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Math Help - Can you integrate this?

  1. #1
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    Can you integrate this?

     <br />
{e}^{ {x}^{2 } }  <br />
?

    with respect to x

    The full question is \iint  {e}^{{ x}^{ 2} }  dx dy by choosing the best order of integration

    Limits for outside integral is  2\sqrt{ln2} and 0
    Inside integral is \sqrt{ln 2} and \frac{y}{ 2}
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by adam_leeds View Post
     <br />
{e}^{ {x}^{2 } }  <br />
?

    with respect to x

    The full question is \iint  {e}^{{ x}^{ 2} }  dx dy by choosing the best order of integration

    Limits for outside integral is  2\sqrt{ln2} and 0
    Inside integral is \sqrt{ln 2} and \frac{y}{ 2}
    This integral has no closed form solution. You'll have to do it numerically.

    -Dan

    Yup. I was wrong. I didn't get back in time to edit my post. Sorry!
    Last edited by topsquark; May 2nd 2011 at 09:11 AM.
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  3. #3
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    Quote Originally Posted by topsquark View Post
    This integral has no closed form solution. You'll have to do it numerically.

    -Dan
    How? do you mean re-write the exponential as a ln?
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  4. #4
    A Plied Mathematician
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    I can't make out your LaTeX there. Is this your integral?

    \int_{0}^{2\sqrt{\ln(2)}}\int_{y/2}^{\sqrt{\ln(2)}}e^{x^{2}}\,dx\,dy?

    In this case, reverse the order of integration, to obtain

    \int_{0}^{\sqrt{\ln(2)}}\int_{0}^{2x}e^{x^{2}}\,dy  \,dx,

    and go from there. To get the limits, draw the region in the xy plane, and figure it out.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    I can't make out your LaTeX there. Is this your integral?

    \int_{0}^{2\sqrt{\ln(2)}}\int_{y/2}^{\sqrt{\ln(2)}}e^{x^{2}}\,dx\,dy?

    In this case, reverse the order of integration, to obtain

    \int_{0}^{\sqrt{\ln(2)}}\int_{0}^{2x}e^{x^{2}}\,dy  \,dx,

    and go from there. To get the limits, draw the region in the xy plane, and figure it out.
    Thanks yes it is.

    The order is not my problem its integrating that with repsect to x? You cant do it.
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  6. #6
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    Quote Originally Posted by adam_leeds View Post
    Thanks yes it is.
    The order is not my problem its integrating that with repsect to x? You cant do it.
    Once you have \int_0^{2x} {e^{x^2 } dy}  = \left. {ye^{x^2 } } \right|_{y = 0}^{2x}  = 2xe^{x^2 } the rest is trivial.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Once you have \int_0^{2x} {e^{x^2 } dy}  = \left. {ye^{x^2 } } \right|_{y = 0}^{2x}  = 2xe^{x^2 } the rest is trivial.
    ???

    We've still got to integrate 2xe^{x^2 } with respect to x
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  8. #8
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    Quote Originally Posted by adam_leeds View Post
    ???

    We've still got to integrate 2xe^{x^2 } with respect to x
    My bad i've been a numpty

    u = x^2
    du = 2xdx

    integrate e^u du

    = e^u = e ^ x^ 2

    put the limits in =

    e (ln2)^0.5 - e^0 = 1
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