## Calculus with Applications (10th Edition)

$$\frac{{25}}{4}\left( {{e^2} - {e^{0.4}}} \right)$$
\eqalign{ & \int_1^5 {\frac{5}{2}{e^{0.4x}}} dx \cr & = \frac{5}{2}\int_1^5 {{e^{0.4x}}} dx \cr & {\text{use substitution}}{\text{. Let }}u = 0.4x,{\text{ so that }}du = 0.4dx,\,\,\,\,\frac{5}{2}du = dx \cr & {\text{the new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = 2,{\text{ then }}u = 0.4\left( 5 \right) = 2 \cr & \,\,\,\,\,\,{\text{If }}x = 1,{\text{ then }}u = 0.4\left( 1 \right) = 0.4 \cr & {\text{Then}} \cr & \frac{5}{2}\int_1^5 {{e^{0.4x}}} dx = \frac{5}{2}\int_{0.4}^2 {{e^u}\left( {\frac{5}{2}du} \right)} \cr & = \frac{{25}}{4}\int_{0.4}^2 {{e^u}du} \cr & {\text{use }}\int_a^b {{e^u}} du = \left( {{e^u}} \right)_a^b \cr & = \frac{{25}}{4}\left( {{e^u}} \right)_{0.4}^2 \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{{25}}{4}\left( {{e^2} - {e^{0.4}}} \right) \cr}