I come back again only to confirm (or not) a generalization. In my post on yesterday the integral
$$ \int_{0}^{6}(x^2+[x])d(|3-x|) $$
was worked out based on a change of variable. I tried to get the same solution in another way - with integration by parts .
After cancelling the absolute value and considering the second integral(due only to simplicity purpose) we have :
$$\int_{3}^{6}(x^2+[x])d(x-3)$$ whose solution is 66
Working the integral....
$$ =\int_{3}^{6}x^2d(x-3) + \int_{3}^{6}[x]d(x-3)$$
$$ =\int_{3}^{6} x^2 dx + ( f(6)a(6) - f(3)a(3)) - \int_{3}^{6} (x-3)d[x] $$
(integration by parts with $f(x)=[x]$ and $a(x)=x$ )
$$ = (72-9) + (18-0) - \int_{3}^{6} x d([x] ) + \int_{3}^{6} (-3)d([x] )$$
$$= (72-9) + (18-0) - (4+5+6) + \int{3}^{6} (-3)d([x] )=66 + 0 =66 $$
It seems reasonable for the last integral to be 0, as floor function [x] doesn't have derivative. Hence, as a generalization we may say that
$$\int_{a}^{b} K d([x-w]) = 0 $$
for every k, w real numbers.
Am I right? And if so how can we prove that?
Thks for cooperation Regards João Pereira