Answer
$\frac{2}{9}e^{3}-\frac{4}{45}$
Work Step by Step
Given: $\int_{0}^{1}\int_{x}^{e^{x}}3xy^{2}dydx$
Let us solve first $\int_{x}^{e^{x}}3xy^{2}dy=xe^{3x}-x^{4}$
Now, $\int_{0}^{1}xe^{3x}-x^{4}dx=(\frac{1}{9}(3x-1)e^{3x})|_{0}^{1}-\frac{x^{5}}{5}|_{0}^{1}$
$=\frac{2}{9}e^{3}+\frac{1}{9}-\frac{1}{5}$
$=\frac{2}{9}e^{3}-\frac{4}{45}$
