Evaluate a definite integral of an exponential function via u-substitution

Let $u = 2x$, so $du = 2\,dx$, giving $dx = \frac{1}{2}\,du$. Converting the bounds: when $x=0$, $u=0$; when $x=1$, $u=2$. The integral becomes $\int_0^1 5^{2x}\,dx = \frac{1}{2}\int_0^2 5^u\,du.$

Using the antiderivative $\int 5^u\,du = \frac{5^u}{\ln 5} + C$, we get $\frac{1}{2}\left[\frac{5^u}{\ln 5}\right]_0^2 = \frac{1}{2}\left(\frac{25}{\ln 5} - \frac{1}{\ln 5}\right) = \frac{24}{2\ln 5} = \frac{12}{\ln 5}.$ This is approximately $7.457$.