Rectangle with double integrals, Has Answer?

The problem can be expressed as the following double integral

$\int_0^1 \! \int_1^2 \! (x + y) \, dx \, dy$

Evaluating the inner integral

$\int_0^1 \! \int_1^2 \! (x + y) \, dx \, dy = \int_0^1 \! [ \frac{1}{2} x^{2} + yx ]_1^2 \, dy$ = $\int_0^1 (\frac{3}{2} + y) \, dy$

Evaluating the outer integral

$\int_0^1 (\frac{3}{2} + y) \, dy = [ \frac{3}{2} y + \frac{1}{2} y^{2} ]_0^1 = \frac{3}{2} + \frac{1}{2} = \boxed{2}$