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Why indefinite integral has such a notation ? What part "dx" has to do with the indefinite integral, and why f(x) is multiplied by it ?

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  • $\begingroup$ Have you searched the site for related questions? $\endgroup$ – Mariano Suárez-Álvarez Jan 12 '18 at 16:50
  • $\begingroup$ @MarianoSuárez-Álvarez yes, a bit, but I haven't found why f(x) is multiplied by dx. $\endgroup$ – Юрій Ярош Jan 12 '18 at 16:52
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There are two good reasons:

  • when there are several variable, the differential indicates on which variable you want to integrate;

  • when you apply a change of variable, say $x=\phi(t)$, you have to multiply the integrand by $\dfrac{dx}{dt}$ and this is expressed in a very natural way by

$$\int f(x)\,dx=\int f(x(t))\frac{dx}{dt}dt=\int g(t)\,dt.$$


In a much less rigorous way, you can consider that the $d$ and $\int$ "operators" are inverses of each other, so that

$$\frac d{dx}\int f(x)\,dx=\frac{f(x)\,dx}{dx}=f(x).$$

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The notation is chosen to match the notation for definite integrals. That notation, in turn, reflects the view of the definite integral as a limit of Riemann sums. The $\int$ symbol is an elongated S, standing for "sum" (and taking the place of $\sum$ in a discrete sum), and the $dx$ stands for an infinitesimally small difference in $x$ (corresponding to the discrete $\Delta x$). The idea is that the Riemann sum $\sum f(x) \Delta x$ takes the height of the function (the value of $f(x)$), multiplied by the length of the interval to the next point (i.e., the difference in $x$), at a discrete set of points; the integral represents doing the same thing at a continuous set of points with infinitesimally small intervals. So, it is as if at each point, the height of the function $f(x)$ is multiplied by the (infinitesimal) length of interval $dx$, and then all the (uncountably infinite) values are summed.

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