Answer
\[A=12\]
Work Step by Step
\[\begin{align}
& \text{From the graph, the area is given by:} \\
& A=\int_{-2}^{0}{\int_{-x-4}^{x}{dy}dx}+\int_{0}^{4}{\int_{2x-4}^{x}{dy}dx} \\
& \text{Integrating} \\
& A=\int_{-2}^{0}{\left( x+x+4 \right)dx}+\int_{0}^{4}{\left( x-2x+4 \right)dx} \\
& A=\int_{-2}^{0}{\left( 2x+4 \right)dx}+\int_{0}^{4}{\left( 4-x \right)dx} \\
& A=\left[ {{x}^{2}}+4x \right]_{-2}^{0}+\left[ 4x-\frac{1}{2}{{x}^{2}} \right]_{0}^{4} \\
& A=\left[ 0 \right]-\left[ {{\left( -2 \right)}^{2}}+4\left( -2 \right) \right]+\left[ 4\left( 4 \right)-\frac{1}{2}{{\left( 4 \right)}^{2}} \right]-\left[ 0 \right] \\
& A=4+8 \\
& A=12 \\
\end{align}\]
