Answer
$${f_{{\text{ave}}}} = \frac{\pi }{{12\left( {\sqrt 3 - 1} \right)}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{{1 + {x^2}}};\,\,\,\left[ {1,\sqrt 3 } \right] \cr
& {\text{Find the average value using the definition }}{f_{{\text{ave}}}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)dx} \cr
& {f_{{\text{ave}}}} = \frac{1}{{\sqrt 3 - 1}}\int_1^{\sqrt 3 } {\frac{1}{{1 + {x^2}}}dx} \cr
& {\text{Integrating }} \cr
& {f_{{\text{ave}}}} = \frac{1}{{\sqrt 3 - 1}}\left( {{{\tan }^{ - 1}}x} \right)_1^{\sqrt 3 } \cr
& {f_{{\text{ave}}}} = \frac{1}{{\sqrt 3 - 1}}\left( {{{\tan }^{ - 1}}\sqrt 3 - {{\tan }^{ - 1}}1} \right) \cr
& {\text{Simplifying}} \cr
& {f_{{\text{ave}}}} = \frac{1}{{\sqrt 3 - 1}}\left( {\frac{\pi }{3} - \frac{\pi }{4}} \right) \cr
& {f_{{\text{ave}}}} = \frac{1}{{\sqrt 3 - 1}}\left( {\frac{\pi }{{12}}} \right) \cr
& {f_{{\text{ave}}}} = \frac{\pi }{{12\left( {\sqrt 3 - 1} \right)}} \cr} $$
