Question

The value of $\underset{-2}{\overset{2}{\int }}|3x^2-3x-6|dx$ is

Answer 1

$x^2-x-2=(x-2)(x+1)$
$\underset{-2}{\overset{2}{\int }}|3x^2-3x-6|dx=3\underset{-2}{\overset{2}{\int }}|x^2-x-2|dx$
$=3\left\{\underset{-2}{\overset{-1}{\int }}(x^2-x-2)dx+\underset{-1}{\overset{2}{\int }}(-x^2+x+2)dx\right\}$
$=3[{(\frac{x^3}{3}-\frac{x^2}{2}-2x)}_{-2}^{-1}-{(\frac{x^3}{3}-\frac{x^2}{2}-2x)}_{-1}^2]$
$=19$