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I have this confusion while studying indefinite integrals. Is dx a derivative of x or a notation?

If it is just a notation,how can we explain that the transformation from dx to du satisfies the operational law of differentiation?

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    $\begingroup$ No, the $dx$ is part of the "overall" integral notation. It specifies the variable used for the integration (maybe the function has more than one variable). $\endgroup$ Dec 6, 2023 at 10:24
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    $\begingroup$ It is just a notation to indicate what is the variable used for integration. For example if $f$ is a function of 2 variables, you could write $F(y)=\int_a^b f(x,y) dx$ $\endgroup$ Dec 6, 2023 at 10:25
  • $\begingroup$ It's the same as for definite integrals. $\endgroup$
    – stange
    Dec 6, 2023 at 10:26
  • $\begingroup$ It means you integrate with respect to the variable $x$. Beyond other things, this notation is very useful for variable change. If you change variables as $x=f(t)$ then the derivative by Leibniz notation is $\frac{dx}{dt}=f'(t)$, and you can actually turn the $dx$ in the integral to $f'(t)dt$. (of course this is justified by a theorem) $\endgroup$
    – Mark
    Dec 6, 2023 at 10:29
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    $\begingroup$ I like to think of $\int()dx$ as an operator $\endgroup$
    – DatBoi
    Dec 6, 2023 at 12:38

3 Answers 3

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Let's backtrack a bit and think about this object we call $$A = \int f(x)\ \mathrm dx,$$ where $f$ is a given function. If you are studying integration at the moment you must have seen a picture that looks something like this:

enter image description here

That is, when we find the integral, we are essentially chopping off all of the little bits that lie under the curve, finding their individual areas, and summing them all up. Each of those little bits have length $\mathrm dx$; the symbol $\mathrm d$ really just specifies that together with the thing after it (in this case $x$, hence $\mathrm dx$), is an indefinitely small quantity - as tiny as practically possible. Clearly, if we break $x$ into very little bits $\mathrm dx$, and sum them all up together, we get $x$. How do we express this sum? The symbol $\int$. So we do write $$\int \mathrm dx = x.$$

Here's a hopefully helpful example. Think of your entire life; let's say you're exactly 13 years old at the moment; not a second older. This means that you've lived for 409,968,000 seconds in total. If you like, you can think of every passing second as a little bit, and your entire life as being composed of 409,968,000 such little bits.

Now let's go back to $A$. Think about the curve above. We have broken the line from $a$ to $b$ into little bits each of length $\mathrm dx$, and by doing so we have formed the little rectangles going from the $x$ axis up to the curve $f$. Thus the height of these rectangles is whatever $f$ is for that $x$; i.e. $f(x)$. The area of a rectangle is the product of its height and width, so the area of each of the really thin rectangles we have created is $\mathrm dA = f(x) \cdot \mathrm dx$, where we've again written $\mathrm dA$ because we expect this area to be really small. The total area under the curve is the sum of each of these tiny little areas, namely $$A = \int \mathrm dA = \int_a^b f(x)\ \mathrm dx.$$


As you learn more mathematics, you will see various notions and interpretations for the quantity $\mathrm dx$, which is more formally called the differential element for some $x$ associated with a certain space. In fact, if you know some linear algebra, it is true that the differentials $\left(\mathrm dx_1\right)_p, \ldots, \left(\mathrm dx_n\right)_p$ at a given point together form a basis for the vector space of linear maps from $\mathbb{R}^n$ to $\mathbb{R}$, and so if a function $f :\mathbb{R}^n \to \mathbb{R}$ is differentiable on all of $\mathbb{R}^n$, the differential at $p$, $\mathrm df_p$, is a linear combination of these basis elements, that is, $$\mathrm df_p = \sum_{i=1}^n D_if(p)\left(\mathrm dx_i\right)_p \implies \mathrm df = \frac{\partial f}{\partial x_1} \cdot \mathrm dx_1 + \ldots + \frac{\partial f}{\partial x_n} \cdot \mathrm dx_n.$$ This is a bit of a roundabout way to show that $$\mathrm df = \frac{\mathrm df}{\mathrm dx} \cdot \mathrm dx.$$

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It is merely a notation. What is meant by this expression is just a function such that its derivative is $f$, and this function should be evaluated at $x$. If you include other variables, it is important to distinguish which one is the argument of the function, in a sense what variable is being integrated.

The notation for indefinite integral is rather weird and confusing in my opinion, and I try not to use it. But it is popular in calculus courses.

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    $\begingroup$ What exactly is confusing about that notation? $\endgroup$
    – stange
    Dec 6, 2023 at 10:27
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    $\begingroup$ The definite integral $∫_a^b f(x)\,\mathrm d x$ does not depend on $x$, however, $∫f(x)\,\mathrm dx$ should depend on $x$! $\endgroup$ Dec 6, 2023 at 10:31
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    $\begingroup$ Consider that $∫ f(x)\,\mathrm dx = ∫_a^x f(y)\,\mathrm dy$ for any $a$ in the domain of $f$. $\endgroup$ Dec 6, 2023 at 10:35
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    $\begingroup$ First time I hear about this somehow being confusing. $\endgroup$
    – Mark
    Dec 6, 2023 at 10:36
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    $\begingroup$ @ZehanLi, the point is that if you have an infinitesimal rectangle of width $dx$ and you apply a change of variables $u=u(x)$, the new width $du$ will be related to the old width $dx$ by the relation $du=\frac{du}{dx}dx$. Correspondingly, the area of the rectangle will also be multiplied by $\frac{du}{dx}$; see my answer. $\endgroup$ Dec 6, 2023 at 13:26
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The $dx$ occurring in the integral is not merely a notation. In the historical infinitesimal calculus, one would partition the domain of integration $[a,b]$ into, say, equal subintervals of infinitesimal length $dx$, with partition points $a=x_0, x_1,\ldots, x_H=b$ where $H$ is an infinite (in modern terminology, unlimited) integer, and $dx=\frac{b-a}{H}$. Then form the corresponding sum $S_H=\sum_{i=1}^H f(x_i) dx$ of infinitely thin rectangles of varying height $f(x_i)$. Such a sum is not exactly the integral; the integral is obtained by discarding an infinitesimal term, or more precisely by taking the standard part of the sum $S_H$: $$\int_a^b f(x) dx = \text{st}\Big(\sum_{i=1}^H f(x_i) dx\Big).$$ As you see, retaining the $dx$ as part of the notation for the definite integral is quite appropriate. Naturally, it is retained for indefinite integrals, as well. It is also essential if one is to get reasonable-looking formulas for change of variables in integrals.

Today's non-infinitesimal approaches cannot explain the notation in this way, and therefore tend to resort to explanations in terms of 1-forms. In modern infinitesimal calculus, $dx$ is naturally an infinitesimal occurring in the Riemann sum $S_H$ (and therefore naturally also in the integral). For details, see Keisler's textbook for infinitesimal calculus.

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    $\begingroup$ The OP isn't defining the Riemann integral in terms of the infinitesimal calculus, and it wasn't defined that way historically. $\endgroup$
    – anomaly
    Dec 6, 2023 at 19:32
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    $\begingroup$ The OP didn't mention "Riemann integral but rather "integral". Leibniz and Newton already had integrals, and Cavalieri computed areas via indivisibles even earlier. @anomaly, if you brush up on your history you will discover that Cavalieri was two centuries before Riemann. $\endgroup$ Dec 7, 2023 at 10:41
  • $\begingroup$ I am aware of the history of integration. It's still irrelevant to an OP asking a question about basic, intro-level calculus. $\endgroup$
    – anomaly
    Dec 7, 2023 at 12:04
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    $\begingroup$ Infinitesimals are helpful in understanding "basic, intro-level calculus". See this publication. I can suggest some others, as well. @anomaly $\endgroup$ Dec 7, 2023 at 12:07
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    $\begingroup$ @anomaly, I suggest you consult some works by the leading educator David Tall. He argues that infinitesimals are a natural product of the human imagination. Trained in the Weierstrassian tradition, you have been taught to think that $\frac{dy}{dx}$ is "just notation", inspite of much evidence to the contrary, such as the standard procedure for converting differential equations into integrals. Calculus is much easier understood via infinitesimals than via epsilon-delta, man. $\endgroup$ Dec 7, 2023 at 12:23

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