Without calculus or integrals, how can you prove that the area under 1/x stays the same, if decomposed into increasingly large segments?

Question

Can someone please elaborate the embolded sentence below? How can you prove that "the area under the curve stayed the same" without integrals?

I'm going to go back in time to the development of the logarithm. The log was a way of performing multiplication (in some base) by simply using addition. Meaning, every time you add one number, your output is proportional to the last output.

At some point in time shortly afterwards, mathematicians were working on integration techniques. They noticed that the basic methods of finding anti derivatives worked for everything except for 1/x. However it was realized, that if you break down the graph of 1/x, into increasingly large segments (1-2, 2-4, 4-8, 8-16...) That the area under the curve stayed the same. this is actually very easily provable! But now look what they found..... They found that the area under the curve increases by simple addition as you double the interval. (Meaning adding the area interval (1-2) + (2-4) + (4-8), is simply the same area added together 3 times).

This should stick out to you, because this is exactly how we defined a logarithm to work. Addition carrying out multiplication with some base number.

Because this logarithm seemed to be so "natural" (it just came about all by itself), it was called the "natural logarithm". So the relation was known that ln(x) = integral of 1/x, before it was known what that base number was. Euler was actually able to calculate the number, and the proof for that is very interesting as well! (See my paper for the proof).

Answer 1

We can try to use the definition of the integral, not the fundamental theorem of algebra (which tells us that we can calculate integrals by using antiderivatives). Let $A= \{x=x_0 < x_1 < \ldots < x_n=y\}$ be any subdivision of $[x, y]$. Then $2A = \{2x_0 < \ldots < 2x_n \}$ is a subdivision of $[2x, 2y]$. The corresponding lower sums are $$ S_A = \sum_{k=0}^{n-1} \frac{x_{k+1} - x_k}{x_{k+1}} $$ and $$ S_{2A} = \sum_{k=0}^{n-1} \frac{2x_{k+1} - 2x_k}{2x_{k+1}} = \sum_{k=0}^{n-1} \frac{x_{k+1} - x_k}{x_{k+1}} = S_A $$ The same holds for the upper sums. Taking the limit for finer subdivisions, we get that the area under $x \mapsto x^{-1}$ over the intervals $[x,y]$ and $[2x,2y]$ is the same.