Answer
$$A = \frac{{64}}{3}$$
Work Step by Step
$$\eqalign{
& y = 6 - \frac{1}{2}{x^2},{\text{ }}y = \frac{3}{4}x,{\text{ }}x = - 2,{\text{ }}x = 2 \cr
& {\text{From the graph we can note that }}6 - \frac{1}{2}{x^2} \geqslant \frac{3}{4}x{\text{ on }}\left[ { - 2,2} \right] \cr
& {\text{Then, the area is given by:}} \cr
& A = \int_{ - 2}^2 {\left( {6 - \frac{1}{2}{x^2} - \frac{3}{4}x} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = \left[ {6x - \frac{1}{6}{x^3} - \frac{3}{8}{x^2}} \right]_{ - 2}^2 \cr
& A = \left[ {6\left( 2 \right) - \frac{1}{6}{{\left( 2 \right)}^3} - \frac{3}{8}{{\left( 2 \right)}^2}} \right] - \left[ {6\left( { - 2} \right) - \frac{1}{6}{{\left( { - 2} \right)}^3} - \frac{3}{8}{{\left( { - 2} \right)}^2}} \right] \cr
& A = \frac{{55}}{6} + \frac{{73}}{6} \cr
& A = \frac{{64}}{3} \cr} $$
