The fundamental theorem of calculus

Theorem

The fundamental theorem of calculus

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Statement

Let \( a \lt b \) be real numbers and let \( f : [a,b] \to \mathbb{R} \) be Riemann integrable (in particular, bounded). Define the indefinite integral \( F : [a,b] \to \mathbb{R} \) by \( F(x) = \int_a^x f(t)\,\mathrm{d}t \). Part I. \( F \) is Lipschitz continuous on \( [a,b] \); and at every point \( c \in [a,b] \) at which \( f \) is continuous, \( F \) is differentiable with \( F'(c) = f(c) \) (one-sided derivatives at the endpoints). In particular, if \( f \) is continuous on all of \( [a,b] \), then \( F \) is an antiderivative of \( f \). Part II (evaluation form). If \( G : [a,b] \to \mathbb{R} \) is continuous on \( [a,b] \), differentiable at every point of \( (a,b) \) with \( G'(x) = f(x) \) for all \( x \in (a,b) \), and \( f \) is Riemann integrable on \( [a,b] \), then \[ \int_a^b f(t)\,\mathrm{d}t = G(b) - G(a). \]

Why it matters

The theorem welds together the two pillars of calculus, which are defined by entirely unrelated limit processes: the derivative (a local limit of difference quotients) and the Riemann integral (a global limit of sums over partitions). Part I manufactures antiderivatives — it is the only general existence theorem for antiderivatives of continuous functions, and it is how the logarithm, the arctangent, and the error function are rigorously defined. Part II converts the computation of an integral, a priori an infinite limiting procedure, into the evaluation of a single function at two points.

Almost every downstream tool of analysis — integration by parts, substitution, the integral form of Taylor's remainder, the theory of ordinary differential equations, and ultimately Stokes' theorem on manifolds — is a corollary or a descendant of this result. Its failure modes (Cantor's function, Volterra's function) drove the twentieth-century rebuilding of integration theory by Lebesgue.

Hypotheses

\( f \) is Riemann integrable on \( [a,b] \).For the Dirichlet function \( f = \mathbf{1}_{\mathbb{Q}} \) on \( [0,1] \), every upper sum is \( 1 \) and every lower sum is \( 0 \), so \( f \) is not integrable and \( F(x) = \int_0^x f \) is not even defined. Boundedness is part of the definition: \( f(t) = 1/\sqrt{t} \) on \( (0,1] \), \( f(0)=0 \), is unbounded, hence not Riemann integrable, and only an improper integral survives.

\( f \) is continuous at \( c \) (needed for \( F'(c) = f(c) \) in Part I).Take the Heaviside step \( f(t) = 0 \) for \( t \lt 0 \), \( f(t) = 1 \) for \( t \geq 0 \), on \( [-1,1] \). Then \( F(x) = \max(x,0) \), which has a corner at \( 0 \): left derivative \( 0 \), right derivative \( 1 \). \( F \) is still Lipschitz, but differentiability fails exactly at the discontinuity of \( f \).

\( G \) is continuous on the closed interval \( [a,b] \) (Part II).Let \( G(x) = 0 \) for \( 0 \leq x \lt 1 \) and \( G(1) = 5 \). Then \( G' = 0 \) on \( (0,1) \), so \( f \equiv 0 \) and \( \int_0^1 f = 0 \), but \( G(1) - G(0) = 5 \). A single jump at an endpoint destroys the telescoping argument.

\( G'(x) = f(x) \) at every point of \( (a,b) \), not merely almost every point.The Cantor–Lebesgue function \( c : [0,1] \to [0,1] \) is continuous, nondecreasing, and satisfies \( c'(x) = 0 \) for every \( x \) outside the Cantor set — a set of measure zero. If "almost everywhere" sufficed, we would get \( c(1) - c(0) = \int_0^1 0 = 0 \); but \( c(1) - c(0) = 1 \).

\( f = G' \) is itself Riemann integrable (Part II) — this is not automatic.Volterra (1881) constructed \( V : [0,1] \to \mathbb{R} \) differentiable at every point with \( V' \) bounded, yet \( V' \) is discontinuous on a fat Cantor set of positive measure, so by Lebesgue's criterion \( V' \) is not Riemann integrable. The formula \( \int_0^1 V' = V(1) - V(0) \) is then meaningless within the Riemann theory.

Proof

Throughout, \( F(x) = \int_a^x f(t)\,\mathrm{d}t \). We prove Part I in Steps 1–6 and Part II in Steps 7–10.

\[ \exists\, M \geq 0 : \ |f(t)| \leq M \quad \forall\, t \in [a,b]. \]

Riemann integrability presupposes boundedness: an unbounded function has some upper or lower Darboux sums infinite, so the Darboux integrals cannot coincide. This \( M \) is fixed for the rest of the proof. A

\[ a \leq x \lt y \leq b \ \Longrightarrow\ F(y) - F(x) = \int_x^y f(t)\,\mathrm{d}t. \]

Additivity of the Riemann integral over adjacent intervals: \( \int_a^y f = \int_a^x f + \int_x^y f \), valid because \( f \) is integrable on every subinterval of \( [a,b] \) (restriction lemma: refinements of partitions of \( [a,b] \) containing \( x \) induce partitions of \( [a,x] \) and \( [x,y] \)). A

Triangle inequality for integrals (\( \pm f \leq |f| \) and monotonicity of the integral), then the bound of Step 1. Hence \( F \) is Lipschitz with constant \( M \), so uniformly continuous on \( [a,b] \). This proves the continuity claim of Part I. B

Fix \( c \in [a,b] \) with \( f \) continuous at \( c \). For \( h \neq 0 \) with \( c + h \in [a,b] \): \[ \frac{F(c+h) - F(c)}{h} - f(c) \;=\; \frac{1}{h} \int_c^{c+h} \bigl( f(t) - f(c) \bigr)\,\mathrm{d}t. \]

By Step 2, \( F(c+h) - F(c) = \int_c^{c+h} f \) (with the orientation convention \( \int_c^{c+h} = -\int_{c+h}^{c} \) when \( h \lt 0 \)), and the constant integrates exactly: \( \int_c^{c+h} f(c)\,\mathrm{d}t = h\,f(c) \). Subtract and divide by \( h \). This algebraic normalisation is the key idea: the difference quotient error is an average of \( f(t) - f(c) \). B

\[ \forall \varepsilon \gt 0\ \exists \delta \gt 0 : \ 0 \lt |h| \lt \delta \ \Longrightarrow\ \left| \frac{F(c+h) - F(c)}{h} - f(c) \right| \leq \varepsilon. \]

Continuity of \( f \) at \( c \): choose \( \delta \gt 0 \) with \( |f(t) - f(c)| \lt \varepsilon \) whenever \( t \in [a,b] \), \( |t - c| \lt \delta \). For \( 0 \lt |h| \lt \delta \), every \( t \) between \( c \) and \( c + h \) satisfies \( |t - c| \leq |h| \lt \delta \), so by the triangle inequality for integrals — valid in both orientations, since \( \bigl| \int_c^{c+h} g \bigr| \leq |h| \sup |g| \) whether \( h \gt 0 \) or \( h \lt 0 \) — \[ \left| \frac{1}{h} \int_c^{c+h} \bigl(f(t) - f(c)\bigr)\,\mathrm{d}t \right| \;\leq\; \frac{1}{|h|} \cdot |h| \cdot \varepsilon \;=\; \varepsilon. \] Note the estimate needs no integrability assumption beyond Step 1: \( f - f(c) \) is integrable on the small interval by Step 2's restriction lemma. C

\[ F'(c) \;=\; \lim_{h \to 0} \frac{F(c+h) - F(c)}{h} \;=\; f(c). \]

Definition of the derivative: Step 5 is precisely the \( \varepsilon\text{–}\delta \) statement that the limit exists and equals \( f(c) \). At \( c = a \) or \( c = b \) the same argument runs with \( h \gt 0 \) or \( h \lt 0 \) only, giving the one-sided derivative. Part I is proved. A

Part II. Let \( P : a = x_0 \lt x_1 \lt \cdots \lt x_n = b \) be an arbitrary partition. Then for each \( i \) there exists \( t_i \in (x_{i-1}, x_i) \) with \[ G(x_i) - G(x_{i-1}) \;=\; f(t_i)\,(x_i - x_{i-1}). \]

Lagrange's Mean Value Theorem applied to \( G \) on \( [x_{i-1}, x_i] \): \( G \) is continuous on this closed subinterval (hypothesis) and differentiable on its interior, which lies inside \( (a,b) \), where \( G' = f \). This is the pivotal move — it converts increments of \( G \) into sampled values of \( f \). B

\[ G(b) - G(a) \;=\; \sum_{i=1}^{n} \bigl( G(x_i) - G(x_{i-1}) \bigr) \;=\; \sum_{i=1}^{n} f(t_i)\,(x_i - x_{i-1}). \]

Telescoping of the finite sum, then substitution from Step 7. Thus \( G(b) - G(a) \) is a Riemann sum of \( f \) for the partition \( P \) — for every partition \( P \), with tags chosen by the MVT. A

With \( m_i = \inf_{[x_{i-1},x_i]} f \) and \( M_i = \sup_{[x_{i-1},x_i]} f \): \[ L(f,P) \;=\; \sum_{i=1}^n m_i \,\Delta x_i \;\leq\; G(b) - G(a) \;\leq\; \sum_{i=1}^n M_i\,\Delta x_i \;=\; U(f,P). \]

Since \( t_i \in (x_{i-1}, x_i) \), we have \( m_i \leq f(t_i) \leq M_i \) by definition of infimum and supremum; multiply by \( \Delta x_i = x_i - x_{i-1} \gt 0 \) and sum. So the fixed number \( G(b) - G(a) \) is squeezed between the lower and upper Darboux sums of every partition. A

\[ \sup_P L(f,P) \;\leq\; G(b) - G(a) \;\leq\; \inf_P U(f,P), \qquad \sup_P L(f,P) = \inf_P U(f,P) = \int_a^b f, \] \[ \therefore\quad G(b) - G(a) = \int_a^b f(t)\,\mathrm{d}t. \]

Step 9 holds for all \( P \), so \( G(b) - G(a) \) is an upper bound for all lower sums and a lower bound for all upper sums; pass to the supremum and infimum. The middle equality is exactly the Darboux definition of Riemann integrability of \( f \) — this is where the hypothesis "\( f \) integrable" is spent, and without it (Volterra) the two outer quantities need not meet. A real number trapped between two equal numbers equals both. C

Result

\[ \textbf{I: } \ \frac{\mathrm{d}}{\mathrm{d}x} \int_a^x f(t)\,\mathrm{d}t = f(x) \ \text{ at each point of continuity of } f; \qquad \textbf{II: } \ \int_a^b G'(t)\,\mathrm{d}t = G(b) - G(a). \]

Reading. Integrating \( f \) and then differentiating gives back \( f \) wherever \( f \) is continuous; differentiating \( G \) and then integrating gives back the total change of \( G \), provided the derivative is honest on the whole open interval and integrable. Differentiation and integration are mutually inverse — but each direction carries its own fine print.

Scope. Riemann integration on a compact interval \( [a,b] \subseteq \mathbb{R} \). Part I needs only integrability of \( f \) plus pointwise continuity where the derivative is claimed. Part II needs \( G \) continuous on \( [a,b] \), \( G' = f \) at every interior point, and \( f \) Riemann integrable. For improper integrals, unbounded intervals, or the Lebesgue theory, separate theorems (with absolute continuity replacing differentiability-plus-integrability) are required.

Corollaries & converses

Existence of antiderivatives. Every continuous \( f : [a,b] \to \mathbb{R} \) has an antiderivative, namely \( F(x) = \int_a^x f \). This is how \( \ln x = \int_1^x t^{-1}\mathrm{d}t \) and \( \arctan x = \int_0^x (1+t^2)^{-1}\mathrm{d}t \) are constructed.
Uniqueness up to a constant. If \( F_1' = F_2' = f \) on an interval, then \( (F_1 - F_2)' = 0 \), so \( F_1 - F_2 \) is constant (corollary of the MVT). Hence Part II is independent of which antiderivative is used.
Integration by parts. For \( u, v \in C^1[a,b] \): \( \int_a^b u v' = [uv]_a^b - \int_a^b u' v \) — apply Part II to \( G = uv \) with the product rule.
Substitution. For \( \varphi \in C^1[\alpha,\beta] \) and \( f \) continuous on \( \varphi([\alpha,\beta]) \): \( \int_\alpha^\beta f(\varphi(s))\varphi'(s)\,\mathrm{d}s = \int_{\varphi(\alpha)}^{\varphi(\beta)} f \) — apply Part II to \( G = F \circ \varphi \) with the chain rule and Part I.
Converse of Part I fails. Not every Riemann integrable \( f \) is a derivative: \( \operatorname{sgn} \) on \( [-1,1] \) is integrable but is no function's derivative, because derivatives satisfy the Darboux intermediate value property (Darboux's theorem) and \( \operatorname{sgn} \) skips every value in \( (-1,0) \cup (0,1) \).
Converse of Part II fails. "\( G \) continuous and \( G' \) exists almost everywhere with \( G' \) integrable" does not imply \( \int_a^b G' = G(b) - G(a) \): the Cantor function is the standard counterexample. The correct equivalence (Lebesgue): the identity holds for the Lebesgue integral exactly when \( G \) is absolutely continuous.

Fails without

Drop continuity of \( f \) at \( c \) (Part I): \( f = \operatorname{sgn} \) on \( [-1,1] \) gives \( F(x) = \int_{-1}^x \operatorname{sgn}(t)\,\mathrm{d}t = |x| - 1 \), which is continuous everywhere but not differentiable at \( 0 \); no version of \( F'(0) = f(0) \) can be rescued, whatever value \( f(0) \) is assigned.
Weaken "\( G' = f \) everywhere on \( (a,b) \)" to "almost everywhere" (Part II): the Cantor–Lebesgue function \( c \) satisfies \( c' = 0 \) off a measure-zero set, is continuous and monotone, yet \( c(1) - c(0) = 1 \neq 0 = \int_0^1 0 \). All the increase happens on the null set where differentiability fails.
Drop integrability of \( f = G' \) (Part II): Volterra's function \( V \) is differentiable everywhere on \( [0,1] \) with \( |V'| \leq 1 \), but \( V' \) is discontinuous on a positive-measure set, hence not Riemann integrable (Lebesgue's criterion); the left side of the formula does not exist. The Henstock–Kurzweil integral repairs exactly this failure.
Drop continuity of \( G \) at an endpoint (Part II): \( G = 0 \) on \( [0,1) \), \( G(1) = 5 \): then \( \int_0^1 G' = 0 \neq 5 = G(1) - G(0) \).

Common errors

Chain-rule amnesia. Writing \( \frac{\mathrm{d}}{\mathrm{d}x} \int_0^{x^2} f(t)\,\mathrm{d}t = f(x^2) \). Correct: \( f(x^2) \cdot 2x \). The FTC differentiates with respect to the upper limit; a composed limit needs the chain rule.
Integrating across a singularity. \( \int_{-1}^{1} t^{-2}\,\mathrm{d}t = \bigl[-t^{-1}\bigr]_{-1}^{1} = -2 \): a negative "value" for a positive integrand. The function is unbounded on \( [-1,1] \), so it is not Riemann integrable and Part II never applied.
"Integrable \( \Rightarrow \) has an antiderivative." False: \( \operatorname{sgn} \) is integrable but is not a derivative (Darboux's theorem). Part I only gives \( F' = f \) at points of continuity.
"Differentiable \( \Rightarrow \) Part II applies." False without checking integrability of the derivative (Volterra). The hypothesis \( f \in \mathcal{R}[a,b] \) must be verified, not assumed.
Dummy-variable collision. Writing \( F(x) = \int_a^x f(x)\,\mathrm{d}x \). The variable of integration must differ from the limit: \( \int_a^x f(t)\,\mathrm{d}t \). The collision breeds genuine errors when differentiating under the integral sign later.
Endpoint carelessness. Claiming \( F'(a) = f(a) \) as a two-sided derivative, or applying the MVT step to a \( G \) not continuous at \( a \) or \( b \).

Discussion

Historically the theorem crystallised in stages. Isaac Barrow's geometric lemma (Lectiones Geometricae, 1670) already contains the inverse relationship between tangent and quadrature problems for monotone curves; Newton and Leibniz turned it into an algorithmic calculus in the 1660s–1680s, treating it as essentially self-evident. The first proof meeting modern standards is Cauchy's (1823), for continuous integrands, using his definition of the integral as a limit of sums. Riemann (1854) widened the class of integrable functions, and it was precisely this widening that made the theorem's hypotheses delicate: Darboux showed derivatives can be discontinuous yet still obey the intermediate value property, and Volterra (1881) produced a bounded derivative that is not Riemann integrable, splitting Part I from Part II irreparably within Riemann's framework.

Conceptually, the two parts are asymmetric. Part I is a smoothing statement: integration raises regularity (a merely integrable \( f \) yields a Lipschitz \( F \); a continuous \( f \) yields a \( C^1 \) function \( F \)). Part II is a rigidity statement: knowledge of the derivative at every interior point, plus continuity at the boundary, pins down the total change. The MVT is the engine of Part II, and it is worth internalising that the proof shows something slightly stronger than convergence of Riemann sums: for every partition there exist tags making the Riemann sum exactly equal to \( G(b) - G(a) \), with no limit taken.

In the Lebesgue theory the correct statement is a characterisation: for \( G : [a,b] \to \mathbb{R} \), the identity \( G(x) - G(a) = \int_a^x G'\,\mathrm{d}\mu \) (Lebesgue integral) holds for all \( x \) if and only if \( G \) is absolutely continuous — Lebesgue (1904) and Vitali. One direction uses the Lebesgue differentiation theorem, which guarantees that an absolutely continuous \( G \) is differentiable almost everywhere with integrable derivative; the Cantor function is continuous and of bounded variation but not absolutely continuous, marking the exact boundary of the theorem. Going further, the Henstock–Kurzweil (gauge) integral integrates every derivative and restores the clean statement "\( \int_a^b G' = G(b) - G(a) \) whenever \( G \) is differentiable on \( [a,b] \)," at the price of losing absolute integrability. In several variables the theorem generalises to \( \int_M \mathrm{d}\omega = \int_{\partial M} \omega \) (Stokes), of which FTC is the case \( M = [a,b] \), \( \partial M = \{a, b\} \) with signs.

Common misconceptions. The theorem does not say "differentiation and integration are inverse on the nose." Integration then differentiation recovers \( f \) only at continuity points (Part I); differentiation then integration recovers \( G \) only under everywhere-differentiability plus integrability of \( G' \) (Part II). Nor does it say every integrable function has an antiderivative, or that every derivative is integrable — both are false, and the counterexamples (sgn, Volterra) are part of the theorem's content, not curiosities.

Worked examples

Example 1. Differentiate \( F(x) = \displaystyle\int_0^{x^2} e^{-t^2}\,\mathrm{d}t \) for \( x \in \mathbb{R} \).

Define \( \Phi(u) = \displaystyle\int_0^{u} e^{-t^2}\,\mathrm{d}t \), so that \( F = \Phi \circ g \) with \( g(x) = x^2 \).

Decomposition into an indefinite integral and a smooth inner function; the integrand \( t \mapsto e^{-t^2} \) is continuous on all of \( \mathbb{R} \), so \( \Phi \) is defined on every interval. A

\[ \Phi'(u) = e^{-u^2} \quad \forall\, u \in \mathbb{R}. \]

FTC Part I: the integrand is continuous everywhere, so the indefinite integral is differentiable everywhere with derivative equal to the integrand. B

\[ F'(x) = \Phi'(x^2)\cdot \frac{\mathrm{d}}{\mathrm{d}x}(x^2) = e^{-x^4}\cdot 2x. \]

Chain rule, legitimate since \( \Phi \) is differentiable at \( g(x) = x^2 \) (Step 2) and \( g \) is differentiable at \( x \). A

\[ \frac{\mathrm{d}}{\mathrm{d}x} \int_0^{x^2} e^{-t^2}\,\mathrm{d}t \;=\; 2x\,e^{-x^4}. \]

Reading. The FTC supplies the derivative of the outer integral; the chain rule transports it through the variable upper limit. Note that \( F \) is differentiable even though no elementary antiderivative of \( e^{-t^2} \) exists — the theorem is an existence statement, not a formula-finding device.

Example 2. Evaluate \( \displaystyle\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n}{n^2 + k^2} \).

\[ \sum_{k=1}^{n} \frac{n}{n^2 + k^2} \;=\; \frac{1}{n} \sum_{k=1}^{n} \frac{1}{1 + (k/n)^2}. \]

Algebra: divide numerator and denominator by \( n^2 \). The right side is the Riemann sum of \( f(x) = \dfrac{1}{1+x^2} \) on \( [0,1] \) for the uniform partition with right-endpoint tags \( t_k = k/n \). B

\[ \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \frac{1}{1+(k/n)^2} \;=\; \int_0^1 \frac{\mathrm{d}x}{1+x^2}. \]

\( f \) is continuous on \( [0,1] \), hence Riemann integrable, and for an integrable function every sequence of tagged Riemann sums with mesh \( \to 0 \) converges to the integral (equivalence of the Darboux and Riemann formulations). B

\[ \int_0^1 \frac{\mathrm{d}x}{1+x^2} \;=\; \arctan(1) - \arctan(0) \;=\; \frac{\pi}{4}. \]

FTC Part II with \( G = \arctan \): \( G \) is continuous on \( [0,1] \), differentiable on \( (0,1) \) with \( G'(x) = \frac{1}{1+x^2} = f(x) \), and \( f \) is integrable. All hypotheses check. A

\[ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{n}{n^2 + k^2} \;=\; \frac{\pi}{4}. \]

Reading. A limit of sums with no visible closed form is converted into an integral, and the integral is then dispatched by exhibiting one antiderivative — the two halves of the FTC working in tandem.

Problems

Compute \( \dfrac{\mathrm{d}}{\mathrm{d}x} \displaystyle\int_x^{x^3} \cos(t^2)\,\mathrm{d}t \).
Solution

Let \( \Phi(u) = \int_0^u \cos(t^2)\,\mathrm{d}t \). The integrand is continuous on \( \mathbb{R} \), so by FTC Part I, \( \Phi'(u) = \cos(u^2) \) for all \( u \). By additivity, \[ \int_x^{x^3} \cos(t^2)\,\mathrm{d}t = \Phi(x^3) - \Phi(x). \] Differentiating with the chain rule: \[ \frac{\mathrm{d}}{\mathrm{d}x}\bigl( \Phi(x^3) - \Phi(x) \bigr) = \cos\bigl((x^3)^2\bigr)\cdot 3x^2 - \cos(x^2) = 3x^2 \cos(x^6) - \cos(x^2). \]
Evaluate \( \displaystyle\lim_{x \to 0} \frac{1}{x} \int_0^x e^{t^2}\,\mathrm{d}t \) without L'Hôpital.
Solution

Let \( F(x) = \int_0^x e^{t^2}\,\mathrm{d}t \). Then \( F(0) = 0 \), so \[ \frac{1}{x}\int_0^x e^{t^2}\,\mathrm{d}t = \frac{F(x) - F(0)}{x - 0}, \] the difference quotient of \( F \) at \( 0 \). The integrand \( e^{t^2} \) is continuous at \( 0 \), so FTC Part I gives \( F'(0) = e^{0^2} = 1 \). By definition of the derivative the limit of the difference quotient is \( F'(0) \). Hence the limit is \( 1 \).
Let \( f : \mathbb{R} \to \mathbb{R} \) be continuous with period \( T \gt 0 \), i.e. \( f(t + T) = f(t) \) for all \( t \). Prove that \( \displaystyle\int_a^{a+T} f(t)\,\mathrm{d}t \) does not depend on \( a \).
Solution

Define \( g(a) = \int_a^{a+T} f(t)\,\mathrm{d}t \). Writing \( \Phi(x) = \int_0^x f(t)\,\mathrm{d}t \), additivity gives \( g(a) = \Phi(a+T) - \Phi(a) \). Since \( f \) is continuous on \( \mathbb{R} \), FTC Part I makes \( \Phi \) differentiable everywhere with \( \Phi' = f \), so \( g \) is differentiable and \[ g'(a) = f(a+T)\cdot 1 - f(a) = f(a+T) - f(a) = 0 \] by periodicity. A function with vanishing derivative on the interval \( \mathbb{R} \) is constant (corollary of the Mean Value Theorem: for \( a_1 \lt a_2 \), \( g(a_2) - g(a_1) = g'(\xi)(a_2 - a_1) = 0 \)). Hence \( g \) is constant, which is the claim.
On \( [-1,1] \) let \( f = \operatorname{sgn} \) (that is, \( f(t) = -1 \) for \( t \lt 0 \), \( f(0) = 0 \), \( f(t) = 1 \) for \( t \gt 0 \)) and let \( F(x) = \int_{-1}^x f(t)\,\mathrm{d}t \). Compute \( F \) explicitly, determine exactly where \( F' \) exists, and reconcile the answer with the FTC.
Solution

\( f \) is bounded with a single discontinuity, hence Riemann integrable. For \( x \leq 0 \): \( F(x) = \int_{-1}^x (-1)\,\mathrm{d}t = (-1)\bigl(x - (-1)\bigr) = -x - 1 \) (changing the integrand's value at the single point \( 0 \) does not affect the integral). For \( x \geq 0 \): \( F(x) = F(0) + \int_0^x 1\,\mathrm{d}t = -1 + x \). Both formulas combine to \[ F(x) = |x| - 1. \] \( F \) is continuous on all of \( [-1,1] \) (indeed Lipschitz with constant \( 1 \), as Part I promises with \( M = 1 \)). For \( x \neq 0 \), \( f \) is continuous at \( x \), and indeed \( F'(x) = \operatorname{sgn}(x) = f(x) \), confirming Part I. At \( x = 0 \) the left derivative of \( F \) is \( -1 \) and the right derivative is \( +1 \), so \( F'(0) \) does not exist. This is consistent with the theorem: Part I claims differentiability only at points where \( f \) is continuous, and \( 0 \) is not such a point. The example shows the continuity hypothesis in Part I is sharp.
(First Mean Value Theorem for integrals.) Let \( f : [a,b] \to \mathbb{R} \) be continuous. Prove there exists \( c \in (a,b) \) with \[ \int_a^b f(t)\,\mathrm{d}t = f(c)\,(b-a). \]
Solution

Define \( F(x) = \int_a^x f(t)\,\mathrm{d}t \). Since \( f \) is continuous on \( [a,b] \), FTC Part I gives: \( F \) is continuous on \( [a,b] \) and differentiable on \( (a,b) \) with \( F'(x) = f(x) \) there (in fact on all of \( [a,b] \), but the open interval suffices). These are exactly the hypotheses of Lagrange's Mean Value Theorem for \( F \) on \( [a,b] \), so there exists \( c \in (a,b) \) with \[ F(b) - F(a) = F'(c)\,(b-a) = f(c)\,(b-a). \] But \( F(a) = 0 \) and \( F(b) = \int_a^b f(t)\,\mathrm{d}t \), so \[ \int_a^b f(t)\,\mathrm{d}t = f(c)\,(b-a), \] as required. Note the conclusion places \( c \) strictly inside \( (a,b) \), which is finer than what one gets from the crude bounds \( m(b-a) \leq \int_a^b f \leq M(b-a) \) plus the intermediate value theorem (that argument only yields \( c \in [a,b] \)).