Answer
$$A = e + {e^{ - 1}} - 2$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can note that the area is given by:}} \cr
& A = \int_{ - 1}^0 {\left( {{e^x} - 1} \right)dx} - \int_0^1 {\left( {{e^x} - 1} \right)dx} \cr
& {\text{Integrating}} \cr
& A = \left[ {{e^x} - x} \right]_{ - 1}^0 - \left[ {{e^x} - x} \right]_0^1 \cr
& A = \left( {{e^0} - 0} \right) - \left( {{e^{ - 1}} + 1} \right) - \left( {{e^1} - 1} \right) + \left( {{e^0} - 0} \right) \cr
& {\text{Simplifying}} \cr
& A = 1 - \frac{1}{e} - 1 - e + 1 + 1 \cr
& A = e + {e^{ - 1}} - 2 \cr} $$
