Answer
$$A = \frac{\pi }{2} - \frac{1}{3}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{1 + {x^2}}},{\text{ }}g\left( x \right) = \frac{1}{2}{x^2} \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }} - {\text{1}} \leqslant x \leqslant 1 \cr
& {\text{From the graph shown below, we can define the area enclosed}} \cr
& {\text{as:}} \cr
& A = \int_{ - 1}^1 {\left( {\frac{1}{{1 + {x^2}}} - \frac{1}{2}{x^2}} \right)} dx \cr
& {\text{By symmetry}} \cr
& A = 2\int_0^1 {\left( {\frac{1}{{1 + {x^2}}} - \frac{1}{2}{x^2}} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = 2\left[ {\arctan \left( x \right) - \frac{1}{6}{x^3}} \right]_0^1 \cr
& A = 2\left[ {\arctan \left( 1 \right) - \frac{1}{6}{{\left( 1 \right)}^3}} \right] - 2\left[ {\arctan \left( 0 \right) - \frac{1}{6}{{\left( 0 \right)}^3}} \right] \cr
& A = 2\left( {\frac{\pi }{4} - \frac{1}{6}} \right) \cr
& A = \frac{\pi }{2} - \frac{1}{3} \cr} $$
