Answer
$\dfrac{e-1}{6}$
Work Step by Step
The domain $D$ for the given region can be expressed as:
$0 \leq x \leq 1$ and $x \leq y \leq 1$
The iterated integral can be calculated as:
$\iint_{D} x^2y d A=\int_0^1 \int_{x}^{1} xe^{y^3} dy dx\\=\int_0^1 \int_0^y x e^{y^3} dx dy\\=\int_0^1 (\dfrac{y^2}{2}e^{y^3}) \ dy$
Suppose that
$y^3=t \implies y^2 dy=\dfrac{dt}{3}$
Now, $\int_0^1 \dfrac{y^2}{2} \times e^{y^3} \ dy=[\dfrac{e^{y^3}}{6}]_0^1\\=\dfrac{1}{6}(e)[e^1-e^0]\\=\dfrac{e-1}{6}$
