Answer
$12$
Work Step by Step
Step 1. Given the function $y=3x^2-3=3(x+1)(x-1)$ and the interval $[-2,2]$, find the zeros at $x=-1,1$; thus we need to divide the integration into three regions: $[-2,-1],[-1,1],[1,2]$
Step 2. We have $A_1=\int_{-2}^{-1}(3x^2-3)dx=(x^3-3x)|_{-2}^{-1}=((-1)^3-3(-1))-((-2)^3-3(-2))=4$
Step 3. Similarly, $A_2=\int_{-1}^{1}(3x^2-3)dx=(x^3-3x)|_{-1}^{1}=((1)^3-3(1))-((-1)^3-3(-1))=-4$
Step 4. For the third part, $A_3=\int_{1}^{2}(3x^2-3)dx=(x^3-3x)|_{1}^{2}=((2)^3-3(2))-((1)^3-3(1))=4$
Step 5. The total area is given by $A=|A_1|+|A_2+|A_3|=12$
