Answer
$$A = \frac{1}{2}\ln \left| {\cosh 4} \right|$$
Work Step by Step
$$\eqalign{
& y = \tanh 2x \cr
& {\text{The area of the region is given by }} \cr
& A = \int_a^b {\left( {f\left( x \right) - g\left( x \right)} \right)} dx,{\text{ }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right],{\text{ then}} \cr
& A = \int_0^2 {\left( {\tanh 2x - 0} \right)} dx \cr
& A = \int_0^2 {\tanh 2x} dx \cr
& A = \int_0^2 {\frac{{\sinh 2x}}{{\cosh 2x}}} dx \cr
& A = \frac{1}{2}\int_0^2 {\frac{{2\sinh 2x}}{{\cosh 2x}}} dx \cr
& {\text{Integrating}} \cr
& A = \frac{1}{2}\left[ {\ln \left| {\cosh 2x} \right|} \right]_0^2 \cr
& A = \frac{1}{2}\left[ {\ln \left| {\cosh 4} \right| - \ln \left| {\cosh 0} \right|} \right] \cr
& A = \frac{1}{2}\ln \left| {\cosh 4} \right| \cr} $$
