Answer
$$A = \frac{{16}}{3}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = {x^2} - 4x + 2,{\text{ }}g\left( x \right) = - {x^2} + 4x - 4 \cr
& {\text{From the graph shown below}} \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}0 \leqslant x \leqslant 1 \cr
& g\left( x \right) \geqslant f\left( x \right){\text{ on the interval 1}} \leqslant x \leqslant 3 \cr
& {\text{The area is given by}} \cr
& A = \int_0^1 {\left[ {\left( {{x^2} - 4x + 2} \right) - \left( { - {x^2} + 4x - 4} \right)} \right]} dx \cr
& {\text{ }} + \int_1^3 {\left[ {\left( { - {x^2} + 4x - 4} \right) - \left( {{x^2} - 4x + 2} \right)} \right]} dx \cr
& A = \int_0^1 {\left( {{x^2} - 4x + 2 + {x^2} - 4x + 4} \right)} dx \cr
& {\text{ }} + \int_1^3 {\left( { - {x^2} + 4x - 4 - {x^2} + 4x - 2} \right)} dx \cr
& A = \int_0^1 {\left( {2{x^2} - 8x + 6} \right)} dx + \int_1^3 {\left( {8x - 6 - 2{x^2}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{{2{x^3}}}{3} - 4{x^2} + 6x} \right]_0^1 + \left[ {4{x^2} - 6x - \frac{{2{x^3}}}{3}} \right]_1^3 \cr
& A = \left[ {\frac{{2{{\left( 1 \right)}^3}}}{3} - 4{{\left( 1 \right)}^2} + 6\left( 1 \right)} \right] + \left[ {4{{\left( 3 \right)}^2} - 6\left( 3 \right) - \frac{{2{{\left( 3 \right)}^3}}}{3}} \right] \cr
& - \left[ {4{{\left( 1 \right)}^2} - 6\left( 1 \right) - \frac{{2{{\left( 1 \right)}^3}}}{3}} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{8}{3} + 0 + \frac{8}{3} \cr
& A = \frac{{16}}{3} \cr} $$
