Answer
$$A = \frac{3}{5}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_0^{2/5} {\left( {4x - x} \right)} dx + \int_{2/5}^1 {\left( { - x + 2 - x} \right)} dx \cr
& A = \int_0^{2/5} {3x} dx + \int_{2/5}^1 {\left( {2 - 2x} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ {\frac{3}{2}{x^2}} \right]_0^{2/5} + \left[ {2x - {x^2}} \right]_{2/5}^1 \cr
& {\text{Evaluating}} \cr
& A = \frac{3}{2}{\left( {\frac{2}{5}} \right)^2} + \left[ {2\left( 1 \right) - {{\left( 1 \right)}^2}} \right] - \left[ {2\left( {\frac{2}{5}} \right) - {{\left( {\frac{2}{5}} \right)}^2}} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{6}{{25}} + 1 - \frac{{16}}{{25}} \cr
& A = \frac{3}{5} \cr} $$
