## Calculus with Applications (10th Edition)

$$A = \frac{1}{3}{e^6} + \frac{1}{3} + \frac{1}{3}{e^9}$$
\eqalign{ & x = 0,\,\,\,\,\,x = 3,\,\,\,\,\,y = 2{e^{3x}},\,\,\,\,\,\,y = {e^{3x}} + {e^6} \cr & {\text{Find the points of intersection of the graphs}} \cr & 2{e^{3x}} = {e^{3x}} + {e^6} \cr & 2{e^{3x}} - {e^{3x}} = {e^6} \cr & {e^{3x}} = {e^6} \cr & 3x = 6 \cr & x = 2 \cr & {\text{Then with the given points and the critical value }}x = 0{\text{ we have the intervals}} \cr & \left( {0,2} \right){\text{ and }}\left( {2,3} \right) \cr & {\text{For the interval }}\left( {0,2} \right){\text{ we have that }}{e^{3x}} + {e^6} > 2{e^{3x}}{\text{ and }} \cr & {\text{For the interval }}\left( {2,3} \right){\text{ }}2{e^{3x}} > {e^{3x}} + {e^6} \cr & \cr & {\text{The area is given by}}: \cr & A = \int_0^2 {\left( {{e^{3x}} + {e^6} - 2{e^{3x}}} \right)} dx + \int_2^3 {\left( {2{e^{3x}} - {e^{3x}} - {e^6}} \right)} dx \cr & A = \int_0^2 {\left( {{e^6} - {e^{3x}}} \right)} dx + \int_2^3 {\left( {{e^{3x}} - {e^6}} \right)} dx \cr & {\text{Integrating}} \cr & A = \left( {{e^6}x - \frac{{{e^{3x}}}}{3}} \right)_0^2 + \left( {\frac{{{e^{3x}}}}{3} - {e^6}x} \right)_2^3 \cr & A = \left( {{e^6}\left( 2 \right) - \frac{{{e^{3\left( 2 \right)}}}}{3}} \right) - \left( {{e^6}\left( 0 \right) - \frac{{{e^{3\left( 0 \right)}}}}{3}} \right) + \left( {\frac{{{e^{3\left( 3 \right)}}}}{3} - {e^6}\left( 3 \right)} \right) - \left( {\frac{{{e^{3\left( 2 \right)}}}}{3} - {e^6}\left( 2 \right)} \right) \cr & A = 2{e^6} - \frac{{{e^6}}}{3} + \frac{1}{3} + \frac{{{e^9}}}{3} - 3{e^6} - \frac{{{e^6}}}{3} + 2{e^6} \cr & A = \frac{1}{3}{e^6} + \frac{1}{3} + \frac{1}{3}{e^9} \cr}