Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.4 Graphs And Applications Involving Logarithmic And Exponential Functions - Exercises Set 6.4 - Page 440: 43

$$A = \frac{1}{2}$$

$$\eqalign{ & {\text{From the graph we can note that the area is given by:}} \cr & A = \int_0^{\ln 2} {\left( {{e^{2x}} - {e^x}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left[ {\frac{1}{2}{e^{2x}} - {e^x}} \right]_0^{\ln 2} \cr & {\text{Simplifying}} \cr & A = \left[ {\frac{1}{2}{e^{2\left( {\ln 2} \right)}} - {e^{\ln 2}}} \right] - \left[ {\frac{1}{2}{e^{2\left( 0 \right)}} - {e^0}} \right] \cr & A = \frac{1}{2}\left( 4 \right) - 2 - \frac{1}{2} + 1 \cr & A = \frac{1}{2} \cr} $$