Learning goals and preparation

We assume Chapter 1's derivative rules, exponential/logarithmic functions, and elementary continuity. This chapter derives accumulation from sums, connects it with antiderivatives, and applies substitution, parts, and numerical quadrature. Business examples use hypothetical mathematical models.

2.1 From a marginal quantity to a total

If $c(q)$ is marginal cost in dollars per unit, then $c(q)\Delta q$ approximates the cost of a small additional quantity.

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

The integral is signed. A negative net cash-flow rate contributes negatively to the total; a probability density, by contrast, must be nonnegative. The meaning follows from the model and units, not the integral symbol alone.

If $C'=c$ and $c$ is continuous, the fundamental theorem gives

$$C(b)-C(a)=\int_a^bc(q)dq.$$

An initial or fixed cost is needed to recover the absolute cost function. Integrating a derivative does not reveal its missing constant.

Worked example 1: reconstruct total cost. Let marginal cost be $c(q)=4+0.6q$, $q\ge0$, and fixed cost $C(0)=80$. Then

$$C(q)=80+\int_0^q(4+0.6s)ds=80+4q+0.3q^2.$$

The cost increase from $10$ to $15$ is $\int_{10}^{15}(4+0.6q)dq=57.5$. This is not exactly $c(10)\cdot5=50$, which uses one initial marginal rate for the whole interval.

2.2 Fundamental theorem and variable upper bounds

For differentiable bounds $u(x),v(x)$ and continuous $f$, the chain rule gives

$$\frac d{dx}\int_{u(x)}^{v(x)}f(t)dt=f(v(x))v'(x)-f(u(x))u'(x).$$

The dummy variable $t$ distinguishes the integration variable from the input controlling the bounds.

Worked example 2: a moving accumulation window. Let $B(t)=\int_{t-1}^{t}e^{-s}ds$. Then $B'(t)=e^{-t}-e^{-(t-1)}$. There is both an entering boundary and a leaving boundary; differentiating only the upper endpoint would miss the second term.

2.3 Substitution reverses the chain rule

When $F'=f$,

$$\int f(g(x))g'(x)dx=F(g(x))+C.$$

For a definite integral transform the bounds too. The formula is justified by differentiating $F(g(x))$, so it retains signs and inner derivative factors.

Worked example 3: a rational marginal law. Suppose a rate is $r(t)=6t/(1+t^2)$ for $0\le t\le2$. Set $u=1+t^2$, so $du=2t\,dt$ and the bounds are $1,5$. The total is $3\int_1^5du/u=3\ln5$. The rate's units and the time variable determine the accumulated output unit.

For an indefinite integral $\int2x e^{x^2}dx=e^{x^2}+C$. The expression $e^{x^2}/(2x)$ would be an invalid guess: differentiating it introduces an extra term. A derivative check is a decisive way to test proposed antiderivatives.

2.4 Parts and discounted accumulations

Integrating the product rule yields $\int u\,dv=uv-\int v\,du$. For $k>0$,

$$\int_0^Tte^{-kt}dt=\frac{1-e^{-kT}(1+kT)}{k^2},$$

using $u=t$, $dv=e^{-kt}dt$. It follows that a hypothetical flow $c(t)=a+bt$ with a chosen continuous discount weight $e^{-kt}$ has weighted accumulation

$$\int_0^T(a+bt)e^{-kt}dt =\frac a k(1-e^{-kT})+\frac b{k^2}(1-e^{-kT}(1+kT)).$$

The exponential weight must be dimensionless, so $k$ has inverse-time units.

Worked example 4: a simple exact total. For $c(t)=2t$, weight $e^{-t}$, and horizon $T=1$, the weighted amount is $2(1-2/e)$. Replacing a whole stream with its final flow times the horizon would produce a different calculation and ignore both timing and changing rates.

2.5 Areas and surplus models

If continuous demand price $p_D(q)$ lies above a transaction price $p_*$ on $0\le q\le Q$, a geometric consumer-surplus model is

$$S=\int_0^Q(p_D(q)-p_*)dq.$$

This is an area under a hypothetical willingness-to-pay curve; its interpretation requires the model's assumptions.

Worked example 5: a triangular surplus. Let $p_D(q)=20-q$ and let $p_*=12$, $Q=8$. Then $S=\int_0^8(8-q)dq=32$, agreeing with the triangle of base eight and height eight. If two curves cross within an integration interval, a signed difference can cancel; an unsigned area requires splitting where their order changes.

The average value of a continuous function over $[a,b]$, $a<b$, is $(b-a)^{-1}\int_a^bf$. Its units match the function, since integration multiplies by the input unit and division removes that interval length. An average cost per unit is not automatically the average value of a marginal-cost function unless the precise cost difference and interval are being compared.

2.6 Numerical quadrature and what an error estimate needs

For equally spaced points $x_i=a+ih$, $h=(b-a)/n$, the composite trapezoidal rule is

$$T_n=h\left(\frac{f(x_0)+f(x_n)}2+\sum_{i=1}^{n-1}f(x_i)\right).$$

It integrates the piecewise straight-line interpolant. For a twice continuously differentiable function with $|f''|\le M$, a standard interpolation-error integration gives

$$\left|\int_a^bf-T_n\right|\le\frac{(b-a)h^2M}{12}.$$

For one subinterval, the interpolation error has the form $f''(\xi)(x-x_i)(x-x_{i+1})/2$ under the usual smoothness hypothesis. Integrating its magnitude gives $Mh^3/12$; summing $n$ pieces gives the displayed bound. The unknown $\xi$ is not fixed across the interval, but the uniform bound is valid.

Why the interpolation error has this form

Let $a<b$. Suppose $f$ is continuous on $[a,b]$, twice differentiable on $(a,b)$, and $|f''(t)|\le M$ there, with $M\ge0$. Continuity of the second derivative is sufficient but is not required for the following argument. The straight-line interpolant through $(a,f(a))$ and $(b,f(b))$ is $\ell(t)=[(b-t)f(a)+(t-a)f(b)]/(b-a)$.

For a fixed $x\in(a,b)$, define

$$\lambda=\frac{f(x)-\ell(x)}{(x-a)(x-b)},\qquad g(t)=f(t)-\ell(t)-\lambda(t-a)(t-b).$$

The function $g$ vanishes at $a,x,b$. Rolle's theorem on $[a,x]$ and $[x,b]$ gives $u\in(a,x)$ and $v\in(x,b)$ with $g'(u)=g'(v)=0$. A second application to $g'$ on $[u,v]$ gives $\xi\in(u,v)$ with $g''(\xi)=0$. The needed continuity of $g'$ follows from its differentiability inside $(a,b)$. Therefore $2\lambda=f''(\xi)$ and

$$f(x)-\ell(x)=\frac{f''(\xi)}2(x-a)(x-b),\qquad |f(x)-\ell(x)|\le\frac M2(x-a)(b-x).$$

At either endpoint the error is zero directly. The point $\xi$ can depend on $x$; we use the uniform bound $M$, rather than integrate an unspecified moving $\xi$.

Integrating the error and adding the panels

Put $h=b-a$. Direct integration of $\ell$ gives $\int_a^b\ell=h[f(a)+f(b)]/2$. Since $f-\ell$ is continuous, the integral triangle inequality and the preceding bound give

$$\left|\int_a^bf(t)\,dt-\frac h2[f(a)+f(b)]\right|\le\frac M2\int_a^b(t-a)(b-t)\,dt=\frac{Mh^3}{12}.$$

For a positive integer $n$ and $A<B$, put $h=(B-A)/n$, $x_i=A+ih$, and $T_n=h[f(A)/2+\sum_{i=1}^{n-1}f(x_i)+f(B)/2]$. Assume the same bound $|f''|\le M$ throughout the open interval and continuity on its closure. Apply this argument to each panel, then add the absolute error bounds:

$$\left|\int_A^Bf-T_n\right|\le\sum_{i=1}^n\frac{Mh^3}{12}=\frac{(B-A)h^2M}{12}.$$

The two endpoint half-weights and the interior full weights arise by adding the panel interpolant integrals. The proof uses the Rolle theorem already established in this note. It bounds exact quadrature error; rounding and uncertainty in sampled data require their own analysis.

Worked example 6: a checked approximation. For $f(x)=x^2$ on $[0,1]$, using $n=2$ gives $T_2=(1/2)(0/2+1/4+1/2)=3/8$. The exact integral is $1/3$, with error $1/24$. Since $M=2$ and $h=1/2$, the bound is $1\cdot(1/4)\cdot2/12=1/24$, attained in this case. A mesh comparison alone would be weaker evidence than this explicit bound.

2.7 Improper totals

An infinite horizon requires a limit:

$$\int_0^\infty f(t)dt=\lim_{T\to\infty}\int_0^Tf(t)dt.$$

For $k>0$, $\int_0^\infty e^{-kt}dt=1/k$ and $\int_0^\infty te^{-kt}dt=1/k^2$, because exponential decay makes the boundary terms vanish. A flow tending to zero does not by itself guarantee finite total: $1/(1+t)$ tends to zero, but its integral grows as $\ln(1+T)$ without bound.

2.8 Graded exercises

Foundation

1. Marginal cost is $C'(q)=3+q$ and $C(0)=50$. Find $C(q)$ and the cost increase from $q=2$ to $q=5$.
2. Evaluate $\int_0^1(2x+3)dx$ and its average value on that interval.
3. Differentiate $A(x)=\int_1^{x^2}\ln t\,dt$ for $x>0$.

Core

4. Evaluate $\int_0^1 4x e^{2x^2}dx$ by substitution.
5. Evaluate $\int_0^2te^{-t}dt$ by parts.
6. For demand $p_D(q)=30-2q$, price $10$ and quantity $10$, compute the surplus-area model.
7. Use the trapezoidal rule with $n=2$ to approximate $\int_0^1x^3dx$. Bound the error using $M=6$.
8. Differentiate $B(t)=\int_t^{2t}(1+s^2)ds$ by the variable-bound formula and by explicit integration.

Synthesis

9. Determine the convergence and value of $\int_0^\infty(2+3t)e^{-2t}dt$.
10. A student claims $\int_a^bf(x)dx=(b-a)f(b)$ for every continuous $f$. Give a counterexample, identify what approximation it represents, and state a condition under which it is exact.

2.9 Full solutions

1. $C(q)=50+3q+q^2/2$. The increase is $3(5-2)+(25-4)/2=39/2$.
2. The integral is $[x^2+3x]_0^1=4$. Since the interval length is one, its average value is also $4$.
3. The chain rule and FTC give $A'(x)=2x\ln(x^2)$ for $x>0$. Positivity ensures the integral's upper bound stays in the logarithm's domain.
4. Set $u=2x^2$, $du=4xdx$; bounds are zero and two. The answer is $\int_0^2e^udu=e^2-1$.
5. An antiderivative is $-(t+1)e^{-t}$. Thus the result is $1-3e^{-2}$.
6. Surplus is $\int_0^{10}(20-2q)dq=200-100=100$.
7. $T_2=(1/2)(0+1/8+1/2)=5/16$. The exact value is $1/4$, so error is $1/16$. The theorem gives upper bound $(1)(1/4)(6)/12=1/8$, a valid but not sharp bound.
8. The boundary formula gives $2(1+4t^2)-(1+t^2)=1+7t^2$. Explicitly, $B=[s+s^3/3]_t^{2t}=t+7t^3/3$, with the same derivative.
9. Split the integral. The constant part gives $2(1/2)=1$, and the time-weighted part gives $3(1/2^2)=3/4$, for total $7/4$. Both improper limits converge because the relevant exponential boundary terms vanish.
10. Take $f(x)=x$ on $[0,1]$. The integral is $1/2$, while the claimed formula gives $1$. It is a single right-endpoint rectangle approximation, not a general integration identity. It is exact if $f$ is constant on the interval, and can be exact for some other special functions/intervals by cancellation, but continuity alone does not guarantee it.

Summary and onward map

An integral reconstructs total change from a marginal quantity and requires an initial constant for an absolute total. Variable bounds contribute through both endpoints. Substitution and parts follow differentiation rules. Weighted streams and surplus are model-dependent accumulations. Trapezoidal quadrature integrates an interpolant and needs derivative bounds for a certified error. Infinite horizons remain limit problems.