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The derivative $\frac{dy}{dx}$ of a function $y=f(x)$ tells us how has the function $y=f(x)$ changes with the change in $x$ at the point $(x,y)$. What is the physical meaning of the integral of the function $y=f(x)$ i.e., $I(a,b)=\int\limits_{a}^{b} f(x)dx$ except the fact that it represents the area under the curve bounded by $x=a$, $x=b$ and $y=f(x)$?

To be specific the work done under a force, in one-dimension, is given by $\int F(x)dx$. Why should it be called a continuous sum?

How does the area interpretation work out if the function being integrated is a function of several variables?

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    $\begingroup$ $\int_a^b f(t)dt$ shows the total distance traveled by the particle from $a$ to $b$. $\endgroup$
    – Nosrati
    Oct 24, 2017 at 13:54
  • $\begingroup$ @MyGlasses: not at all ! That distance would be $\int_a^b\sqrt{1+f'^2(t)}\,dt$, from $(a,f(a))$ to $(b,f(b))$. $\endgroup$
    – user65203
    Oct 24, 2017 at 13:57
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    $\begingroup$ You understand, do you not, that a mathematical operation (addition, multiplication, differentiation, integration, etc.) does NOT HAVE "physical meaning"? A mathematical operation can be used in many different ways, applied to "physical" problems (or non-physical problems). There is NO one "physical" meaning. $\endgroup$
    – user247327
    Oct 24, 2017 at 14:00
  • $\begingroup$ Instead of considering $\int F(x)dx$, try considering $\frac{d}{dx}W(x)$. $\endgroup$
    – John Joy
    Oct 24, 2017 at 14:39

3 Answers 3

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Here is a nice video by 3Blue1Brown with visuals

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  • $\begingroup$ I was just about to recommend the same thing, that video is genius $\endgroup$
    – Harry Alli
    Oct 24, 2017 at 13:54
  • $\begingroup$ ikr, the whole channel is awesome :) $\endgroup$
    – Xhaka
    Oct 24, 2017 at 13:56
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    $\begingroup$ Link only answers are discouraged here. Please edit to summarize what the video does as well as provide the link. $\endgroup$ Oct 24, 2017 at 13:57
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To the extent that one interprets a differential form representing the "infinitesimal" variations of a function, the fundamental theorem of calculus can be directly seen as taking the infinitesimal variation and producing the total variation:

$$\int_{x=a}^{x=b} \mathrm{d}f(x) = f(b) - f(a) $$

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  • $\begingroup$ This always makes sense in 1D, but it is not guaranteed to make sense in 2D or 3D, which is where we do physics. $\endgroup$
    – Ian
    Oct 24, 2017 at 14:31
  • $\begingroup$ @Ian: In physics, one computes path integrals, which are 1D. There are higher dimensional versions of the FTAC as well; e.g. Stokes' theorem, Green's theorem, or simply the generalized Stokes' theorem. $\endgroup$
    – user14972
    Oct 24, 2017 at 14:31
  • $\begingroup$ Yes, except that not all line integrals are of the form $\int df$ when the underlying space has dimension greater than 1. $\endgroup$
    – Ian
    Oct 24, 2017 at 17:01
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When integrating the displacement function, call it $f(x),$ you compute the "absement.$ In kinematics, the absement is the measure of sustained displacement. (Wikipedia)

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