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It is commonly taught that to integrate a function f(x), with respect to x, from x = a to x = b, one calculates:

$$\int_a^b f(x) \ dx$$

We add dx at the end of the integral to show that we are summing up the area of an infinite number of thin blocks, where the height of each block is given by f(x) and the width is given by dx. In this way, we can calculate the area bounding by the x axis, f(x), x = a, and x = b.

However, what about integrals that were meant to calculate certain other quantities, such as arc length? The formula for the arc length of a function ends with dx, but dx exists for another reason. This reason is that when deriving the arc length formula, $$(\Delta\ x)^2$$ appears under a square root, leaving $$\Delta\ x$$

Then, as $$\Delta\ x$$ becomes the infinitesimal dx, and the summation becomes an integral, we see the formula become an integral that ends with dx. This is not the same as an integral that ends with dx in order to find the area under a curve. Yet somehow, all integrals, whatever their purposes, end with dx. What is the general meaning of dx in an integral, seeing that all integrals must end in dx?

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  • $\begingroup$ Are you sure this question does not already exist ? $\endgroup$ – LL 3.14 May 17 '20 at 20:29
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    $\begingroup$ They don't end with $dx$. They contain $dx$. Granted, it's common to write it at the end, but it isn't mandatory. $\endgroup$ – Arthur May 17 '20 at 20:31
  • $\begingroup$ The simple answer is a notational inheritance from $\sum f(x)\Delta x$ to $\int f(x)\mathrm dx$. The longer answer has to delve in to differential forms where a statement like $\int_M \mathrm d\omega=\int_{\partial M}\omega$ makes sense. $\endgroup$ – Hagen von Eitzen May 17 '20 at 20:33
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    $\begingroup$ In mathematics, usually, $\mathrm{d}x$ is put at the end of the integral. But actually, several notations are rather equivalent. You could write also $∫ f = \int f(x)\,\mathrm{d} x$. $\endgroup$ – LL 3.14 May 17 '20 at 20:34
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    $\begingroup$ @HagenvonEitzen Some physicists would beg to differ :) $\endgroup$ – cmk May 17 '20 at 20:39
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So, if you have a function $f$, the arc length formula tells you that the length of $(x,f(x))$ for $x$ going from $a$ to $b$ is $$ \int_a^b \sqrt{1+f'(x)^2}\,\mathrm{d}x $$ And actually, it is also the area under the function $\sqrt{1+f'(x)^2}$, this is the remarkable fact that the theorem tells you. Then, not all integrals end by $\mathrm{d}x$, which correspond to integrals with Lebesgue measure. More generally you can have $$ \int f(x)\,\mu(\mathrm{d}x), $$ and in the case of the Lebesgue measure, you have the common notation $∫ f$ instead of $∫ f(x) \mathrm{d}x$.

Moreover, in the case when $g$ is a function of bounded variations, then $μ = g'$ is a measure and there is the Stieltjes integral notation $$ ∫f(x)\,\mu(\mathrm{d}x) = ∫f(x)\,\mathrm{d}g(x) = ∫f\,\mathrm{d}g $$

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