Answer
$$A = - 1 + \frac{1}{{2{e^2}}} + \frac{1}{2}{e^4}$$
Work Step by Step
$$\eqalign{
& x = - 1,\,\,\,\,\,x = 2,\,\,\,\,\,y = 2{e^{2x}},\,\,\,\,\,\,y = {e^{2x}} + 1 \cr
& {\text{Find the points of intersection of the graphs}} \cr
& 2{e^{2x}} = {e^{2x}} + 1 \cr
& 2{e^{2x}} - {e^{2x}} = {e^{2x}} + 1 - {e^{2x}} \cr
& {e^{2x}} = 1 \cr
& \ln {e^{2x}} = \ln 1 \cr
& 2x = 0 \cr
& x = 0 \cr
& {\text{Then with the given points and the critical value }}x = 0{\text{ we have the intervals}} \cr
& \left( { - 1,0} \right){\text{ and }}\left( {0,2} \right) \cr
& {\text{For the interval }}\left( { - 1,0} \right){\text{ we have that }}{e^{2x}} + 1 > 2{e^{2x}}{\text{ and }} \cr
& {\text{For the interval }}\left( {0,2} \right){\text{ }}2{e^{2x}} > {e^{2x}} + 1 \cr
& \cr
& {\text{The area is given by}}: \cr
& A = \int_{ - 1}^0 {\left( {{e^{2x}} + 1 - 2{e^{2x}}} \right)} dx + \int_0^2 {\left( {2{e^{2x}} - {e^{2x}} - 1} \right)} dx \cr
& A = \int_{ - 1}^0 {\left( {1 - {e^{2x}}} \right)} dx + \int_0^2 {\left( {{e^{2x}} - 1} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {x - \frac{1}{2}{e^{2x}}} \right)_{ - 1}^0 + \left( {\frac{1}{2}{e^{2x}} - x} \right)_0^2 \cr
& A = \left( {0 - \frac{1}{2}{e^0}} \right) - \left( { - 1 - \frac{1}{2}{e^{ - 2}}} \right) + \left( {\frac{1}{2}{e^{2\left( 2 \right)}} - 2} \right) - \left( {\frac{1}{2}{e^0} - 0} \right) \cr
& A = - \frac{1}{2} + 1 + \frac{1}{{2{e^2}}} + \frac{1}{2}{e^4} - 2 - \frac{1}{2} \cr
& A = - 1 + \frac{1}{{2{e^2}}} + \frac{1}{2}{e^4} \cr} $$