Answer
2
Work Step by Step
We are given that $f(x)=x^3$, interval $[0,2]$
Apply formula: $\overline{f}=\dfrac{1}{b-a}\int_a^b f(x) \ dx$
So, we have: $\overline{f}=\dfrac{1}{2-0}\int_0^2 x^3 \ dx=\dfrac{1}{2}\int_0^2 x^3 \ dx$
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 $\overline{f}=\dfrac{1}{2}\int_0^2 x^3 \ dx=\dfrac{1}{2}[\dfrac{x^4}{4}]_0^2$
or, $=\dfrac{1}{2}(4-0)$
or, $=2$
