TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Line integrals under reversal and concatenation

Statement

Let $γ$ be a piecewise- $C^{1}$ path, and let $f$ be a continuous scalar field and $F$ a continuous vector field on a set containing its trace. Then

$\int_{γ^{-}} f d s = \int_{γ} f d s, \int_{γ^{-}} F \cdot d r = - \int_{γ} F \cdot d r .$

If piecewise- $C^{1}$ paths $α, β : [0, 1] \to R^{n}$ satisfy $α (1) = β (0)$ , and $f$ and $F$ are continuous on a set containing both traces, then

$\int_{α * β} f d s = \int_{α} f d s + \int_{β} f d s,$ $\int_{α * β} F \cdot d r = \int_{α} F \cdot d r + \int_{β} F \cdot d r .$

Facts & Assumptions

Given: The paths and fields in the Statement.

[L1]

Reversal is $γ^{-} (t) = γ (a + b - t)$ , and concatenation uses $α (2 t)$ and $β (2 t - 1)$ on the two halves of $[0, 1]$ (Reversal, concatenation, closed paths, and oriented piecewise-C1 reparametrizations).

[L2]

Scalar line integrals are unchanged by oriented reparametrization; vector line integrals are unchanged under preservation and negated under reversal (Scalar line integrals are parametrization-independent; vector line integrals retain orientation and change sign when it reverses).

[L3]

Line-integral sums are independent of the admissible partition (The piecewise-C1 line-integral sums do not depend on the admissible partition).

[L4]

Oriented one-variable integrals are additive across every intermediate point (For $a < c < b$ : $f$ is integrable on $[a, b]$ if and only if it is integrable on $[a, c]$ and on $[c, b]$ , and then $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

[L5]

Scalar and vector line integrals are sums of their defining one-variable integrals over smooth pieces (Scalar line integrals with respect to arc length and vector-field line integrals).

Proof

technique · direct

1.1L1L2L5algebra

If $a = b$ , [L5] makes the two line integrals over both $γ$ and $γ^{-}$ zero. If $a < b$ , the affine map $t \mapsto a + b - t$ is orientation-reversing, so applying [L2] to the reversal in [L1] proves the two formulas.

1.2L1algebra

Split the concatenation at $1 / 2$ . The first half is the orientation-preserving affine reparametrization $t \mapsto 2 t$ of $α$ , and the second is the orientation-preserving affine reparametrization $t \mapsto 2 t - 1$ of $β$ .

2.1step 1.2L2L3L4L5

By [L2], each half-integral in step 1.2 equals the corresponding integral over $α$ or $β$ . By [L3], [L4], and [L5], the sum of the two half-integrals is the integral over $α * β$ . This proves both concatenation formulas.

3.1L1L5algebra∎

The join point is an allowed partition point, so no derivative match is required there. If either path is constant, its derivative and both of its line-integral contributions are zero, and the formulas remain valid.

Depends on

Dependency tree · 2 levels

Sources

J. Lebl, Basic Analysis II, section 9.2