Line integral and dependenece on direction I have been doing some mistake for quite a while now and I don't see it. Therefore, I wanted to ask you whether anybody here could tell me what I have been doing wrong:
$\gamma:[0,1] \mapsto \mathbb{R}^n$, $t \mapsto x+t(y-x)$
$\int_{\gamma} f(s) ds= \int_0^1 f(\gamma(t))||\gamma'(t)|| dt  = -\int_1^0 f(\gamma(1-u))||y-x||_2 dt =\int_0^1 f(\gamma(1-u))||y-x||_2 dt.$
Obviously this is not what one would expect after the substitution ( a minus sign is missing, cause now I am going along the curve from the other direction). So please, could anybody here tell me where my stupid mistake is?

When you do the line integral with respect to arclength (as opposed to a work integral), the answer is independent of the orientation on the curve. If you think of $f$ as a density function on a thin wire, for example, the total mass doesn't depend on which direction you travel.
