I came up with this alternate Riemann sum that correctly gives the value of a definite integral (at least for some simple polynomial and trig functions I tested with wolfram alpha):
$$\lim\limits_{n \to\infty}\sum_{i=1}^{l\cdot n}\frac{f\left(\frac{i}{n}\right)}{n}$$
where $l$ is the value we want from the antiderivative.
That is,
$$\lim\limits_{n \to\infty}\sum_{i=1}^{l\cdot n}\frac{f\left(\frac{i}{n}\right)}{n} = \int_{0}^l f(x)$$
Here is the "classic" version from Calculus textbooks I've seen:
$$\lim\limits_{n \to\infty}\sum_{i=1}^{n}f\left(a+i\frac{b-a}{n}\right)\left(\frac{b-a}{n}\right)$$
And here is my version on some interval $[a,b]$ instead of $[0,l]$:
$$\lim\limits_{n \to\infty}\sum_{i=1}^{\left(b-a\right)\cdot n}\frac{f\left(a+\frac{i}{n}\right)}{n}$$
I tried looking for more information on it with no success. I want to know is it valid in all cases, is it really somehow just a trivial rewrite of the classic version and does it have a name? Is there some textbook or other resource that mentions it somewhere or somewhere to learn more about it? To me I see it as very different from the "average value" that the classic Riemann sum does, although I think you can still view it that way. Sorry if this is a trivial rewrite of the other version, I am new to calculus.
I like this version better because it is more easy to understand in my opinion because to me it shows the "reversal" of the derivative definition more clearly:
$$\lim\limits_{h \to 0}\frac{f(x+h)-f(x)}{h}$$
It is more clear to me in the version I proposed that multiplying by $\frac{1}{n}$ in the Riemann sum is canceling out the $h$ from the derivative. And then we are just adding up all the original values of $f$ that $f'$ is derived from. As opposed to the classic version where we multiply by $\frac{b-a}{n}$