\(A₁= \int\limits_{0}^{1}x*(x-1)*(3-x)*dx=\int\limits_{0}^{1}(4x^2-x^3-3x)*dx=[\frac{4}{3}x^3-\frac{1}{4}x^4-\frac{3}{2}x^2]\)
\(A₁=[\frac{4}{3}-\frac{1}{4}-\frac{3}{2}]-0=-\frac{5}{12}\)
\(A₁=|-\frac{5}{12}|=\frac{5}{12}\)
\(A₂=\int\limits_{1}^{3}(4x^2-x^3-3x)*dx=[\frac{4}{3}x^3-\frac{1}{4}x^4-\frac{3}{2}x^2]\)
\(A₂=[36-\frac{81}{4}-\frac{27}{2}]-[-\frac{5}{12}]\)
\(A₂=36-\frac{81}{4}-\frac{27}{2}+\frac{5}{12}=\frac{8}{3}\)
\(A=A₁+A₂=\frac{5}{12}+\frac{32}{12}=\frac{37}{12}\)