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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.

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    $\begingroup$ quora.com/… $\endgroup$ – D.R. Oct 27 '18 at 21:47
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    $\begingroup$ I suggest you get a different book for it appears the current book is misleading you. Infinitesmals are only an intuitive way of understanding integration. $\endgroup$ – William Elliot Oct 27 '18 at 22:23
  • $\begingroup$ Sorry, I don't get how this is related to the question. @D.R. $\endgroup$ – Samuel Shokry Oct 27 '18 at 22:38
  • $\begingroup$ The limits weren't intuitive to me, the infinitesimals work better for me. @WilliamElliot $\endgroup$ – Samuel Shokry Oct 27 '18 at 22:39
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    $\begingroup$ You should have a look at this question and the answers and comments therein: math.stackexchange.com/q/2929795/72031 The fact you ask is dependent on uniform continuity. But this aspect is not explicitly mentioned by Keisler. $\endgroup$ – Paramanand Singh Oct 28 '18 at 9:56
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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.

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  • $\begingroup$ I guess this would be a stupid question, but why do we have to substite $dx$ by $g'(u)du$, Shouldn't the answer be independent of the subdivision as you have mentioned? of course that would yield a different answer, but I can't figure out why, Is it because of the parametrizing step? could you explain a little bit more?. $\endgroup$ – Samuel Shokry Nov 6 '18 at 23:57
  • $\begingroup$ You get a different subdivision in the above formulas by setting $g(u)=u$. However, that equality already used the more fundamental fact that the integral value is independent of the subdivision. You want to have substitution available as tool in case that the product $f(g(u))g'(u)$ dramatically simplifies. $\endgroup$ – LutzL Nov 7 '18 at 0:07

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