Olhando o gráfico da função y, devemos separar a integral entre dois intervalos de modo a tomarmos o módulo da função no intervalo onde ela é negativa.
[tex]\int_{-1}^{3}[x(x+1)(x-3)]dx\\\\\int_{-1}^{3}[x(x^2-3x+x-3)]dx\\\\\int_{-1}^{3}(x^3-2x^2-3x)dx\\\\\int_{-1}^{0}(x^3-2x^2-3x)dx+\int_{0}^{3}|x^3-2x^2-3x|dx\\\\\frac{x^4}{4}-\frac{2}{3}x^3-\frac{3}{2}x^2\:\: \Big|_{-1}^{0}+|\frac{x^4}{4}-\frac{2}{3}x^3-\frac{3}{2}x^2|\:\: \Big|_{0}^{3}\\\\(\frac{0}{4}-\frac{2}{3}.0-\frac{3}{2}.0)-(\frac{(-1)^4}{4}-\frac{2}{3}(-1)^3-\frac{3}{2}(-1)^2)+(|\frac{3^4}{4}-\frac{2}{3}3^3-\frac{3}{2}3^2-|\frac{0}{4}-\frac{2}{3}.0-\frac{3}{2}.0|)[/tex]
[tex]-(\frac{1}{4}+\frac{2}{3}-\frac{3}{2})+|\frac{3^4}{4}-2.3^2-\frac{3^3}{2}|\\\\-(\frac{3+8-18}{12})+|\frac{3^5-24.3^2-6.3^3}{12}|\\\\\frac{7}{12}+|\frac{-135}{12}|\\\\\frac{142}{12}=\frac{71}{6}[/tex]
