Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13
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False: vector line integrals are invariant under reversing a path

Statement

For every continuous vector field F and piecewise-C1 path γ on whose trace it is defined, ∫γ−F⋅dr=∫γF⋅dr.

Facts & Assumptions

Given: The constant field F=(1,0) and the segment γ(t)=(t,0) on [0,1].

[L1]

On a C1 path η, the vector line integral is the integral of ⟨F(η(t)),η′(t)⟩ (Scalar line integrals with respect to arc length and vector-field line integrals).

[L2]

An orientation-reversing reparametrization negates a vector line integral but leaves a scalar ds integral unchanged (Scalar line integrals are parametrization-independent; vector line integrals retain orientation and change sign when it reverses).

Refutation

technique · direct
1.1

Since γ′(t)=(1,0), [L1] and [L3] give ∫γF⋅dr=∫011 dt=1.

givenL1L3algebra
2.1

The reversal is orientation-reversing, so [L2] gives ∫γ−F⋅dr=−1.

givenstep 1.1L2algebra
3.1

Since −1≠1, the statement is false. The scalar analogue is true because [L2] says that ds removes the orientation sign.

step 1.1step 2.1L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources