$f$ is differentiable in $[0,1]$ ,$\sup_{x\in[0,1]}|f'(x)| \le M\lt+\infty $

Proof. $\blacktriangleleft$ Rewrite the equation as follows: \begin{align*} S_n := \int_0^1 f(x) \mathrm dx - \sum_0^{n-1} \frac {f(j/n)}n &= \sum_0^{n-1} \int_{j/n}^{(j+1)/n} f(x) \mathrm dx - \frac {f(j/n)}n\\ &= \sum_0^{n-1} \int_{j/n}^{(j+1)/n} (f(x) - f(j/n)) \mathrm dx\\ &= \sum_0^{n-1} \int_{j/n}^{(j+1)/n} f’(c_j) (x - j/n) \mathrm dx. \end{align*}

Now by assumption, $-M \leqslant f’(x) \leqslant M$ on $[0,1]$. Since $(x - j/n)$ is always nonnegative on $[j/n, (j+1)/n]$, we have $$ - M \sum_0^{n-1} \int_{j/n}^{(j+1)/n} (x - j/n) \mathrm dx \leqslant S_n \leqslant M \sum_0^{n-1} \int_{j/n}^{(j+1)/n} (x-j/n)\mathrm dx. $$ Since $$ \sum_0^{n-1} \int_{j/n}^{(j+1)/n} (x - j/n)\mathrm dx = \frac 12 \sum_0^{n-1} \left( \frac {j+1}n - \frac jn\right)^2 = \frac 1{2n}, $$ the desired estimate follows. $\blacktriangleright$