Integration with bounded derivative

Question

I try to solve the following problem, but I have no idea how to link bounded derivative to integral: IN Riemann Sence

Let $f:[0,1]\rightarrow \mathbb{R}$ be a differentiable function such that $|f'(x)|\leq M$ for all $x \in (0,1).$ Show that $$\Big|\int_0^1 f - \frac{1}{n} \sum_{k=1}^n f(\frac{k}{n})\Big| \leq \frac{M}{n}.$$

Answer 1

Hint: divide $[0,1]$ in $n$ subintervals of length $1/n$. Compare integrals in each subinterval with summands of sum. MVT will be useful.

Answer 2

Note that $f' \leq M$ implies that $f$ is Lipschitz continous on $[0,1]$ with Lipschitz constant $M$ (i.e. $\forall x,y, |f(x)-f(y)| \leq M|x-y|$).

So

$$\left| \int_0^1 f(t)dt - \frac{1}{n} \sum_{k=1}^n f \left( \frac{k}{n} \right) \right|= \left| \sum_{k=1}^n \int_{\frac{k-1}{n}}^{\frac{k}{n}} \left( f(t) -f\left( \frac{k}{n} \right) \right) dt \right| \leq $$

$$\leq \sum_{k=1}^n \int_{\frac{k-1}{n}}^{\frac{k}{n}} \left| f(t) -f \left( \frac{k}{n} \right) \right| dt \leq \sum_{k=1}^n \int_{\frac{k-1}{n}}^{\frac{k}{n}} M \left|t-\frac{k}{n} \right| dt =$$

$$= M \sum_{k=1}^n \frac{1}{2n^2} = \frac{M}{2n}$$