Answer
$$A = \frac{{95}}{{12}}$$
Work Step by Step
$$\eqalign{
& x = {y^3} - 4{y^2} + 3y,{\text{ }}x = {y^2} - y \cr
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_0^1 {\left[ {\left( {{y^3} - 4{y^2} + 3y} \right) - \left( {{y^2} - y} \right)} \right]} dy \cr
& {\text{ }} + \int_1^3 {\left[ {\left( {{y^2} - y} \right) - \left( {{y^3} - 4{y^2} + 3y} \right)} \right]} dy \cr
& {\text{Integrate}} \cr
& A = \int_0^1 {\left( {{y^3} - 5{y^2} + 4y} \right)} dy + \int_1^3 {\left( {5{y^2} - {y^3} - 4y} \right)} dy \cr
& {\text{Evaluating}} \cr
& A = \left[ {\frac{{{y^4}}}{4} - \frac{{5{y^3}}}{3} + 2{y^2}} \right]_0^1 + \left[ {\frac{{5{y^3}}}{3} - \frac{{{y^4}}}{4} - 2{y^2}} \right]_1^3 \cr
& {\text{Simplifying}} \cr
& A = \left[ {\frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{5{{\left( 1 \right)}^3}}}{3} + 2{{\left( 1 \right)}^2}} \right] + \left[ {\frac{{5{{\left( 3 \right)}^3}}}{3} - \frac{{{{\left( 3 \right)}^4}}}{4} - 2{{\left( 3 \right)}^2}} \right] \cr
& {\text{ }} - \left[ {\frac{{5{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( 1 \right)}^4}}}{4} - 2{{\left( 1 \right)}^2}} \right] \cr
& A = \frac{7}{{12}} + \frac{{27}}{4} + \frac{7}{{12}} \cr
& A = \frac{{95}}{{12}} \cr} $$
