Answer
$$A = \frac{7}{3}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph}} \cr
& {\text{We let: }}f\left( x \right) = - \left| x \right|{\text{ and }}g\left( x \right) = {x^2} - 2 \cr
& {\text{ }}f\left( x \right) \geqslant {\text{ }}g\left( x \right)\,{\text{ on the interval }}\left[ { - 1,1} \right] \cr
& A = \int_{ - 1}^1 {\left( { - \left| x \right| - \left[ {{x^2} - 2} \right]} \right)dx} \cr
& {\text{Using the symmetry of the graph we can define the area as}} \cr
& A = 2\int_0^1 {\left( { - x - \left[ {{x^2} - 2} \right]} \right)} dx \cr
& A = 2\int_0^1 {\left( { - x - {x^2} + 2} \right)} dx \cr
& {\text{Integrate}} \cr
& A = 2\left[ { - \frac{{{x^2}}}{2} - \frac{{{x^3}}}{3} + 2x} \right]_0^1 \cr
& A = 2\left[ { - \frac{{{{\left( 1 \right)}^2}}}{2} - \frac{{{{\left( 1 \right)}^3}}}{3} + 2\left( 1 \right)} \right] - 2\left[ { - \frac{{{{\left( 0 \right)}^2}}}{2} - \frac{{{{\left( 0 \right)}^3}}}{3} + 2\left( 0 \right)} \right] \cr
& A = 2\left[ {\frac{7}{6}} \right] - 2\left[ 0 \right] \cr
& A = \frac{7}{3} \cr} $$
