CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11

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Omitting the absolute value from the Jacobian gives negative length under the reflection $x \mapsto 1 - x$

Statement refuted

False claim. In the unoriented change-of-variables formula, one may replace $∣ φ^{'} ∣$ by $φ^{'}$ .

Facts & Assumptions

Given: The reflection $φ (x) = 1 - x$ on $[0, 1]$ and the constant function $f = 1$ on its image.

[L1]

The one-dimensional unoriented formula uses the absolute derivative (In one dimension the compact-Jordan formula is substitution over the unoriented image interval with the absolute derivative).

[L2]

Oriented substitution records orientation through the order of its endpoint limits (Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ ).

Counterexample

technique · direct

1.1

The image is $[0, 1]$ , so its unoriented length integral is $\int_{0}^{1} 1 d y = 1$ .

given

2.1

Since $φ^{'} = - 1$ , the proposed un-absolute right side is $\int_{0}^{1} (- 1) d x = - 1$ , not $1$ .

givenstep 1.1

3.1

With the required absolute value, [L1] gives $\int_{0}^{1} ∣ - 1 ∣ d x = 1$ . The negative value in step 2.1 instead belongs to the oriented formula [L2], whose image endpoints occur in reverse order. Hence the claim is false.

L1L2step 2.1∎

Depends on

Used by

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Dependency tree · two levels

19 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.

Omitting the absolute value from the Jacobian gives negative length under the reflection x↦1−xx↦1−x