---
tag: 9SZXY65416
kind: proof
name: Proof that reversing a path changes the sign of a line integral
msc: ["26B12"]
proves: 0FSCXYG856
uses: [0YS8RX87F4, 0M50E4WYWJ, 0C28RT5BG7, 069P20KYQT, 0JF1XXGWB4, 0NHGY0N0JB, 0RFWW56Y22, 03HF5PMY5T, 0WS1A8E2P9]
sources:
  - url: https://en.wikipedia.org/wiki/Line_integral
    locator: Line integral of a vector field
---

## Proof

We work directly with the sums in the definitions. Reversal maps each partition of $[a, b]$ to another partition of the same mesh.

1. Let $\rho(t) = a + b - t$. It is a decreasing bijection of $[a, b]$ onto itself, and $\gamma^- = \gamma \circ \rho$.
2. Since $\gamma$ is piecewise $C^1$, there is a [[partition|069P20KYQT]] $a = p_0 < \dots < p_m = b$ such that $\gamma$ is a [[$C^1$ curve|0JF1XXGWB4]] on each $[p_{l-1}, p_l]$. By the [[extreme value theorem|0WS1A8E2P9]] there is $M \ge 0$ with $\norm{\gamma'} \le M$ on each piece, using one-sided derivatives at the $p_l$.
3. Let $s < t$ lie in one piece. By the [[fundamental theorem of calculus|0NHGY0N0JB]] in each component, $\gamma(t) - \gamma(s) = \int_s^t \gamma'(u) \, du$. With $w = \gamma(t) - \gamma(s)$, the [[Cauchy–Schwarz inequality|0RFWW56Y22]] gives $\norm{w}^2 = \int_s^t w \cdot \gamma'(u) \, du \le \norm{w} M (t - s)$, so $\norm{w} \le M (t - s)$.
4. For any $s < t$ in $[a, b]$, insert the points $p_l$ between them. The [[triangle inequality|03HF5PMY5T]] and step 3 give $\norm{\gamma(t) - \gamma(s)} \le M (t - s)$.
5. Hence each inscribed polygon of $\gamma$ has length at most $M (b - a)$. So $\gamma$ is [[rectifiable|0C28RT5BG7]], and the line integrals over $\gamma$ exist.
6. Let $[c, d] \subseteq [a, b]$, and let $P$ be a partition $c = t_0 < \dots < t_k = d$. Put $u_j = \rho(t_{k-j})$. Then $P^* = (u_0, \dots, u_k)$ is a partition of $[\rho(d), \rho(c)]$, and $P \mapsto P^*$ is a bijection between the partitions of the two intervals.
7. The terms $\norm{\gamma^-(t_i) - \gamma^-(t_{i-1})} = \norm{\gamma(u_{k-i+1}) - \gamma(u_{k-i})}$ are the terms of the polygon length of $\gamma$ for $P^*$, in reverse order. So the two polygon lengths are equal.
8. Taking suprema, the length of $\gamma^-$ on $[c, d]$ equals the length of $\gamma$ on $[\rho(d), \rho(c)]$. In particular $\gamma^-$ is rectifiable.
9. Now let $a = t_0 < \dots < t_k = b$ with points $\tau_i \in [t_{i-1}, t_i]$. Let $u_j = \rho(t_{k-j})$ and, for $j = k + 1 - i$, let $\sigma_j = \rho(\tau_i)$. Then $\sigma_j \in [u_{j-1}, u_j]$, and the two partitions have the same mesh.
10. For $j = k + 1 - i$ we have $\gamma^-(t_i) = \gamma(u_{j-1})$, $\gamma^-(t_{i-1}) = \gamma(u_j)$ and $\gamma^-(\tau_i) = \gamma(\sigma_j)$.
11. So $\sum_i F(\gamma^-(\tau_i)) \cdot \big( \gamma^-(t_i) - \gamma^-(t_{i-1}) \big) = -\sum_j F(\gamma(\sigma_j)) \cdot \big( \gamma(u_j) - \gamma(u_{j-1}) \big)$.
12. By step 8, the length of $\gamma^-$ on $[t_{i-1}, t_i]$ equals the length $\Delta s_j$ of $\gamma$ on $[u_{j-1}, u_j]$. So $\sum_i f(\gamma^-(\tau_i)) \Delta s^-_i = \sum_j f(\gamma(\sigma_j)) \Delta s_j$, where $\Delta s^-_i$ is the length of $\gamma^-$ on $[t_{i-1}, t_i]$.
13. Let the mesh tend to $0$. The right sides of steps 11 and 12 tend to $-\int_\gamma F \cdot d\mathbf{r}$ and $\int_\gamma f \, ds$. This holds by the definitions of the [[vector line integral|0YS8RX87F4]] and the [[scalar line integral|0M50E4WYWJ]].
14. Hence the left sides converge to the same values, so $\int_{\gamma^-} F \cdot d\mathbf{r} = -\int_\gamma F \cdot d\mathbf{r}$ and $\int_{\gamma^-} f \, ds = \int_\gamma f \, ds$.

## Idea

A partition of the reversed curve is a partition of the original curve read backwards. Each chord $\gamma(t_i) - \gamma(t_{i-1})$ turns into its negative, so the vector sums change sign. Lengths of pieces do not change, so the scalar sums stay the same.