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In this example,

If we use the relation $F'(x) = f(x)$, this consequence of the fundamental theorem may be written in the form

$$ f(b) - f(a) = \int_a^b F'(x) dx = \color{#c66}{\boxed{\color{black}{ \int_a^b \frac{dF(x)}{\color{red}{dx}}\color{red}{dx} = \int_a^b dF(x)}}}, $$

In the last equation of this example, obviously, the two $dx$ are canceled in the left before we get the right, this indicates that the relationship between $f(x)$ and $dx$ within $\int_a^b f(x)\,\mathrm{d}x$ is multiplication , but I cannot find a reasonable explanation for this. Using infinitely small quantities to explain it is wrong since infinitely small has no place in modern mathematics saying here .

Anyone can explain why the relationship between $f(x)$ and $dx$ in $\int_a^b f(x)\,\mathrm{d}x$ is multiplication? what does $dx$ mean here ?

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    $\begingroup$ So first of all, what exactly do you mean when say that the relationship between $f(x)$ and $\mathrm{d}x$ is multiplication? As for your question, I suggest to study the proof of the fundamental theorem of calculus. Also it might help to study an analysis course which covers integral without the notation $\mathrm{d}x$, which only seems to cause problems and misunderstandings. $\endgroup$ – Mathematician 42 Apr 24 '16 at 6:25
  • $\begingroup$ That book in the link is not the one-stop definitive source on what's right and what's wrong in mathematics. There is nothing wrong with infinitesimal quantities. Also is that image with (31) even from the same book or author(s)? If not, then you're trying to reconcile something from one book using the rationale contained in a different book or put forth by a different person. Not always a fruitful endeavor. $\endgroup$ – tilper Apr 24 '16 at 7:11
  • $\begingroup$ @tilper yes, from the same author ,just another book here books.google.ca/… $\endgroup$ – iMath Apr 24 '16 at 7:30
  • $\begingroup$ @Mathematician42 never saw an analysis course or calculus without $dx$ or $dy$, could you give an example ? $\endgroup$ – iMath Apr 24 '16 at 10:03
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    $\begingroup$ @iMath where did you get the idea that infinitesimals have no place in modern mathematics? $\endgroup$ – Mikhail Katz May 2 '16 at 9:27
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It's not really multiplication. It's actually largely just notational.

$dx$ can be interpreted as the differential of $x$.

$dF(x)$ can be interpreted as the differential of $F(x)$, which is actually just $F'(x)\, dx.$ This comes from the fact that $\dfrac{dF(x)}{dx} = F'(x).$ We can move the $dx$ to the other side of the equation to get $dF(x) = F'(x) \, dx.$ This is known as separating the variables. This looks like we're multiplying both sides by $dx$ but we're not. We can't multiply by $dx$ because $dx$ doesn't actually represent a number. It represents an infinitely small quantity, somewhat similar to how $\infty$ represents an infinitely large one.

So, when we simplify $\dfrac{dF(x)}{dx} \, dx$ to just $dF(x)$ it's just a matter of convenient notation.

Here's a concrete example following the same pattern as (31) in the provided image, where in this example I use $F(x) = \sin x.$ So then $F'(x) = \cos x$ and we have:

$$\sin 2 - \sin 1 = \int_1^2 \cos x \, dx = \int_1^2 \frac{d \sin x}{dx} \, dx = \int_1^2 d\sin x$$

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  • $\begingroup$ If you think dx as an infinitely small quantity to explain this question , I recommend you to read this before you answer this question books.google.com.hk/… $\endgroup$ – iMath Apr 24 '16 at 6:40
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    $\begingroup$ @iMath If you have also read what you recommend, then you should "be careful not to attach too much significance to what is, after all, a pure convention as to how the limit shall be denoted". $\endgroup$ – lastresort Apr 24 '16 at 6:51
  • $\begingroup$ yes, I did read it, but in calculation, the relationship between f(x) and dx in $\int_a^b f(x)\,\mathrm{d}x$ is treated as if it is multiplication, I just wonder why $\endgroup$ – iMath Apr 24 '16 at 6:54
  • $\begingroup$ @iMath, I opened the link before answering but decided it was pointless to guess exactly which parts of that you thought were relevant to the question, especially since I couldn't find anything relevant. My answer is more than "dx is an infinitely small quantity, the end." If my post doesn't help answer your question then please ask specific follow-up questions on what isn't clear to you rather than giving reading recommendations. $\endgroup$ – tilper Apr 24 '16 at 6:58
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    $\begingroup$ @iMath, what makes you feel that infinitesimals don't have a legal place? Which law have they violated? $\endgroup$ – Mikhail Katz May 2 '16 at 10:30
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Your query could be better understood in geometrical manner as follow:

In the integral

$\int_{a}^{b} f(x)\, dx$

The product $f(x)\,dx $ is nothing but multiplication of height and breadth which would actually constitute area of elementary strip below curve $f(x)$ and combination of areas of these strips from $a$ to $b$ constitute total area under $f(x)$ .

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In Keisler's calculus book it is explained how the expression $f(x)dx$ can indeed be viewed as a product. The dismissive comment about infinitesimals that you quote seem to come from the book by Courant and Robbins. This was written before Abraham Robinson introduced modern infinitesimals in 1961. Some of Galois' contemporaries also laughed at him when he introduced group theory.

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