Answer
\[\frac{128}{3}\]
Work Step by Step
\[\begin{align}
& y=\left| x \right|\to y=x,\text{ }x\ge 0,\text{ }y=-x,\text{ }x<0 \\
& x=y,\text{ }x\ge 0,\text{ }x=-y,\text{ }x<0 \\
& \text{From the graph, the region }R\text{ is} \\
& R=\left\{ \left( x,y \right):-y\le x\le y,\text{ 0}\le y\le 4\text{ } \right\} \\
& \text{Then,} \\
& \iint_{R}{\left( x+y \right)}dA=\int_{0}^{4}{\int_{-y}^{y}{\left( x+y \right)dx}dy} \\
& \text{Integrating} \\
& =\int_{0}^{4}{\left[ \frac{1}{2}{{x}^{2}}+xy \right]_{-y}^{y}dy} \\
& =\int_{0}^{4}{\left[ \left( \frac{1}{2}{{y}^{2}}+{{y}^{2}} \right)-\left( \frac{1}{2}{{y}^{2}}-{{y}^{2}} \right) \right]dy} \\
& =\int_{0}^{4}{\left( \frac{1}{2}{{y}^{2}}+{{y}^{2}}-\frac{1}{2}{{y}^{2}}+{{y}^{2}} \right)dy} \\
& =\int_{0}^{4}{2{{y}^{2}}dy} \\
& =\left[ \frac{2{{y}^{3}}}{3} \right]_{0}^{4} \\
& =\frac{2{{\left( 4 \right)}^{3}}}{3}-\frac{2{{\left( 0 \right)}^{3}}}{3} \\
& =\frac{128}{3} \\
\end{align}\]
