Answer
$32$
Work Step by Step
\[\begin{align}
& \text{From the graph we can define the area} \\
& A=\int_{0}^{2}{\left( 5-{{x}^{2}}-{{x}^{2}}+3 \right)}dx+\int_{2}^{4}{\left( {{x}^{2}}-3-5+{{x}^{2}} \right)dx} \\
& A=\int_{0}^{2}{\left( 8-2{{x}^{2}} \right)}dx+\int_{2}^{4}{\left( 2{{x}^{2}}-8 \right)dx} \\
& \text{Integrating} \\
& A=\left[ 8x-\frac{2}{3}{{x}^{3}} \right]_{0}^{2}+\left[ \frac{2}{3}{{x}^{3}}-8x \right]_{2}^{4} \\
& \text{Evaluating we obtain} \\
& A=\left[ 8\left( 2 \right)-\frac{2}{3}{{\left( 2 \right)}^{3}} \right]+\left[ \frac{2}{3}{{\left( 4 \right)}^{3}}-8\left( 4 \right) \right]-\left[ \frac{2}{3}{{\left( 2 \right)}^{3}}-8\left( 2 \right) \right] \\
& A=\frac{32}{3}+\frac{32}{3}+\frac{32}{3} \\
& A=32 \\
\end{align}\]