Question

The value of ${\int }_{-2}^2\left|3x^2-3x-6\right|dx$ is

Answer 1

Evaluate the integral.

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

Hence, the required value of the integral is $19$.