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This question already has an answer here:

Jerk, $j(t)=\dddot x(t)$, acceleration, $a(t)=\ddot x(t)$ and velocity, $v(t)=\dot x(t)$ are derivatives of position, $x(t)$. What is the meaning of the integral of position?

For example, one could write

$$\int x \mathrm{d}t = \int x \frac{\mathrm{d}t}{\mathrm{d}x}\mathrm{d}x = \int x\frac{1}{v} \mathrm{d}x$$

which reminds the equation for expected value.

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marked as duplicate by sammy gerbil, stafusa, user259412, M. Enns, Kyle Kanos Apr 23 '18 at 10:21

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  • $\begingroup$ if you perform the integral over a bounded domain (say $0\leq t\leq T$) than this gives the mean-position ($T$-times it). There is no further meaning to my knowledge $\endgroup$ – Bort Dec 14 '15 at 14:45
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I might be wrong, but I guess you need a more primitive concept than position, call it S, that changes at a constant rate if position were constant: dS=xdt. One way to build such a variable is to assume that space expands, and this expansion makes everything expand, including rulers, so that positions will remain the same as space expands. Your new variable S=∫xdt, is the total "amount of space" created during the interval Δt. Of course, unless you could somehow measure this amount of space, this variable would remain useless from a practical point of view

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