The value of the infinitesimal in integral doesn't matter?

Question

I am studying calculus by the infinitesimal approach using "Elementary Calculus: An Infinitesimal Approach" textbook. in page 187, the author proved that the value of the infinitesimal we integrate with respect to doesn't matter as long as it is an infinitesimal. But it looks like he substituted $x$ with $u$, and $dx$ by $du$, it is like a change of variable where $x = u$, so $dx$ and $du$ might have different values. Generally, we have to substitute $du$ with $g'(x)dx$ . I am confused between the two notions, especially that first wasn't discussed in college.

Answer 1

If you define the integral of continuous functions as $$ {\rm std}\left(\sum_{x\in I}f(x)dx\right) $$ where

• $I$ is some subdivision of the interval $[a,b]$ with an inf-large number of inf-small sub-intervals,
• $dx=x^+-x\approx 0$ for all $x\in I$, with $x^+$ the successor of $x$ in $I$,

then indeed the value of the integral is independent of the subdivision.

You can parametrize $I$ over some other subdivision $J$ using a monotonously increasing differentiable function $g$ and define $x=g(u)$ for $u\in J$ so that $$dx=x^+-x=g(u^+)-g(u)=g'(u)du+O(du^2).$$ As $\max_{u\in J}|du|\approx 0$ the last term remains inf-small in the sum so that indeed $$ {\rm std}\left(\sum_{x\in I}f(x)dx\right)={\rm std}\left(\sum_{u\in J}f(g(u))g'(u)du\right) $$ in this notation.