Answer
\[A=\frac{9}{2}\]
Work Step by Step
\[\begin{align}
& \text{From the graph the region R is given by} \\
& x+2\le y\le -{{x}^{2}}+4 \\
& -2\le x\le 1 \\
& \text{Using an iterated integral we can define the area as:} \\
& A=\int_{-2}^{1}{\int_{x+2}^{-{{x}^{2}}+4}{dy}dx} \\
& \text{Integrating} \\
& A=\int_{-2}^{1}{\left( -{{x}^{2}}+4-x-2 \right)dx} \\
& A=\int_{-2}^{1}{\left( -{{x}^{2}}-x+2 \right)dx} \\
& A=\left[ -\frac{1}{3}{{x}^{3}}-\frac{1}{2}{{x}^{2}}+2x \right]_{-2}^{1} \\
& A=\left[ -\frac{1}{3}{{\left( 1 \right)}^{3}}-\frac{1}{2}{{\left( 1 \right)}^{2}}+2\left( 1 \right) \right]-\left[ -\frac{1}{3}{{\left( -2 \right)}^{3}}-\frac{1}{2}{{\left( -2 \right)}^{2}}+2\left( -2 \right) \right] \\
& A=\frac{7}{6}+\frac{10}{3} \\
& A=\frac{9}{2} \\
\end{align}\]
