## University Calculus: Early Transcendentals (3rd Edition)

$4$
We can calculate the area, by integrating the function from -2 to 2. $\int_{-2}^{2} (3x^2-3)dx =[x^3-3x]_{-2}^{2}=[2^3-3\cdot2]-[(-2)^3-3\cdot(-2)]=[8-6]-[-8+6]=2-(-2)=4$