1. Simplify the expression inside modulus

We have

$$I=\int_{-2}^{2} |3x^2-3x-6|\,dx$$

Factor the quadratic:

$$3x^2-3x-6=3(x^2-x-2)=3(x-2)(x+1)$$

So,

$$I=\int_{-2}^{2} 3|(x-2)(x+1)|\,dx$$

2. Find where the expression changes sign

The roots are

$$x=-1,\quad x=2$$

On the interval $[-2,2]$:

• For $x\in[-2,-1]$, $(x-2)(x+1)>0$
• For $x\in[-1,2]$, $(x-2)(x+1)<0$

Hence,

$$|3x^2-3x-6|= \begin{cases} 3x^2-3x-6, & -2\le x\le -1 \\ -(3x^2-3x-6)=-3x^2+3x+6, & -1\le x\le 2 \end{cases}$$

3. Split the integral

Thus,

$$I=\int_{-2}^{-1} (3x^2-3x-6)\,dx+\int_{-1}^{2} (-3x^2+3x+6)\,dx$$

4. Evaluate the first integral

An antiderivative of $3x^2-3x-6$ is

$$x^3-\frac{3x^2}{2}-6x$$

So,

$$I_1=\left[x^3-\frac{3x^2}{2}-6x\right]_{-2}^{-1}$$

At $x=-1$:

$$-1-\frac{3}{2}+6=\frac{7}{2}$$

At $x=-2$:

$$-8-6+12=-2$$

Therefore,

$$I_1=\frac{7}{2}-(-2)=\frac{11}{2}$$

5. Evaluate the second integral

An antiderivative of $-3x^2+3x+6$ is

$$-x^3+\frac{3x^2}{2}+6x$$

So,

$$I_2=\left[-x^3+\frac{3x^2}{2}+6x\right]_{-1}^{2}$$

At $x=2$:

$$-8+6+12=10$$

At $x=-1$:

$$1+\frac{3}{2}-6=-\frac{7}{2}$$

Therefore,

$$I_2=10-\left(-\frac{7}{2}\right)=\frac{27}{2}$$

6. Add both parts

$$I=I_1+I_2=\frac{11}{2}+\frac{27}{2}=\frac{38}{2}=19$$

Therefore,

$$\int_{-2}^{2}|3x^2-3x-6|\,dx=19$$