Answer
$$A = \frac{{71}}{6}$$
Work Step by Step
$$\eqalign{
& {\text{The area of the region is given by}} \cr
& A = \int_0^1 {\left[ {\left( {{y^3} - 4{y^2} + 3y} \right) - \left( {{y^2} - y} \right)} \right]d} y \cr
& \,\,\,\,\, + \int_1^4 {\left[ {\left( {{y^2} - y} \right) - \left( {{y^3} - 4{y^2} + 3y} \right)} \right]} dy \cr
& {\text{Simplifying the integrand}} \cr
& A = \int_0^1 {\left( {{y^3} - 4{y^2} + 3y - {y^2} + y} \right)} dy + \int_1^4 {\left( {5{y^2} - y - {y^3} - 3y} \right)} dy \cr
& A = \int_0^1 {\left( {{y^3} - 5{y^2} + 4y} \right)} dy + \int_1^4 {\left( {5{y^2} - {y^3} - 4y} \right)} dy \cr
& {\text{Integrating}} \cr
& A = \left[ {\frac{{{y^4}}}{4} - \frac{{5{y^3}}}{3} + 2{y^2}} \right]_0^1 + \left[ {\frac{{5{y^3}}}{3} - \frac{{{y^4}}}{4} - 2{y^2}} \right]_1^4 \cr
& A = \left[ {\frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{5{{\left( 1 \right)}^3}}}{3} + 2{{\left( 1 \right)}^2}} \right] - 0 \cr
& \,\,\,\,\,\, + \left[ {\frac{{5{{\left( 4 \right)}^3}}}{3} - \frac{{{{\left( 4 \right)}^4}}}{4} - 2{{\left( 4 \right)}^2}} \right] - \left[ {\frac{{5{{\left( 1 \right)}^3}}}{3} - \frac{{{{\left( 1 \right)}^4}}}{4} - 2{{\left( 1 \right)}^2}} \right] \cr
& A = \frac{7}{{12}} - 0 + \frac{{32}}{3} + \frac{7}{{12}} \cr
& A = \frac{{71}}{6} \cr} $$