Answer
$\dfrac{e-1}{8}$
Work Step by Step
The domain $D$ for given region can be expressed as: $0 \leq x \leq 1$ and $x^{2/3} \leq y \leq 1$
The iterated integral can be calculated as:
$\iint_{D} f(x,y) d A=\int_0^1 \int_{x^{2/3}}^{1} xe^{y^4} dy dx\\=\int_0^1 \int_{0}^{y^{3/2}} xe^{y^4} dy \\=\int_0^1 (\dfrac{y^3e^{y^4}}{2}) \ dy$
Suppose that $y^4=t \implies 4y^3 dy=dt$
Now, $\int_0^1 \dfrac{y^3}{2} \times e^{y^4} \ dy=[\dfrac{e^{y^4}}{8}]_0^1\\=\dfrac{e}{8}-\dfrac{1}{8}\\=\dfrac{e-1}{8}$
