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