Answer
$$A = \frac{9}{2}$$
Work Step by Step
$$\eqalign{
& x = {y^2} + 1,{\text{ }}x = y + 3,{\text{ }}y = 0 \cr
& {\text{From the graph we can note that }}y + 3 > {y^2} + 1{\text{ on }}\left[ { - 1,2} \right] \cr
& {\text{Then, the area is given by:}} \cr
& A = \int_{ - 1}^2 {\left[ {\left( {y + 3} \right) - \left( {{y^2} + 1} \right)} \right]} dy \cr
& A = \int_{ - 1}^2 {\left( {y + 3 - {y^2} - 1} \right)} dy \cr
& A = \int_{ - 1}^2 {\left( {y - {y^2} + 2} \right)} dy \cr
& {\text{Integrate and evaluate}} \cr
& A = \left[ {\frac{1}{2}{y^2} - \frac{1}{3}{y^3} + 2y} \right]_{ - 1}^2 \cr
& A = \left[ {\frac{1}{2}{{\left( 2 \right)}^2} - \frac{1}{3}{{\left( 2 \right)}^3} + 2\left( 2 \right)} \right] - \left[ {\frac{1}{2}{{\left( { - 1} \right)}^2} - \frac{1}{3}{{\left( { - 1} \right)}^3} + 2\left( { - 1} \right)} \right] \cr
& A = \left( {\frac{{10}}{3}} \right) - \left( { - \frac{7}{6}} \right) \cr
& A = \frac{9}{2} \cr} $$
