Line integral: $\int _{\gamma_1} F . d\gamma_1 =\pm \int_{\gamma_2} F . d\gamma_2$? Let $F$ be a continuous vector field in $\Omega$ and, $\gamma_1: [a,b] \to \Omega$ and $\gamma_2:[c,d] \to \Omega$ two curves of class $C^1$ such that $\text{Im}(\gamma_1) = \text{Im}(\gamma_2)$. Is the statement 
$$\int _{\gamma_1} F \cdot d\gamma_1 = \pm \int_{\gamma_2} F \cdot d\gamma_2$$
true or false? Justify.
I didn't think in anything to prove or disprove the statement. I would like a hint, if possible.
Thanks in advance!
 A: If we assume that $\gamma_1$ and $\gamma_2$ are $C^1$, trace out the same path, are one-one and onto and simple, then we can construct a function $x(t)$ such that $$\gamma_1\big(x(t)\big)=\gamma_2(t)$$ This implies also that $$\gamma_1'(x(t))x'(t)=\gamma_2'(t)$$ If they have the same directions then $x$ will have a strictly positive derivative, and otherwise a strictly negative derivative. We have $$\int_{\gamma_1}F\cdot d\gamma_1=\int_a^b{F(\gamma_1(u))\cdot\gamma_1'(u)\,du}$$ Now let $u=x(t),\;du=x'(t)\,dt$ and assume they have the same direction. We get the new integral $$\int_c^d{F(\gamma_2(t))\cdot \gamma_1'(x(t))\,x'(t)\,dt}=\int_c^d{F(\gamma_2(t))\cdot \gamma_2'(t)\,dt}=\int_{\gamma_2}{F\cdot d\gamma_2}$$ This is positive because $\gamma_2(c)$ will be the start of the path and $\gamma_2(d)$ the end - if they had opposite directions, we would have to switch the bounds and hence add a negative sign to the integral.
The only part of this proof which you may want to elaborate on is why there must exist such a function $x(t)$ and why it would be differentiable and strictly increasing/decreasing.
