ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$\int_{0}^{1} x^{m} = 1 / ι (m + 1)$ , computed by the fundamental theorem and checked against the definition

Example

Let $m \in N$ and let $p_{m} : [0, 1] \to R$ be $p_{m} (x) = x^{m}$ (Integer powers $a^{m}$ ). Then $p_{m}$ is integrable on $[0, 1]$ and

$\int_{0}^{1} x^{m} d x = \frac{1}{ι (m + 1)},$

where $ι (m + 1)$ is the canonical natural of $m + 1$ in $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field) and is positive because $m + 1 \geq 1$ (Canonical naturals are positive and strictly increasing).

The $ι$ is not decoration. A natural number is a von Neumann natural, that is a set, so $m + 1$ is not an element of $R$ and $1 / (m + 1)$ is not an expression of the field; what the display says is $1 / ι (m + 1)$ , and that is why the reader meets $ι$ here at all.

Two independent checks are carried out below: the value at $m = 0$ against the published formula for the integral of a constant (If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ ), and the monotonicity of the answer in $m$ against the pointwise inequality $x^{m + 1} \leq x^{m}$ on $[0, 1]$ .

Facts & Assumptions

Given: A natural number $m$ and the function $p_{m} (x) = x^{m}$ on $[0, 1]$ .

[L1]

For $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable at every real $c$ with derivative $ι (n) c^{n - 1}$ (For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, claim 2); the constant $x \mapsto x^{0} = 1$ has derivative $0$ (claim 1).

[L2]

A scalar multiple of a differentiable function is differentiable with the scaled derivative (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , claim 2).

[L3]

Every polynomial function is continuous on every subset of $R$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point), and a continuous function on $[0, 1]$ is integrable there (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ).

[L4]

If $H$ is differentiable at every point of $[p, q]$ with $H^{'}$ integrable there, then $\int_{p}^{q} H^{'} = H (q) - H (p)$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ); a continuous function on an interval has a primitive (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[L5]

Powers: $1^{n} = 1$ for every $n$ , $0^{n} = 0$ for $n \geq 1$ , $0^{0} = 1$ , and $0 \leq x \leq 1$ gives $0 \leq x^{n + 1} \leq x^{n} \leq 1$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , claims 1 and 3).

[L6]

$ι (0) = 0$ , $ι (n + 1) = ι (n) + 1$ , $ι (n) > 0$ for $n \geq 1$ , and $ι$ is increasing on the naturals $\geq 1$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L7]

$\int_{p}^{q} c = c (q - p)$ for a constant $c$ , and if $u \leq w$ pointwise with both integrable then $\int_{p}^{q} u \leq \int_{p}^{q} w$ (If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L8]

Ordered-field arithmetic: a positive real has a positive inverse, and $s \leq t$ with $s, t > 0$ gives $1 / t \leq 1 / s$ (Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The integral with oriented limits: $\int_{a}^{a} f : = 0$ and $\int_{b}^{a} f : = - \int_{a}^{b} f$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

Verification

technique · direct

1.1

$p_{m}$ is continuous on $[0, 1]$ , hence integrable there, by [L3].

1.2

Define $H : [0, 1] \to R$ by $H (x) : = x^{m + 1} / ι (m + 1)$ ; this is legitimate because $ι (m + 1) > 0$ by [L6].

L6L8construct

2.1

By [L1] with $n : = m + 1 \geq 1$ and [L2], $H$ is differentiable at every point of $[0, 1]$ with $H^{'} (x) = ι (m + 1) x^{m} / ι (m + 1) = x^{m} = p_{m} (x)$ .

step 1.2L1L2L6

3.1

By [L4] applied to $H$ on $[0, 1]$ , whose derivative $p_{m}$ is integrable by step 1.1, $\int_{0}^{1} x^{m} = H (1) - H (0)$ .

step 1.1step 2.1L4

4.1

By [L5], $H (1) = 1^{m + 1} / ι (m + 1) = 1 / ι (m + 1)$ and $H (0) = 0^{m + 1} / ι (m + 1) = 0$ , since $m + 1 \geq 1$ ; so $\int_{0}^{1} x^{m} = 1 / ι (m + 1)$ .

step 1.2step 3.1L5L6

5.1

First check, at $m = 0$ . There $p_{0}$ is the constant function $1$ by [L5], so $\int_{0}^{1} p_{0} = 1 \cdot (1 - 0) = 1$ by [L7], while the formula gives $1 / ι (1) = 1 / 1 = 1$ . The two agree.

step 4.1L5L6L7

6.1

Second check, monotonicity in $m$ . By [L5], $x^{m + 1} \leq x^{m}$ for every $x \in [0, 1]$ , so $\int_{0}^{1} x^{m + 1} \leq \int_{0}^{1} x^{m}$ by [L7]; the formula gives $1 / ι (m + 2) \leq 1 / ι (m + 1)$ , which holds by [L6] and [L8]. The two agree.

step 4.1L5L6L7L8∎

Remarks

The primitive is the only input, and the index range inside it is the trap. For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term states its power rule for $n \geq 1$ only, because at $n = 0$ the formula $ι (0) c^{- 1}$ mentions $c^{- 1}$ , which is undefined at $c = 0$ . The exponent used above is $m + 1$ , which is $\geq 1$ for every $m \in N$ including $m = 0$ , so claim 2 of that lemma applies with no case split and the example is correct at $m = 0$ as well.
The same computation from the definition is a different exercise. The previous page's companion evaluates $\int_{0}^{1} x^{2}$ from Darboux sums on uniform partitions and a closed form for $\sum_{k < n} ι (k)^{2}$ ; that is the value $1 / 3$ , which is $1 / ι (3)$ and agrees with the formula above at $m = 2$ . The point of computing by the fundamental theorem is that no closed form for a power sum is needed at any $m$ .
Linearity extends this to every polynomial. For $P (x) = \sum_{j < n} a_{j} x^{j}$ one gets $\int_{0}^{1} P = \sum_{j < n} a_{j} / ι (j + 1)$ from Integrable functions on $[a, b]$ form a set closed under sums and scalar multiples, and $\int_{a}^{b} (λ f + μ g) = λ \int_{a}^{b} f + μ \int_{a}^{b} g$ ; that is a routine consequence and is not stated as a separate claim.

Depends on

Used by

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Dependency tree · two levels

75 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

Fundamental theorem of calculus (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 6 (standard reference, not scraped)
UC Berkeley Math 128A, Integration notes (standard reference, not scraped)

∫01xm=1/ι(m+1)∫01​xm=1/ι(m+1), computed by the fundamental theorem and checked against the definition