Answer
${g_{avg}} = \frac{9}{2}{\tan ^{ - 1}}\left( 2 \right)$
Work Step by Step
$$\eqalign{
& g\left( t \right) = \frac{9}{{1 + {t^2}}},{\text{ }}\left[ {0,2} \right] \cr
& {\text{Use The Mean Value Theorem for Integrals }} \cr
& {f_{avg}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)} dx \cr
& {\text{Therefore}}{\text{,}} \cr
& {g_{avg}} = \frac{1}{{2 - 0}}\int_0^2 {\frac{9}{{1 + {t^2}}}} dt \cr
& {g_{avg}} = \frac{9}{2}\int_0^2 {\frac{1}{{1 + {t^2}}}} dt \cr
& {\text{Integrate using }}\int {\frac{1}{{1 + {x^2}}}} dx = {\tan ^{ - 1}}x,{\text{ so}} \cr
& {g_{avg}} = \frac{9}{2}\left[ {{{\tan }^{ - 1}}t} \right]_0^2 \cr
& {\text{Evaluate and simplify}} \cr
& {g_{avg}} = \frac{9}{2}\left[ {{{\tan }^{ - 1}}\left( 2 \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right] \cr
& {g_{avg}} = \frac{9}{2}{\tan ^{ - 1}}\left( 2 \right) \cr} $$
