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)
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