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 ? 

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 $g(x)=2x$
