Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - Review Exercises - Page 394: 55

$$A = - 2{e^{ - 2}} + 2$$

$$\eqalign{ & {\text{The area of the region bounded by the graphs is given by}} \cr & A = \int_0^2 {2{e^{ - x}}} dx \cr & {\text{Integrating and evaluating}} \cr & A = 2\left[ { - {e^{ - x}}} \right]_0^2 \cr & A = - 2\left[ {{e^{ - x}}} \right]_0^2 \cr & A = - 2\left[ {{e^{ - 2}} - {e^0}} \right] \cr & A = - 2{e^{ - 2}} + 2 \approx 1.729 \cr & {\text{Graph}} \cr} $$