Let $f(x)$ be a real positive continous function on $[0,1]$ such that $$ \int_{0}^{1}f(x)\ dx=2019,\text{and}\int_{0}^{1}f^2(x)=20\ 220\ 821. $$ Prove that, there exist a unique partition $0\leq x_0<x_1<\cdots<x_n=1$ such that $$\int_{x_i}^{x_{i+1}}f(x)=\frac 1n\int_{0}^{1}f(x)\ dx.$$and evaluate the limit $$\dfrac 1n\sum_{i=0}^{n}f\left(x_i\right).$$
The proving part is trivial and I have no problem with that. However, the second evaluation part seems tricky. The hint is to use "Integration by parts" but as $f(x)$ is not differentiable, I suppose it is referring to the second mean-value theorem on the integral, as is called in V.A. Zorich's book. I applied the theorem and got something like this:
Consider the integral $$\int_{x_i}^{x_{i+1}}f^2(x)\ dx.$$ By the second mean-value, we can find $\xi_i\in[x_i,x_i+1]$ such that $$\int_{x_i}^{x_{i+1}}f^2(x)\ dx=f(x_i)\int_{x_i}^{\xi}f(x)dx+f(x_{i+1})\int_{\xi}^{x_i}f(x)\ dx.$$ Summing in terms of $i$ from $1$ to $n-1$, one has $$\int_{0}^{1}f^2(x)\ dx=f(x_0)\int_{0}^{\xi_0}f(x_0)\ dx+\sum_{i=1}^{n-1}f(x_i)\int_{\xi_{i-1}}^{\xi_{i}}f(x)dx+f(x_0)\int_{\xi_n}^{1}f(x_0)\ dx.$$
I have a hard time estimating each of the integrals $$\Big|\int_{\xi_i}^{\xi_{i+1}}f(x)\ dx-\int_{x_i}^{x_{i+1}}f(x)\ dx\Big|=\Big|\int_{x_{i+1}}^{\xi_{i+1}}f(x)\ dx-\int_{x_n}^{\xi_{n}}f(x)\ dx\Big|.$$Intuitively, as the values of $f$ can be very small in some of the intervals, I don't have any idea how to prove that the length of the intervals is $o\left(\dfrac 1{n}\right)$ when $n\to \infty,$ which, from my perspective is the only way to show that the difference above is an infinitesimal when summed $n$ times and $n\to \infty.$ And only by showing this can I evaluate the limit in a rigorous and accurate way in my opinion.
I guess that the answer is $\frac{20\ 220\ 821}{2019}$. It's completely trivial when I'm convinced that the estimate above is correct and appropriate. This problem seems to be an old one since I've encountered it from many sources. So feel free to provide the original problem and close this one if you must (I don't want it, but it should be as in the regulations of Math StackExchange.)