$\int_0^1 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f(i/n)$ is not valid for non-monotone function. How to find such a counterexample?
As is well-known, for monotone function $f$, we have $\int_0^1 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f(i/n)$. then for non-monotone function, is there any simple example such that $\int_0^1 f(x)dx$ exists (or for any $\delta>0$, $f$ is integrable on $[\delta, 1]$, and $\int_0^1 f(x)dx=\lim_{\delta\to 0}\int_\delta^1 f(x)dx$ exists), but $\int_0^1 f(x)dx=\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n f(i/n)$ is not valid.
For a counterexample, take $f:[0,1] \to \mathbb{R}$ such that for $m = 0,1,2,\ldots$
$$f(x) = \begin{cases}2^m,&x = 2^{-m}\\0, &\text{otherwise} \end{cases}$$
Then
$$S_n=\sum_{k=1}^{2^n}f\left( \frac{k}{2^n}\right)\left(\frac{k}{2^n}- \frac{k-1}{2^n}\right) = 2^{-n}\sum_{k=1}^{2^n}f\left( \frac{k}{2^n}\right) = 2^{-n}\sum_{r=0}^{n}\frac{2^n}{2^r}= 2 - \frac{2}{2^{n+1}},$$
and
$$\lim_{n \to \infty}S_n = 2 \neq 0 = \int_0^1f(x) \, dx$$