Answer
$ \dfrac{1754}{15}$
Work Step by Step
The iterated integral can be calculated as:
$\iint_{D} f(x,y) d A= \int_{1}^{3} \int_{x}^{2x+1} x^2y \ dy \ dx \\= \int_{1}^{3} [\dfrac{x^2y^2}{2}]_x^{2x+1} \ dy\\=(\dfrac{1}{2}) \int_{1}^{3} [x^2(4x^2+4x+1)-x^2(x^2)] \ dx\\=(\dfrac{1}{2})[\dfrac{3x^5}{5}+x^4+\dfrac{x^3}{3}]_1^3 \\= \dfrac{1754}{15}$
