---
tag: 0FSCXYG856
kind: statement
role: proposition
name: Reversing a path changes the sign of a line integral
aliases: [orientation-reversal-of-line-integrals]
msc: [26B12]
uses: [0YS8RX87F4, 0M50E4WYWJ, 0W3W1VYWBP]
sources:
  - url: https://en.wikipedia.org/wiki/Line_integral
    locator: Line integral of a vector field
  - url: https://en.wikipedia.org/wiki/Line_integral
    locator: Line integral of a scalar field
---

## 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|0W3W1VYWBP]], $(\gamma^-)'(t) = -\gamma'(a + b - t)$. The substitution $u = a + b - t$ then gives both identities.
The [[line integral of a vector field|0YS8RX87F4]] depends on the orientation of the curve. The [[line integral of a scalar field|0M50E4WYWJ]] 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.