Prove the following inequality for all $n \in \mathbb{N}$

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

If I'm not mistaken, we can get a slightly sharper bound. Write

$$\int_0^1\,f(x)\,dx = \sum_{k=1}^n\int_{(k-1)/n}^{k/n}\,f(x)\,dx$$

so then we need to estimate

\begin{align} &\left| \sum_{k=1}^n\left[\left(\int_{(k-1)/n}^{k/n}\,f(x)\,dx\right)-\frac1n f(k/n)\right] \right| \\=\,\,& \left| \sum_{k=1}^n\left(\int_{(k-1)/n}^{k/n}\,f(x) - f(k/n)\,dx\right) \right|. \end{align}

Now, use the mean value theorem. For each $x\in\Big((k-1)/n,\, k/n\Big)$, there is some $\eta_x \in (x,\,k/n)$ with

$$f(x) - f(k/n) = f'(\eta_x) \big(x-k/n\big).$$

Then our estimate becomes

\begin{align} &\left| \sum_{k=1}^n\left(\int_{(k-1)/n}^{k/n}\,f(x) - f(k/n)\,dx\right) \right| \\\leqslant \,\,& \sum_{k=1}^n\int_{(k-1)/n}^{k/n}\,|f'(\eta_x)| \big(k/n - x\big)\,dx \\\leqslant \,\,& M\,\sum_{k=1}^n\int_{(k-1)/n}^{k/n}\,\big(k/n - x\big)\,dx \\= \,\,& M\,\sum_{k=1}^n\left[\frac{kx}n - \frac{x^2}2\right]_{(k-1)/n}^{k/n} \\= \,\,& M\,\sum_{k=1}^n\frac1{2n^2} = \frac M{2n}. \end{align}

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As a bonus, it's easy to verify that the bound is sharp when $f$ is affine (that is, when $f'$ is constant).