CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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If $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $f^{'}$ extends continuously to $[a, b]$ , then the graph of $f$ has length $\int_{a}^{b} \sqrt{1 + f^{'} (t)^{2}} d t$

Statement

Let $a < b$ , let $f : [a, b] \to R$ be continuous and differentiable on $(a, b)$ , and suppose $f^{'}$ extends continuously to $q : [a, b] \to R$ . Then the graph path $γ (t) = (t, f (t))$ is rectifiable and

$L (γ) = \int_{a}^{b} \sqrt{1 + q (t)^{2}} d t .$

Facts & Assumptions

Given: The scalar function $f$ , extension $q$ , and graph path $γ$ .

[L1]

Vector differentiation is componentwise (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[L2]

In $R^{2}$ , $∥ (x, y) ∥_{2} = \sqrt{x^{2} + y^{2}}$ (The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[L3]

A path with continuous derivative extension $v$ has length $\int ∥ v ∥_{2}$ (If $γ : [a, b] \to R^{n}$ is continuous, differentiable on $(a, b)$ , and $γ^{'}$ extends continuously to $[a, b]$ , then $L (γ) = \int_{a}^{b} ∥ γ^{'} (t) ∥_{2} d t$ ).

Proof

technique · specialization

1.1

By [L1], the graph path has interior derivative $(1, f^{'} (t))$ and continuous extension $v (t) = (1, q (t))$ .

givenL1

2.1

By [L2], $∥ v (t) ∥_{2} = \sqrt{1 + q (t)^{2}}$ .

step 1.1L2

3.1

Apply [L3] and substitute the speed from step 2.1.

step 1.1step 2.1L3∎

Depends on

Used by

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Sources

A. R. Shastri, Metric Spaces, Section 5 (standard reference, not scraped)

If ff is continuous on [a,b][a,b], differentiable on (a,b)(a,b), and f′f′ extends continuously to [a,b][a,b], then the graph of ff has length ∫ab1+f′(t)2 dt∫ab​1+f′(t)2​dt