Answer
\[A=\frac{9}{2}\]
Work Step by Step
\[\begin{align}
& \text{From the graph} \\
& R=\left\{ \left( x,y \right):{{x}^{2}}\le y\le x+2,\text{ }-\text{1}\le x\le 2\text{ } \right\} \\
& \text{Then,} \\
& A=\int_{-1}^{2}{\int_{{{x}^{2}}}^{x+2}{dy}dx} \\
& \text{Integrate} \\
& A=\int_{-1}^{2}{\left[ y \right]_{{{x}^{2}}}^{x+2}dx} \\
& A=\int_{-1}^{2}{\left( x+2-{{x}^{2}} \right)dx} \\
& A=\left[ \frac{1}{2}{{x}^{2}}+2x-\frac{1}{3}{{x}^{3}} \right]_{-1}^{2} \\
& A=\left[ \frac{1}{2}{{\left( 2 \right)}^{2}}+2\left( 2 \right)-\frac{1}{3}{{\left( 2 \right)}^{3}} \right]-\left[ \frac{1}{2}{{\left( -1 \right)}^{2}}+2\left( -1 \right)-\frac{1}{3}{{\left( -1 \right)}^{3}} \right] \\
& A=\frac{10}{3}+\frac{7}{6} \\
& A=\frac{9}{2} \\
\end{align}\]
