Answer
$$A = \frac{1}{3}$$
Work Step by Step
$$\eqalign{
& x = {y^2} - 2y,{\text{ }}x = - 1,{\text{ }}y = 0 \cr
& {\text{From the graph we can note that }}{y^2} - 2y > - 1{\text{ on }}\left[ {0,1} \right] \cr
& {\text{Then, the area is given by:}} \cr
& A = \int_0^1 {\left( {{y^2} - 2y + 1} \right)} dy \cr
& {\text{Factor the trinomial}} \cr
& A = \int_0^1 {{{\left( {y - 1} \right)}^2}} dy \cr
& {\text{Integrate and evaluate}} \cr
& A = \left[ {\frac{{{{\left( {y - 1} \right)}^3}}}{3}} \right]_0^1 \cr
& A = \frac{1}{3}\left[ {{{\left( {1 - 1} \right)}^3} - {{\left( {0 - 1} \right)}^3}} \right] \cr
& A = \frac{1}{3}\left[ 1 \right] \cr
& A = \frac{1}{3} \cr} $$
