9. Applications and improper integrals

An integral solves a model only after the quantity and its units have been identified. A thin slice can approximate area or volume; summing slices and passing to a limit explains the integral, not just the formula.

Definition

Improper integral

For $f$ continuous on every finite interval $[a,b]$ with $b>a$, define $\int_a^\infty f(x)dx=\lim_{b\to\infty}\int_a^b f(x)dx$ when the limit exists as a finite real number. For an endpoint singularity, use the appropriate one-sided limit, for example $\int_0^1f=\lim_{\varepsilon\downarrow0}\int_\varepsilon^1f$. If there are several infinite endpoints or singularities, split at interior points and require each resulting improper integral to converge separately; cancellation in a symmetric limit is not enough.

Theorem

Power benchmark and comparison

For real $p$, $\int_1^\infty x^{-p}dx$ converges exactly when $p>1$, while $\int_0^1x^{-p}dx$ converges exactly when $p<1$. On a tail $[A,\infty)$, suppose $0\le f(x)\le g(x)$ and both functions are integrable on every finite subinterval. If the integral of $g$ over that tail converges, so does the integral of $f$; if the integral of $f$ diverges, so does the integral of $g$. This comparison controls only the tail: check any other improper endpoints or singularities separately.

Proof

The power test

For $p\ne1$, integrate to obtain $[x^{1-p}/(1-p)]$ and take the relevant endpoint limit. At infinity the endpoint term tends to zero exactly when $1-p<0$; at zero it tends to zero exactly when $1-p>0$. For $p=1$, the antiderivative is $\ln x$ and either endpoint limit diverges. Comparison follows by bounding nondecreasing partial integrals of nonnegative functions and using completeness.

Worked example

Area and volume by slices

On $[0,1]$, the region between $y=x$ and $y=x^2$ has area $\int_0^1(x-x^2)dx=1/6$, because $x\ge x^2$ there. Rotating the region under $y=x$ above the axis about the $x$-axis gives cross-sectional disks of radius $x$ and area $\pi x^2$, hence volume $\pi\int_0^1x^2dx=\pi/3$. For an annular slice, subtract the inner disk area from the outer one.

Worked example

Convergence before evaluation

For $\int_1^\infty dx/(1+x^2)$, the integrand is nonnegative and at most $1/x^2$ for $x\ge1$, so the power test and comparison imply convergence. Its exact value is $\lim_{b\to\infty}(\arctan b-\pi/4)=\pi/4$. In contrast, $\int_1^\infty dx/x$ diverges; an antiderivative existing on every finite interval does not imply convergence.

If $f$ has a continuous derivative on $[a,b]$, the arc length of its graph is $\int_a^b\sqrt{1+[f'(x)]^2}dx$. The formula comes from approximating short arcs by straight segments. A rate $r(t)$ accumulated over time yields net change $\int_a^b r(t)dt$; the physical interpretation requires matching units and signs.

Washers and sign changes

Suppose $0\le r(x)\le R(x)$ are continuous on $[a,b]$. Rotating the region between them about the horizontal axis gives volume $\pi\int_a^b(R(x)^2-r(x)^2)dx$. The two squares represent areas of cross-sectional disks; $\pi(R-r)^2$ would describe a different disk. If the order of curves changes or the region crosses the axis, split into intervals and reassess the geometry rather than applying one formula blindly.

Worked example

An integrable endpoint singularity

Although $x^{-1/2}$ is unbounded near zero, $\int_0^1x^{-1/2}dx=\lim_{\varepsilon\downarrow0}[2\sqrt{x}]_\varepsilon^1=2$. In contrast $\int_0^1x^{-1}dx$ diverges logarithmically. An unbounded integrand does not by itself decide convergence.

Choosing a method

Draw the region and decide which curve is above or which radius is outer before integrating. Use symmetry only after verifying it. For an improper integral, identify every problematic endpoint, write explicit limits, and test each piece. Use comparison when a benchmark controls size; use antiderivatives only after writing the limit.

Exercises

1. Determine convergence and value, when finite, of $\int_1^\infty x^{-3}dx$.
2. Find the area between $y=2x$ and $y=x^2$.
3. Does $\int_{-1}^1 dx/x$ converge as an improper integral?

Full solutions

1. $\int_1^b x^{-3}dx=[-1/(2x^2)]_1^b=1/2-1/(2b^2)$, which tends to $1/2$.
2. The curves meet at $x=0$ and $x=2$. On $[0,2]$, $2x\ge x^2$, so the area is $\int_0^2(2x-x^2)dx=[x^2-x^3/3]_0^2=4/3$.
3. No. The required split at zero yields $\int_{-1}^0dx/x=-\infty$ and $\int_0^1dx/x=+\infty$. Their symmetric principal value is zero, but that is a different notion and does not establish improper convergence.

Connection

Improper integration returns to the limits of the opening chapters. The complete toolkit is a loop: functions describe a model, derivatives analyse local change, and integrals accumulate that change while limits justify the passage to arbitrarily small or large scales.

Proofs supporting the methods used in this note

Finite real limit laws and geometric sequences: the missing proofs

Work over the ordered real numbers, with the Archimedean property, positive-integer induction and the usual absolute-value triangle inequality. A sequence converges to a finite real $L$ when every $\varepsilon>0$ admits an $N$ that works for every $n\ge N$.

Uniqueness, boundedness and subsequences. If $a_n\to L,M$ and $L\ne M$, require both errors below $|L-M|/3$ at the same sufficiently large index. The triangle inequality gives $|L-M|<2|L-M|/3$, a contradiction. For boundedness, require $|a_n-L|<1$ in the tail, and take a maximum of $|L|+1$ and the finitely many initial absolute values. For a subsequence $n_1<n_2<\cdots$, induction gives $n_k\ge k$; hence every tail threshold for the original sequence also works eventually for the subsequence.

Sum, scalar and product. If $a_n\to a$, $b_n\to b$, choose both input errors below $\varepsilon/2$; then

$$|(a_n+b_n)-(a+b)|\le |a_n-a|+|b_n-b|<\varepsilon.$$

For a fixed scalar $c\ne0$ use input tolerance $\varepsilon/|c|$; $c=0$ is immediate. Choose $B\ge1$ bounding all $|b_n|$. Require

$$|a_n-a|<\frac{\varepsilon}{2B},\qquad |b_n-b|<\frac{\varepsilon}{2(|a|+1)}.$$

The common later threshold gives $|a_nb_n-ab|\le B|a_n-a|+|a||b_n-b|<\varepsilon$.

Reciprocal and quotient. If $b\ne0$, first require $|b_n-b|<|b|/2$, ensuring $|b_n|>|b|/2$ and nonzero denominators. Then

$$\left|\frac1{b_n}-\frac1b\right|\le\frac{2|b_n-b|}{|b|^2}.$$

Requiring also $|b_n-b|<\varepsilon|b|^2/2$ proves reciprocal convergence. The product law now proves $a_n/b_n\to a/b$. Any undefined initial quotients may be omitted; convergence is a tail assertion.

Order and squeeze. If $a_n\le b_n$ eventually and $a_n\to a$, $b_n\to b$, then $a\le b$: otherwise errors below $(a-b)/3$ would force $a_n>b_n$. If $a_n\le z_n\le b_n$ eventually and both outer limits are $L$, require both errors below $\varepsilon$. Then $L-\varepsilon<z_n<L+\varepsilon$, proving $z_n\to L$. This proof does not require a previously known limit of $z_n$.

For the corresponding function-limit laws, fix a common allowed approach domain $D\subseteq\mathbb R$ and an accumulation point $a$ of $D$: every punctured neighborhood of $a$ contains a point of $D$. Let $f,g:D\to\mathbb R$ have the stated finite relative-domain limits at $a$. Replace each tail threshold in the displayed estimates by a punctured-domain radius, and take the minimum of finitely many positive radii. For the product proof only local boundedness is needed: $g(x)\to b$ gives $|g(x)|\le|b|+1$ on a sufficiently small punctured $D$-neighborhood, even if $g$ is unbounded elsewhere on $D$. The same estimates prove sum, scalar, product, reciprocal and quotient laws with their stated nonzero limiting-denominator condition; the same order and squeeze estimates apply on the common approach domain. Accumulation supplies actual points when proving uniqueness or excluding an incompatible order of limiting values.

For composition let $B\subseteq\mathbb R$, let $b\in B$, and let $F:B\to\mathbb R$ be continuous at $b$ relative to $B$. Suppose $g:D\to\mathbb R$ tends to $b$ and $g(x)\in B$ throughout some sufficiently small punctured $D$-neighborhood of $a$. Given $\varepsilon>0$, continuity supplies $\eta>0$ such that $u\in B$ and $|u-b|<\eta$ imply $|F(u)-F(b)|<\varepsilon$. The limit of $g$ and the composition-domain condition supply a common positive radius forcing both conditions for every allowed approaching $x$. Thus $F(g(x))\to F(b)$. This includes inputs with $g(x)=b$ and does not divide by an inner increment.

Geometric classification. Bernoulli's bound $(1+t)^n\ge1+nt$ for $t\ge0$ follows by induction: multiply the case $n$ by the positive $1+t$ and retain the nonnegative remainder $nt^2$. If $0<q<1$, set $t=q^{-1}-1>0$. Thus

$$0<q^n\le\frac1{1+nt}\longrightarrow0.$$

For any requested tolerance, the Archimedean property supplies $N$ making the right side smaller. Hence $r^n\to0$ for $0<|r|<1$ by $|r^n|=|r|^n$; $r=0$ gives zero for every positive index. At $r=1$ the sequence is constantly one. At $r=-1$, the even and odd subsequences are one and minus one, so uniqueness and subsequence inheritance exclude convergence. If $r>1$, Bernoulli with $t=r-1$ gives $r^n\ge1+nt\to+\infty$. If $r<-1$, $|r^n|\to+\infty$ by the same argument, contradicting boundedness of any convergent real sequence. These cases exhaust all real ratios.

Subsequences

Definition

Subsequence

A subsequence of $(x_n)$ is $(x_{n_k})$ where $n_1<n_2<\cdots$ are positive integers. If $x_n\to L$, then every subsequence converges to $L$ because $n_k\ge k$ and therefore eventually lies beyond any prescribed index.

Theorem

Bolzano–Weierstrass

Every bounded real sequence has a convergent subsequence.

Proof

Nested-interval proof

Place every term in a closed bounded interval $I_1$. Bisect it. At least one half contains infinitely many terms; choose that half as $I_2$. Continue, obtaining nested closed intervals $I_k$ of lengths tending to zero, each containing infinitely many terms. Choose indices $n_k$ strictly increasing with $x_{n_k}\in I_k$; this is possible because each interval contains infinitely many terms. Completeness gives a common point $L$ in all intervals: the supremum of their left endpoints lies in each interval because every left endpoint is at most every right endpoint. Since $x_{n_k},L\in I_k$, their distance is at most the interval length and tends to zero.

The conclusion is about a subsequence, not necessarily the whole sequence. For example, $(-1)^n$ has constant subsequences converging to $1$ and $-1$, but does not converge.

Real closed sets: the sequence criterion and the open complement

Work on the real line in the usual classical set-theoretic setting, with countable witness selections available. A set $U$ is open when every $y$ in $U$ has some $r>0$ with $(y-r,y+r)$ contained in $U$. We use the Archimedean consequence that $1/n$ tends to zero. For every $K$ subset of the real line, the following conditions are equivalent:

1. Every sequence of points of $K$ that converges in the real line has its limit in $K$.
2. The complement of $K$ is open.

Proof of 2 implies 1. Suppose $x_n$ belongs to $K$ and $x_n$ tends to $x$. If $x$ were outside $K$, openness would give $r>0$ with $(x-r,x+r)$ disjoint from $K$. Convergence gives a threshold after which $|x_n-x|<r$, contradicting $x_n$ in $K$. Therefore $x$ belongs to $K$.

Proof of 1 implies 2. Fix $y$ outside $K$. If no positive-radius interval around $y$ avoided $K$, then for every positive integer $n$ there would be a point $x_n$ in $K$ with $|x_n-y|<1/n$. Choose these witnesses as a sequence. Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$, $|x_n-y|<1/n\le1/N<\varepsilon$, so $x_n$ tends to $y$. Condition 1 would force $y$ into $K$, a contradiction. Hence some interval around $y$ avoids $K$. This holds for every point of the complement, proving it open. The witness-sequence step uses the stated classical setting; this proof is not offered as a theorem about choice-free foundations. If $K$ is empty there is no sequence in $K$, and its complement is the whole open line. If $K$ is the whole line, its empty complement is open vacuously. Both edge cases also satisfy the equivalence.

Learning goals and definitions

A set $K\subseteq\mathbb R$ is compact if every family of open sets whose union contains $K$ has a finite subfamily still covering $K$. A set is closed if it contains all limits of convergent sequences of its points, equivalently if its complement is open. Sequential compactness means every sequence in $K$ has a subsequence converging to a member of $K$.

The empty set is compact, closed and bounded, and sequentially compact vacuously. In proofs using a selected point or an infinite sequence, treat that case separately. We may assume the real completeness property and its nested-interval and Bolzano–Weierstrass consequences from the earlier sequence units.

Theorem: Heine–Borel in the real line

For $K\subseteq\mathbb R$, the following are equivalent: compactness by open covers; closedness and boundedness; sequential compactness.

Compact implies closed and bounded. The intervals $(-n,n)$, $n\ge1$, cover $\mathbb R$, hence $K$. A finite subcover gives a bound. To show closedness, take $y\notin K$. For each $x\in K$, the interval $B(x,|x-y|/2)$ avoids $y$. A finite subfamily covers $K$. Let $\delta$ be the minimum of their positive half-distances to $y$. Every point in the chosen interval centered at $x$ has distance from $y$ strictly greater than $|x-y|/2\ge\delta$. Thus $B(y,\delta)$ misses $K$, showing its complement open. If $K$ is empty this is immediate.

Closed and bounded implies compact. Enclose nonempty $K$ in a closed bounded interval $I_0$. Suppose an open cover of $K$ has no finite subcover. Bisect $I_0$. At least one closed half meets $K$ in a set not finitely coverable; otherwise two finite collections would cover $K$. Choose such a half and repeat. We obtain nested closed intervals $I_n$ with lengths tending to zero, and $K\cap I_n$ not finitely coverable. Each intersection is nonempty; choose $x_n\in K\cap I_n$. The nested-interval property gives a unique common point $p$, and $x_n\to p$ because both lie in $I_n$. Closedness gives $p\in K$. Some member $O$ of the cover contains $p$, and openness gives a ball around $p$ inside $O$. For large $n$, the whole interval $I_n$ lies in that ball, so $O$ alone covers $K\cap I_n$, a contradiction.

Closed and bounded implies sequentially compact. Every sequence in $K$ is bounded, so Bolzano–Weierstrass supplies a convergent subsequence. Closedness retains its limit in $K$.

Sequentially compact implies closed and bounded. If unbounded, choose $x_n\in K$ with $|x_n|>n$; no subsequence can converge to a finite real number, a contradiction. If $p$ is a limit of points in $K$, choose $x_n\in K$ with $|x_n-p|<1/n$. A subsequence converges to a point of $K$ by sequential compactness, but it also converges to $p$; uniqueness of real limits gives $p\in K$. Thus $K$ is closed. This completes all implications.

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Uniform continuity

Definition

One delta for the entire domain

A function $f:A\to\mathbb R$ is uniformly continuous on $A$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in A$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$. The chosen $\delta$ can depend on $\varepsilon$ but not on $x$ or $y$.

Theorem

Heine–Cantor theorem

Every continuous real-valued function on a compact set is uniformly continuous.

Proof

Contradiction using compactness

If uniform continuity failed, some $\varepsilon_0>0$ would permit, for every $n$, points $x_n,y_n\in K$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge\varepsilon_0$. Compactness supplies a subsequence $x_{n_k}\to x\in K$. Since the paired distances tend to zero, also $y_{n_k}\to x$. Continuity gives both $f(x_{n_k})\to f(x)$ and $f(y_{n_k})\to f(x)$, contradicting the output separation.

Worked example

Pointwise but not uniform on an unbounded domain

The function $f(x)=x^2$ is continuous on $\mathbb R$ but not uniformly continuous there. Take $x_n=n$ and $y_n=n+1/n$. Then $|x_n-y_n|=1/n\to0$, but $|f(y_n)-f(x_n)|=2+1/n^2\not\to0$. On any closed bounded interval, Heine–Cantor does give uniform continuity. It is also direct: if $|x|,|y|\le M$, then $|x^2-y^2|\le2M|x-y|$.

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Proof

Why continuity suffices

A continuous function on a compact interval is uniformly continuous. Given $\varepsilon>0$ and $b>a$, choose $\delta$ so that points less than $\delta$ apart have output difference less than $\varepsilon/(2(b-a))$. A partition with mesh below $\delta$ has oscillation on every subinterval at most that bound, hence $U-L<\varepsilon$. The Darboux criterion yields integrability. For $a=b$ the integral is zero by convention.

Bounded proper Riemann integrals: mesh equivalence and elementary properties

Let $a<b$ be finite, and let $f:[a,b]\to\mathbb R$ be bounded. For a partition $P$, write $L(f,P),U(f,P)$ for the lower and upper Darboux sums. We use real completeness and the proved Darboux criterion: equality of lower and upper integrals is equivalent to arbitrarily small $U-L$. This packet does not assume continuity of $f$.

Darboux implies every fine tagged sum converges. Let the Darboux integral be $I$. Given $\varepsilon>0$, choose a fixed partition $P_0$ with gap below $\varepsilon/2$. Choose $M>0$ with $|f|\le M$, and let $k$ be the number of interior partition points of $P_0$. For any other partition $P$, mark as bad every cell whose closure contains one of those $k$ points. Their total length is at most $2k\|P\|$, because each point belongs to at most two cells. Every remaining cell lies within a single $P_0$ cell. The contribution of all good cells to the oscillation sum $U(f,P)-L(f,P)$ is at most the gap for $P_0$; the bad cells contribute at most $2M(2k\|P\|)$. Thus

$$U(f,P)-L(f,P)<\varepsilon/2+4Mk\|P\|.$$

Take $\|P\|<\varepsilon/[8M\max(1,k)]$. Every tagged sum $S(P)$ and $I$ lie between $L(f,P)$ and $U(f,P)$, so $|S(P)-I|<\varepsilon$, independently of tags.

The converse. Suppose all sufficiently fine tagged sums are within $\eta$ of a common finite $I$. Fix one such fine partition. Each cell supremum and infimum can be approximated by actual function values with error below $\eta/(b-a)$. The resulting upper-approximating tagged sum exceeds $U(f,P)-\eta$, and the lower-approximating sum is below $L(f,P)+\eta$. Therefore $U(f,P)<I+2\eta$ and $L(f,P)>I-2\eta$. Since $\eta$ can be arbitrarily small, the Darboux criterion gives integrability. Lower and upper bounds on tagged sums then identify its integral as $I$. Suprema need not be attained for this argument.

Linearity, constants and order. For integrable $f,g$ and real $\alpha,\beta$, on any common tagged partition the exact finite identity is $S(\alpha f+\beta g)=\alpha S(f)+\beta S(g)$. Its limit exists, so the equivalence just proved gives integrability and the corresponding linear integral identity. The constant function $c$ has every sum equal to $c(b-a)$. If $f\le g$, their tagged sums with the same tags obey the same order; their limits do too. Consequently $|\int f|\le M(b-a)$ whenever $|f|\le M$.

Absolute values. The reverse triangle inequality gives $\operatorname{osc}_J|f|\le\operatorname{osc}_J f$ on every cell $J$: apply $||f(x)|-|f(y)||\le|f(x)-f(y)|$ and approximate the suprema/infima. Thus every Darboux gap for $|f|$ is at most the gap for $f$, and $|f|$ is integrable. Since $-|f|\le f\le|f|$, order and linearity give $|\int_a^b f|\le\int_a^b|f|$.

Restriction and interval addition. For $a<c<b$, refine any small-gap partition by $c$. The nonnegative gaps on $[a,c]$ and $[c,b]$ sum to the refined whole gap; each can therefore be arbitrarily small. Both restrictions are integrable. Join partitions of the two pieces with meshes tending to zero. The exact sum on the joined partition is the sum of the two tagged sums, so passage to their three limits gives $\int_a^b f=\int_a^c f+\int_c^b f$. Conversely, if both restrictions are bounded integrable, join two partitions with gaps below $\varepsilon/2$; the resulting whole gap is below $\varepsilon$, proving whole integrability and the same identity. At a shared endpoint both restrictions use the same value of $f$; no endpoint term has positive interval length. Set equal-bound integrals to zero and reversed-bound integrals to the negative. Splitting a finite interval and cancellation now extend addition to every ordering of its three endpoints.

These proofs justify the estimates used in accumulation, continuous FTC, integration by parts and proper substitution. They assert bounded proper Riemann statements only, and do not replace the separate convergence definitions at singularities or infinite endpoints.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Mean value theorem. Under the same continuity and differentiability hypotheses but without equal endpoint values, some $c\in(a,b)$ satisfies $f(b)-f(a)=f'(c)(b-a)$.

Subtract the secant line $f(a)+[f(b)-f(a)](x-a)/(b-a)$ from $f$ and apply Rolle. This proves the theorem without assuming $f'$ is continuous. Consequences include: derivative zero on an interval makes the function constant; nonnegative derivative makes it nondecreasing; and $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for points in it. An interval-wide derivative bound cannot be inferred from one derivative value.

4.5 Integrals as accumulated change

For $a<b$, take a partition $a=x_0<x_1<\cdots<x_n=b$, with $\Delta x_i=x_i-x_{i-1}$, tags $\xi_i\in[x_{i-1},x_i]$, and mesh $\|P\|=\max_i\Delta x_i$. A finite number $I$ is the definite integral $\int_a^bf(x)dx$ if, for every $\varepsilon>0$, there is $\delta>0$ such that every partition with $\|P\|<\delta$ and every choice of tags satisfy $|\sum_i f(\xi_i)\Delta x_i-I|<\varepsilon$. In that case $I$ is the common limit. Continuous functions on a closed interval are integrable.

Let $f$ be continuous on $[a,b]$ and $A(x)=\int_a^xf(t)dt$. For $x\in(a,b)$ and nonzero $h$ with $x+h\in[a,b]$, the integral estimate gives

$$\left|\frac{A(x+h)-A(x)}h-f(x)\right|\le\sup_{t\in[\min(x,x+h),\max(x,x+h)]}|f(t)-f(x)|\longrightarrow0\quad(h\to0).$$

Thus $A'(x)=f(x)$ in the interior; at the endpoints the same argument gives one-sided derivatives. If $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $F'=f$ there, the mean value theorem makes $F-A$ constant on $[a,b]$. Therefore $\int_a^bf=F(b)-F(a)$.

For example, $\int x^2dx=x^3/3+C$, and $\int_a^bx^2dx=(b^3-a^3)/3$. The indefinite integral denotes a family of antiderivatives; a definite integral is a number. Negative rate values contribute negatively to a net accumulation.

A complete length proof for a finite C1 graph

Let $a<b$ be finite real numbers and $f\in C^1([a,b],\mathbb R)$, with one-sided endpoint derivatives. Put $\gamma(t)=(t,f(t))\in\mathbb R^2$. Define the graph's length to be the supremum of all inscribed polygonal lengths

$$L(P)=\sum_i\|\gamma(t_i)-\gamma(t_{i-1})\|,\qquad a=t_0<\cdots<t_n=b.$$

Then this supremum is finite and equals $\int_a^b\sqrt{1+f'(t)^2}\,dt$. The integral is a proper continuous integral. The degenerate interval $a=b$ has length and integral zero.

Here are the norm facts used in the proof. For vectors $u,v\in\mathbb R^2$, expanding gives $\|u\|^2\|v\|^2-(u\cdot v)^2=(u_1v_2-u_2v_1)^2\ge0$, hence $|u\cdot v|\le\|u\|\|v\|$. Expanding $\|u+v\|^2$ now gives the triangle inequality and thus $|\|u\|-\|v\||\le\|u-v\|$. For continuous $w$ on a finite interval, define $v=\int w$ componentwise. If $v\ne0$, let $e=v/\|v\|$; scalar linearity and order give $\|v\|=\int e\cdot w\le\int\|w\|$. If $v=0$, the same integral norm inequality is immediate. These statements use only proper continuous integrals.

Since $\gamma'=(1,f')$ is continuous, its norm is continuous and integrable. The componentwise fundamental theorem gives

$$\gamma(v)-\gamma(u)=\int_u^v\gamma'(t)\,dt\quad(a\le u\le v\le b).$$

The integral norm inequality on each cell proves $L(P)\le I:=\int_a^b\|\gamma'\|$ for every partition, so the supremum is at most the finite $I$.

Conversely fix $\varepsilon>0$. Uniform continuity of $\gamma'$ supplies $\delta>0$ such that $\|\gamma'(s)-\gamma'(t)\|<\varepsilon/[2(b-a)]$ when $|s-t|<\delta$. For any partition of mesh below $\delta$ and any tags $\xi_i$, the fundamental theorem and integral norm bound yield

$$\left\|\gamma(t_i)-\gamma(t_{i-1})-\gamma'(\xi_i)\Delta t_i\right\|\le\frac{\varepsilon\Delta t_i}{2(b-a)}.$$

The reverse triangle inequality and summation therefore give

$$\left|L(P)-\sum_i\|\gamma'(\xi_i)\|\Delta t_i\right|\le\varepsilon/2.$$

The tagged sums of the continuous speed tend to $I$ on all sufficiently fine meshes. Thus $|L(P)-I|<\varepsilon$ for all sufficiently fine meshes, after also controlling that tagged-sum error. In particular, uniform partitions give polygonal lengths tending to $I$, so their supremum is at least $I$. Together with the earlier upper bound, this proves equality. This establishes the C1 graph formula from the polygonal-supremum definition, rather than assuming that short chords already equal short arcs.