Answer
See explanation.
Work Step by Step
Since the graph of $f$ is a linear function, the area between $f$ and the $x$ -axis has a trapezoidal shape and is $\frac{(-a+b) \cdot (f(b)+f(a)) }{2}$
The average value of $f$ on $[a, b]$ is $f_{a v \epsilon}= \frac{(f(a)+f(b)) \cdot(-a+b)}{2} \cdot \frac{1}{-a+b} =$
$ \frac{(f(b)+f(a)) }{2}$
Since $f$ is a linear function, we have $f_{a v g}=\frac{f(b)+f(a)}{2}=f\left(\frac{b+a}{2}\right)$