Answer
$$\int_{-1}^2\int_{1}^3(x+y^2)\hspace{0.5mm}dxdy=18$$
Work Step by Step
We will begin with integral with respect to x:
$\int_{-1}^2\int_{1}^3(x+y^2)\hspace{0.5mm}dxdy=\int_{-1}^2\left(\frac{x^2}{2}+y^2x\right)\bigg\vert_1^3dy$
$=\int_{-1}^2\left(\left(\frac{9}{2}+3y^2\right)-\left(\frac{1}{2}+y^2\right)\right)dy$
$=\int_{-1}^2\left(\frac{9}{2}+3y^2-\frac{1}{2}-y^2\right)dy=\int_{-1}^2\left(2y^2+4\right)dy$
$=\left(\frac{2}{3}y^3+4y\right)\bigg\vert_{-1}^2=\left
(\frac{16}{3}+8\right)-\left(-\frac{2}{3}-4\right)$
$=\frac{16}{3}+8+\frac{2}{3}+4=\frac{18}{3}+12=18$
