I've always just accepted that the value of the line integral
$$ \int\limits_C f(x,y)\ \mathrm{d}s $$
is independent of the direction in which $C$ is traversed, but when I actually try to satisfy myself that this is true I can't make it work.
For example, if I use the parabola $y=x^2$ between $(0,0)$ and $(1,1)$ as my curve, and just integrate arc length for simplicity I get
$$ x=t, \phantom{5pt} y=t^2, \phantom{5pt} t=0...1 $$ $$ \int\limits_C \mathrm{d}s = \int\limits_0^1 \sqrt{1+4t^2}\ \mathrm{d} t. $$
However, if I choose to traverse $C$ from $(1,1)$ to $(0,0)$ then my instinct is telling me the parameterisation should be $$ x=t, \phantom{5pt} y=t^2, \phantom{5pt} t=1...0 $$ and then I guess the sign of $y'(t)$ would change, with $t$ going in the opposite direction, but it doesn't matter because it's squared anyway so $\mathrm{d} s$ remains unchanged but now the bounds have switched:
$$ \int\limits_{-C} \mathrm{d}s = \int\limits_1^0 \sqrt{1+4t^2}\ \mathrm{d} t = - \int\limits_C \mathrm{d} s, $$
which contradicts the assumption that the direction of traversal doesn't matter.
In my mind I think I had justified this assumption intuitively by thinking if you traverse a curve backwards then the sign of $\mathrm{d} s$ must change which counteracts the change of bounds but everything I've read seems to state that $\mathrm{d} s$ remains unchanged.
Can anybody offer some insight into where my reasoning has gone astray?
I think you parameterisation of the opposite direction is not valid. it should be : $x=1-t$, $y = (1-t)^2$ , $t = 0..1$.
I don't think it is valid to have the boundaries where the first is larger than the second. i.e $t = b..a$ where $ b > a$
If you will think on the length of an arc, you will see it pretty similar. you will end with negative length, thus, in order to use opposite boundaries , as you want you should multiply the norm by $(-1)$.