Answer
$$4$$
Work Step by Step
\begin{align*}
\int_{0}^{1} \int_{0}^{2}\left(x+4 y^{3}\right) d x d y&=\int_{0}^{1} \int_{0}^{2} x d x d y+\int_{0}^{1} \int_{0}^{2} 4 y^{3} d x d y\\
&= \left(\int_{0}^{2} x d x\right)\left(\int_{0}^{1} 1 d y\right)+
\left(\int_{0}^{1} 4 y^{3} d y\right)\left(\int_{0}^{2} 1 d x\right)
\\
&=\left(\left.\frac{1}{2} x^{2}\right|_{0} ^{2}\right)\left(\left.y\right|_{0} ^ { 1 }\right)
+
\left(\left.y^{4}\right|_{0} ^{1 }\right)\left(\left.x\right|_{0} ^{2}\right)\\
&=4
\end{align*}
