If $f$ is Lipschitz-continuous then $\left\lvert\int_0^1 f(x)\,dx-\frac1n\sum\limits_{k=1}^{n}f \left(\frac kn\right)\right\rvert<\frac{M}{2n}$

Question

Suppose $f$ is Lipschitz-continuous on $[0,1]$, so that $|f(x)-f(y)|<M|x-y|$ for all $x,y\in[0,1]$; prove $$\left\lvert\int_0^1 f(x) \, dx-\frac 1 n \sum_{k=1}^n f\left(\frac k n\right)\right\rvert<\frac{M}{2n}$$

I get RHS $\frac{M}{2n^2}$ instead. Book seems to give that answer then confusingly states $\frac{M}{2n}$ at very end. Just want to confirm which is correct.

EDIT: Nevermind its $\frac{M}{2n}$. question is from Intro to Calculus and Analysis Vol 1 By Courant, S2.3.3

Answer 1

\begin{align*} \left|\int_{0}^{1}f(x)dx-\dfrac{1}{n}\sum_{k=1}^{n}f(k/n)\right|&=\left|\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}(f(x)-f(k/n))dx\right|\\ &< M\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}|x-k/n|dx\\ &=M\sum_{k=1}^{n}\int_{(k-1)/n}^{k/n}(k/n-x)dx\\ &=M\left(-\int_{0}^{1}xdx+\dfrac{1}{n^{2}}\sum_{k=1}^{n}k\right)\\ &=M\left(-\dfrac{1}{2}+\dfrac{1}{n^{2}}\cdot\dfrac{n^{2}+n}{2}\right)\\ &=\dfrac{M}{2n}. \end{align*}