Answer
$\dfrac{23}{84}$
Work Step by Step
We can define the domain $D$ in the Type-1 as follows: $
D=\left\{ (x, y) | x^3 \leq y \leq x, \ 0 \leq x \leq 1 \right\}
$
Therefore, $\iint_{D} (x^2+2y) \ dA=\int_{0}^{1} \int_{x^3}^{x} (x^2+2y) \ dy \ dx \\ =\int_0^1 [x^2y+y^2]_{x^3}^{x} \ dx \\= \int_0^1 [x^2 (x)+x^2-x^2 \times x^3 -(x^3)^2] d x \\ = [\dfrac{x^4}{4} +\dfrac{x^3}{3} -\dfrac{x^6}{6}-\dfrac{x^7}{7}]_0^1 \\= \dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{7} \\=\dfrac{23}{84}$
