Evaluate double integral of triangular region Evaluate $$ \int\int_D e^{y^2} dA  $$ where D is the triangular region with vertices (0,0), (0,1) and (2,1)
My attempts:
$$ \int^{2}_0 \int^{1}_\frac{x}{2} e^{y^2} dy dx $$
or
$$ \int^{0}_1\int^{2y}_2 e^{y^2} dx dy $$
but I couldn't evaluate the integral so I think I must've done something wrong when finding the region D.

The integral $$\int_0^2\int_{x/2}^1 e^{y^2} dy dx$$ is hard to evaluate, so you need to get the $dx$ inside. The following will do:
$$\int_0^1\int_0^{2y} e^{y^2} dx dy.$$
Evaluation is simple:
\begin{align}
\int_0^1\int_0^{2y} e^{y^2} dx dy &= \int_0^1 2y e^{y^2} dy \\
                                  &= \left[e^{y^2}\right]_0^1 \\
                                  &= e-1.
\end{align}
