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

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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

Statement

Let $a < b$ be reals, let $f : [a,b] \to \mathbb{R}$ be integrable (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 $\int_a^b f$ ), let $F$ be its integral function (The integral function $F(x) := \int_a^x f$ of an integrable $f$ ), and let $c \in [a,b]$ be a point at which $f$ is continuous (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). Then $F$ is differentiable at $c$ as a function on $[a,b]$ (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) and

$F'(c) \;=\; f(c) .$

At $c = a$ and $c = b$ this is the one-sided statement, which is what 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 means at those points: every point of a nondegenerate interval is a limit point of it, so $F'(c)$ is a meaningful symbol at every $c \in [a,b]$ , and the difference quotient is taken over $[a,b] \setminus \{c\}$ .

Consequently, if $f$ is continuous on the whole of $[a,b]$ , then $F$ is a primitive of $f$ there: $F' = f$ at every point of $[a,b]$ .

Continuity at $c$ is a hypothesis and it cannot be dropped. For an integrable $f$ that is discontinuous at $c$ , $F'(c)$ may fail to exist, and it may exist and differ from $f(c)$ ; both are exhibited on the companion page, by an integrable function with no primitive and by a false statement about the integral function.

Facts & Assumptions

Given: Reals $a < b$ , an integrable $f : [a,b] \to \mathbb{R}$ , its integral function $F(x) = \int_a^x f$ , a point $c \in [a,b]$ at which $f$ is continuous, and a real $\varepsilon > 0$ .

[L1]

$F(x) - F(c) = \int_c^x f$ for all $x, c \in [a,b]$ , in either order (The integral function $F(x) := \int_a^x f$ of an integrable $f$ , 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 $a,b,c$ ).

[L2]

Continuity at $c$ : for every real $\eta > 0$ there is a real $\delta > 0$ such that every $t \in [a,b]$ with $|t - c| < \delta$ satisfies $|f(t) - f(c)| < \eta$ (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).

[L3]

Every point of a nondegenerate interval is a limit point of it, so $F'(c) = \lim_{x\to c} \bigl(F(x)-F(c)\bigr)/(x-c)$ is a meaningful symbol, the limit being taken over $[a,b]\setminus\{c\}$ (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, The $\varepsilon$ - $\delta$ limit $\lim_{x \to c} f(x) = L$ of $f : A \to \mathbb{R}$ at a limit point $c$ of $A$ , Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

For $p < q$ in $[a,b]$ : $f$ and $|f|$ and every constant are integrable on $[p,q]$ ; $\int_p^q \lambda = \lambda(q-p)$ ; sums and scalar multiples of integrable functions are integrable with the corresponding identity; and $\bigl|\int_p^q u\bigr| \le \int_p^q |u|$ (A function integrable on $[a,b]$ is integrable on every closed subinterval, If $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_a^b c = c(b-a)$ , Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and $\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g$ , If $f,g$ are integrable on $[a,b]$ then so are $\lvert f\rvert$ , $f^{2}$ , $fg$ , $\max(f,g)$ and $\min(f,g)$ , and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$ ).

[L5]

If $u \le w$ pointwise on $[p,q]$ and both are integrable then $\int_p^q u \le \int_p^q w$ (If $f \le g$ on $[a,b]$ and both are integrable then $\int_a^b f \le \int_a^b g$ ; and $m(b-a) \le \int_a^b f \le M(b-a)$ ).

[L6]

With oriented limits, $\int_q^p u = -\int_p^q u$ and $\int_p^p u = 0$ (The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ ).

[L7]

Absolute value and ordered-field arithmetic: $|{-t}| = |t|$ , $|t\,s| = |t||s|$ , $|t| \le c$ follows from $-c \le t \le c$ , a positive real has a positive inverse, and the order is total and transitive (Basic properties of the absolute value, Absolute value in an ordered field, Ordered field, Complete ordered field (least-upper-bound property)). The nonstrict forms of the order facts follow from the strict ones by adjoining equality.

[L8]

For every real $\eta > 0$ there is a real $\eta' > 0$ with $\eta' < \eta$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Every complete ordered field is Archimedean).

Proof

technique · direct

1.1

By [L2] with $\eta := \varepsilon \cdot 2^{-1}$ , fix a real $\delta > 0$ such that $|f(t)-f(c)| < \varepsilon\cdot 2^{-1}$ for every $t \in [a,b]$ with $|t-c| < \delta$ .

givenL2L7L8choose

1.2

For $x \in [a,b]$ with $x \ne c$ , [L1] and [L4] give $F(x) - F(c) - f(c)(x-c) = \int_c^x f - \int_c^x f(c) = \int_c^x \bigl(f(t) - f(c)\bigr)\,\mathrm{d}t$ , the constant $f(c)$ having integral $f(c)(x-c)$ over the oriented interval from $c$ to $x$ by [L4] and [L6].

L1L4L6

2.1

The estimate for $c < x < c+\delta$ . Every $t \in [c,x]$ has $|t-c| < \delta$ , so $-\varepsilon\cdot2^{-1} \le f(t)-f(c) \le \varepsilon\cdot 2^{-1}$ there by step 1.1, whence $\bigl|\int_c^x (f - f(c))\bigr| \le \varepsilon\cdot2^{-1}\,(x-c)$ by [L4] and [L5].

step 1.1step 1.2L4L5L7

2.2

The estimate for $c-\delta < x < c$ . By [L6], $\int_c^x (f-f(c)) = -\int_x^c (f-f(c))$ , and every $t \in [x,c]$ has $|t-c| < \delta$ , so the same argument gives $\bigl|\int_c^x (f-f(c))\bigr| \le \varepsilon\cdot 2^{-1}\,(c-x)$ .

step 1.1step 1.2L4L5L6L7

3.1

In both cases $\bigl|\int_c^x(f-f(c))\bigr| \le \varepsilon\cdot 2^{-1}\,|x-c|$ , so dividing by the nonzero $|x-c|$ and using step 1.2 gives $\Bigl|\dfrac{F(x)-F(c)}{x-c} - f(c)\Bigr| \le \varepsilon\cdot 2^{-1} < \varepsilon$ for every $x \in [a,b]$ with $0 < |x-c| < \delta$ .

step 1.2step 2.1step 2.2L7

4.1

Since $\varepsilon > 0$ was arbitrary, the limit of the difference quotient of $F$ at $c$ exists and equals $f(c)$ by [L3]; that is, $F'(c) = f(c)$ .

step 1.1step 3.1L3

5.1

If $f$ is continuous at every point of $[a,b]$ then step 4.1 applies at every $c \in [a,b]$ , so $F' = f$ on $[a,b]$ and $F$ is a primitive of $f$ .

step 4.1L3∎

Remarks

The estimate is written out for $x < c$ as well, and that is not redundancy. For $x < c$ the factor $1/(x-c)$ is negative and the naive chain $m(x-c) \le \int_c^x \le M(x-c)$ reverses; what makes the argument uniform is taking absolute values before dividing, which is what steps 2.1, 2.2 and 3.1 do. This is the single most common error in this proof.
The route is the definition of the derivative, not the mean value theorem for integrals. Deducing $F'(c) = f(c)$ from If $f$ is continuous on $[a,b]$ and $g$ is integrable with $g \ge 0$ , there is $\xi \in [a,b]$ with $\int_a^b fg = f(\xi)\int_a^b g$ would need $f$ continuous on a whole subinterval around $c$ , which is a strictly stronger hypothesis than continuity at the single point $c$ . The theorem as stated is the sharp one.
What is proved at a point is proved at a point. Nothing here says $F$ is differentiable anywhere else, and nothing says $F' = f$ off the continuity set of $f$ . Where $f$ is merely integrable, all that survives is The integral function of a bounded integrable $f$ is Lipschitz, hence uniformly continuous.
Forward references, orientation only. The two failures at a discontinuity are worked out on the companion page as The sign function is Riemann integrable on $[-1,1]$ and has no primitive there ↗ and FALSE: for every integrable $f$ on $[a,b]$ , the integral function $F(x)=\int_a^x f$ satisfies $F' = f$ on $[a,b]$ ↗; nothing above depends on either.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 88 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)
Carnegie Mellon 21-269, Riemann integration notes (standard reference, not scraped)
J. Lebl, Basic Analysis I, Fundamental theorem of calculus (standard reference, not scraped)

The first fundamental theorem: if fff is integrable on [a,b][a,b][a,b] and continuous at ccc, then F′(c)=f(c)F'(c) = f(c)F′(c)=f(c); in particular a continuous fff has FFF as a primitive