Answer
$$ \iint \limits_{D} x^3d A=\frac{3e^4}{16}+\frac{1}{16}$$
Work Step by Step
Given $$ \iint \limits_{D} x^3d A$$
$$D=\{(x, y) |1\leq x \leq e,0\leq y \leq \ln x\}$$
So, we have
\begin{aligned} A&=\iint \limits_{D} x^3 \ d A\\
&=\int_{0}^{e} \int_{0}^{\ln x} x^3 \ d y \ d x\\
&=\int_{0}^{e} x^3 \left[ y\right]_{0}^{\ln x} \ d x\\&
= \int_{0}^{e}x^3 \ln x \ d x\\
\end{aligned}
So, by the partition technique, let $$u=ln x \Rightarrow du= \frac{1}{x}dx$$ $$dv=x^3 \ dx \Rightarrow v=\frac{x^4}{4} $$ So, we get
\begin{aligned} A &=\left[uv \right]_{1}^{e}- \int_{1}^{e} vdu\\ &
=\left[\frac{x^4 \ln x}{4} \right]_{1}^{e}- \int_{1}^{e}\frac{x^3}{4} \ d x\\ &=\left[\frac{x^4 \ln x}{4} \right]_{1}^{e}-\left[ \frac{x^4}{16} \right]_{1}^{e} \\
&=\left[\frac{e^4 \ln e}{4} -0 \right] -\left[\frac{e^4}{16} -\frac{1}{16} \right] \\
&= \frac{e^4}{4}-\frac{e^4}{16}+\frac{1}{16}\\
&=\frac{3e^4}{16}+\frac{1}{16}
\end{aligned}