Answer
(a) $f_{avg} = \frac{1}{3}$
(b) $c=\sqrt 3$
Work Step by Step
$f(t) = \frac{1}{t^2}$ at interval $[1,3]$
(a). Finding the Average
$f_{avg} = \frac{1}{b-a} $ $\int $ $f(x) dx$
$f_{avg} = \frac{1}{3-1}$ $\int_{1}^{3}\frac{1}{t^2}\,dt$
= $\frac{1}{2}$ $\int_{1}^{3}\frac{-1}{t}\, dt$
= $\frac{1}{2} [ \frac{-1}{3} - [-1]] $
$f_{avg} = \frac{1}{3}$
(b) Find $c$ in the given interval such that $f_{avg} = f(c)$
$\frac{1}{3} = (\frac{1}{c})^2$
$\frac{1}{3}$ = $\frac{1}{c^2}$
$3 = c^2$
$c = \sqrt 3$ ($c=-\sqrt 3$ is not in $[1,3]$)
