# Compute integral

## The Attempt at a Solution

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 :

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

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

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