Solution
= Solution
The integrand is continuous on the compact rectangle, so <Fubini's theorem> allows integrating in $y$ first. For $x\ne0$, $\int_0^1xe^{xy}dy=e^x-1$; the same formula holds by continuity at $x=0$. Hence
$$
\boxed{\int_0^1\int_0^2xe^{xy}dx\,dy=\int_0^2(e^x-1)dx=e^2-3.}
$$