I recently asked a question (During U-substitution, is the $du$ that we use as the infinitesimal in $\int f(u) \ du$ changing size?) and was explained that there is sort of a "freedom" with how one chooses to subdivide the area underneath a given curve so long as the way you define your dx (in the context of Riemann sum) will converge to zero as n→∞ .
This immediately made me realize that you could subdivide a given area under the curve an infinitely many number of ways (even "more so" when you consider that the subdivisions do not even have to be uniform).
As I understand it, these changes in subdivisions do not alter the resulting value of the integral. And perhaps that is where my misunderstanding arrives.
Consider the picture below:
In the first picture, we would have an integral of the following form: $\int_A^B f(x) d(x)$
In the second picture, we would have an integral of the following form: $\int_A^B f(x) 2dx$ (at least I believe that is how the notation would work)
My confusion is the following: using either one of these versions of uniform subdivisions will result in the same value. However, I feel like there must be a different interpretation that captures the fact that the subdivision for the lower graph is actually going to "fill up" the area under the curve more quickly.
Could someone please explain?
Edit 1: It seems like a clarification that I need in my question (based on below comments) is my misunderstanding of why a change in our “dx of choice” necessarily redefines the interval we are discussing. I do not see why I can’t change the subdivision without changing the interval.
Edit 2: As I know understand, the picture that I have is incorrect in that the bottom image should have its interval redefined to $2A$ and $2B$. However, I still do not understand why the change in interval demands that a corresponding change in subdivision must occur.
In the Riemann sum, if you widen the intervals , say from $\Delta x$ to $ 2 \Delta x$, then the number of intervals is half, to cover the same interval in $x$, therefore the sum in the limit does not change.
If I got properly your comment, you are concerned with the doubt :
" up to which point I can modify the subdivision of the $x$ interval from a standard linear (equal $\Delta x$) into an arbitrary subdivision (let's call it non-linear), without altering the limit of the Riemann sum ?".
To that, I am going to answer to you just intuitively ; others more expert in the field might give a more rigorous answer.
And the intuition is that given in the sketch: until the function that determines the partition is bi-jective the limit of the sum will not change.
That is because, where you are taking smaller $\Delta x'_{\,k} $, you are taking more bars in the histogram, so that the weighted sums $f(x'_{\,k}) \cdot \Delta x'_{\,k} $ and $f(x_{\,k}) \cdot \Delta x_{\,k} $ are going to coincide in the limit.