Answer
$ av(f)=\dfrac{1}{(b-a)} \int_a^b f(x) dx$
Work Step by Step
When the function $f(x)$ has minimum and maximum values defined on $[p,q]$ then the maximum-minimum inequality tells that
$min f(q-p) \leq \int_p^q f(x) dx \leq max f(q-p)$
Need to apply formula such as: $ \int_a^b c dx =c (b-a)$
The left hand side suggests that
$ av(f) (b-a)=\int_a^b f(x) dx$
This can be re-formed as:
$ av(f)=\dfrac{1}{(b-a)} \int_a^b f(x) dx$
So, the result is proved by the definition.
