Let $X$ and $Y$ be continuous random variables with joint density function $$ f(x,y)=0.25$$ for $$0 \leq x \leq 2$$ and $$x-2 \leq y \leq x$$ What's $$E[X^3Y]$$
The solution is $$ 0.25 \int_{0}^{2} \! \int_{x-2}^{x} x^3yf(x,y) \,dy\,dx$$
Can I switch to $dxdy$?
$$ 0.25 \int_{0}^{2} \! \int_{y}^{2} x^3yf(x,y) \,dx\,dy$$+$$ 0.25 \int_{0}^{2} \! \int_{-2}^{y+2} x^3yf(x,y) \,dx\,dy$$
The result is different.
When you switch the order of integration you get $\int_{-2}^{2} \int_{y\vee 0}^{y+2} x^{3}yf(x,y)\ dxdy$ where $y \vee 0 $ is the maximum of $y$ and $0$. Now see if you get the same answer. [Split the intergal as $\int_0^{2}\int_y^{y+2} x^{3}y\ dxdy +\int_{-2}^{0}\int_0^{y+2} x^{3}y\ dxdy $].