Answer
$$A = \frac{{37}}{{12}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x\left( {{x^2} - 3x + 3} \right),{\text{ }}g\left( x \right) = {x^2} \cr
& {\text{From the graph shown below}} \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}0 \leqslant x \leqslant 1 \cr
& g\left( x \right) \geqslant f\left( x \right){\text{ on the interval }}1 \leqslant x \leqslant 3 \cr
& {\text{The enclosed area is given by }} \cr
& A = \int_0^1 {\left[ {x\left( {{x^2} - 3x + 3} \right) - {x^2}} \right]dx + \int_1^3 {\left[ {{x^2} - x\left( {{x^2} - 3x + 3} \right)} \right]} } dx \cr
& A = \int_0^1 {\left( {{x^3} - 3{x^2} + 3x - {x^2}} \right)dx + \int_1^3 {\left( {{x^2} - {x^3} + 3{x^2} - 3x} \right)} } dx \cr
& A = \int_0^1 {\left( {{x^3} - 4{x^2} + 3x} \right)dx + \int_1^3 {\left( {4{x^2} - {x^3} - 3x} \right)} } dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{{{x^4}}}{4} - \frac{{4{x^3}}}{3} + \frac{{3{x^2}}}{2}} \right]_0^1 + \left[ {\frac{{4{x^3}}}{3} - \frac{{{x^4}}}{4} - \frac{{3{x^2}}}{2}} \right]_1^3 \cr
& A = \left[ {\frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{4{{\left( 1 \right)}^3}}}{3} + \frac{{3{{\left( 1 \right)}^2}}}{2}} \right] + \left[ {\frac{{4{{\left( 3 \right)}^3}}}{3} - \frac{{{{\left( 3 \right)}^4}}}{4} - \frac{{3{{\left( 3 \right)}^2}}}{2}} \right] \cr
& - \left[ {\frac{{4{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{3{{\left( 1 \right)}^2}}}{2}} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{5}{{12}} + \frac{9}{4} + \frac{5}{{12}} \cr
& A = \frac{{37}}{{12}} \cr} $$
