Answer
$${\text{Area}} = \ln \left( {\frac{{\sqrt 2 + 1}}{{\sqrt 2 }}} \right)$$
Work Step by Step
$$\eqalign{
& \sec x \geqslant \tan x{\text{ on the interval }}\left[ {0,\frac{\pi }{4}} \right] \cr
& {\text{The area is given by}} \cr
& {\text{Area}} = \int_0^{\pi /4} {\left( {\sec x - \tan x} \right)dx} \cr
& {\text{Integrating}} \cr
& {\text{Area}} = \left[ {\ln \left| {\sec x + \tan x} \right| - \ln \left| {\sec x} \right|} \right]_0^{\pi /4} \cr
& {\text{Area}} = \left[ {\ln \left| {\sqrt 2 + 1} \right| - \ln \left| {\sqrt 2 } \right|} \right] - \left[ {\ln \left| {1 + 0} \right| - \ln \left| 1 \right|} \right] \cr
& {\text{Area}} = \ln \left( {\frac{{\sqrt 2 + 1}}{{\sqrt 2 }}} \right) \cr} $$
