Answer
${f_{avg}} = 1 + \frac{4}{\pi }\ln \left( {\frac{{\sqrt 2 }}{2}} \right)$
Work Step by Step
$$\eqalign{
& f\left( x \right) = x{\sec ^2}x{\text{ on the interval }}\underbrace {\left[ {0,\frac{\pi }{4}} \right]}_{\left[ {a,b} \right]} \cr
& {\text{The average value is given by}} \cr
& {f_{avg}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)} dx \cr
& {\text{Therefore}}{\text{,}} \cr
& {f_{avg}} = \frac{1}{{\pi /4 - 0}}\int_0^{\pi /4} {x{{\sec }^2}x} dx \cr
& {f_{avg}} = \frac{4}{\pi }\int_0^{\pi /4} {x{{\sec }^2}x} dx \cr
& \cr
& {\text{Integrating by parts}} \cr
& u = x,{\text{ }}du = dx \cr
& dv = {\sec ^2}xdx,{\text{ }}v = \tan x \cr
& \int {x{{\sec }^2}x{\text{ }}} dx = x\tan x - \int {\tan x} dx \cr
& \int {x{{\sec }^2}x{\text{ }}} dx = x\tan x + \ln \left| {\cos x} \right| + C \cr
& \cr
& {f_{avg}} = \frac{4}{\pi }\int_0^{\pi /4} {x{{\sec }^2}x} dx \cr
& {f_{avg}} = \frac{4}{\pi }\left[ {x\tan x + \ln \left| {\cos x} \right|} \right]_0^{\pi /4} \cr
& {\text{Evaluating}} \cr
& {f_{avg}} = \frac{4}{\pi }\left[ {\left( {\frac{\pi }{4}} \right)\tan \left( {\frac{\pi }{4}} \right) + \ln \left| {\cos \frac{\pi }{4}} \right|} \right] - \frac{4}{\pi }\left[ {\left( 0 \right)\tan \left( 0 \right) + \ln \left| {\cos 0} \right|} \right] \cr
& {\text{Simplifying}} \cr
& {f_{avg}} = \frac{4}{\pi }\left[ {\frac{\pi }{4} + \ln \left( {\frac{{\sqrt 2 }}{2}} \right)} \right] - \frac{4}{\pi }\left[ 0 \right] \cr
& {f_{avg}} = 1 + \frac{4}{\pi }\ln \left( {\frac{{\sqrt 2 }}{2}} \right) \approx 0.5587 \cr} $$
