TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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

1.1

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.

L1L2L5algebra

1.2

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 $β$ .

L1algebra

2.1

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.

step 1.2L2L3L4L5

3.1

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.

L1L5algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 87 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. Lebl, Basic Analysis II, section 9.2 (standard reference, not scraped)