Answer
$A = \int_1^2 {\left( {3x - {x^2} - 2} \right)} dx$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = 2 - x{\text{ and }}y = 2x - {x^2} \cr
& {\text{Find the intersection points between the curves}} \cr
& y = y \cr
& 2 - x = 2x - {x^2} \cr
& {x^2} - x - 2x + 2 = 0 \cr
& {x^2} - 3x + 2 = 0 \cr
& {\text{Factoring}} \cr
& \left( {x - 2} \right)\left( {x - 1} \right) = 0 \cr
& {x_1} = 1{\text{ and }}{x_2} = 2 \cr
& {\text{We obtain the interval }}\left[ {1,2} \right] \cr
& {\text{The enclosed area is shown in the graph below}}{\text{.}} \cr
& 2x - {x^2} \geqslant 2 - x{\text{ on the interval }}\left[ {1,2} \right] \cr
& {\text{Therefore}} \cr
& {\text{The area is given by}} \cr
& A = \int_1^2 {\left[ {\left( {2x - {x^2}} \right) - \left( {2 - x} \right)} \right]} dx \cr
& A = \int_1^2 {\left( {2x - {x^2} - 2 + x} \right)} dx \cr
& A = \int_1^2 {\left( {3x - {x^2} - 2} \right)} dx \cr} $$
