Learning goals and preparation

We assume definite integrals, the fundamental theorem, substitution, integration by parts, and derivatives. We now derive integral models from small pieces. You will distinguish signed totals from positive totals; select slices, washers or shells for a volume; and formulate length, mass, work and pressure calculations with consistent units.

The most important step is often constructing the integrand. An antiderivative of an incorrectly modeled quantity solves the wrong problem perfectly. Every derivation below identifies what a small piece represents, what its measure is, and which assumptions allow summation.

4.1 Rates, displacement, and total distance

Let $a<b$ and let position $s\in C^1([a,b])$, with continuous velocity $v=s'$. Then displacement over $[a,b]$ is $s(b)-s(a)=\int_a^bv(t)dt$. Total distance is $\int_a^b|v(t)|dt$, because movement in opposite directions should not cancel. The same distinction applies to a net flow rate versus total material moved.

Worked example 1: a reversal. Let $v(t)=t-2$ m/s for $0\le t\le5$. Displacement is

$$\int_0^5(t-2)dt=\frac52\text{ m}.$$

Velocity is negative up to $2$ and positive afterward. The total distance is

$$\int_0^2(2-t)dt+\int_2^5(t-2)dt=2+\frac92=\frac{13}{2}\text{ m}.$$

A position offset does not affect either change. An initial position is needed only to recover the absolute final position from displacement.

4.2 Area between curves and changing orientation

If $f\ge g$ are continuous on $[a,b]$, thin vertical strips have area approximately $(f(x)-g(x))\Delta x$. Their Riemann-sum limit is

$$A=\int_a^b(f(x)-g(x))dx.$$

If the order changes, split at crossings or integrate $|f-g|$. Horizontal strips are often preferable when the left and right boundaries are simpler as functions of $y$.

Continuous strips give Jordan area

Let $a<b$ and let $f,g:[a,b]\to\mathbb R$ be continuous with $g\le f$. Use ordinary Jordan area: finite unions of rectangles have their usual area, and a bounded set is measurable when inner and outer rectangular areas can be made arbitrarily close. On a partition interval $I$, put $f_- =\min_I f$, $f_+=\max_I f$, and similarly $g_-,g_+$. A rectangle of height $f_+-g_-$ covers the part of the region over $I$, while the rectangle between $g_+$ and $f_-$ is contained in it when $f_-\ge g_+$; otherwise use an empty inner rectangle. The height gap is at most $\operatorname{osc}_I f+\operatorname{osc}_I g$. Uniform continuity makes the total area gap at most $(b-a)(\omega_f(\delta)+\omega_g(\delta))\to0$, where $\delta$ is the mesh and $\omega_f(\delta)=\sup_{|x-y|\le\delta}|f(x)-f(y)|$. Boundary line segments have zero area. The strip set is therefore Jordan measurable. Every tagged sum for $f-g$ lies between those two area sums, proving $A=\int_a^b(f-g)$.

Worked example 2: choose the easy direction. The region between $x=y^2$ and $x=2-y$ has intersection values satisfying $y^2=2-y$, hence $y=-2,1$. Across this range the right boundary is $2-y$ and the left is $y^2$. The area is

$$\int_{-2}^1(2-y-y^2)dy =\left[2y-\frac{y^2}{2}-\frac{y^3}{3}\right]_{-2}^1=\frac92.$$

Vertical strips would need separate upper and lower branch descriptions. Choosing the direction of slicing reduces both algebra and the risk of missing a boundary piece.

4.3 Volumes by slices, washers and shells

If a solid has cross-sectional area $A(x)$ perpendicular to the $x$ axis, slices of thickness $\Delta x$ have approximate volume $A(x)\Delta x$. Under appropriate continuity assumptions,

$$V=\int_a^bA(x)dx.$$

For a solid of revolution around the $x$ axis, an outer radius $R(x)$ and inner radius $r(x)$ give washer area $\pi(R^2-r^2)$, provided $R\ge r\ge0$. Square the radii before subtracting: $(R-r)^2$ is generally not the correct cross-sectional area.

A precise sufficient theorem for cross-sectional volume

Let $a<b$, $c<d$, $e<\ell$, and let $E\subseteq[a,b]\times[c,d]\times[e,\ell]$ be a bounded Jordan-measurable set. Define its volume as the proper Riemann integral of $\mathbf1_E$ over that box. Assume each section $E_x=\{(y,z):(x,y,z)\in E\}$ is Jordan measurable, and write $A(x)$ for its area. Then $A$ is Riemann integrable and $V(E)=\int_a^bA(x)dx$.

Here is the rectangular Darboux argument, which avoids assuming a general slicing theorem. On a product grid, let $m_{ijk}$ and $M_{ijk}$ be the infimum and supremum of $\mathbf1_E$ on cell $I_i\times J_j\times K_k$. For every $x\in I_i$, section integrability gives

$$\sum_{j,k}m_{ijk}|J_j||K_k|\le A(x)\le\sum_{j,k}M_{ijk}|J_j||K_k|.$$

Taking the infimum and supremum over $x\in I_i$, multiplying by $|I_i|$ and adding yields $L_3\le L(A)\le U(A)\le U_3$. Jordan measurability gives grids with $U_3-L_3$ arbitrarily small. Thus the Darboux criterion proves integrability of $A$. Both $V(E)$ and $\int A$ lie in these same arbitrarily small brackets, so they are equal. This is a sufficient domain condition; continuity of a named section-area function alone does not establish that an arbitrary set is Jordan measurable or that every section is measurable.

Continuous washers have a direct volume proof

Let $a<b$ and let $R,r$ be continuous on $[a,b]$ with $0\le r\le R$. Rotate the radial interval $r(x)\le\sqrt{y^2+z^2}\le R(x)$ about the $x$ axis. Over a partition interval $I$, annular cylinders with squared-radius bounds give outer cross-sectional area $\pi(\max_I R^2-\min_I r^2)$ and inner area $\pi\max\{\min_I R^2-\max_I r^2,0\}$. Their difference is at most $\pi(\operatorname{osc}_I R^2+\operatorname{osc}_I r^2)$. Multiply by $|I|$ and sum. Uniform continuity makes the inner/outer volume gap tend to zero, proving Jordan measurability and

$$V=\pi\int_a^b(R(x)^2-r(x)^2)dx.$$

This uses the elementary area of an annulus and volume of a constant cylinder. Annuli are Jordan measurable: each boundary circle is the union of two continuous graphs on a compact interval, and the preceding strip argument with equal upper and lower functions shows those graphs have zero Jordan area. Extruding rectangular inner and outer area approximations through a fixed height proves that a constant cylinder is Jordan measurable with volume equal to base area times height. Equal radii and zero radii are allowed and contribute zero. The formula does not use the square of the difference of radii.

Continuous shells retain radius and multiplicity

Let $0\le a<b$ and let continuous $g\le f$ on $[a,b]$ bound a region with nonnegative radial coordinate $x$. Rotate it about the $y$ axis and put $H=f-g$. On a radial interval $I_i=[x_{i-1},x_i]$, use a cylindrical annulus of area $\pi(x_i^2-x_{i-1}^2)$ and heights $\max\{\min_{I_i}f-\max_{I_i}g,0\}$ and $\max_{I_i}f-\min_{I_i}g$ for inner and outer volumes. Their gap tends to zero by uniform continuity, bounded by $\pi(b^2-a^2)(\omega_f(\delta)+\omega_g(\delta))$. Thus the solid is Jordan measurable. With midpoint $m_i=(x_i+x_{i-1})/2$, the intermediate sums are exactly

$$\sum_i\pi(x_i^2-x_{i-1}^2)H(m_i)=\sum_i2\pi m_iH(m_i)(x_i-x_{i-1}).$$

They converge to $2\pi\int_a^bxH(x)dx$. Each radial annulus occurs once under $x\ge0$. If a generating region also uses negative $x$, overlapping rotated pieces must first be removed or separated; blindly adding both sides can double-count the same volume.

Worked example 3: a washer calculation. Rotate the region $x^2\le y\le x$, $0\le x\le1$, about the $x$ axis. The outer radius is $x$, inner radius $x^2$, so

$$V=\pi\int_0^1(x^2-x^4)dx=\frac{2\pi}{15}.$$

For rotation around the $y$ axis, a vertical strip generates a thin cylindrical shell with circumference $2\pi x$, height $x-x^2$, and thickness $dx$. Thus the volume about this different axis is

$$V=2\pi\int_0^1x(x-x^2)dx=\frac\pi6.$$

The different results are not contradictory: the axes of rotation are different. Shells require a nonnegative radius equal to distance from the axis; rotation about a shifted axis changes that radius.

A cone of height $H>0$ and base radius $R\ge0$ supplies an independent slicing check (with $R=0$ understood as the degenerate zero-volume case). Measuring $x$ from the tip gives radius $Rx/H$, hence

$$V=\int_0^H\pi(Rx/H)^2dx=\frac13\pi R^2H.$$

4.4 Arc length and surfaces of revolution

Why the real numbers need more than arithmetic

The rational numbers support the familiar ordered arithmetic, yet the set $\{q\in\mathbb Q:q>0,\ q^2<2\}$ has no rational least upper bound. The real number system adds a completeness property: bounded nonempty sets have least upper bounds. This is the source of many existence results later in analysis, not simply another rule for manipulating inequalities.

Definition

Upper bound and supremum

For a nonempty $A\subset\mathbb R$, $u$ is an upper bound if $a\le u$ for every $a\in A$. A number $s$ is $\sup A$ if it is an upper bound and $s\le u$ for every upper bound $u$. A maximum is an element of $A$ that is also an upper bound. The maximum, when it exists, is the supremum; a supremum need not belong to $A$. Lower bounds, infimum and minimum are defined by reversing inequalities.

Theorem

Completeness axiom

Every nonempty subset of $\mathbb R$ that is bounded above has a real supremum. Equivalently, every nonempty subset bounded below has a real infimum, via $\inf A=-\sup(-A)$.

For example, $\sup(0,1)=1$ although $1$ is not a maximum of that set. Conversely, $\sup[0,1]=1$ and it is a maximum. The empty set and an unbounded-above set do not have a real supremum under this definition.

Theorem

Supremum approximation criterion

An upper bound $s$ of a nonempty set $A$ is $\sup A$ if and only if, for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a\le s$.

Proof

Proving both directions

If $s=\sup A$ and no element exceeds $s-\varepsilon$, then $s-\varepsilon$ is a smaller upper bound, contradiction. Conversely, suppose $s$ is an upper bound and the approximation property holds. If another upper bound $u$ were smaller than $s$, choose $\varepsilon=s-u>0$. The property gives $a>u$, contradicting that $u$ is an upper bound. Thus $s$ is the least upper bound.

Worked example

Use both halves of a supremum proof

Let $A=\{n/(n+1):n\in\mathbb N, n\ge1\}$. Every element is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>1/\varepsilon$ by the Archimedean property below. Then $1-n/(n+1)=1/(n+1)<\varepsilon$. Thus $\sup A=1$. No element equals $1$, so $A$ has no maximum. An upper-bound calculation alone would not prove leastness: $2$ is also an upper bound.

Archimedean property and density

Theorem

Archimedean property

For every real $x$ there exists a positive integer $n>x$. Consequently, for every $\varepsilon>0$ there is $n\in\mathbb N$ such that $1/n<\varepsilon$.

Proof

Completeness forces integers to be unbounded

If the positive integers were bounded above, completeness would give $s=\sup\mathbb N$. Since $s-1$ is not an upper bound, some $n\in\mathbb N$ satisfies $n>s-1$. But $n+1>s$ and $n+1\in\mathbb N$, contradicting that $s$ bounds the set. Apply this to $1/\varepsilon$ to obtain the second statement.

Between any real numbers $x<y$ there is a rational number. Indeed choose $n$ with $n(y-x)>1$; the Archimedean property and the integer-part property yield an integer $m$ with $nx<m<ny$, so $x<m/n<y$. The integer-part property follows from the unboundedness of integers in both directions and the well-ordering of the natural numbers. An irrational also lies between $x$ and $y$: choose a rational $q$ with $x-\sqrt2<q<y-\sqrt2$; then $q+\sqrt2$ is irrational and lies in the interval. Thus density does not say rationals and irrationals are equal sets.

Worked example

An infimum via reflection

For $B=\{2+1/n:n\ge1\}$, every element exceeds $2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$ to obtain $2<2+1/n<2+\varepsilon$. Therefore $2$ is the infimum, but not a minimum. The maximum is $3$, attained at $n=1$.

Quantifiers in convergence

Definition

Convergence of a real sequence

A sequence $(x_n)$ converges to $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for every $n\ge N$, $|x_n-L|<\varepsilon$. Write $x_n\to L$. The index $N$ may depend on $\varepsilon$ but must work for all later indices.

To negate convergence to $L$, there must be some $\varepsilon_0>0$ such that for every $N$ there is an $n\ge N$ with $|x_n-L|\ge\varepsilon_0$. A few early terms do not affect convergence; a recurring failure arbitrarily late does.

Worked example

Construct an epsilon–N proof

For $x_n=(3n+1)/(n+2)$, guess $L=3$. Since

$$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}<\frac5n,$$

choose $N>5/\varepsilon$. Then every $n\ge N$ gives $|x_n-3|<\varepsilon$. The guess is motivation; the inequality and quantified choice establish the proof.

Theorem

Uniqueness, boundedness and limit laws

A convergent real sequence has a unique limit and is bounded. If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$, $x_ny_n\to xy$, and $x_n/y_n\to x/y$ when $y\ne0$ and the quotient is defined eventually. If $x_n\le z_n\le y_n$ eventually and both outer sequences converge to $L$, then $z_n\to L$.

Proof

Uniqueness and boundedness

If $x_n\to L$ and $x_n\to M$, then for any $\varepsilon>0$, sufficiently late terms satisfy both $|x_n-L|<\varepsilon/2$ and $|x_n-M|<\varepsilon/2$. The triangle inequality yields $|L-M|<\varepsilon$. Since this holds for every positive $\varepsilon$, $L=M$. For boundedness choose $N$ so that $|x_n-L|<1$ for $n\ge N$. The finite initial terms and the bound $|x_n|\le|L|+1$ for the tail together give a finite bound for the whole sequence.

The converse of boundedness is false: $(-1)^n$ is bounded but does not converge. If it converged to $L$, its even subsequence would converge to $1$ and its odd subsequence to $-1$, violating uniqueness.

Proof

Product law mechanism

A convergent sequence $(y_n)$ is bounded, say $|y_n|\le M$. Write

$$|x_ny_n-xy|\le|y_n||x_n-x|+|x||y_n-y|.$$

Both terms tend to zero. For the quotient, $y_n\to y\ne0$ implies $|y_n|\ge|y|/2$ eventually; this keeps the denominator away from zero and gives $1/y_n\to1/y$.

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.

Compact subsets of the real line

A set $K\subset\mathbb R$ is compact if every open cover has a finite subcover. In $\mathbb R$, Heine–Borel says this is equivalent to being closed and bounded; it is also equivalent to every sequence in $K$ having a subsequence converging to a point of $K$. The closedness requirement keeps the subsequential limit in the set; boundedness permits extraction. We use the sequential form in the proofs below.

Theorem

Extreme value theorem

If $K\subset\mathbb R$ is nonempty and compact and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both a maximum and a minimum on $K$.

Proof

Extract a maximizing subsequence

If $f$ were unbounded above, choose $x_n\in K$ with $f(x_n)>n$. A subsequence $x_{n_k}\to x\in K$ exists; continuity would give $f(x_{n_k})\to f(x)$, contradicting divergence to infinity. The same reasoning applied to $-f$ gives boundedness below. Let $s=\sup f(K)$. Choose $x_n\in K$ with $s-1/n<f(x_n)\le s$. A convergent subsequence tends to some $x\in K$, and continuity implies $f(x)=s$. Apply the argument to $-f$ for a minimum.

Worked example

Which hypothesis fails?

The continuous function $f(x)=x$ on $(0,1)$ is bounded but has no maximum or minimum: $0$ and $1$ are not in the domain. The set is bounded but not closed, hence not compact. On $[0,1]$, both endpoints are present and the extrema are attained. Compactness is a sufficient global hypothesis, not a condition required for every individual function to attain extrema.

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.

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 $\varepsilon$.

Theorem: finitely many discontinuities are permitted

If $f$ is bounded and continuous except possibly at finitely many points, it is integrable.

Proof. Let $|f|\le M$, taking $M>0$. Surround the exceptional points by finitely many small intervals, with endpoints included in the partition, having total length less than $\varepsilon/(4M)$. At endpoints of $[a,b]$ use one-sided intervals. Their contribution to the upper-lower gap is at most $2M$ times their total length, hence less than $\varepsilon/2$. On the remaining finitely many closed intervals, $f$ is continuous and therefore uniformly continuous. Subdivide each sufficiently finely that its oscillation on every resulting subinterval is less than $\varepsilon/[2(b-a)]$. Their total gap is less than $\varepsilon/2$. Combining all partition points gives gap below $\varepsilon$. If an exceptional interval endpoint would itself be exceptional, move it slightly so all actual exceptional points lie inside the selected intervals or at the domain endpoints. This keeps the remaining closed intervals continuous.

Continuous functions and all tagged meshes

Let $f:[a,b]\to\mathbb R$ be continuous, where $a<b$. The extreme value theorem makes $f$ bounded, and Heine–Cantor makes it uniformly continuous. For a partition $P$, write $L(f,P)$ and $U(f,P)$ for its lower and upper Darboux sums. The proved Darboux integrability theorem for continuous functions supplies the common integral $I$.

Fix $\varepsilon>0$. Uniform continuity supplies $\delta>0$ such that $|f(x)-f(y)|<\varepsilon/[2(b-a)]$ whenever $|x-y|<\delta$. If the mesh of $P$ is less than $\delta$, every subinterval has oscillation at most $\varepsilon/[2(b-a)]$. Therefore

$$U(f,P)-L(f,P)\le\frac{\varepsilon}{2}<\varepsilon.$$

For every selection of tags, its tagged sum $S(P)$ lies between $L(f,P)$ and $U(f,P)$; so does $I$. Thus $|S(P)-I|\le U(f,P)-L(f,P)<\varepsilon$. The same $\delta$ works for every partition of sufficiently small mesh and every choice of tags. This proves the tagged-mesh definition for continuous functions on the stated closed interval.

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 $\int_a^bf=F(b)-F(a)$.

Componentwise integration and uniform continuity

Let $a<b$ and let $r=(r_1,\ldots,r_n):[a,b]\to\mathbb R^n$ be $C^1$, with $n\ge1$. Define the integral of a continuous vector function componentwise. For $a\le u<v\le b$, the scalar fundamental theorem applied to each coordinate gives

$$\int_u^v r'(t)\,dt=\bigl(r_1(v)-r_1(u),\ldots,r_n(v)-r_n(u)\bigr)=r(v)-r(u).$$

Each $r_i'$ is continuous on the compact interval and hence uniformly continuous. Given $\eta>0$, choose $\delta_i>0$ such that $|r_i'(t)-r_i'(s)|<\eta/\sqrt n$ whenever $t,s\in[a,b]$ and $|t-s|<\delta_i$. With $\delta=\min_i\delta_i$, every such pair with $|t-s|<\delta$ satisfies

$$\|r'(t)-r'(s)\|^2=\sum_{i=1}^n|r_i'(t)-r_i'(s)|^2<\eta^2.$$

Thus $r'$ is uniformly continuous in the Euclidean norm. These two facts apply even when $r'$ vanishes.

1. Define length before replacing it by an integral

For $r:[a,b]\to\mathbb R^n$, define its length as the supremum over finite partitions $a=t_0<\cdots<t_m=b$ of the polygonal sums $\sum_i\|r(t_i)-r(t_{i-1})\|$. The supremum allows arbitrarily fine polygonal approximations; one endpoint chord is generally only a lower bound.

Theorem: $C^1$ length formula

For a $C^1$ curve on a compact interval,

$$L(r)=\int_a^b\|r'(t)\|\,dt.$$

Upper bound. On a subinterval, the vector fundamental theorem gives displacement $d=\int r'(t)dt$. If $d\ne0$, dot with its unit direction to obtain $\|d\|\le\int\|r'(t)\|dt$; if $d=0$, that inequality is immediate. Summing over any partition gives polygonal length at most the speed integral.

Lower bound. Uniform continuity of $r'$ says that for any $\eta>0$, a sufficiently fine partition makes $\|r'(t)-r'(\xi_i)\|<\eta$ on every subinterval, where $\xi_i$ is any chosen tag there. Thus its displacement differs from $r'(\xi_i)\Delta t_i$ by a vector of norm at most $\eta\Delta t_i$. The reverse triangle inequality gives chord length at least $\|r'(\xi_i)\|\Delta t_i-\eta\Delta t_i$. The speed itself varies by at most $\eta$ there, so its integral differs from that tagged value by at most $\eta\Delta t_i$. Summing yields polygonal length at least the speed integral minus $2\eta(b-a)$. Since arbitrarily small $\eta$ are possible, the supremum reaches the integral. This proves both directions rather than only the familiar chord upper bound.

Regularity is not needed for this formula: a $C^1$ curve may stop. Regularity becomes essential when we invert arc length to use it as a coordinate.

For a continuously differentiable graph $y=f(x)$, a short chord has length approximately

$$\sqrt{(\Delta x)^2+(f'(x)\Delta x)^2}=\sqrt{1+f'(x)^2}\,\Delta x.$$

Taking the limit of polygonal lengths yields

$$L=\int_a^b\sqrt{1+f'(x)^2}dx.$$

The square root is essential: $\int f'$ gives net vertical change, not curve length. For a parametrized curve $(x(t),y(t))$ traversed once with continuous derivative, the corresponding formula is $\int\sqrt{x'(t)^2+y'(t)^2}dt$. Repeated traversal counts length repeatedly.

Worked example 4: a nonstraight graph. For $f(x)=2x^{3/2}/3$ on $[0,1]$, $f'=\sqrt x$, so

$$L=\int_0^1\sqrt{1+x}dx=\frac23(2\sqrt2-1).$$

This exceeds the straight-line distance between endpoints, as expected.

If a nonnegative graph is rotated about the $x$ axis, small frusta have lateral area approaching circumference $2\pi f(x)$ times arc-length element. Thus

$$S=2\pi\int_a^bf(x)\sqrt{1+f'(x)^2}dx.$$

This is lateral surface area, not including end disks unless they are separately added. For $f(x)=x$, $0\le x\le1$, it gives $S=2\pi\sqrt2\int_0^1xdx=\pi\sqrt2$, the lateral area of the corresponding cone.

The frustum limit for a nonnegative C1 graph

Let $a<b$ and $f\in C^1([a,b])$ with $f\ge0$. For a partition $P$, rotate each chord of the graph about the $x$ axis and sum its frustum lateral area:

$$S(P)=\sum_i\pi(f(x_{i-1})+f(x_i))\sqrt{(x_i-x_{i-1})^2+(f(x_i)-f(x_{i-1}))^2}.$$

For this class of surfaces, define lateral area by the common limit of these rotational frustum sums as the mesh tends to zero. The following proof establishes that this limit exists and is independent of the chosen partitions. The mean value theorem supplies $\xi_i\in(x_{i-1},x_i)$ with $f(x_i)-f(x_{i-1})=f'(\xi_i)\Delta x_i$. If $M=\max|f'|$, the difference between $S(P)$ and the tagged sum for $2\pi f\sqrt{1+f'^2}$ at these tags is bounded by

$$2\pi\sqrt{1+M^2}(b-a)\omega_f(\|P\|)\longrightarrow0.$$

That integrand is continuous, so all such tagged sums have the same integral limit. Therefore $S=2\pi\int_a^bf(x)\sqrt{1+f'(x)^2}dx$. Zero radii are harmless in this construction. End disks are excluded. This establishes the formula for the stated rotational approximation; it does not assert that arbitrary triangulations of arbitrary surfaces have the same area limit.

4.5 Density, mass and centres of mass

For a rod on $[a,b]$ with $a<b$ and continuous nonnegative linear density $\rho$, assume total mass $M=\int_a^b\rho(x)dx>0$. A small piece has mass approximately $\rho(x)dx$, so

$$M=\int_a^b\rho(x)dx,\qquad \bar x=\frac{\int_a^bx\rho(x)dx}{M}\quad(M>0).$$

The numerator is the first moment about zero. Constant density cancels in a centroid calculation, but variable density cannot be discarded.

Worked example 5: a nonuniform rod. Let $0\le x\le2$ m and $\rho(x)=1+x$ kg/m, with the numerical coefficient understood in compatible units. Then $M=[x+x^2/2]_0^2=4$ kg. The first moment is $\int_0^2(x+x^2)dx=14/3$ kg·m, so $\bar x=7/6$ m. It lies to the right of the midpoint because the rod is denser there.

For a uniform lamina bounded by $0\le y\le f(x)$ on $[a,b]$, assume $a<b$, $f$ continuous and nonnegative, and area $A=\int_a^bf(x)dx>0$. The centroid formulas are

$$\bar x=\frac{\int xf(x)dx}{A},\qquad \bar y=\frac{\int f(x)^2/2\,dx}{A}.$$

The second numerator follows because a vertical strip's own midpoint has height $f(x)/2$ and its area is $f(x)dx$. For the triangular lamina under $f(x)=x$ on $[0,1]$, this gives $(\bar x,\bar y)=(2/3,1/3)$.

Continuous strip moments and centroids

Let $a<b$, and let $f:[a,b]\to[0,\infty)$ be continuous. For the lamina $0\le y\le f(x)$ with $f\ge0$, the same rectangles also prove the moment formulas. A rectangle over a short interval with height $h$ has vertical first moment $h^2\Delta x/2$ and horizontal first moment $h\int_I x\,dx$. Uniform convergence of the lower/upper step heights to $f$, together with boundedness of $x$ and $f$, makes these moments converge to $\int_a^b f(x)^2/2\,dx$ and $\int_a^bxf(x)dx$. Divide by the positive area to obtain the centroid; this argument does not divide by zero for a zero-area lamina.

4.6 Work and pressure require a force model

Work along a straight direction is $W=\int F(x)dx$ when $F(x)$ is the signed force component in that direction. With displacement in metres and force in newtons, the result is joules. A changing magnitude cannot generally be replaced by its endpoint value times the full distance.

Worked example 6: a spring. Under Hooke's-law model $F(x)=kx$ for extension $x\ge0$, the work to extend from $a$ to $b$, $0\le a<b$, is

$$W=\frac k2(b^2-a^2).$$

If $k=200$ N/m, $a=0.1$ m and $b=0.3$ m, the work is $100(0.09-0.01)=8$ J. This model assumes the spring remains within the range where force is proportional to extension.

For hydrostatic force in a stationary fluid of constant density $\rho$, gauge pressure at depth $h$ below the surface is $\rho gh$. A horizontal strip of a vertical plate with width $w(h)$ and thickness $dh$ experiences force $\rho gh\,w(h)dh$. Thus

$$F=\rho g\int h\,w(h)dh.$$

For a rectangular plate of width $3$ m extending from depth $1$ m to $3$ m, $F=3\rho g\int_1^3h\,dh=12\rho g$, with units determined by the stated SI inputs. The average depth works in this rectangular example because pressure is affine and width is constant; the integral formulation handles more general shapes.

4.7 Graded exercises

Foundation

1. For velocity $v(t)=3-2t$ on $[0,3]$, find displacement and total distance.
2. Find the area between $y=2x$ and $y=x^2$ on their bounded region.
3. A solid has cross-sectional area $A(x)=\pi(1+x)^2$ for $0\le x\le2$. Find its volume.

Core

4. Rotate the region under $y=\sqrt x$, $0\le x\le4$, about the $x$ axis. Find its volume.
5. Rotate the region under $y=1-x$, $0\le x\le1$, about the $y$ axis using shells. Check against the cone formula.
6. Find the arc length of the straight segment $y=3x+1$ from $x=0$ to $x=2$ and check with endpoint distance.
7. A rod on $[0,1]$ has density $\rho(x)=2x$. Find its mass and centre of mass.
8. Under $F(x)=100x$, find work extending a spring from zero to $0.2$ m and from $0.2$ to $0.4$ m. Explain why they differ despite equal extension distances.

Synthesis

9. Find the centroid of a uniform lamina under $y=x^2$ on $[0,1]$.
10. A vertical rectangular plate of width $w>0$ extends from depth $a\ge0$ to $a+H$, $H>0$, in a stationary constant-density fluid. Derive its total gauge force and show that it equals pressure at its midpoint depth times its area. State which assumptions make this simplification valid.

4.8 Full solutions

1. The signed integral is $[3t-t^2]_0^3=0$. Velocity changes sign at $3/2$. Each half contributes distance $9/4$, so total distance is $9/2$.
2. Intersections satisfy $2x=x^2$, so $x=0,2$. On $[0,2]$ the line is above the parabola. Area is $\int_0^2(2x-x^2)dx=[x^2-x^3/3]_0^2=4/3$.
3. $V=\pi[(1+x)^3/3]_0^2=26\pi/3$.
4. Disk radius is $\sqrt x$, so $V=\pi\int_0^4x\,dx=8\pi$. Squaring the radius, rather than integrating the radius itself, is crucial.
5. Shell radius is $x$ and height $1-x$, so $V=2\pi\int_0^1x(1-x)dx=\pi/3$. The solid is a cone of radius and height one, also giving $\pi/3$.
6. The length is $\int_0^2\sqrt{1+9}dx=2\sqrt{10}$. Endpoint differences are horizontal $2$ and vertical $6$, giving $\sqrt{4+36}=2\sqrt{10}$.
7. Mass is $\int_0^12xdx=1$. The first moment is $\int_0^12x^2dx=2/3$, so the centre of mass is $2/3$ from the left endpoint.
8. Work is $50(b^2-a^2)$. The two values are $2$ J and $6$ J. The force is larger during the second interval because extension is larger; work depends on force over the whole path, not distance alone.
9. Area is $A=1/3$. The horizontal first moment is $\int_0^1x^3dx=1/4$, giving $\bar x=3/4$. The vertical first moment is $\frac12\int_0^1x^4dx=1/10$, giving $\bar y=3/10$.
10. Integrate $F=\rho gw\int_a^{a+H}h\,dh=\rho gw(aH+H^2/2)=\rho g(a+H/2)(wH)$. The first factor is gauge pressure at midpoint depth and $wH$ is area. It works because the plate is vertical and rectangular, width is constant, density and gravitational acceleration are constant, and hydrostatic gauge pressure is linear in depth. It is not an unrestricted rule for arbitrary plate shapes or variable fluid density.

Summary and onward map

Construct integrals by identifying a small contribution and its measure. Velocity integrates to displacement; speed to distance. Area requires correct boundary order. Slices, washers and shells use different geometries, and radii are distances from the chosen axis. Arc length introduces a slope-dependent element. Density weights mass and moments; force models weight work or pressure strips. Units and independent geometric checks often expose errors before any difficult integration is attempted.