Answer
$A\approx 0.9133$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sec \frac{{\pi x}}{4}\tan \frac{{\pi x}}{4},{\text{ }}g\left( x \right) = \left( {\sqrt 2 - 4} \right)x + 4,{\text{ }}x = 0 \cr
& {\text{Find the intersection points between the graphs}} \cr
& f\left( x \right) = g\left( x \right) \cr
& \sec \left( {\frac{{\pi x}}{4}} \right)\tan \left( {\frac{{\pi x}}{4}} \right) = \left( {\sqrt 2 - 4} \right)x + 4 \cr
& {\text{Solving}}{\text{, we obtain }}x = 1{\text{ }} \cr
& {\text{From the graph shown below}}{\text{, we can define the enclosed area}} \cr
& {\text{as:}} \cr
& A = \,\int_0^1 {\sec \left( {\frac{{\pi x}}{4}} \right)\tan \left( {\frac{{\pi x}}{4}} \right)} dx + \,\int_1^{\pi /2} {\left[ {\left( {\sqrt 2 - 4} \right)x + 4} \right]} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = \,\frac{4}{\pi }\left[ {\sec \left( {\frac{{\pi x}}{4}} \right)} \right]_0^1 + \left[ {\frac{{\left( {\sqrt 2 - 4} \right)}}{2}{x^2} + 4x} \right]_1^{\pi /2} \cr
& A = \,\frac{4}{\pi }\left[ {\sec \left( {\frac{\pi }{4}} \right) - \sec \left( 0 \right)} \right] \cr
& + \left[ {\frac{{\left( {\sqrt 2 - 4} \right)}}{2}{{\left( {\frac{\pi }{2}} \right)}^2} + 4\left( {\frac{\pi }{2}} \right)} \right] - \left[ {\frac{{\left( {\sqrt 2 - 4} \right)}}{2}{{\left( 1 \right)}^2} + 4\left( 1 \right)} \right] \cr
& {\text{Simplifying}} \cr
& A = \,\frac{4}{\pi }\left( {\sqrt 2 - 1} \right) + \frac{{\left( {\sqrt 2 - 4} \right)}}{8}{\pi ^2} + 2\pi - \frac{{\left( {\sqrt 2 - 4} \right)}}{2} - 4 \cr
& A = \,\frac{4}{\pi }\left( {\sqrt 2 - 1} \right) + \frac{{\left( {\sqrt 2 - 4} \right)}}{8}{\pi ^2} + 2\pi - \frac{{\sqrt 2 }}{2} - 2 \cr
& A \approx 0.9133 \cr} $$
