Is it possible to calculate for example $\int_{0}^{1} x \mathrm{d}2x$ My question is just for fun, but I want also to verify if I understand something in variation calculus... 
I want to know if it is possible to calculate this : 
$$
\int_{0}^{1} x \mathrm{d}2x
$$
A geometric argument is enough to conclude the area is an half, but here I have a two in my integral and I hope I can't find a result of 1... like this 
$$
\int_{0}^{1} x \mathrm{d}2x = 2\int_{0}^{1} x\mathrm{d}x = 1
$$
So it looks like a false result ! Can you explain me why I'm wrong ? 
 A: Well it is not a contradiction, actually the real value is 1, because in your case you can apply $$ \int f(x) dg = \int f(x)\cdot g'(x) dx. $$This gives exactly $$ \int_{0}^1 x d(2x) = \int_{0}^1 x\cdot 2 dx = 2\int_{0}^1 x dx =1 $$
OBS: Since you are working on a Riemann-Stieltjes integral you are NOT computing areas,also the change of parameters needs to be justified to be a valid approach 
A: using the integration by parts of Riemann–Stieltjes integral we get $$\int_{a}^{b}f(x)dg(x)=-f(a)g(a)+f(b)g(b)-\int_{a}^{b}g(x)df(x)$$ now substitute $a=0$ , $b=1$ and $f(x)=x$ and $g(x)=2x$
A: May I propose you another, more geometric, approach?:
$$\int_0^1 x\,d(2x)=\frac12\int_0^1 2x\,d(2x)$$
Think of the above integral as usually done in high school: that gives the area between the identity function $\;f(2x)=2x\;$ and the $\;2x$-axis . 
In this "scaled up" axis system, we have the area of a straight angle triangle with vertices at the origin $\;(0,0)\;$, at $\;\left(0,2\right)\;$ and at $\;(2,2)\;$ , and thus its area is $\;\frac{2\cdot2}2=2\;$ . This, times $\;\frac12\;$ , gives us the correct solution of $\;1\;$ .
A: $$\int\limits_0^1x\,\mathrm d(2x)\stackrel{u=2x}=\frac{1}{2}\int\limits_0^2u\,\mathrm du$$
How are the limits decided? This way:
$$u=2x\implies \begin{cases}x=0\implies u=0\\ x=1\implies u=2\end{cases}$$
Can you continue?
