## Calculus with Applications (10th Edition)

$$A = \frac{1}{e} + {e^2} - 3$$
\eqalign{ & f\left( x \right) = {e^x} - 1;{\text{ }}\left[ { - 1,2} \right] \cr & {\text{First check if the graph crosses the }}x - {\text{axis in the given interval}} \cr & {e^x} - 1 = 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) = {e^{ - 1}} - 1 < 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( {{e^x} - 1} \right)} dx} \right| + \int_0^2 {\left( {{e^x} - 1} \right)} dx \cr & A = \left| {\left( {{e^x} - x} \right)_{ - 1}^0} \right| + \left( {{e^x} - x} \right)_0^2 \cr & A = \left| {\left( {{e^0} - 0} \right) - \left( {{e^{ - 1}} - \left( { - 1} \right)} \right)} \right| + \left( {{e^2} - 2} \right) - \left( {{e^0} - 0} \right) \cr & {\text{simplifying}} \cr & A = \left| {1 - {e^{ - 1}} - 1} \right| + {e^2} - 2 - 1 \cr & A = \frac{1}{e} + {e^2} - 3 \cr}