What is the geometrical meaning of a Riemann-Stieltjes integral? Geometrical meaning of Reimann integration is  to find the area under the curve of function and $x$-axis. Due to this geometrical interpretation, all theorems on Riemann integration are easily understandable. I want to know what is the geometrical meaning of Riemann-Stieltjes integration. How can one see its graphical representation? Is it similar to finding the area between the curve and $x$-axis, or is it something else? Please someone explain with an example (without mathematical solution) with geometrical meaning.
 A: Since, Stieltjes integral is the generalization of Riemann integral so, basic ideas on R integral is must before getting into the R-S integral or simply Stieltjes integral.
Ok, i would like to give the geometrical interpretation of Steiltjes integral so that you will get the intuitive meaning of R-S integral.
GEOMETRICAL INTERPRETATION OF R-S INTEGRAL:
Here in R-S integral, let’s consider we are integrating the function $f(\xi )$ with respect to the monotonic function $\alpha (\xi )$within the interval$[a,b]$.
mathematically we represent as below,
$$\int_{a}^{b}f(\xi )d\alpha (\xi )$$
let’s get into the geometrical interpretation of above integral.
we generally take the area of the $f(\xi )$with respect to the x-axis within certain interval in Reimann integral. Similarly we are calculating the area but a bit complex than the Reimann integral.
let’s take three axes where we keep $\xi $ and the functions $f(\xi )$ and $\alpha (\xi )$ at x-axis,y-axis and z-axis respectively. Now, we erect the wall from the curve traced by those functions then we can have the junction curve(since the intersection of the planes is curve). let’s take a light source emitting the parallel rays of light along the x-axis which results in the formation of shadow behind the wall on $f− \alpha$ plane as shown in image description,
Hence, the R-S integral gives the area of the shadow.
A: Let us take a simple functions $f(x) = x$ and $g(x) = x^2$, and simple bounds of integration.
So, our integral takes the form $\int_{0}^{2} f(x) \text{d} g(x) = \int_ {0}^{2} x \text{d} x^2$.
There are, at least, two ways to solve it.
1) Figure out that $\text{d} x^2 = 2x \text{d}x$:
$\int_ {0}^{2} x \text{d} x^2 = \int_ {0}^{2} 2x^2\text{d}x = \frac{2 x^3}{3} |_{0}^{2} = \frac {16}{3}$.
2) Change of variable, say, $t = x^2$ and so $ x = t^{1/2}$, which gives us:
$\int_ {0}^{2} x \text{d} x^2 = \int_ {0}^{2^2} t^{1/2} \text{d} t = \frac{2}{3} t^{3/2} |_{0}^{4} = \frac{16}{3}$. 
(notice the change in upper bound of integration.)

In order to visualise this example let us consider the code below (in Mathematica):
Show[
 ContourPlot3D[{f == x, g == x^2}, {g, 0, 4}, {x, 0, 2}, {f, 0, 2}, 
  AxesLabel -> 
   Flatten[Style[#, 12, Italic] & /@ {"g(x)", "x", "f(x)"}], 
  Mesh -> None, ContourStyle -> Opacity[0.5],
  ViewPoint -> {1.88, -2.45, 1.29}, 
  ViewVertical -> {0.45, -0.57, 0.67}],

 ListPointPlot3D[Table[{t^2, 0, t}, {t, 0, 4, 0.001}], 
  Filling -> Bottom, PlotStyle -> PointSize[Tiny]]
 ]

You may play around with the code to change point of view and other elements.
