Answer
Yes, see explanations.
Work Step by Step
Step 1. By definition, the average value of an integrable function $f(x)$ over an interval of $[a,b]$ can be written as $\bar f=av(f)=\frac{1}{b-a}\int_a^b f(x)dx$.
Step 2. Given that the interval length is $2$, we have $b-a=2$ and $\bar f=av(f)=\frac{1}{2}\int_a^b f(x)dx$ which equals to half of the function's integral over the interval.
