In differentiation, we denote $dx$ as in infinitesimal change in $x$ and $dy$ the corresponding change.But in integration, we divide $x$ into infinitely many equal intervals and name them $dx$. Then $f(x)dx$ represents the area of an infinitely thin rectangle. From differentiation we know, $dy = f(x)dx$.Why do we substitute then $dy$ in place of $f(x)dx$? I mean $dy$ is change in differentiation. But here in integration we take $dy$ to be the area of each infinitesimal rectangle. Why do we mix them up? Please someone explain with clarity.
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There are some wonderful answers on this thread. – jlammy Jan 29 '21 at 14:06
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Yeah i went through this before but my question is entirely different. – a_i_r Jan 29 '21 at 14:10
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1We don't put $dy$ in place of $f(x)dx$. We put $dy$ in place of $f'(x)dx$. If we set $y = f(x)$, then $dy = f(x + dx) - f(x)$ (by definition if you like). But $f(x + dx) - f(x) = f'(x) dx$ (again, by definition if you like). Thus $dy = f'(x) dx$. This is all in the language of infinitesimal calculus. There are many ways to look at these symbols. As purely formal objects (this is the traditional approach), as one-forms, and so on. – Charles Hudgins Jan 29 '21 at 14:28
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Related, possibly helpfukL https://math.stackexchange.com/questions/1991575/why-cant-the-second-fundamental-theorem-of-calculus-be-proved-in-just-two-lines/1991585#1991585 – Ethan Bolker Jan 29 '21 at 15:31
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Without being formal, $dy = f'(x)dx$ not $dy = f(x)dx$
So $dy$ does actually stand for a small change in $y$. Since you're integrating $f'(x)dx$, it would give you the change in the antiderivative of $f'(x)$, which is $f$ which is equal to $y$.
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$dx$ and $dy$ are not infinitesimal changes, this is an old, obsolete legend. Here is the truth: https://en.wikipedia.org/wiki/Differential_of_a_function.
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1According to the link, "Traditionally, the variables dx and dy are considered to be very small (infinitesimal), and this interpretation is made rigorous in non-standard analysis." – johnnyb Jan 29 '21 at 15:20
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@johnnyb While that is true, it has often been my experience that many people who invoke the use of infinitesimals either (1) don't fully understand the framework needed to make the use of such objects rigorous, or (2) they understand the framework but will resort to standard definitions/ways of thinking when convenient. If you're going to use infinitesimals to teach/develop Calculus, the way you think about other objects also has to change accordingly. For instance, in non-standard analysis, we no longer have the limit definition of a derivative (limits are no longer necessary). – Michael Jan 29 '21 at 15:48
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@johnnyb Instead, the derivative of a function $f$ at $x$ as defined in non-standard analysis is $$f'(x) = \text{st}\left(\frac{f(x+h) - f(x)}{h}\right)$$ where st is what's known as the "standard part" function. – Michael Jan 29 '21 at 15:56
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@johnnyb: the article explains that the differential is the linear part of the variation and is a finite, arbitrary quantity. Nothing non-standard. – Jan 29 '21 at 16:01
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I agree that "many people who invoke the use of infinitesimals either (1) don't fully understand the framework needed to make the use of such objects rigorous", but the fact is that the same thing is true of real numbers. The question is whether or not the simplifications in thought that are likely to be made will be in the correct direction or not. I've found that basic infinitesimal thinking generally leads to clearer understanding of the processes than those based on the reals, and prevents far more common mistakes than it causes. – johnnyb Jan 29 '21 at 16:11
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@johnnyb: I don't think we are in phase. The article has no connection to infinitesimals. – Jan 29 '21 at 16:13
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You said differentials as infinitesimals were an obsolete legend, but pointed to an article that, while not specifically focused on infinitesimals, said that differentials as infinitesimals were perfectly sound. – johnnyb Jan 29 '21 at 16:23
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@johnnyb Fair enough - I agree that we don't always rigorously understand everything that we work with. However, while thinking of $dx$ as an infinitesimal might make for a good pedagogical tool to develop heuristics in students seeing Calculus for the first time, this thinking doesn't ultimately result in sound Mathematics (in the usual framework in which Calculus is developed) under careful scrutiny. – Michael Jan 29 '21 at 16:25
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But that's the thing - it now does. Prior to the 60s, it was just heuristics. Now it is a heuristic that can be continued into a theoretically sound mathematics. Which is how we teach the rest of mathematics. – johnnyb Jan 29 '21 at 16:39
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No that's a drastic over-simplification of the situation. Sure, Non-standard analysis rigorously develops Calculus based on infinitesimals, but this development of Calculus looks VERY different from Calculus taught to students using limits and $\epsilon-\delta$ arguments. If you're using limit definitions of things in your Calculus, the differentials are not infinitesimals. Likewise, if your differentials are infinitesimals, then you are implicitly in the realm of non-standard analysis and you will certainly not be using limit definitions in THAT development of Calculus. It's one or the other. – Michael Jan 29 '21 at 16:49
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The answer is "the fundamental theorem of calculus". The fundamental theorem of calculus shows the deep connection between slopes and areas. Because of the fundamental theorem of calculus, $dy$ is both the $y$-component of the slope of $f(x)$ and the area of the individual boxes $f'(x)\,dx$. That's the fundamental theorem of calculus.
However, I think viewing calculus as fundamentally about slopes and areas is a wrong way of going about it. I think a simpler way to think about it is this:
- Differentiation is about infinitesimal changes
- Integration is about adding together infinitesimal changes
Here, you can easily see why integration works: you are adding up changes to get the total change (and the total change will be off from the `true value' by a constant, $+C$).
That is, if I differentiate $x^2$, I will get $2x\,dx$. This means that the value of the change in $x^2$ depends on both (a) the current value of $x$, and (b) the amount of (infinitesimal) change in $x$. If I add up all of those changes (I integrate it), I will get the total change. In other words,
$$\text{total change in }x^2 = \int 2x\,dx$$
If I make a graph of this, it will be the same as the original graph, but perhaps off by a constant - which is precisely what integration does! So integration gives us a mechanism to add together an infinite number of infinitely small values.
So, how does this relate to areas?
The area under a curve is simply adding together all of the infinitely small rectangles under it. Each rectangle is $y$ high and $dx$ wide, or $y\,dx$. Therefore, the sum of all the individual rectangles is $\int y\,dx$.
Now, we know enough to find the relationship between slopes and areas. To find the slope of an equation, we take the differential and divide by $dx$. To find the area under the curve of an equation for $y$, we multiply by $y$ and integrate. Since integration and differentiation are opposites of each other, the process of finding areas is the reverse of the process of finding slopes.
johnnyb
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By the way, I give a more full presentation of these ideas in chapters 16-18 of "Calculus from the Ground Up". – johnnyb Jan 29 '21 at 14:32