compute the double integral

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

how can I compute sin(y^2)

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- May 16th 2009, 10:45 PMyzc717integral
compute the double integral

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

how can I compute sin(y^2) - May 17th 2009, 12:35 AMCaptainBlack
- May 17th 2009, 12:41 AMMoo
Hello,

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

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

Reverse the order of the integrals (which is possible since the integrand is positive) :

$\displaystyle 0\leq x\leq y\leq \sqrt{\pi} \Rightarrow $ y ranges from 0 to $\displaystyle \sqrt{\pi}$

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

So the integral is now :

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

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

And that's easier :p - May 17th 2009, 12:47 AMyzc717
- May 17th 2009, 01:11 AMMoo
The inner integral will give y.

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

For this one : substitution !