Post
    okay idid that but then i would have to integrate e^-(u)^2how can i do that

    \frac{d}{dx}e^{-x^2}=-2x e^{-2x}

    so:

    \int x e^{-2x} \ dx= -\frac{1}{2} e^{-x^2}

    and so:

     <br />
\int_{-\infty}^{\infty}x e^{-2x} \ dx=\lim_{a\to -\infty,\ b\to \infty}\int_{a}^{b}x e^{-2x} \ dx=\lim_{a\to -\infty,\ b\to \infty}\left(e^{-a^2}-e^{-b^2}\right)=0<br />

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