Answer
$$A = \frac{{32}}{3}$$
Work Step by Step
$$\eqalign{
& \left( i \right){\text{From the graph we can see that the net area of the region is:}} \cr
& A = \int_{ - 2}^2 {\left( {4 - {x^2}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {4x - \frac{1}{3}{x^3}} \right]_{ - 2}^2 \cr
& {\text{Evaluating}} \cr
& A = \left[ {4\left( 2 \right) - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - \left[ {4\left( { - 2} \right) - \frac{1}{3}{{\left( { - 2} \right)}^3}} \right] \cr
& A = \frac{{16}}{3} + \frac{{16}}{3} \cr
& A = \frac{{32}}{3} \cr} $$
