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Math Help - In my notes i have this equals this

  1. #1
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    In my notes i have this equals this

    But i have no idea how it does im getting a different answer, any help please?

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  2. #2
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    Quote Originally Posted by adam_leeds View Post
    But i have no idea how it does im getting a different answer, any help please?


    \frac{2}{2\pi}\int^a_0xe^{-x^2}\,dx=-\frac{1}{2\pi}\int^a_0(-2x\,dx)e^{-x^2}= -\frac{1}{2\pi}\left[e^{-x^2}\right]^a_0=-\frac{1}{2\pi}\left[e^{-a^2}-e^0\right] =\frac{1-e^{-a^2}}{2\pi} , so there's a mistake in your notes.

    Tonio
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  3. #3
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by tonio View Post
    \frac{2}{2\pi}\int^a_0xe^{-x^2}\,dx=-\frac{1}{2\pi}\int^a_0(-2x\,dx)e^{-x^2}= -\frac{1}{2\pi}\left[e^{-x^2}\right]^a_0=-\frac{1}{2\pi}\left[e^{-a^2}-e^0\right] =\frac{1-e^{-a^2}}{2\pi} , so there's a mistake in your notes.

    Tonio
     \frac{2}{2 \pi } \int_0^a xe^{-x^2} dx

    Let  u=x^2 and  du = 2xdx


     \frac{1}{2 \pi } \int_0^{a^2} e^{-u} du = \frac{1}{2 \pi } [ e^{-a^2} - 1 ]

    I think you dont carry through your negative on the third step?

    But the above is NOT equal to

    \frac{2}{ \sqrt{ \pi }} [ e^{-a^2} - 1 ]

    Your professor has messed up the constants somewhere along the way.
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  4. #4
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    Quote Originally Posted by AllanCuz View Post
     \frac{2}{2 \pi } \int_0^a xe^{-x^2} dx

    Let  u=x^2 and  du = 2xdx


     \frac{1}{2 \pi } \int_0^{a^2} e^{-u} du = \frac{1}{2 \pi } [ e^{-a^2} - 1 ]

    I think you dont carry through your negative on the third step?


    \int e^{-u}\,du=-e^{-u} , and not only e^{-u} . You forgot the minus sign here.

    Tonio



    But the above is NOT equal to

    \frac{2}{ \sqrt{ \pi }} [ e^{-a^2} - 1 ]

    Your professor has messed up the constants somewhere along the way.
    .
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  5. #5
    Senior Member AllanCuz's Avatar
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    So what I got from this thread was trust the judgment of MFH helpers over my own, always! lol thanks
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