Answer
$ \dfrac{4}{3} $
Work Step by Step
We can define the domain $D$ as follows: $
D=\left\{ (x, y) | 1 \leq x \leq 2, \ y-1 \leq y \leq 1 \right\}
$
Therefore, $\iint_{D} (2x+y) \ dA=\int_{1}^{2} \int_{y-1}^{1} (2x+y) \ dx \ dy \\ =\int_1^2 [x^2+xy]_{y-1}^{1} \ dy \\= \int_1^2 (1+y) -[(y^2 -2y+1) +(y^2-y) ] dy \\ =\int_1^2 4y -2y^2 dy \\ [2y^2 -\dfrac{2y^3}{3} ]_1^2 \\= [2(2)^2 -\dfrac{2(2)^3}{3}] - [2(1)^2 -\dfrac{2(1)^3}{3}] \\=6-\dfrac{14}{3} \\= \dfrac{4}{3} $
