Answer
$$A = 32$$
Work Step by Step
$$\eqalign{
& {\text{From the graph}} \cr
& {\text{We let: }}f\left( x \right) = 2{x^2} + 10x - 5{\text{ and }}g\left( x \right) = - {x^2} + 4x + 4 \cr
& {\text{ }}g\left( x \right) \geqslant f\left( x \right){\text{ on the interval }}\left[ { - 3,1} \right] \cr
& {\text{The area enclosed can define as}} \cr
& A = \int_{ - 3}^1 {\left[ {g\left( x \right) - f\left( x \right)} \right]} dx \cr
& A = \int_{ - 3}^1 {\left[ {\left( { - {x^2} + 4x + 4} \right) - \left( {2{x^2} + 10x - 5} \right)} \right]} dx \cr
& A = \int_{ - 3}^1 {\left( { - {x^2} + 4x + 4 - 2{x^2} - 10x + 5} \right)} dx \cr
& A = \int_{ - 3}^1 {\left( { - 3{x^2} - 6x + 9} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ { - {x^3} - 3{x^2} + 9x} \right]_{ - 3}^1 \cr
& A = \left[ { - {{\left( 1 \right)}^3} - 3{{\left( 1 \right)}^2} + 9\left( 1 \right)} \right] - \left[ { - {{\left( { - 3} \right)}^3} - 3{{\left( { - 3} \right)}^2} + 9\left( { - 3} \right)} \right] \cr
& A = \left( { - 1 - 3 + 9} \right) - \left( {81 - 27 - 27} \right) \cr
& A = 5 + 27 \cr
& A = 32 \cr} $$
