Solution
= Solution
The region is $0\leq y\leq1$, $2y\leq x\leq2$. Reversing the order describes the same triangle as $0\leq x\leq2$, $0\leq y\leq x/2$. Since the <integrand> is continuous on this compact region, <Fubini's theorem> permits
$$
\int_0^1\int_{2y}^2 e^{x^2}\,dx\,dy=\int_0^2\int_0^{x/2}e^{x^2}\,dy\,dx=\frac12\int_0^2xe^{x^2}\,dx=\boxed{\frac{e^4-1}{4}}.
$$
The new inner <integral> introduces the factor $x$ needed for the <change of variables> $u=x^2$.