Answer
$0.0665$
Work Step by Step
\[\begin{align}
& y={{e}^{x}}\text{ and }y=-{{x}^{2}}-2x \\
& \text{Find the intersection points let }y=y \\
& {{e}^{x}}=-{{x}^{2}}-2x \\
& \text{Using a graphing calculator we obtain:} \\
& {{x}_{1}}=-1.9241,\text{ }{{x}_{2}}=-0.4164 \\
& -{{x}^{2}}-2x\ge {{e}^{x}}\text{ on the interval }\left( -1.9241,-0.4164 \right) \\
& \text{The area between the curves is:} \\
& A=\int_{-0.4164}^{-1.9241}{\left( -{{x}^{2}}-2x-{{e}^{x}} \right)}dx \\
& \text{Integrating} \\
& A=\left[ -\frac{{{x}^{3}}}{3}-{{x}^{2}}-{{e}^{x}} \right]_{-0.4164}^{-1.9241} \\
& A=\left[ -\frac{{{\left( -1.9241 \right)}^{3}}}{3}-{{\left( -1.9241 \right)}^{2}}-{{e}^{-1.9241}} \right]- \\
& \left[ -\frac{{{\left( -0.4164 \right)}^{3}}}{3}-{{\left( -0.4164 \right)}^{2}}-{{e}^{-0.4164}} \right] \\
& \text{Simplifying} \\
& A\approx 0.0665 \\
\end{align}\]
