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I know that he Riemann Stieltjes integral is a generalization of the Riemann integral, i.e. we integrate one function say $f$ with respect to another say $g$ and then we write $\int f(x)d(g(x))$.

But I'd like to see it geometrically, or at least to try.

Because I can see the geometric of the Riemann integral which is the area under the function $f$, but how is for the Stieltjes-integral??

user441848
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  • Short answer is you can think of the $g$ as doing a kind of weighted average on the interval over which you're integrating in a possibly non-uniform way (which is the case of the standard Riemann integral, putting $g = \mathrm{id}_{\Bbb{R}}$). – YoTengoUnLCD Jun 12 '17 at 01:24
  • I am not sure that visualising is the right term, but suppose $g$ is $C^1$ then the integral is $\int f g' $. – copper.hat Jun 12 '17 at 05:05

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