TITLE: Prove that $ \int_0^1 f(x)dx - \frac{1}{n}\sum_{k=1}^n f(\frac{k}{n}) = O(\frac{1}{n}) $
QUESTION [5 upvotes]: Let the function $ f(x) $ is bounded and monotone on $[0,1]$. Prove that
$ \int_0^1 f(x)dx - \frac{1}{n}\sum_{k=1}^n f(\frac{k}{n}) = O(\frac{1}{n}) $
It is clear that $ \int_0^1 f(x)dx - \frac{1}{n}\sum_{k=1}^n f(\frac{k}{n}) \to 0 $ if $ n \to \infty $
But I have no idea how to prove that it is exactly $ O(\frac 1n) $.
Thanks for the help! Do not judge strictly I am just starting to study integral calculus.

REPLY [3 votes]: You have, as f is increasing,
$$ \int_0^1 f(x) dx=  \sum_{k=1}^n \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(x) dx \leq \sum_{k=1}^n \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(\frac{k}{n}) dx \leq \frac{1}{n} \sum_{k=1}^n  f(\frac{k}{n}) $$
and
$$ \int_0^1 f(x) dx=  \sum_{k=1}^n \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(x) dx \geq 
\sum_{k=1}^n \int_{\frac{k-1}{n}}^{\frac{k}{n}} f(\frac{k-1}{n}) dx $$
$$\geq \frac{1}{n} \sum_{k=1}^n  f(\frac{k-1}{n}) \geq \frac{f(0)}{n}-\frac{f(1)}{n} + \frac{1}{n} \sum_{k=1}^n  f(\frac{k}{n}) $$
Hence 
$$\frac{f(0)-f(1)}{n} \leq \int_0^1 f(x) dx - \frac{1}{n} \sum_{k=1}^n  f(\frac{k}{n}) \leq 0$$