Answer
$$A = \frac{{23}}{3}$$
Work Step by Step
$$\eqalign{
& x = - 2,\,\,\,\,\,x = 1,\,\,\,\,\,y = 2x,\,\,\,\,\,\,y = {x^2} - 3 \cr
& {\text{Find the points of intersection of the graphs}} \cr
& 2x = {x^2} - 3 \cr
& {x^2} - 2x - 3 = 0 \cr
& \left( {x - 3} \right)\left( {x + 1} \right) = \cr
& {x_1} = - 1{\text{ and }}{x_2} = 3 \cr
& {\text{Then with the given points and the critical values we have the intervals}} \cr
& \left( { - 2, - 1} \right){\text{ and }}\left( { - 1,1} \right) \cr
& {\text{For the interval }}\left( { - 2,1} \right){\text{ we have that }}{x^2} - 3 \geqslant 2x{\text{ and for the interval }}\left( { - 1,1} \right) \cr
& 2x \geqslant {x^2} - 3 \cr
& {\text{The area is given by}}: \cr
& A = \int_{ - 2}^{ - 1} {\left[ {\left( {{x^2} - 3} \right) - 2x} \right]} dx + \int_{ - 1}^1 {\left[ {2x - \left( {{x^2} - 3} \right)} \right]} dx \cr
& A = \int_{ - 2}^{ - 1} {\left( {{x^2} - 2x - 3} \right)} dx + \int_{ - 1}^2 {\left( {2x - {x^2} + 3} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {\frac{{{x^3}}}{3} - {x^2} - 3x} \right)_{ - 2}^{ - 1} + \left( {{x^2} - \frac{{{x^3}}}{3} + 3x} \right)_{ - 1}^1 \cr
& A = \left( {\frac{{{{\left( { - 1} \right)}^3}}}{3} - {{\left( { - 1} \right)}^2} - 3\left( { - 1} \right)} \right) - \left( {\frac{{{{\left( { - 2} \right)}^3}}}{3} - {{\left( { - 2} \right)}^2} - 3\left( { - 2} \right)} \right) + \left( {{{\left( 1 \right)}^2} - \frac{{{{\left( 1 \right)}^3}}}{3} + 3\left( 1 \right)} \right) \cr
& \,\,\,\,\,\, - \,\left( {{{\left( { - 1} \right)}^2} - \frac{{{{\left( { - 1} \right)}^3}}}{3} + 3\left( { - 1} \right)} \right) \cr
& A = \frac{5}{3} + \frac{2}{3} + \frac{{11}}{3} + \frac{5}{3} \cr
& A = \frac{{23}}{3} \cr} $$
