Answer
$$A = \frac{9}{2}$$
Work Step by Step
$$\eqalign{
& {x^2} = y,{\text{ }}x = y - 2 \Rightarrow {\text{ }}y = x + 2 \cr
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_{ - 1}^2 {\left( {x + 2 - {x^2}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{1}{2}{x^2} + 2x - \frac{1}{3}{x^3}} \right]_{ - 1}^2 \cr
& {\text{Evaluate}} \cr
& A = \left[ {\frac{1}{2}{{\left( 2 \right)}^2} + 2\left( 2 \right) - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - \left[ {\frac{1}{2}{{\left( { - 1} \right)}^2} + 2\left( { - 1} \right) - \frac{1}{3}{{\left( { - 1} \right)}^3}} \right] \cr
& {\text{Simplify}} \cr
& A = \frac{{10}}{3} + \frac{7}{6} \cr
& A = \frac{9}{2} \cr} $$
