Answer
$$\int_0^1\int_0^2 (x+y)\hspace{0.5mm}dydx=3$$
Work Step by Step
We will start with the integral with respect to $y$:
$\int_0^1\int_0^2 (x+y)\hspace{0.5mm}dydx=\int_0^1(xy+\frac{y^2}{2})\bigg\vert_0^2dx=\int_0^1\left((2x+\frac{2^2}{2})-(0-0)\right)dx=\int_0^1(2x+2)dx=\left(x^2+2x\right)\bigg\vert_0^1=(1^2+2)-(0+0)=3$
