Find the probability that the electron in the ground-state $\mathrm{H}$ atom is less than a distance $a$ from the nucleus.
We want the probability that the radial coordinate lies between 0 and $a$. This is found by taking the infinitesimal probability (6.116) of being between $r$ and $r+d r$ and summing it over the range from 0 to $a$. This sum of infinitesimal quantities is the definite integral
$$
\begin{aligned}
\int_0^a R_{n l}^2 r^2 d r & =\frac{4}{a^3} \int_0^a e^{-2 r / a} r^2 d r=\left.\frac{4}{a^3} e^{-2 r / a}\left(-\frac{r^2 a}{2}-\frac{2 r a^2}{4}-\frac{2 a^3}{8}\right)\right|_0 ^a \\
& =4\left[e^{-2}(-5 / 4)-(-1 / 4)\right]=0.323
\end{aligned}
$$