Answer
$\dfrac{-1}{4}$
Work Step by Step
We are given that $f(x)=x^3-x$, interval $[0,1]$
Apply formula: $\overline{f}=\dfrac{1}{b-a}\int_a^b f(x) \ dx$
So, we have: $\overline{f}=\dfrac{1}{1-0}\int_{0}^1 (x^3-x) \ dx=\dfrac{1}{1}\int_{0}^1 (x^3-x) \ dx$
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 $\overline{f}=\dfrac{1}{1}\int_{0}^1 (x^3-x) \ dx=[\dfrac{x^4}{4}-\dfrac{x^2}{2}]_{0}^1$
or, $=(\dfrac{1}{4}-\dfrac{1}{2})-0$
or, $=\dfrac{-1}{4}$
