Let $f :$ [0:1] $\to \mathbb {R}$ continuously differentiable. Prove:
$|\int_0^1 f(x)dx - {1 \over n} \sum_{k=0}^{n-1} f({k \over n})| \leq {M \over n}$ with $M = \int_0^1|f^{'}(x)|dx$.
So I have sketched down some of the things I know but I haven't been able to make a whole proof out of it, and maybe what I wrote down is actually useless.
We know $f$ is continous therfore we have $f \in $ R([0,1]). Let $\prod_n$ be a series of partitions that part [0,1] into $n$ equal segments, and ${t_i^{(n)}}$ a set of numbers s.t $t_i := x_{i-1} \in $[$x_{i-1}, x_i$].
Now notice: $S(f, \prod_n,{t_i^{(n)}}) = \sum_{k=0}^{n-1}f(t_i)\Delta x_i = {1 \over n}\sum_{k=0}^{n-1}f({k \over n})$ and since $f \in$ R([0,1]) & $\lambda (\prod_n) \to 0$ then:
$\lim_{n \to \infty}S(f, \prod_n,{t_i^{(n)}}) = \int_0^1f(x)dx$.
From Newton-Leibnitz thereom we have $|f(1)-f(0)| = |\int_0^1f^{'}(x)dx| \leq \int_0^1|f^{'}(x)|dx = M$.
That's it this is all I have hoping for help!