Question

The value of $\int_{-2}^{2} |3x^2-3x-6| \,dx$ is ________ .

Hint:

When integrating an absolute value function, $|f(x)|$, the first step is always to find the roots of $f(x)=0$. Then, split the integral at these roots and remove the absolute value signs by considering the sign of $f(x)$ in each sub-interval.

Answer 1

Correct Answer: 19

\[ \int_{-2}^{2}|3x^2-3x-6|dx \] Roots: \[ 3(x-2)(x+1)=0 \Rightarrow x=-1,2 \] Split integral: \[ =\int_{-2}^{-1}(3x^2-3x-6)dx+\int_{-1}^{2}-(3x^2-3x-6)dx \] Antiderivative: \[ x^3-\frac32x^2-6x \] Evaluating: \[ =\frac{11}{2}+\frac{27}{2}=19 \] \[ \boxed{19} \]