Answer
\[\overline{f}=2\]
Work Step by Step
\[\begin{align}
& \text{From the graph we can define the region }R\text{ as:} \\
& R=\left\{ \left( x,y \right):0\le y\le 1-{{x}^{2}},\text{ }-1\le x\le 1\text{ } \right\} \\
& \text{The area of the region is } \\
& A=\int_{-1}^{1}{\left( 1-{{x}^{2}} \right)}dx \\
& A=\frac{4}{3} \\
& \text{The average of }f(x,y)\text{ on the region }R\text{ is} \\
& \overline{f}=\frac{1}{A}\iint_{R}{f\left( x,y \right)}dydx \\
& \overline{f}=\frac{3}{4}\int_{-1}^{1}{\int_{0}^{1-{{x}^{2}}}{\left( 2+x \right)}dy}dx \\
& \text{Integrate} \\
& \overline{f}=\frac{3}{4}\int_{-1}^{1}{\left[ \left( 2+x \right)y \right]_{0}^{1-{{x}^{2}}}}dx \\
& \overline{f}=\frac{3}{4}\int_{-1}^{1}{\left[ \left( 2+x \right)\left( 1-{{x}^{2}} \right) \right]}dx \\
& \overline{f}=\frac{3}{4}\int_{-1}^{1}{\left( 2-2{{x}^{2}}+x-{{x}^{3}} \right)}dx \\
& \overline{f}=\frac{3}{4}\left[ 2x-\frac{2}{3}{{x}^{3}}+\frac{1}{2}{{x}^{2}}-\frac{1}{4}{{x}^{4}} \right]_{-1}^{1} \\
& \overline{f}=\frac{3}{4}\left[ 2\left( 1 \right)-\frac{2}{3}{{\left( 1 \right)}^{3}}+\frac{1}{2}{{\left( 1 \right)}^{2}}-\frac{1}{4}{{\left( 1 \right)}^{4}} \right] \\
& -\frac{3}{4}\left[ 2\left( 1 \right)-\frac{2}{3}{{\left( 1 \right)}^{3}}+\frac{1}{2}{{\left( 1 \right)}^{2}}-\frac{1}{4}{{\left( 1 \right)}^{4}} \right] \\
& \overline{f}=\frac{19}{16}+\frac{13}{16} \\
& \overline{f}=\frac{32}{16} \\
& \overline{f}=2 \\
\end{align}\]
