Answer
$$A = \frac{9}{2}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_{ - 1}^2 {\left( {{x^2} + 1 - x} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{1}{3}{x^3} - \frac{1}{2}{x^2} + x} \right]_{ - 1}^2 \cr
& {\text{Evaluate }} \cr
& A = \left[ {\frac{1}{3}{{\left( 2 \right)}^3} - \frac{1}{2}{{\left( 2 \right)}^2} + \left( 2 \right)} \right] - \left[ {\frac{1}{3}{{\left( { - 1} \right)}^3} - \frac{1}{2}{{\left( { - 1} \right)}^2} + \left( { - 1} \right)} \right] \cr
& {\text{Simplify}} \cr
& A = \frac{8}{3} + \frac{{11}}{6} \cr
& A = \frac{9}{2} \cr} $$
