Good evening, I'm trying to solve an exercise in introductory statistics.
After applying the condition that $\int_{\mathbb R^2} f(x,y) \, \mathrm{d}x \,\mathrm{d}y = 1$, I find that $c = 3/2$.Then $$\mathbb{P}\left(X \leqslant-\frac{1}{2}, Y \leqslant \frac{1}{2}\right)=\int_{-\infty}^{-\frac{1}{2}}\left(\int_{-\infty}^{\frac{1}{2}} f(x, y) \,\mathrm{d}y \right) \,\mathrm{d}x$$
The solution given in my lecture note is $$=\int_{-1}^{-\frac{1}{\sqrt{2}}}\left(\int_{0}^{\frac{1}{2}} \frac{3}{2} d y\right) d x+\int_{-\frac{1}{\sqrt{2}}}^{-\frac{1}{2}}\left(\int_{0}^{x^{2}} \frac{3}{2} d y\right) d x$$
I could not figure out how to split in such way. Could you please elaborate on this issue? Thank you so much!
Update: Here's the picture of the support, but I still don't know how to use the graph of the support to calculate the integral.
Can you understand the integrals limits from here?