Answer
$$A = e - 1 + \frac{1}{{{e^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 1 - {e^{ - x}};{\text{ }}\left[ { - 1,2} \right] \cr
& {\text{First check if the graph crosses the }}x - {\text{axis in the given interval}} \cr
& 1 - {e^{ - x}} = 0 \cr
& {\text{solve for }}x \cr
& {e^{ - x}} = 1 \cr
& x = 0 \cr
& {\text{the graph crosses the }}x{\text{ - axis at }}x = 0{\text{ and the given interval }}\left[ { - 1,2} \right],{\text{ }} \cr
& f\left( { - 1} \right) = 1 - {e^{ - \left( { - 1} \right)}} < 0,{\text{ }}f\left( x \right){\text{ is negative for }}\left( { - 1,0} \right) \cr
& {\text{then}} \cr
& {\text{the definite integral of the total area will be}}: \cr
& A = \left| {\int_{ - 1}^0 {\left( {1 - {e^{ - x}}} \right)} dx} \right| + \int_0^2 {\left( {1 - {e^{ - x}}} \right)} dx \cr
& A = \left| {\left( {x + {e^{ - x}}} \right)_{ - 1}^0} \right| + \left( {x + {e^{ - x}}} \right)_0^2 \cr
& A = \left| {\left( {0 + {e^0}} \right) - \left( { - 1 + {e^1}} \right)} \right| + \left( {2 + {e^{ - 2}}} \right) - \left( {0 + {e^0}} \right) \cr
& {\text{simplifying}} \cr
& A = \left| {1 + 1 - e} \right| + 2 + {e^{ - 2}} - 1 \cr
& A = e - 2 + 1 + {e^{ - 2}} \cr
& A = e - 1 + \frac{1}{{{e^2}}} \cr} $$
