Answer
$$A = 8\arctan {e^2} - 2\pi $$
Work Step by Step
$$\eqalign{
& y = \operatorname{sech} \frac{x}{2} \cr
& {\text{The area of the region is given by }} \cr
& A = \int_a^b {\left( {f\left( x \right) - g\left( x \right)} \right)} dx,{\text{ }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right],{\text{ then}} \cr
& A = \int_{ - 4}^4 {\left( {\operatorname{sech} \frac{x}{2}} \right)} dx \cr
& A = \int_{ - 4}^4 {\operatorname{sech} \left( {\frac{x}{2}} \right)} dx \cr
& {\text{By definition sech}}u = \frac{2}{{{e^u} + {e^{ - u}}}} \cr
& A = \int_{ - 4}^4 {\frac{2}{{{e^{x/2}} + {e^{ - x/2}}}}} dx \cr
& {\text{Multiply numerator and denominator by }}{e^{x/2}} \cr
& A = 2\int_{ - 4}^4 {\frac{{{e^{x/2}}}}{{{{\left( {{e^{x/2}}} \right)}^2} + 1}}} dx \cr
& {\text{By the property of even integrals}} \cr
& A = 4\int_0^4 {\frac{{{e^{x/2}}}}{{{{\left( {{e^{x/2}}} \right)}^2} + 1}}} dx \cr
& A = 8\int_0^4 {\frac{{\left( {1/2} \right){e^{x/2}}}}{{{{\left( {{e^{x/2}}} \right)}^2} + 1}}} dx \cr
& {\text{Integrate }} \cr
& A = 8\left[ {\arctan {e^{x/2}}} \right]_0^4 \cr
& A = 8\left[ {\arctan {e^{4/2}} - \arctan {e^{0/2}}} \right] \cr
& A = 8\left[ {\arctan {e^2} - \arctan 1} \right] \cr
& A = 8\left[ {\arctan {e^2} - \frac{\pi }{4}} \right] \cr
& A = 8\arctan {e^2} - 2\pi \cr} $$
