Answer
$$A = \frac{{37}}{{12}}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_0^1 {\left( {{x^3} - 4{x^2} + 3x} \right)} dx + \int_1^3 {\left( {4{x^2} - 3x - {x^3}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{1}{4}{x^4} - \frac{4}{3}{x^3} + \frac{3}{2}{x^2}} \right]_0^1 + \left[ {\frac{4}{3}{x^3} - \frac{1}{4}{x^4} - \frac{3}{2}{x^2}} \right]_1^3 \cr
& {\text{Evaluating}} \cr
& A = \left( {\frac{1}{4} - \frac{4}{3} + \frac{3}{2}} \right) + \left[ {\frac{4}{3}{{\left( 3 \right)}^3} - \frac{1}{4}{{\left( 3 \right)}^4} - \frac{3}{2}{{\left( 3 \right)}^2}} \right] \cr
& {\text{ }} - \left[ {\frac{4}{3}{{\left( 1 \right)}^3} - \frac{1}{4}{{\left( 1 \right)}^4} - \frac{3}{2}{{\left( 1 \right)}^2}} \right] \cr
& A = \frac{5}{{12}} + \frac{9}{4} + \frac{5}{{12}} \cr
& A = \frac{{37}}{{12}} \cr} $$
