Average approximates definite integral.

Question

In my studies, I read a result similar to the following.

Let $g:[0,1]\rightarrow \mathbb{R}$ be twice continously differentiable on $[0,1]$ (this is probably stronger than necessary). Then \begin{equation} \bigg\lvert 1/K_s \sum_{k=1}^{K_s} g\{x_i\} - \int_{0}^{1}g(u)du \bigg\rvert \leq \dfrac{C}{K_s}, \end{equation} for some constant $C>0$, where $\{x_1,\dotsc,x_{K_s}\}$ is a set of equidistant points in $[0,1]$ and $K_s$ its cardinality. By set of equidistant points I mean that for some $a\in \mathbb{R},\exists d>0:\{x_1,\dotsc,x_{K_s}\}=\{a+d\mathbb{Z}\}\cap [0,1]$ and $a+d(x_{K_s}+1) \notin [0,1]$. Here,$\{a+d\mathbb{Z}\}=\{a+dx: x\in \mathbb{Z}\}$. This precision is due to @Sangchul Lee comment.

I gave a cumbersome proof to this problem using the definition (upper/lower Riemann sum and constructing an appropriate partition). Though I think it is correct, I wonder if there is an easy way to prove it.

Question: what is the easiest way to prove it?

Thanks in advance!

Answer 1

The proof should be using the following properties:

1. The area under the curve that is above the average value of $g$ in range $[a..b]$, exactly cancels the area between average and the curve when $g$ is below the average.

2. a random sample $g(x_i)$ picks samples equally often from above the average than below the average.

3. Then there's the link between average and integral: $$ \int_a^b{f(x)dx} = (b-a)\cdot\mathbf{av}_f(a,b)$$