## Calculus with Applications (10th Edition)

$$\frac{{3\left( {1 - {e^{ - 4}}} \right)}}{2}$$
\eqalign{ & \int_0^2 {3{e^{ - 2x}}} dx \cr & {\text{use substitution}}{\text{. Let }}u = - 2x,{\text{ so that }}du = - 2dx,\,\,\,\, - \frac{1}{2}du = dx \cr & {\text{the new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = 2,{\text{ then }}u = - 2\left( 2 \right) = - 4 \cr & \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = - 2\left( 0 \right) = 0 \cr & {\text{Then}} \cr & \int_0^2 {3{e^{ - 2x}}} dx = \int_0^{ - 4} {3{e^u}\left( { - \frac{1}{2}du} \right)} \cr & = - \frac{3}{2}\int_0^{ - 4} {{e^u}du} \cr & {\text{use }}\int_a^b {{e^u}} du = \left( {{e^u}} \right)_a^b \cr & = - \frac{3}{2}\left( {{e^u}} \right)_0^{ - 4} \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & = - \frac{3}{2}\left( {{e^{ - 4}} - {e^{ - 0}}} \right) \cr & = - \frac{3}{2}\left( {{e^{ - 4}} - 1} \right) \cr & = \frac{{3\left( {1 - {e^{ - 4}}} \right)}}{2} \cr}