## Calculus: Early Transcendentals (2nd Edition)

$$\frac{{{5^{25}} - 1}}{{5\ln 5}}$$
\eqalign{ & \int_0^5 {{5^{5x}}} dx \cr & {\text{substitute }}u = 5x,{\text{ }}du = 5dx \cr & {\text{express the limits in terms of }}u \cr & x = 5{\text{ implies }}u = 5\left( 5 \right) = 25 \cr & x = 0{\text{ implies }}u = 5\left( 0 \right) = 0 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^5 {{5^{5x}}} dx = \frac{1}{5}\int_0^{25} {{5^u}du} \cr & {\text{by the formula }}\int {{a^u}} du = \frac{{{a^u}}}{{\ln a}} + C \cr & {\text{letting }}a = 5 \cr & = \frac{1}{5}\left. {\left( {\frac{{{5^u}}}{{\ln 5}}} \right)} \right|_0^{25} \cr & {\text{use the fundamental theorem}} \cr & = \frac{1}{{5\ln 5}}\left( {{5^{25}} - {5^0}} \right) \cr & = \frac{{{5^{25}} - 1}}{{5\ln 5}} \cr}