<s>[INST] Is the double integral equal to the area? I have to compute the double integral $\int_0^{\sqrt{\frac{\pi}{2}}}\int_x^{\sqrt{\frac{\pi}{2}}} 2\sin(y^2)dydx$. 
I drawed the region $0\leq x\leq \sqrt{\frac{\pi}{2}}$ and $x\leq y\leq \sqrt{\frac{\pi}{2}}$. This is a triangle with height and base equal to $\sqrt{\frac{\pi}{2}}$, right? 
The double integral is equal to the are of the triangle, or not? 
So, is the double integral equal to $\frac{1}{2}\cdot \sqrt{\frac{\pi}{2}}\cdot \sqrt{\frac{\pi}{2}}=\frac{\pi}{4}$ ?
 [/INST] Hint. Note that the double integral over that triangle can be written as iterated integrals in two ways. 
\begin{align*}
\int_{x=0}^{\sqrt{\frac{\pi}{2}}}\left(\int_{y=x}^{\sqrt{\frac{\pi}{2}}} 2\sin(y^2)dy\right)dx
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}\left(\int_{x=0}^{y} 2\sin(y^2)dx\right)dy\\
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}2\sin(y^2)\left(\int_{x=0}^{y} dx\right)dy\\
&=\int_{y=0}^{\sqrt{\frac{\pi}{2}}}2y\sin(y^2)dy.
\end{align*}
Can you take it from here? 
P.S. The final result is different from what you have found. It is not the area of the triangle. 
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