Learning goals and preparation

We assume substitution, integration by parts, continuous definite integrals, finite sums, and finite-average ideas. The Monte Carlo section introduces the expectation/variance notation it uses. We develop integration methods: trigonometric/rational/irrational examples, reduction formulas, integral limits and inequalities, Simpson quadrature, and Monte Carlo estimation. The last method is random and does not provide the same kind of deterministic guarantee as a derivative-based quadrature error bound.

8.1 Choosing an integration structure

Substitution reverses the chain rule and parts reverses the product rule. Other techniques reduce an integrand to these familiar structures. Always retain the original domain and verify an indefinite answer by differentiation.

Worked example 1: rational partial fractions. For

$$\frac{3x+4}{(x+1)(x+2)}=\frac A{x+1}+\frac B{x+2},$$

matching coefficients gives $A+B=3$, $2A+B=4$, hence $A=1,B=2$. An antiderivative is $\ln|x+1|+2\ln|x+2|+C$ on each interval avoiding the poles. A singularity inside definite bounds must be treated as an improper integral; an antiderivative cannot be evaluated across it as if the integrand were continuous.

Before partial fractions, divide polynomials if the numerator degree is not smaller than the denominator degree. For example $(x^2+1)/(x-1)=x+1+2/(x-1)$, giving antiderivative $x^2/2+x+2\ln|x-1|+C$ on intervals excluding one. A repeated linear factor needs every denominator power: $1/[x(x+1)^2]=1/x-1/(x+1)-1/(x+1)^2$, with antiderivative $\ln|x|-\ln|x+1|+1/(x+1)+C$. For an irreducible quadratic, split off the denominator derivative: $(2x+3)/(x^2+4)$ integrates to $\ln(x^2+4)+(3/2)\arctan(x/2)+C$. Here $d(\arctan u)/du=1/(1+u^2)$ follows by differentiating $\tan y=u$ on its principal branch.

For an irrational expression, substitute an inner radical or a trigonometric identity. For $\int x/\sqrt{1+x^2}dx$, $u=1+x^2$ gives $\sqrt{1+x^2}+C$. For $\int_0^1\sqrt{1-x^2}dx$, $x=\sin\theta$, $0\le\theta\le\pi/2$, gives $\int_0^{\pi/2}\cos^2\theta d\theta=\pi/4$. The branch makes $\sqrt{\cos^2\theta}=\cos\theta$; without it an absolute value is required.

A direct root substitution can avoid trigonometry entirely. For $x>0$, set $u=\sqrt x$ in $\int dx/[\sqrt x(1+\sqrt x)]$. Since $dx=2u\,du$, the integral is $2\ln(1+u)+C=2\ln(1+\sqrt x)+C$. The original denominator still excludes zero; simplifying after substitution does not add that input back.

For odd trigonometric powers, retain a factor for the differential. For example $\int\sin^3x\cos xdx=\sin^4x/4+C$. For even powers, use $\sin^2x=(1-\cos2x)/2$. A named “trig integral” is not one single recipe; identify its algebraic structure.

8.2 A reduction formula derived by parts

Let $I_n=\int_0^{\pi/2}\sin^n xdx$ for integers $n\ge0$. For $n\ge2$, integrate $\sin^{n-1}x\sin x$ by parts with $u=\sin^{n-1}x$, $dv=\sin xdx$. Endpoint terms vanish, giving

$$I_n=(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2xdx=(n-1)(I_{n-2}-I_n).$$

Hence

$$I_n=\frac{n-1}{n}I_{n-2},\qquad I_0=\pi/2,\quad I_1=1.$$

The formula reduces the exponent by two and needs separate even and odd starting cases. For example $I_4=(3/4)(1/2)(\pi/2)=3\pi/16$ and $I_5=(4/5)(2/3)=8/15$.

For an indefinite example, let $J_n=\int x^ne^xdx$. Integration by parts gives $J_n=x^ne^x-nJ_{n-1}$ for integers $n\ge1$, with $J_0=e^x+C$. Therefore $J_2=e^x(x^2-2x+2)+C$. At each recurrence the arbitrary constants are absorbed into the one final constant; a definite use instead keeps endpoint terms. Differentiating the final expression checks the reduction independently.

An indefinite reduction formula must retain any boundary term and arbitrary constant. Using a definite formula outside its stated limits or forgetting its base cases is not justified by the word “reduction.”

8.3 Riemann sums as limits to evaluate

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

As a special case, if $f$ is continuous on $[a,b]$, then with $h=(b-a)/n$,

$$h\sum_{k=1}^nf(a+kh)\longrightarrow\int_a^bf(x)dx.$$

This can evaluate a sequence limit without finding a closed finite sum. Identify both the sample point and the width; a sum with the wrong normalization need not have the same limit.

Worked example 2: recognize the width. Since

$$\sum_{k=1}^n\frac{n}{n^2+k^2}=\frac1n\sum_{k=1}^n\frac1{1+(k/n)^2},$$

the limit is $\int_0^1dx/(1+x^2)=\pi/4$. Omitting the prefactor $1/n$ in the rewritten expression would change the scaling completely.

A function's average over $[a,b]$ is its integral divided by $b-a$. A Riemann sum can approximate a total or an average; its normalization determines which.

8.4 Integral inequalities

If continuous functions satisfy $f\le g$ pointwise, their integrals obey the same inequality, by comparing each tagged sum. In particular if $m\le f\le M$, then $m(b-a)\le\int_a^bf\le M(b-a)$. Also

$$\left|\int_a^bf\right|\le\int_a^b|f|.$$

For real continuous $f,g$, the integral Cauchy–Schwarz inequality is

$$\left(\int_a^bfg\right)^2\le\left(\int_a^bf^2\right)\left(\int_a^bg^2\right).$$

If $\int g^2>0$, integrate $(f-tg)^2\ge0$ and minimize the resulting quadratic in $t$; its minimum cannot be negative, yielding the inequality. If $\int g^2=0$, continuity forces $g=0$ throughout, making the result immediate. This proof uses positive integrals rather than an unsupported transfer of a finite-vector formula.

Worked example 3: a simple integral bound. On $[0,1]$, $1\le\sqrt{1+x}\le\sqrt2$, so its integral lies between one and $\sqrt2$. Concavity additionally puts the graph below the tangent $1+x/2$, giving upper bound $5/4$, sharper than $\sqrt2$. Each bound follows from an explicit pointwise inequality.

8.5 Simpson's rule and its conditions

Let $a<b$ and let $n$ be a positive even integer. For $n$ equal subintervals of width $h=(b-a)/n$, Simpson's rule is

$$S_n=\frac h3\left[f(x_0)+f(x_n)+4\sum_{i\text{ odd}}f(x_i)+2\sum_{\substack{i\text{ even}\\0<i<n}}f(x_i)\right].$$

It integrates the quadratic interpolant on each pair of subintervals. On a pair centred at zero with nodes $-h,0,h$, the three Lagrange basis polynomials have integrals $h/3,4h/3,h/3$, producing the familiar weights. Symmetry also makes the cubic term integrate exactly, so Simpson's rule is exact for every polynomial of degree at most three.

For a function with continuous fourth derivative bounded by $M_4$, the standard composite error theorem gives

$$\left|\int_a^bf-S_n\right|\le\frac{(b-a)h^4M_4}{180}.$$

The fourth derivative, smoothness and even subdivision requirement are part of the theorem. Here is a derivation of the constant, rather than an inference from observed mesh differences. On a unit panel $[-1,1]$, define the error functional $E(f)=\int_{-1}^1f(t)dt-[f(-1)+4f(0)+f(1)]/3$. It annihilates polynomials through degree three. Insert the integral Taylor remainder about $-1$ and exchange integrals over the finite triangle (the integrands are continuous). The resulting expression is

$$E(f)=\int_{-1}^1K(s)f^{(4)}(s)ds,\qquad K(s)=-\frac{(1-|s|)^3(1+3|s|)}{72}.$$

For instance, for $0\le s\le1$, the kernel is $(1-s)^4/24-(1-s)^3/18$, equal to the displayed form; symmetry gives the negative half. Its absolute integral is $[2/72]\int_0^1(1-s)^3(1+3s)ds=1/90$. Thus a panel of half-width $h$ has error at most $M_4h^5/90$ after scaling. There are $(b-a)/(2h)$ panels, giving $(b-a)h^4M_4/180$ as claimed.

Worked example 4: compare a known integral. For $f=x^4$ on $[0,1]$ with $n=2$, $S_2=(1/6)(0+4/16+1)=5/24$. Exact integral is $1/5$, so error is $1/120$. Since $M_4=24$ and $h=1/2$, the theorem's bound is $(1)(1/16)(24)/180=1/120$, attained here. For a cubic, the method would be exact under exact arithmetic.

Chapter 2's trapezoidal rule has an $h^2$ derivative bound, while Simpson's has an $h^4$ bound with stronger smoothness. Higher formal order does not remove domain singularities or arithmetic error.

8.6 Monte Carlo integration and random error

Here $E[X]$ denotes the probability-weighted average of a random quantity, and $\operatorname{Var}(X)=E[(X-E[X])^2]$ its expected squared deviation. For a uniform point $U$ on $[a,b]$, the definition is $E[f(U)]=(b-a)^{-1}\int_a^bf$. Independent samples have product joint weights/densities; consequently averages of products from different samples factor into products of averages. Expanding the square then shows the variance of their sum is the sum of variances.

Let $a<b$, let $N$ be a positive integer, and let $U_1,\ldots,U_N$ be independent uniformly distributed points on $[a,b]$. For integrable $f$ with finite second moment, define

$$\widehat I_N=\frac{b-a}{N}\sum_{i=1}^Nf(U_i).$$

Uniformity gives $E[f(U_i)]=(b-a)^{-1}\int_a^bf$, so the estimator is unbiased: $E[\widehat I_N]=I$. Independence gives

$$\operatorname{Var}(\widehat I_N)=\frac{(b-a)^2\operatorname{Var}(f(U))}{N}.$$

Its root-mean-square error is therefore proportional to $N^{-1/2}$, not $N^{-1}$ or $N^{-2}$. This is a statistical error statement; a particular random sample can err in either direction.

For $\varepsilon>0$, Chebyshev's inequality yields the probabilistic bound

$$P(|\widehat I_N-I|\ge\varepsilon)\le\frac{(b-a)^2\operatorname{Var}(f(U))}{N\varepsilon^2}.$$

For completeness, the bound follows from $E[(\widehat I_N-I)^2]\ge\varepsilon^2P(|\widehat I_N-I|\ge\varepsilon)$, because the squared error is at least $\varepsilon^2$ on that event and nonnegative elsewhere. Unbiasedness identifies the expected squared error with the variance. The probability refers to repeated independent random sampling under the model. It is not an absolute guarantee that one computed estimate is within the tolerance. A deterministic grid is not an independent uniform sample, and a pseudorandom generator's output does not prove independence merely because it looks irregular.

Worked example 5: exact variance of a toy estimator. For $f(u)=u^2$ on $[0,1]$, $I=1/3$, $E[f(U)^2]=1/5$, so variance of $f(U)$ is $4/45$. Thus estimator variance is $4/(45N)$. At $N=100$, Chebyshev bounds probability of error at least $0.1$ by $4/45$, about $0.089$. The bound is conservative but mathematically explicit. No fixed sample output is claimed here.

8.7 Choosing an approximation responsibly

First inspect continuity, singularities and scale. Use an exact antiderivative or reduction when practical. For smooth one-dimensional functions, deterministic quadrature with a derivative bound often gives a transparent certificate. Monte Carlo is useful when random sampling is part of the problem or geometry/dimension makes grids inconvenient, but requires a statistical uncertainty statement.

A useful report names the method, assumptions, step/subdivision or sample count, arithmetic precision, and the kind of error bound actually established. “More samples” and “more decimal places” are different controls.

8.8 Graded exercises

Foundation

1. Find $\int 2x/\sqrt{1+x^2}dx$ and verify it by differentiation.
2. Decompose and integrate $1/[x(x+1)]$ on an interval avoiding its poles.
3. Use the reduction formula to calculate $I_6=\int_0^{\pi/2}\sin^6xdx$.

Core

4. Evaluate $\lim_n(1/n)\sum_{k=1}^n(k/n)^2$ as an integral.
5. Bound $\int_0^1e^{-x^2}dx$ between two constants using $e^{-1}\le e^{-x^2}\le1$.
6. Use integral Cauchy–Schwarz to prove $(\int_0^1f)^2\le\int_0^1f^2$ for continuous real $f$.
7. Apply Simpson's rule with $n=2$ to $f(x)=x^3$ on $[0,1]$. Explain exactness.
8. For $f=e^x$ on $[0,1]$ with $M_4\le3$, what Simpson error bound is obtained with $n=4$?

Synthesis

9. For uniform $U$ on $[0,1]$ and $f(U)=U$, derive the Monte Carlo estimator's mean and variance. Give a Chebyshev error probability bound for tolerance $\varepsilon$ and sample size $N$.
10. A student reports “the Monte Carlo estimate is unbiased, so its error is zero.” Explain the difference between expectation and the realized estimate, using the variance formula and a possible one-point sample.

Additional method checks

11. Integrate $(x^2+2)/(x-1)$ by polynomial division.
12. Integrate $dx/[\sqrt x(2+\sqrt x)]$ on its real domain.
13. Use the $J_n$ recurrence to find $\int x^3e^xdx$ and verify its derivative.

8.9 Full solutions

1. Substitution gives $2\sqrt{1+x^2}+C$. Its derivative is $2x/\sqrt{1+x^2}$.

2. $1/[x(x+1)]=1/x-1/(x+1)$. An antiderivative is $\ln|x|-\ln|x+1|+C$, separately on intervals avoiding zero and negative one.

3. $I_6=(5/6)I_4=(5/6)(3\pi/16)=5\pi/32$.

4. The right-endpoint sums of the continuous function $x^2$ on $[0,1]$ converge to $\int_0^1x^2dx=1/3$.

5. Monotonicity of integration gives $e^{-1}\le\int_0^1e^{-x^2}dx\le1$.

6. Apply the inequality with $g=1$; its squared integral factor is $\int_0^11^2dx=1$, giving the result directly.

7. $S_2=(1/6)(0+4(1/8)+1)=1/4$, matching the exact integral. Simpson integrates every cubic exactly; equivalently its fourth derivative is zero.

8. Here $h=1/4$, so the bound is $3/(180\cdot4^4)=1/15360$.

9. $E[U]=1/2$, $E[U^2]=1/3$, so variance is $1/12$. The estimator is the sample mean, with expectation $1/2$ and variance $1/(12N)$. Chebyshev gives $P(|\widehat I_N-1/2|\ge\varepsilon)\le1/(12N\varepsilon^2)$, capped at one if the displayed upper bound exceeds one.

10. Unbiasedness says the average over repeated hypothetical samples equals the true integral; individual samples fluctuate. For $f(u)=u$ and $N=1$, a sample near zero gives an estimate near zero instead of the true $1/2$. Positive variance $1/12$ quantifies that variability. Expected signed error zero does not mean absolute or squared error zero.

11. Division gives $x+1+3/(x-1)$, so the antiderivative is $x^2/2+x+3\ln|x-1|+C$, separately on intervals avoiding one.

12. The real domain is $x>0$. Set $u=\sqrt x$, $dx=2u\,du$, giving $2\int du/(2+u)=2\ln(2+\sqrt x)+C$. The logarithm argument is positive on the domain.

13. $J_3=x^3e^x-3J_2=e^x(x^3-3x^2+6x-6)+C$. The derivative adds the polynomial and its derivative, cancelling lower terms to leave $x^3e^x$.

Summary

Reduction formulas come from actual integration identities and need base cases. Riemann sums can evaluate limits once the width and sampling points are identified. Integral inequalities follow pointwise bounds or positivity arguments. Simpson's rule uses paired quadratic interpolation and a qualified fourth-derivative error theorem. Monte Carlo is unbiased under independent uniform sampling, with random uncertainty rather than a deterministic tolerance guarantee.