CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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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 $G$

Statement

Let $I \subseteq \mathbb{R}$ be order-convex with at least two elements (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length) and let $f : I \to \mathbb{R}$ be continuous on $I$ (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point). Call $G : I \to \mathbb{R}$ a primitive of $f$ on $I$ when $G$ is differentiable at every point of $I$ as a function on $I$ with $G' = f$ there (The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set). Then:

Existence. Fix $c_0 \in I$ . The function $F : I \to \mathbb{R}, \qquad F(x) \;:=\; \int_{c_0}^x f$ is defined at every $x \in I$ (The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ , The integral function $F(x) := \int_a^x f$ of an integrable $f$ ) and is a primitive of $f$ on $I$ .
Uniqueness up to a constant. If $G_1$ and $G_2$ are primitives of $f$ on $I$ then there is a real $k$ with $G_1(x) = G_2(x) + k$ for every $x \in I$ .
Evaluation. If $a, b \in I$ with $a < b$ and $G$ is any primitive of $f$ on $I$ , then $\int_a^b f \;=\; G(b) - G(a) .$

The scope is exactly the continuous case, and that is not a limitation of the proof. An integrable function need not have a primitive, and a function with a primitive need not be integrable; this corollary is precisely the intersection where both fundamental theorems apply, and both witnesses are on the companion page.

Facts & Assumptions

Given: An order-convex $I \subseteq \mathbb{R}$ with at least two elements, a continuous $f : I \to \mathbb{R}$ , a base point $c_0 \in I$ , and a real $\varepsilon > 0$ .

[L1]

A continuous function on a closed bounded interval with distinct endpoints is integrable there; a restriction of a continuous function is continuous (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, 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 $\int_a^b f$ ).

[L2]

Order-convexity: if $p, q \in I$ then every real between $p$ and $q$ lies in $I$ , so the closed interval with endpoints $p$ and $q$ is contained in $I$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

$\int_p^p u = 0$ , $\int_q^p u = -\int_p^q u$ , and for $u$ integrable on a closed bounded interval containing $p,q,r$ one has $\int_p^q u + \int_q^r u = \int_p^r u$ (The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ , 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 $a,b,c$ , claim 3).

[L4]

First fundamental theorem: if $u$ is integrable on $[p,q]$ with $p<q$ and continuous at $c \in [p,q]$ , then $x \mapsto \int_p^x u$ has derivative $u(c)$ at $c$ as a function on $[p,q]$ ; written out, for every real $\varepsilon>0$ there is a real $\delta>0$ with $\bigl|\bigl(\int_p^x u - \int_p^c u\bigr)/(x-c) - u(c)\bigr| < \varepsilon$ for every $x \in [p,q]$ with $0<|x-c|<\delta$ (The first fundamental theorem: if $f$ is integrable on $[a,b]$ and continuous at $c$ , then $F'(c) = f(c)$ ; in particular a continuous $f$ has $F$ as a primitive, The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ ).

[L5]

Second fundamental theorem: 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 $\int_a^b f = G(b)-G(a)$ ).

[L6]

If $u$ is continuous on an order-convex $I$ and differentiable with $u' = 0$ at every interior point of $I$ , then $u$ is constant on $I$ (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

[L7]

A differentiable function is continuous, and the restriction of a function differentiable at $c$ to a subset still having $c$ as a limit point is differentiable at $c$ with the same derivative; every point of a nondegenerate interval is a limit point of it (A function differentiable at $c$ is continuous at $c$ , The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L8]

Ordered-field arithmetic and minima of two reals: the order is total and transitive, and $\min\{s,t\}$ is a real that is $\le$ both (Maximum and minimum of a set, Ordered field, Complete ordered field (least-upper-bound property)).

Proof

technique · direct

1.1

$F$ is defined. For $x \in I$ the closed interval with endpoints $c_0$ and $x$ lies in $I$ by [L2], $f$ is continuous there, hence integrable when $x \ne c_0$ by [L1], and $\int_{c_0}^{c_0} f = 0$ by [L3]; so $F(x)$ names a real for every $x \in I$ .

givenL1L2L3

1.2

A closed neighbourhood inside $I$ . Fix $c \in I$ . If some element of $I$ is $< c$ , choose $u \in I$ with $u < c$ ; otherwise put $u := c$ . If some element of $I$ is $> c$ , choose $v \in I$ with $v > c$ ; otherwise put $v := c$ . Not both $u = c$ and $v = c$ , since $I$ would then have $c$ as its only element; so $u < v$ , and $[u,v] \subseteq I$ by [L2].

givenL2choose

1.3

Claim 2. Let $G_1, G_2$ be primitives of $f$ on $I$ and put $u := G_1 - G_2$ . Then $u$ is differentiable at every point of $I$ with $u' = f - f = 0$ there, in particular at every interior point of $I$ , and $u$ is continuous on $I$ by [L7]; so [L6] gives a real $k$ with $u \equiv k$ .

L6L7

2.1

Put $\eta := \min\{\,c-u,\ v-c\,\}$ if $u<c$ and $c<v$ , $\eta := v-c$ if $u = c$ , and $\eta := c-u$ if $v = c$ ; in every case $\eta > 0$ .

step 1.2L8construct

2.2

$f$ is integrable on $[u,v]$ by [L1], and for $x \in [u,v]$ , [L3] applied to the points $c_0, u, x$ inside the closed interval with endpoints $\min\{c_0,u\}$ and $\max\{c_0,v\}$ , which lies in $I$ by [L2], gives $F(x) = F(u) + \int_u^x f$ .

step 1.1step 1.2L1L2L3

3.1

Every point of $I$ within $\eta$ of $c$ lies in $[u,v]$ . Let $x \in I$ with $|x-c| < \eta$ . If $x < c$ then $I$ has an element below $c$ , so $u < c$ and $\eta \le c-u$ , whence $x > c-\eta \ge u$ . If $x > c$ then symmetrically $x < c+\eta \le v$ . And $u \le c \le v$ covers $x = c$ . So $u \le x \le v$ .

step 1.2step 2.1L8

3.2

Hence for $x \in [u,v]$ with $x \ne c$ , $\bigl(F(x)-F(c)\bigr)/(x-c) = \bigl(\int_u^x f - \int_u^c f\bigr)/(x-c)$ , the constant $F(u)$ cancelling.

step 2.2algebra

3.3

By [L4] applied on $[u,v]$ at the point $c$ , fix a real $\delta > 0$ with $\bigl|\bigl(\int_u^x f - \int_u^c f\bigr)/(x-c) - f(c)\bigr| < \varepsilon$ for every $x \in [u,v]$ with $0<|x-c|<\delta$ , and put $\delta' := \min\{\delta,\eta\} > 0$ .

step 2.2givenL1L4L8choose

4.1

Every $x \in I$ with $0 < |x-c| < \delta'$ lies in $[u,v]$ by step 3.1, so by step 3.2 and step 3.3, $\bigl|\bigl(F(x)-F(c)\bigr)/(x-c) - f(c)\bigr| < \varepsilon$ .

step 3.1step 3.2step 3.3

5.1

As $\varepsilon > 0$ was arbitrary and $c$ is a limit point of $I$ by [L7], $F$ is differentiable at $c$ with $F'(c) = f(c)$ ; since $c \in I$ was arbitrary, $F$ is a primitive of $f$ on $I$ , which is claim 1.

step 1.2step 4.1L7

6.1

Claim 3. Let $a<b$ in $I$ and let $G$ be a primitive of $f$ on $I$ . Then $[a,b] \subseteq I$ by [L2], the restriction of $G$ to $[a,b]$ is differentiable at every point of $[a,b]$ with derivative $f$ there by [L7], and $f$ is integrable on $[a,b]$ by [L1]; so [L5] gives $\int_a^b f = G(b)-G(a)$ .

L1L2L5L7∎

Remarks

Steps 1.2, 2.1 and 3.1 are the only work beyond citing the two fundamental theorems. The first fundamental theorem: if $f$ is integrable on $[a,b]$ and continuous at $c$ , then $F'(c) = f(c)$ ; in particular a continuous $f$ has $F$ as a primitive is stated on a closed bounded interval, while $I$ here may be open, half-open or unbounded, so the derivative it produces is the derivative of a restriction. What those steps supply is a closed subinterval $[u,v] \subseteq I$ that contains all points of $I$ within $\eta$ of $c$ , after which the difference quotients of $F$ and of the restriction agree on a punctured neighbourhood and the $\varepsilon$ - $\delta$ statement transfers verbatim.
"Two primitives differ by a constant" is not re-minted here. A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant already states exactly that, in its second clause, for functions with equal derivatives on an order-convex domain; claim 2 is that statement applied to $G_1 - G_2$ .
Order-convexity of $I$ is essential to claim 2 and harmless elsewhere. On a domain in two pieces a function may be constant on each with different constants, which is why A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant carries the same hypothesis. Claims 1 and 3 use it only to know that closed subintervals spanned by points of $I$ lie in $I$ .
Forward references, orientation only. The two witnesses bounding the scope of this corollary are The sign function is Riemann integrable on $[-1,1]$ and has no primitive there ↗ and A function differentiable on $[0,1]$ whose derivative is unbounded, hence not Riemann integrable ↗ on the companion page; nothing above depends on either.

Depends on

Used by

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Sources

Antiderivative (Wikipedia) (standard reference, not scraped)
Fundamental theorem of calculus (Wikipedia) (standard reference, not scraped)
Encyclopedia of Mathematics, Integral calculus (standard reference, not scraped)
J. Lebl, Basic Analysis I, Fundamental theorem of calculus (standard reference, not scraped)

Every continuous function on an interval has a primitive; two primitives differ by a constant; and ∫abf=G(b)−G(a)\int_a^b f = G(b)-G(a)∫ab​f=G(b)−G(a) for any primitive GGG