Prove the following inequality with one integral and one sum

Question

Let $f(x)$ be a derivable function on the interval $(0,1)$ and continuous on $[0,1]$. Assume that $|f'(x)| \leq M$ for every $x \in (a,b)$. Prove the following inequality for all $n\in\mathbb{N}$, $$ \left| \int_0^1 f(x) dx - \frac{1}{n} \sum_{k=1}^n f(k/n) \right| \leq \frac{M}{n}. $$

Answer 1

$$\left|\int_{\frac {k-1} n}^{\frac k n} [f(x)-f(\frac k n )] dx\right|\leq \int_{\frac {k-1} n}^{\frac k n} M\left|x-\frac k n\right| dx $$ by MVT. Since $\left|x-\frac k n\right|\leq \frac 1n$ this gives $$\left|\int_{\frac {k-1} n}^{\frac k n} [f(x)-f(\frac k n )] dx\right|\leq \int_{\frac {k-1} n}^{\frac k n} M \frac 1 n dx=\frac M {n^{2}} $$ Sum this over all $k$. Can you finish the proof now?