Solution from first principles

$\frac{{\int }_{-3\/2}^1{\int }_{y^2}^{\left(3-y\right)\/2}\left(2 x^2\+3 y^2\right) \mathit{\ⅆ}x \mathit{\ⅆ}y}{{\int }_{-3\/2}^1{\int }_{y^2}^{\left(3-y\right)\/2}1 \mathit{\ⅆ}x \mathit{\ⅆ}y}$ = $\frac{405}{112}$

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{FunctionAverage}\left(2 x^2\+3 y^2\,x\=y^2..\left(3-y\right)\/2\,y\=-3\/2..1\,\mathrm{output}\=\mathrm{integral}\right)$

$\frac{{\∫}_{-\frac{3}{2}}^1{\∫}_{y^2}^{\frac{3}{2}-\frac{1}{2}y}\left(2x^2\+3y^2\right)\ⅆx\ⅆy}{{\∫}_{-\frac{3}{2}}^1{\∫}_{y^2}^{\frac{3}{2}-\frac{1}{2}y}1\ⅆx\ⅆy}$

$\mathrm{Student}:-\mathrm{MultivariateCalculus}:-\mathrm{FunctionAverage}\left(2 x^2\+3 y^2\,x\=y^2..\left(3-y\right)\/2\,y\=-3\/2..1\right)$

$\frac{405}{112}$