The Second Fundamental Theorem of Calculus

Suppose you know a function ff. Suppose you also know some function FF whose derivative is ff. Can you compute ∫abf\int_a^b f without ever building a Riemann sum? This is the question the Second Fundamental Theorem of Calculus answers. The answer is a resounding yes: the integral collapses to the boundary values F(b)−F(a)F(b) - F(a). This single fact converts integration, defined analytically as a limit of approximating sums, into an algebraic bookkeeping problem. That problem is simply "find an antiderivative and plug in endpoints." The result matters because it is the bridge between two theories developed for entirely different purposes. Differentiation measures instantaneous rate of change. Integration measures accumulated area. These are inverse operations. The inverse relationship can also be exploited computationally. This is one of the great unifications in mathematics. Without this theorem, every integral would require the full machinery of Riemann sums, upper and lower integrals, and limiting arguments. With it, computing ∫12x3 dx\int_1^2 x^3\,dx becomes the one-line calculation [x44]12=16−14=154\left[\tfrac{x^4}{4}\right]_1^2 = \tfrac{16-1}{4} = \tfrac{15}{4}. The theorem sits alongside the First Fundamental Theorem of Calculus as one half of a matched pair. The First Fundamental Theorem says the accumulation function x↦∫axf(t) dtx \mapsto \int_a^x f(t)\,dt is itself an antiderivative of ff. This holds when ff is continuous. The Second Fundamental Theorem runs in the reverse direction. Given any antiderivative FF of an integrable ff, the definite integral's value is determined. It is determined by FF evaluated at the endpoints. The two theorems are logically independent statements. They are often merged, though, in calculus courses. Here we isolate the evaluation theorem instead. We prove it carefully using Basic Properties: Linearity and Additivity of the Riemann integral. We also use the Mean Value Theorem. This article gives the precise hypotheses under which the evaluation formula holds. It offers a complete proof via Riemann sums and the Mean Value Theorem. It also includes a worked computation with an asymmetric integrand. A careful counterexample follows, showing why Riemann integrability of ff cannot be dropped from the hypotheses. It closes by placing the theorem in the broader landscape of techniques it makes possible. These include substitution, integration by parts, and beyond.

Key takeaways
  • The theorem states: if F′=fF' = f on [a,b][a,b] and ff is Riemann integrable, then ∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a).
  • It converts integration into antiderivative-finding, replacing limits of Riemann sums with algebra.
  • The proof uses the Mean Value Theorem to match Riemann-sum pieces to FF's increments exactly.
  • Both hypotheses — differentiability of FF everywhere with F′=fF'=f, and integrability of ff — are essential.
  • Dropping integrability of ff produces genuine counterexamples (a derivative that is unbounded, hence not Riemann integrable).
  • Key idea: telescoping the difference F(b)−F(a)F(b)-F(a) against a partition turns it into a Riemann sum for ff, forcing equality in the limit.

Where this lives today

The evaluation theorem is the computational engine behind virtually all closed-form integral evaluation in physics and engineering: computing work done by a force, charge accumulated from current, or displacement from velocity all reduce to finding an antiderivative and subtracting endpoint values. It underlies Integration by Substitution, where a change of variable u=g(x)u = g(x) transforms ∫f(g(x))g′(x) dx\int f(g(x))g'(x)\,dx into ∫f(u) du\int f(u)\,du, and the evaluation theorem is what lets you convert the resulting antiderivative back into a definite integral value. It equally underlies Integration by Parts, derived by applying the evaluation theorem to the product rule (uv)′=u′v+uv′(uv)' = u'v + uv'. Beyond first-year calculus, the theorem generalizes in analysis to the Fundamental Theorem of Calculus for Lebesgue integrals, where FF need only be absolutely continuous (not necessarily differentiable everywhere) — a refinement designed precisely to handle pathologies like the counterexample above. In numerical analysis, quadrature methods (Simpson's rule, Gaussian quadrature) are justified as approximations to the right-hand side of the theorem when no closed-form antiderivative exists. In complex analysis, the theorem's structure reappears as the fundamental theorem of contour integration for holomorphic primitives. In probability and statistics, evaluating cumulative distribution functions from density functions (and vice versa) is a direct application. In control theory and signal processing, computing the state transition from a known rate function again reduces to this evaluation.

:::quiz[In the proof, what does the Mean Value Theorem supply on each subinterval [xᵢ₋₁, xᵢ]?]

  • A bound on the oscillation of f
  • A point tᵢ where F(xᵢ) − F(xᵢ₋₁) = f(tᵢ)Δxᵢ exactly ✓
  • The exact value of the integral over that subinterval
  • A proof that f is continuous at tᵢ

:::quiz[Why does the counterexample F(x) = x² sin(1/x²) (F(0)=0) fail to satisfy the theorem's conclusion on [0,1]?]

  • F is not continuous at 0
  • F is not differentiable at 0
  • F'(x) is unbounded near 0, so f = F' is not Riemann integrable on [0,1] ✓
  • The Mean Value Theorem fails on [0,1]

:::quiz[If f is continuous on [a,b], which prior theorem guarantees an antiderivative automatically exists?]

  • The Second Fundamental Theorem of Calculus itself (circular)
  • The First Fundamental Theorem of Calculus, via F(x) = ∫ₐˣ f(t) dt ✓
  • The Extreme Value Theorem
  • Rolle's Theorem

Ask the tutor: Can you show me why the Mean Value Theorem step actually requires FF to be differentiable at every interior point of every subinterval, not just at the tagged points tit_i?

DefinitionAntiderivative and the Evaluation Theorem's setup

Let f:[a,b]→Rf : [a,b] \to \mathbb{R} be a function. A function F:[a,b]→RF : [a,b] \to \mathbb{R} is called an antiderivative of ff on [a,b][a,b] if FF is continuous on [a,b][a,b], differentiable on (a,b)(a,b) (with one-sided derivatives assumed to exist and match ff at the endpoints, in the strong version), and

F′(x)=f(x)for all x∈[a,b].F'(x) = f(x) \quad \text{for all } x \in [a,b].

We say ff is Riemann integrable on [a,b][a,b] (written f∈R[a,b]f \in \mathcal{R}[a,b]) if the upper and lower Darboux integrals of ff over [a,b][a,b] coincide; their common value is ∫abf(x) dx\int_a^b f(x)\,dx. The evaluation theorem (Second Fundamental Theorem of Calculus) asserts that when both of these hypotheses hold simultaneously — FF is an antiderivative of ff on [a,b][a,b], and ff is Riemann integrable on [a,b][a,b] — then

∫abf(x) dx=F(b)−F(a).\int_a^b f(x)\,dx = F(b) - F(a).

Intuition

Picture the antiderivative FF as a "distance traveled" function and f=F′f = F' as a velocity function. The claim ∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a) says: total displacement equals the integral of velocity over time. This is believable at the level of physical intuition — you cannot get from position F(a)F(a) to position F(b)F(b) except by accumulating the velocity ff along the way — but the theorem promotes this intuition to a theorem about limits of Riemann sums. The proof idea is a clever telescoping trick. Take any partition a=x0<x1<⋯<xn=ba = x_0 < x_1 < \cdots < x_n = b. Then trivially,

F(b)−F(a)=∑i=1n(F(xi)−F(xi−1)),F(b) - F(a) = \sum_{i=1}^n \big(F(x_i) - F(x_{i-1})\big),

because the sum telescopes: every intermediate value of FF cancels. Now, on each tiny subinterval [xi−1,xi][x_{i-1}, x_i], the Mean Value Theorem hands us a point tit_i where

F(xi)−F(xi−1)=F′(ti)(xi−xi−1)=f(ti) Δxi.F(x_i) - F(x_{i-1}) = F'(t_i)(x_i - x_{i-1}) = f(t_i)\,\Delta x_i.

So F(b)−F(a)F(b)-F(a) is exactly equal to ∑if(ti)Δxi\sum_i f(t_i)\Delta x_i — a genuine Riemann sum for ff, for every partition, with a specific (Mean-Value-Theorem-supplied) choice of sample points. Since ff is assumed Riemann integrable, every Riemann sum (with any tags) converges to ∫abf\int_a^b f as the mesh shrinks. In particular these particular sums converge to ∫abf\int_a^b f. But they are all identically equal to F(b)−F(a)F(b) - F(a). Hence F(b)−F(a)=∫abfF(b) - F(a) = \int_a^b f.

Figure
Telescoping F over a partition matches a Riemann sum for f
a=x₀ x₁ x₂ x₃=b F(a) F(x₁) F(x₂) F(b) slope f(t₁) slope f(t₂) slope f(t₃) Each rise F(xᵢ)−F(xᵢ₋₁) = f(tᵢ)Δxᵢ (Mean Value Theorem)

The proof, in outline

The picture above is the entire proof: telescoping, then the Mean Value Theorem on each piece, then invoking integrability of ff to pass to the limit. We now write this out with full rigor.

TheoremSecond Fundamental Theorem of Calculus (Evaluation Theorem)

Let f:[a,b]→Rf : [a,b] \to \mathbb{R} with a<ba < b, and suppose: 1. (Integrability) f∈R[a,b]f \in \mathcal{R}[a,b], i.e. ff is Riemann (Darboux) integrable on [a,b][a,b]; 2. (Antiderivative) F:[a,b]→RF : [a,b] \to \mathbb{R} is continuous on [a,b][a,b], and F′(x)=f(x)F'(x) = f(x) for every x∈(a,b)x \in (a,b), with the one-sided derivatives at the endpoints also equal to f(a)f(a) and f(b)f(b) respectively. Then

∫abf(x) dx=F(b)−F(a).\int_a^b f(x)\,dx = F(b) - F(a).
Proof

Technique: telescoping sum matched to Riemann sums via the Mean Value Theorem. Let P={a=x0<x1<⋯<xn=b}P = \{a = x_0 < x_1 < \cdots < x_n = b\} be an arbitrary partition of [a,b][a,b].

Step 1 (Telescoping). By definition of a telescoping sum,

F(b)−F(a)=∑i=1n(F(xi)−F(xi−1)).(1)F(b) - F(a) = \sum_{i=1}^n \big(F(x_i) - F(x_{i-1})\big). \tag{1}

This is a purely algebraic identity: expanding the right side, every term F(xi)F(x_i) for 1≤i≤n−11 \le i \le n-1 appears once with a ++ sign (from the ii-th summand) and once with a −- sign (from the (i+1)(i+1)-th summand), so it cancels; only F(xn)=F(b)F(x_n) = F(b) and −F(x0)=−F(a)-F(x_0) = -F(a) survive.

Step 2 (Mean Value Theorem on each subinterval). Fix i∈{1,…,n}i \in \{1, \dots, n\}. On the subinterval [xi−1,xi][x_{i-1}, x_i], the function FF is continuous (being continuous on all of [a,b][a,b], hence on the subinterval) and differentiable on the open interval (xi−1,xi)(x_{i-1}, x_i) (since (xi−1,xi)⊆(a,b)(x_{i-1},x_i) \subseteq (a,b), where F′=fF' = f is assumed to exist). By the Mean Value Theorem, there exists a point ti∈(xi−1,xi)t_i \in (x_{i-1}, x_i) such that

F(xi)−F(xi−1)=F′(ti) (xi−xi−1)=f(ti) Δxi,F(x_i) - F(x_{i-1}) = F'(t_i)\,(x_i - x_{i-1}) = f(t_i)\,\Delta x_i,

where Δxi:=xi−xi−1\Delta x_i := x_i - x_{i-1} and we used hypothesis (2) that F′=fF' = f on (a,b)(a,b).

Step 3 (Recognizing a Riemann sum). Substituting Step 2 into the right-hand side of (1) gives

F(b)−F(a)=∑i=1nf(ti) Δxi=:S(f,P,{ti}),(2)F(b) - F(a) = \sum_{i=1}^n f(t_i)\,\Delta x_i =: S(f, P, \{t_i\}), \tag{2}

where the right side is precisely a Riemann sum for ff associated to the partition PP with tags ti∈[xi−1,xi]t_i \in [x_{i-1},x_i] (in fact tit_i lies in the open subinterval, but this is still an admissible tag). Crucially, identity (2) holds exactly, with no error term, for every partition PP — the specific tags tit_i depend on PP (they come from the Mean Value Theorem), but the equality F(b)−F(a)=S(f,P,{ti})F(b)-F(a) = S(f,P,\{t_i\}) is exact, not approximate.

Step 4 (Passing to the limit using integrability of ff). Since f∈R[a,b]f \in \mathcal{R}[a,b], by the Riemann-sum characterization of the integral (equivalent to the Darboux definition): for every ε>0\varepsilon > 0 there exists δ>0\delta > 0 such that for every partition PP with mesh ∥P∥:=max⁡iΔxi<δ\|P\| := \max_i \Delta x_i < \delta, and for every choice of tags ti∈[xi−1,xi]t_i \in [x_{i-1}, x_i],

∣S(f,P,{ti})−∫abf∣<ε.\left| S(f, P, \{t_i\}) - \int_a^b f \right| < \varepsilon.

(This equivalence of the Darboux integral and the tagged-Riemann-sum limit is a standard prior fact about the Riemann integral, following from squeezing Riemann sums between lower and upper Darboux sums; we take it as known, consistent with Basic Properties: Linearity and Additivity having already established the integral's basic apparatus.) Now choose, for each nn, a partition PnP_n with mesh ∥Pn∥→0\|P_n\| \to 0 as n→∞n \to \infty (for instance the uniform partition into nn equal pieces, ∥Pn∥=(b−a)/n→0\|P_n\| = (b-a)/n \to 0). For each PnP_n, Steps 1–3 produce tags {ti(n)}\{t_i^{(n)}\} with

F(b)−F(a)=S(f,Pn,{ti(n)})for every n.F(b) - F(a) = S(f, P_n, \{t_i^{(n)}\}) \quad \text{for every } n.

By Step 4's Riemann-sum convergence (applied with ε→0\varepsilon \to 0, i.e. taking nn large enough that ∥Pn∥<δ(ε)\|P_n\| < \delta(\varepsilon)), the right side satisfies

S(f,Pn,{ti(n)})⟶∫abfas n→∞.S(f, P_n, \{t_i^{(n)}\}) \longrightarrow \int_a^b f \quad \text{as } n \to \infty.

But the left side F(b)−F(a)F(b) - F(a) is a constant, not depending on nn at all. A sequence that is eventually within ε\varepsilon of ∫abf\int_a^b f for every ε>0\varepsilon>0, while also being constantly equal to F(b)−F(a)F(b)-F(a), forces

F(b)−F(a)=∫abf(x) dx.F(b) - F(a) = \int_a^b f(x)\,dx.

This completes the proof. ■\blacksquare

∎
CorollaryAdditivity-compatible form

If FF is an antiderivative of f∈R[a,b]f \in \mathcal{R}[a,b] and c∈(a,b)c \in (a,b), then applying the theorem on [a,c][a,c] and on [c,b][c,b] separately gives

∫acf+∫cbf=(F(c)−F(a))+(F(b)−F(c))=F(b)−F(a)=∫abf,\int_a^c f + \int_c^b f = \big(F(c)-F(a)\big) + \big(F(b)-F(c)\big) = F(b)-F(a) = \int_a^b f,

recovering the additivity property of the integral (compare Basic Properties: Linearity and Additivity) as a consistency check on the theorem.

Worked examples

Example 1: An integral requiring a non-obvious antiderivative, with asymmetric bounds. Compute ∫14(3x2−1x)dx.\displaystyle \int_1^{4} \left(3x^2 - \frac{1}{\sqrt{x}} \right) dx. Step 1 — check integrability. The integrand f(x)=3x2−x−1/2f(x) = 3x^2 - x^{-1/2} is continuous on [1,4][1,4] (no singularity, since 1≤x≤41 \le x \le 4 keeps us away from x=0x=0). A continuous function on a closed bounded interval is Riemann integrable, so hypothesis (1) of the theorem holds. Step 2 — find an antiderivative. Using linearity of differentiation,

F(x)=x3−2xF(x) = x^3 - 2\sqrt{x}

satisfies F′(x)=3x2−1x=f(x)F'(x) = 3x^2 - \dfrac{1}{\sqrt{x}} = f(x) for x>0x > 0, in particular on (1,4)(1,4), and FF is continuous on [1,4][1,4] (and differentiable at the endpoints too, with the correct one-sided derivatives, since x−1/2x^{-1/2} is continuous there). Hypothesis (2) holds. Step 3 — apply the theorem.

∫14(3x2−1x)dx=F(4)−F(1)=(43−24)−(13−21).\int_1^4 \left(3x^2 - \frac{1}{\sqrt x}\right) dx = F(4) - F(1) = \big(4^3 - 2\sqrt4\big) - \big(1^3 - 2\sqrt1\big).

Compute each piece: 43=644^3 = 64, 24=42\sqrt4 = 4, so F(4)=64−4=60F(4) = 64 - 4 = 60. And F(1)=1−2=−1F(1) = 1 - 2 = -1. Hence

∫14(3x2−1x)dx=60−(−1)=61.\int_1^4 \left(3x^2 - \frac{1}{\sqrt x}\right)dx = 60 - (-1) = 61.

As a sanity check, note f(x)>0f(x) > 0 throughout [1,4][1,4] (since 3x2≥33x^2 \ge 3 while 1/x≤11/\sqrt{x} \le 1 there), so a positive answer is consistent with ff being positive over an interval of length 3.

Example 2: A piecewise antiderivative, testing additivity across a break point. Compute ∫03f(x) dx\displaystyle \int_0^{3} f(x)\,dx where

f(x)={2x0≤x≤24+3(x−2)22<x≤3.f(x) = \begin{cases} 2x & 0 \le x \le 2\\ 4 + 3(x-2)^2 & 2 < x \le 3.\end{cases}

Step 1 — check integrability. First check continuity of ff at x=2x=2: from the left, f(2−)=2(2)=4f(2^-) = 2(2) = 4; from the right, f(2+)=4+3(0)2=4f(2^+) = 4 + 3(0)^2 = 4. They match, and ff is a polynomial (hence continuous) on each piece, so ff is continuous on all of [0,3][0,3], hence Riemann integrable there. Step 2 — construct a global antiderivative. On [0,2][0,2], an antiderivative of 2x2x is x2+C1x^2 + C_1. On [2,3][2,3], an antiderivative of 4+3(x−2)24 + 3(x-2)^2 is 4x+(x−2)3+C24x + (x-2)^3 + C_2. To glue these into one continuous, piecewise-differentiable FF we need FF to match at x=2x=2 and have matching derivative there (which is automatic since both pieces reduce to f(2)=4f(2)=4 at the junction from the formulas above, guaranteeing F′F' exists at 22 too, provided the values agree). Choose C1=0C_1 = 0, so the left piece is F(x)=x2F(x) = x^2 for x∈[0,2]x \in [0,2], giving F(2)=4F(2) = 4. For the right piece we need 4(2)+(0)3+C2=44(2) + (0)^3 + C_2 = 4, i.e. 8+C2=48 + C_2 = 4, so C2=−4C_2 = -4. Thus

F(x)={x20≤x≤24x+(x−2)3−42<x≤3.F(x) = \begin{cases} x^2 & 0 \le x \le 2 \\ 4x + (x-2)^3 - 4 & 2 < x \le 3. \end{cases}

One checks directly that FF is continuous at 22 (both formulas give 44) and that F′(2)=4=f(2)F'(2) = 4 = f(2) from both sides (differentiate each piece and take the one-sided limit at x=2x=2), so FF is a genuine antiderivative of ff on all of [0,3][0,3] in the sense of the theorem. Step 3 — apply the theorem.

∫03f(x) dx=F(3)−F(0).\int_0^3 f(x)\,dx = F(3) - F(0).

Compute F(3)=4(3)+(1)3−4=12+1−4=9F(3) = 4(3) + (1)^3 - 4 = 12 + 1 - 4 = 9. Compute F(0)=02=0F(0) = 0^2 = 0. Hence

∫03f(x) dx=9−0=9.\int_0^3 f(x)\,dx = 9 - 0 = 9.

As a check via additivity, ∫022x dx=[x2]02=4\int_0^2 2x\,dx = [x^2]_0^2 = 4, and ∫23(4+3(x−2)2) dx=[4x+(x−2)3]23=(12+1)−(8+0)=5\int_2^3 (4+3(x-2)^2)\,dx = [4x+(x-2)^3]_2^3 = (12+1)-(8+0) = 5; summing gives 4+5=94+5=9, matching. This confirms both the theorem's answer and the corollary on additivity across c=2c=2.

CounterexampleWhy f must be Riemann integrable — an unbounded derivative

The theorem requires both that FF have F′=fF' = f everywhere on (a,b)(a,b) and that ff itself be Riemann integrable. It is tempting to think the second hypothesis is redundant — surely if FF is differentiable, its derivative is automatically integrable? This is false, and the failure is exactly what forces us to state integrability of ff as a separate, necessary hypothesis. Define F:[0,1]→RF : [0,1] \to \mathbb{R} by

F(x)={x2sin⁡ ⁣(1x2)x≠00x=0.F(x) = \begin{cases} x^2 \sin\!\left(\dfrac{1}{x^2}\right) & x \neq 0 \\ 0 & x = 0. \end{cases}

FF is differentiable everywhere on [0,1][0,1]. For x≠0x \neq 0, the product and chain rules give

F′(x)=2xsin⁡ ⁣(1x2)−2xcos⁡ ⁣(1x2).F'(x) = 2x\sin\!\left(\frac{1}{x^2}\right) - \frac{2}{x}\cos\!\left(\frac{1}{x^2}\right).

At x=0x = 0, we compute directly from the definition of the derivative:

F′(0)=lim⁡h→0F(h)−F(0)h=lim⁡h→0h2sin⁡(1/h2)h=lim⁡h→0hsin⁡ ⁣(1h2)=0,F'(0) = \lim_{h \to 0} \frac{F(h) - F(0)}{h} = \lim_{h\to 0} \frac{h^2 \sin(1/h^2)}{h} = \lim_{h \to 0} h\sin\!\left(\frac{1}{h^2}\right) = 0,

since ∣hsin⁡(1/h2)∣≤∣h∣→0|h \sin(1/h^2)| \le |h| \to 0 (squeeze theorem). So FF is differentiable on all of [0,1][0,1], with

f(x):=F′(x)={2xsin⁡(1/x2)−2xcos⁡(1/x2)x≠00x=0.f(x) := F'(x) = \begin{cases} 2x\sin(1/x^2) - \frac{2}{x}\cos(1/x^2) & x \neq 0\\ 0 & x=0.\end{cases}

ff is unbounded near 00, hence NOT Riemann integrable on [0,1][0,1]. Take xn=12πnx_n = \frac{1}{\sqrt{2\pi n}} for n∈Nn \in \mathbb{N}, so that 1/xn2=2πn1/x_n^2 = 2\pi n and cos⁡(1/xn2)=cos⁡(2πn)=1\cos(1/x_n^2) = \cos(2\pi n) = 1. Then

f(xn)=2xnsin⁡(2πn)−2xncos⁡(2πn)=0−2xn=−22πn→−∞f(x_n) = 2x_n \sin(2\pi n) - \frac{2}{x_n}\cos(2\pi n) = 0 - \frac{2}{x_n} = -2\sqrt{2\pi n} \to -\infty

as n→∞n \to \infty, while xn→0+x_n \to 0^+. So ff takes arbitrarily large-magnitude values arbitrarily close to 00; ff is unbounded on [0,1][0,1]. A basic prerequisite fact about the Riemann integral is that every Riemann integrable function on a closed bounded interval must be bounded (unbounded functions make upper Darboux sums infinite, or force the supremum/infimum defining the sums to be undefined on any subinterval containing points where f→±∞f\to\pm\infty). Hence f∉R[0,1]f \notin \mathcal{R}[0,1].

Conclusion. FF is a legitimate antiderivative of ff on [0,1][0,1] — differentiable everywhere with F′=fF' = f — yet ∫01f(x) dx\int_0^1 f(x)\,dx does not exist as a Riemann integral, so the formula ∫01f=F(1)−F(0)\int_0^1 f = F(1) - F(0) is not merely false, it is meaningless (the left side is undefined). This shows the integrability hypothesis on ff cannot be removed or deduced from the existence of FF: differentiability of FF alone is not strong enough to guarantee f=F′f=F' is even bounded, let alone Riemann integrable.

Scope, edge cases & common traps

SituationWhat happens
ff continuous on [a,b][a,b]Theorem applies directly; FF can be built via the First Fundamental Theorem of Calculus as F(x)=∫axfF(x)=\int_a^x f.
ff Riemann integrable but discontinuous at finitely many pointsTheorem still applies as stated, provided a genuine antiderivative FF (differentiable everywhere with F′=fF'=f) exists — such FF may fail to exist if ff has jump discontinuities, since derivatives satisfy the intermediate value property (Darboux's theorem on derivatives) and cannot jump.
FF differentiable everywhere but F′=fF'=f unboundedIntegrability hypothesis fails; conclusion can be literally undefined (see counterexample).
ff integrable, but FF merely continuous with F′=fF'=f only except at finitely many pointsA stronger version of the theorem (using generalized/Lebesgue-style arguments) still often holds, but this exceeds the elementary Riemann-sum proof given here; treat as a distinct, more advanced result.
Improper integrals (a=−∞a=-\infty or unbounded ff near an endpoint)Outside the scope of this theorem; requires the separate theory of improper integrals as a limit of proper ones.

The mathematician behind it

Origins · 1 minIsaac Newton and how this was discovered
IN
Isaac Newton
1642–1727 · Cambridge, England, during the Scientific Revolution

Newton developed the calculus of fluxions in the 1660s–1670s, treating derivatives ("fluxions") and what we now call antiderivatives as inverse operations from the start — his method of quadrature was built around finding a function whose fluxion matched the given curve, then reading off areas from the resulting formula. Independently, Gottfried Wilhelm Leibniz arrived at an equivalent formalism with his differential and integral notation, and the notation ∫\int and dxdx that survives today is his. The rigorous ε\varepsilon–δ\delta underpinnings that make the Mean-Value-Theorem-based proof above airtight came later, chiefly from Augustin-Louis Cauchy and Bernhard Riemann in the 19th century, who replaced the intuitive "infinitesimal" reasoning of Newton and Leibniz with limits of sums.

History
- 1666 — Newton circulates early manuscripts on fluxions, treating integration as the inverse of differentiation.
- 1675 — Leibniz introduces the integral sign ∫\int and systematic differential notation, publishing calculus results shortly after.
- 1823 — Cauchy gives the first rigorous limit-based definition of the definite integral for continuous functions, and proves an early rigorous version of the evaluation theorem.
- 1854 — Riemann's habilitation extends the theory to integrability of more general (possibly discontinuous) functions via what are now called Riemann sums.
- 1902 — Henri Lebesgue's measure-theoretic integral resolves subtler versions of the fundamental theorem, handling functions like the one in the counterexample above via a more general (non-Riemann) framework.
Do I need f to be continuous for the Second Fundamental Theorem to apply?

No. The theorem as proved here only requires ff to be Riemann integrable on [a,b][a,b], together with the existence of some antiderivative FF with F′=fF' = f everywhere on (a,b)(a,b) (and matching one-sided derivatives at the endpoints). Continuity of ff is a common and convenient special case — it guarantees integrability automatically and lets you construct FF via the First Fundamental Theorem of Calculus — but the evaluation formula itself works more generally, for instance when ff has finitely many jump discontinuities, as long as a genuine antiderivative still exists.

How is this different from the First Fundamental Theorem of Calculus?

The First Fundamental Theorem says that if ff is continuous on [a,b][a,b], then the accumulation function A(x)=∫axf(t) dtA(x) = \int_a^x f(t)\,dt is differentiable with A′(x)=f(x)A'(x) = f(x); it produces an antiderivative from an integral. The Second Fundamental Theorem runs the other direction: given any antiderivative FF (found by any means, not necessarily by integrating ff), it lets you evaluate ∫abf\int_a^b f by computing F(b)−F(a)F(b)-F(a). They are converse-flavored statements that are logically distinct, though often taught together and sometimes both called "the Fundamental Theorem of Calculus."

Can two different antiderivatives of the same f give different answers?

No, and this is a useful consistency check. If F1F_1 and F2F_2 are both antiderivatives of ff on [a,b][a,b], then (F1−F2)′=f−f=0(F_1 - F_2)' = f - f = 0 on (a,b)(a,b), so by the Mean Value Theorem F1−F2F_1 - F_2 is constant on [a,b][a,b]. Hence F1(b)−F1(a)=F2(b)−F2(a)F_1(b) - F_1(a) = F_2(b) - F_2(a): the specific antiderivative you choose does not matter, only the endpoint difference does. This is why we speak of "an" antiderivative rather than "the" antiderivative.

What goes wrong if F fails to be differentiable at even one interior point?

The Mean Value Theorem step in the proof requires FF to be differentiable on the open subinterval (xi−1,xi)(x_{i-1}, x_i) for every subinterval in every partition. If FF fails to be differentiable at a single point c∈(a,b)c \in (a,b), then any partition with cc strictly inside some subinterval breaks the Mean Value Theorem application on that piece, and the telescoping trick no longer produces an exact Riemann sum. In practice, one can often still recover the theorem by splitting the integral at cc (via additivity) and checking whether one-sided derivatives at cc still match f(c)f(c), but this requires care and is not automatic.

Is the Second Fundamental Theorem still true for Lebesgue integrals?

A closely related but distinct statement holds: if FF is absolutely continuous on [a,b][a,b], then F′=fF' = f almost everywhere, ff is Lebesgue integrable, and ∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a). Absolute continuity is a strictly stronger condition than mere differentiability everywhere — it rules out pathologies like the Cantor function and like the counterexample in this article. The Riemann-integral version proved here is the classical, calculus-course statement; the Lebesgue version is the appropriate generalization studied in measure-theoretic real analysis.

Why does the proof need the tags tᵢ to come from the Mean Value Theorem specifically, rather than arbitrary tags?

The telescoping identity F(b)−F(a)=∑i(F(xi)−F(xi−1))F(b)-F(a) = \sum_i (F(x_i)-F(x_{i-1})) is exactly true for any partition, but to rewrite each summand as f(ti)Δxif(t_i)\Delta x_i exactly (not approximately), we need a point tit_i where the average rate of change of FF on [xi−1,xi][x_{i-1},x_i] equals the instantaneous rate f(ti)f(t_i) — that is precisely what the Mean Value Theorem guarantees. Without it, we would only have an approximate Riemann sum (using, say, f(xi−1)Δxif(x_{i-1})\Delta x_i), and the argument would need an extra error-bounding step using uniform continuity, which the Mean Value Theorem approach elegantly avoids.

Before this

Where this leads