Answer
$$A = \frac{2}{3}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {\left( {x - 1} \right)^2},{\text{ }}g\left( x \right) = - {\left( {x - 1} \right)^2} \cr
& {\text{From the graph shown below}} \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}0 \leqslant x \leqslant 1 \cr
& {\text{The area is given by}} \cr
& A = \int_0^1 {\left[ {{{\left( {x - 1} \right)}^2} + {{\left( {x - 1} \right)}^2}} \right]} dx \cr
& A = 2\int_0^1 {{{\left( {x - 1} \right)}^2}} dx \cr
& {\text{ Integrate}} \cr
& A = 2\left[ {\frac{{{{\left( {x - 1} \right)}^3}}}{3}} \right]_0^1 \cr
& A = \frac{2}{3}\left[ {{{\left( {x - 1} \right)}^3}} \right]_0^1 \cr
& {\text{Simplifying}} \cr
& A = \frac{2}{3}\left[ {{{\left( {1 - 1} \right)}^3} - {{\left( {0 - 1} \right)}^3}} \right] \cr
& A = \frac{2}{3}\left( 1 \right) \cr
& A = \frac{2}{3} \cr} $$
