Answer
$$A = 39$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = 2{x^2} + 10x - 5,{\text{ }}g\left( x \right) = - {x^2} + 4x + 4 \cr
& {\text{From the graph shown below}} \cr
& g\left( x \right) \geqslant f\left( x \right){\text{ on the interval }} - {\text{3}} \leqslant x \leqslant 1 \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}1 \leqslant x \leqslant 2 \cr
& {\text{The area is given by}} \cr
& A = \int_{ - 3}^1 {\left[ {\left( { - {x^2} + 4x + 4} \right) - \left( {2{x^2} + 10x - 5} \right)} \right]} dx \cr
& {\text{ }} + \int_1^2 {\left[ {\left( {2{x^2} + 10x - 5} \right) - \left( { - {x^2} + 4x + 4} \right)} \right]} dx \cr
& A = \int_{ - 3}^1 {\left( { - {x^2} + 4x + 4 - 2{x^2} - 10x + 5} \right)} dx \cr
& {\text{ }} + \int_1^2 {\left( {2{x^2} + 10x - 5 + {x^2} - 4x - 4} \right)} dx \cr
& A = \int_{ - 3}^1 {\left( { - 3{x^2} - 6x + 9} \right)} dx + \int_1^2 {\left( {3{x^2} + 6x - 9} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ { - {x^3} - 3{x^2} + 9x} \right]_{ - 3}^1 + \left[ {{x^3} + 3{x^2} - 9x} \right]_1^2 \cr
& A = \left[ { - {{\left( 1 \right)}^3} - 3{{\left( 1 \right)}^2} + 9\left( 1 \right)} \right] - \left[ { - {{\left( { - 3} \right)}^3} - 3{{\left( { - 3} \right)}^2} + 9\left( { - 3} \right)} \right] \cr
& + \left[ {{{\left( 2 \right)}^3} + 3{{\left( 2 \right)}^2} - 9\left( 2 \right)} \right] - \left[ {{{\left( 1 \right)}^3} + 3{{\left( 1 \right)}^2} - 9\left( 1 \right)} \right] \cr
& {\text{Simplifying}} \cr
& A = 5 + 27 + 2 + 5 \cr
& A = 39 \cr} $$
