Answer
$${f_{avg}} = \frac{3}{{\ln 4}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {e^x} + {e^{ - x}},{\text{ }}\left[ {\ln \frac{1}{2},\ln 2} \right] \cr
& {\text{The average value is given by}} \cr
& {f_{avg}} = \frac{1}{{\ln 2 - \ln \left( {1/2} \right)}}\int_{\ln \left( {1/2} \right)}^{\ln 2} {\left( {{e^x} + {e^{ - x}}} \right)dx} \cr
& {\text{Integrating}} \cr
& {f_{avg}} = \frac{1}{{\ln 4}}\left[ {{e^x} + {e^{ - x}}} \right]_{\ln \left( {1/2} \right)}^{\ln 2} \cr
& {f_{avg}} = \frac{1}{{\ln 4}}\left[ {{e^{\ln 2}} + {e^{ - \ln 2}}} \right] - \frac{1}{{\ln 4}}\left[ {{e^{\ln 1/2}} + {e^{ - \ln 1/2}}} \right] \cr
& {f_{avg}} = \frac{1}{{\ln 4}}\left( {2 - \frac{1}{2}} \right) - \frac{1}{{\ln 4}}\left( {\frac{1}{2} - 2} \right) \cr
& {f_{avg}} = \frac{1}{{\ln 4}}\left( {\frac{3}{2}} \right) - \frac{1}{{\ln 4}}\left( { - \frac{3}{2}} \right) \cr
& {f_{avg}} = \frac{3}{{\ln 4}} \cr} $$
