Reversing a path changes the sign of a line integral

Statement

Let $\gamma \colon [a, b] \to \mathbb{R}^n$ be a piecewise $C^1$ curve with image $C$, and let $\gamma^- \colon [a, b] \to \mathbb{R}^n$, $\gamma^-(t) = \gamma(a + b - t)$, be the reversed curve. For each continuous vector field $F$ on $C$ and each continuous function $f$ on $C$, $$\int_{\gamma^-} F \cdot d\mathbf{r} = -\int_\gamma F \cdot d\mathbf{r}, \qquad \int_{\gamma^-} f \, ds = \int_\gamma f \, ds.$$

Remarks

Proof: by the parametric formula, $(\gamma^-)'(t) = -\gamma'(a + b - t)$. The substitution $u = a + b - t$ then gives both identities. The line integral of a vector field depends on the orientation of the curve. The line integral of a scalar field does not, because it uses only $\|\gamma'\|$. More generally, under a $C^1$ change of parameter $\varphi$ with $\varphi' > 0$, the vector line integral does not change. With $\varphi' < 0$ it changes sign.

Explanations

L1 human

If a force helps you along a path, it works against you on the way back. So the work, a line integral of a vector field, changes sign when you reverse the path. The length of the path and the mass of a wire do not.