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 ? 

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 
