 Solution

An orientation-preserving reparametrization has the form $\tilde{\gamma }(s)=\gamma (h(s))$, where $h:[\alpha ,\beta ]\to [a,b]$ is an increasing, piecewise smooth change of parameter with $h(\alpha )=a$, $h(\beta )=b$. For regular parametrizations one may take $h$ and its inverse continuously differentiable on each piece. The chain rule gives ${\tilde{\gamma }}^{\prime }(s)={\gamma }^{\prime }(h(s))h^{\prime }(s)$. The change of variables $t=h(s)$ therefore yieldsThis proves reparametrization invariance of a vector line integral. If $h$ is decreasing, the endpoints in the final integral are reversed and the sign changes. Thus the same direction and the same multiplicity of traversal are essential.