Answer
$-1$
Work Step by Step
We are given that $f(x)=x^3-1$, $x=-2$ to $x=2$
Apply formula: $\overline{f}=\dfrac{1}{b-a}\int_a^b f(x) \ dx$
So, we have: $\overline{f}=\dfrac{1}{2-(-2)}\int_{-2}^2 (x^3-1) \ dx=\dfrac{1}{4} [\dfrac{x^4}{4}-x]_{-2}^2$
In order to solve the above integral, we will use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Now, we have $\overline{f}=\dfrac{1}{4} [\dfrac{(2)^4}{4}-2]-\dfrac{1}{4} [\dfrac{(-2)^4}{4}-(-2)]$
or, $=-1$
