## Calculus 8th Edition

$$4$$
$f$ is continuous on $[1,3]$, so by the Mean Value Theorem for Integrals there exists a number $c$ in $[1,3]$ such that $\int^3 _1 f(x) dx = f(c) (3-1), 8 = 2f(c)$; that is, there is a number $c$ such that $f(c) = \frac{8}{2} = 4$