Answer
$$A = 1$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = 1 - \left| {\frac{x}{2} - 1} \right| \cr
& {\text{By the definition of absolute value }} \cr
& \left| {\frac{x}{2} - 1} \right| = - \left( {\frac{x}{2} - 1} \right){\text{ for }}\frac{x}{2} - 1 < 0 \cr
& \left| {\frac{x}{2} - 1} \right| = \frac{x}{2} - 1{\text{ for }}\frac{x}{2} - 1 \geqslant 0 \cr
& {\text{Therefore}} \cr
& 1 - \left| {\frac{x}{2} - 1} \right| = \frac{x}{2}{\text{ for }}x < 2 \cr
& 1 - \left| {\frac{x}{2} - 1} \right| = 2 - \frac{x}{2}{\text{ for }}x \geqslant 2 \cr
& {\text{The graph of the enclosed area is shown below}} \cr
& {\text{The enclosed area is given by }} \cr
& A = \int_0^2 {\left( {\frac{x}{2} - \frac{x}{6}} \right)} dx + \int_2^3 {\left( {2 - \frac{x}{2} - \frac{x}{6}} \right)} dx \cr
& {\text{Simplify the integrands}} \cr
& A = \int_0^2 {\frac{x}{3}} dx + \int_2^3 {\left( {2 - \frac{{2x}}{3}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{{{x^2}}}{6}} \right]_0^2 + \left[ {2x - \frac{{{x^2}}}{3}} \right]_2^3 \cr
& {\text{Evaluate and simplify}} \cr
& A = \left[ {\frac{{{{\left( 2 \right)}^2}}}{6} - \frac{{{{\left( 0 \right)}^2}}}{6}} \right] + \left[ {2\left( 3 \right) - \frac{{{{\left( 3 \right)}^2}}}{3}} \right] - \left[ {2\left( 2 \right) - \frac{{{{\left( 2 \right)}^2}}}{3}} \right] \cr
& A = \frac{2}{3} + 3 - \frac{8}{3} \cr
& A = 1 \cr} $$
