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Prove It I find the easiest way to reverse orders of integration is to look at the inequalities and "solve" them simultaneously.
Notice that
$\displaystyle \tan{\left(\frac{\pi x}{4}\right)} \leq y \leq x$ and $\displaystyle 0 \leq x \leq 1$.
You should be able to see that since $\displaystyle y \leq x$ and $\displaystyle x \leq 1$, then $\displaystyle y \leq 1$.
Also, if $\displaystyle \tan{\left(\frac{\pi x}{4}\right)} \leq y$ and $\displaystyle 0 \leq x$, then $\displaystyle 0 \leq y$.
Finally, if $\displaystyle \tan{\left(\frac{\pi x}{4}\right)} \leq y$, then
$\displaystyle \frac{\pi x}{4} \leq \arctan{y}$
$\displaystyle x \leq \frac{4}{\pi}\arctan{y}$.
So your new inequalities are
$\displaystyle 0 \leq y \leq 1$ and $\displaystyle y \leq x \leq \frac{4}{\pi}\arctan{y}$.
Thus the double integral is
$\displaystyle \int_0^1{\int_{\tan{\left(\frac{\pi x}{4}\right)}}^x{f(x, y)\,dy}\,dx} = \int_0^1{\int_y^{\frac{4}{\pi}\arctan{y}}{f(x, y)\,dx}\,dy}$.