Answer
$4$
Work Step by Step
The Mean Value Theorem for Integrals states: If $ f $ is continuous on $[a,b]$, then there exists a number c in $[a,b]$ such that:
$$
f(c)=f_{\text {ave }}=\frac{1}{b-a} \int_{a}^{b} f(x) d x
$$
Since $ f $ is continuous on $[1,3]$, and $\int_{1}^{3} f(x) d x =8$ then there exists a number $ c $ in $[1,3]$ such that
$$
\int_{1}^{3} f(x) d x=f(c)(3-1)=2f(c)=8
$$
$ \Rightarrow $
$$
f(c)=\frac{8}{2}=4
$$
