Answer
$$A = \frac{1}{3}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph}} \cr
& {\text{We let: }}f\left( x \right) = {x^2} - 2x{\text{ and }}g\left( x \right) = - {x^2} + 4x - 4 \cr
& {\text{ }}g\left( x \right) \geqslant f\left( x \right){\text{ on the interval }}\left[ {1,2} \right] \cr
& {\text{The area enclosed can define as}} \cr
& A = \int_1^2 {\left[ {g\left( x \right) - f\left( x \right)} \right]} dx \cr
& A = \int_1^2 {\left[ {\left( { - {x^2} + 4x - 4} \right) - \left( {{x^2} - 2x} \right)} \right]} dx \cr
& A = \int_1^2 {\left( { - {x^2} + 4x - 4 - {x^2} + 2x} \right)} dx \cr
& A = \int_1^2 {\left( { - 2{x^2} + 6x - 4} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ { - \frac{{2{x^3}}}{3} + 3{x^2} - 4x} \right]_1^2 \cr
& A = \left[ { - \frac{{2{{\left( 2 \right)}^3}}}{3} + 3{{\left( 2 \right)}^2} - 4\left( 2 \right)} \right] - \left[ { - \frac{{2{{\left( 1 \right)}^3}}}{3} + 3{{\left( 1 \right)}^2} - 4\left( 1 \right)} \right] \cr
& A = \left( { - \frac{4}{3}} \right) - \left( { - \frac{5}{3}} \right) \cr
& A = \frac{1}{3} \cr} $$
