Compute

${\displaystyle \underset{0}{\overset{1}{}}}{\displaystyle \underset{3y}{\overset{3}{}}}{e}^{x^2}dxdy$

$\frac{e+1}{2}$

$e^2$

$\frac{2e^6-3}{4}$

$e^3$

$e^5$

$\frac{e^9-1}{6}$

$1$