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Math Help - integral

  1. #1
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    integral

    compute the double integral

    \int\int sin(y^2) dydx from [0,sqrt(pi)]x[x,sqrt(pi)]

    how can I compute sin(y^2)
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  2. #2
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    Quote Originally Posted by yzc717 View Post
    compute the double integral

    \int\int sin(y^2) dydx from [0,sqrt(pi)]x[x,sqrt(pi)]

    how can I compute sin(y^2)
    Can you explain the limits more clearly.

    CB
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  3. #3
    Moo
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    Hello,

    One can guess that 0\leq x \leq \sqrt{\pi} and x\leq y\leq\sqrt{\pi}

    that is \int_0^{\sqrt{\pi}}\int_x^{\sqrt{\pi}} \sin(y^2) ~dy ~dx

    Reverse the order of the integrals (which is possible since the integrand is positive) :
    0\leq x\leq y\leq \sqrt{\pi} \Rightarrow y ranges from 0 to \sqrt{\pi}

    0\leq x\leq y\leq \sqrt{\pi} \Rightarrow x varies from 0 to y.

    So the integral is now :

    \int_0^{\sqrt{\pi}}\left(\int_0^y \sin(y^2) ~dx\right)~dy

    =\int_0^{\sqrt{\pi}}\left(\sin(y^2)\int_0^y dx\right)~dy

    And that's easier
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  4. #4
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    Quote Originally Posted by Moo View Post

    =\int_0^{\sqrt{\pi}}\left(\sin(y^2)\int_0^y dx\right)~dy

    And that's easier

    that's what I got, I suddenly forget how to compute the sin(y^2)...
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  5. #5
    Moo
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    The inner integral will give y.

    Which finally gives \int_0^{\sqrt{\pi}} y\sin(y^2)~dy

    For this one : substitution !
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