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Math Help - Integrating an exponential function

  1. #1
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    Integrating an exponential function

    How do you go about integrating 2e^(x^2) between y and 1? According to wolfram alpha it gives an error though I would assume there can be an answer expressed to this in one form or another.

    Thanks
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  2. #2
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    Re: Integrating an exponential function

    Quote Originally Posted by UnbeatableZoomy View Post
    How do you go about integrating 2e^(x^2) between y and 1? According to wolfram alpha it gives an error though I would assume there can be an answer expressed to this in one form or another.

    Thanks
    There is no closed form solution. Though I wonder...did you mean to write \int_1^y 2x e^{x^2} dx instead? This one can be done.

    -Dan
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  3. #3
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    Re: Integrating an exponential function

    Yep, though between 1 and y
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    Re: Integrating an exponential function

    If you mean integrating 2e^(x^2) from 1 to y, then there is no closed form like topsquark mentioned.

    However, if you are integrating 2xe^(x^2) from 1 to y, then use a direct substution. Let u(x) = x^2. Then du = 2x dx. Then u(1) = 1 and u(y) = y^2.

    Integrating 2xe^(x^2) from 1 to y is the same as integrating e^u from 1 to y^2.

    Thus, the integral is e^u | from 1 to y^2.

    Thus, e^(y^2) - e^1.

    Thus, e^(y^2) - e.
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    Re: Integrating an exponential function

    I didn't mention this but the question was actually a double integral which finished with dxdy so I think it involved a change of order
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    Re: Integrating an exponential function

    Well, that's a interesting development. What is the entire question asking?
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    Re: Integrating an exponential function

    UNBEATABLE....

    you are posting an exponential function and then you are talking for a double integral? What really do you want?
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  8. #8
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    Re: Integrating an exponential function

    The question asks you to evaluate the integral between 1 and 0 of the integral between 1 and y of 2e^(x^2) dxdy.

    Word isn't working and I haven't yet learned Latex thus why I've just described it in a sentence.
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  9. #9
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    Re: Integrating an exponential function

    I would suggest reversing the order of integration. By the bounds of your integral, we have \displaystyle 0 \leq y \leq 1 and  \displaystyle 0 \leq x \leq y, which implies \displaystyle 0 \leq x \leq y \leq 1.

    From these we can determine the new bounds are \displaystyle 0 \leq x \leq 1 and \displaystyle x \leq y \leq 1. So

    \displaystyle \begin{align*} \int_0^1{ \int_0^y{ 2e^{x^2} \,dx } \,dy } &= \int_0^1{ \int_x^1{ 2e^{x^2}  \,dy } \,dx } \end{align*}

    This should be able to be evaluated.
    Last edited by Prove It; April 16th 2013 at 03:46 AM.
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  10. #10
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    Re: Integrating an exponential function

    Actually, no. The first integral will give \int_0^1 (1- x)e^{x^2}dx= \int_0^1 e^{x^2}dx- \int_0^1 xe^{x^2}dx. The second integral can be integrated as an exponential but the first cannot.
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  11. #11
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    Re: Integrating an exponential function

    That's why I said SHOULD :P haha
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