Solution: Look at the graph of the function \[ f(x) = \begin{cases} \frac{1}{n}, &\text{ if } \frac{1}{n+1} \leq x \leq \frac{1}{n} ;\\ 0, &\text{ if } x = 0. \end{cases} \]

Since $0\leq f(x) \leq 1$, so $f$ is bounded and it is continuous on $[0,1]$ except at $x = \frac{1}{n},$ for $n = 2,3,\dots$. The set of discontinuity of $f$ is countable, so $f$ is Riemann integrable (we could have also used the fact that the set of discontinuity of $f$ is infinite with one limit point $0$, so $f$ is integrable).

Now we need to find the integral. Let us write \[ f_n(x) = \frac{1}{n},\quad \frac{1}{n+1} \leq x \leq \frac{1}{n}, n \in \mathbb{N} . \] Then, \begin{align*} \int _0^1 f(x) & = \lim_{n \to \infty} \left[ \int _{\frac{1}{2}}^1 f_1(x) + \int _{\frac{1}{3}} ^{\frac{1}{2}} f_2(x) + \dots + \int_{\frac{1}{n+1}^{\frac{1}{n}}}f_n(x) \right] \\[2ex] & = \lim_{n \to \infty} \left[ 1 \left( 1-\frac{1}{2} \right) + \frac{1}{2}\left( \frac{1}{2} - \frac{1}{3} \right) + \dots + \frac{1}{n}\left( \frac{1}{n} - \frac{1}{n+1} \right) \right] \\[2ex] & = \lim_{n \to \infty} \left[ \sum_{i=1}^{n} \frac{1}{i^2} - \sum_{i=1}^{n} \frac{1}{i\cdot (i+1)} \right] \tag{$\star $}. \end{align*}

We note that \begin{align*} \sum_{i=1}^{n} \frac{1}{i\cdot (i + 1)} & = \sum_{i=1}^{n} \left( \frac{1}{i} - \frac{1}{i+1} \right) = 1 - \frac{1}{n+1}. \end{align*} Substituting this in $(\star)$, we get \begin{align*} \int _0^1 f(x) & = \lim_{n \to \infty} \left[ \sum_{i=1}^{n} \frac{1}{i^2} - \sum_{i=1}^{n} \frac{1}{i\cdot (i+1)} \right] \\ & = \lim_{n \to \infty} \left[ \sum_{i=1}^{n} \frac{1}{i^2} - \left( 1 - \frac{1}{n+1} \right) \right] \\ & = \lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{i^2} - 1 \\ & = \frac{\pi ^2}{6} - 1. \end{align*} Thus, \[ \textcolor{teal}{\boxed{ \int _0^1 f(x) = \frac{\pi ^2}{6} - 1 }} \]