Answer
$a^2 b+\dfrac{3}{2} ab^2$
Work Step by Step
We can define the domain $D$ in the rectangle as follows: $
D=\left\{ (x, y) | 0 \leq x \leq 9, \ 0 \leq y \leq b \right\}
$
Therefore, $$\iint_{D} f(x,y) dA=\iint_{D} (2x+3y) dA \\=\int_{0}^{a} \int_{0}^{b} f(x,y) (2x+3y) \ dy \ dx \\=\int_0^a [2xy +\dfrac{3y^2}{2}]_0^b \ dx \\= \int_0^a 2bx +\dfrac{3b^2}{2} \ dx \\= [bx^2 +\dfrac{3b^2 x}{2}]_0^a \\= a^2 b+\dfrac{3}{2} ab^2$$
