Answer
\[A=\frac{9}{8}\]
Work Step by Step
\[\begin{align}
& \text{Let }y=1+x-{{x}^{2}}\text{ and }y={{x}^{2}} \\
& \text{From the graph, the area is given by:} \\
& A=\int_{-1/2}^{1}{\int_{{{x}^{2}}}^{1+x-{{x}^{2}}}{dy}dx} \\
& \text{Integrating} \\
& A=\int_{-1/2}^{1}{\left( 1+x-{{x}^{2}}-{{x}^{2}} \right)dx} \\
& A=\int_{-1/2}^{1}{\left( 1+x-2{{x}^{2}} \right)dx} \\
& A=\left[ x+\frac{1}{2}{{x}^{2}}-\frac{2}{3}{{x}^{3}} \right]_{-1/2}^{1} \\
& A=\left[ \left( 1 \right)+\frac{1}{2}{{\left( 1 \right)}^{2}}-\frac{2}{3}{{\left( 1 \right)}^{3}} \right]-\left[ \left( -\frac{1}{2} \right)+\frac{1}{2}{{\left( -\frac{1}{2} \right)}^{2}}-\frac{2}{3}{{\left( -\frac{1}{2} \right)}^{3}} \right] \\
& A=\frac{5}{6}+\frac{7}{24} \\
& A=\frac{9}{8} \\
\end{align}\]
