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Home/Calculus/Chapter 6/The Riemann Integral
CALC-6.1
4-5 hours

The Riemann Integral

Build the definite integral rigorously from Riemann sums and Darboux sums, establishing precise conditions for when a function is integrable.

Learning Objectives
Understand partitions and their refinements
Define Riemann sums and their properties
Master Darboux upper and lower sums
Prove the definition of the Riemann integral
Establish necessary and sufficient conditions for integrability
Identify classes of integrable functions
Apply integrability criteria to specific functions
Understand the Du Bois-Reymond criterion

1. The Area Problem

Motivating Question

How do we compute the area under the curve y=x2y = x^2 from x=0x = 0 to x=1x = 1?

The ancient Greeks approximated areas using inscribed/circumscribed shapes. We formalize this by dividing [a,b][a, b] into subintervals and summing rectangle areas.

2. Partitions and Riemann Sums

Definition 6.1: Partition

Let f(x)f(x) be defined on [a,b][a, b]. A partition of [a,b][a, b] is a finite set of points:

Δ:a=x0<x1<x2<<xn1<xn=b\Delta: a = x_0 < x_1 < x_2 < \cdots < x_{n-1} < x_n = b

This divides [a,b][a, b] into nn subintervals [xk1,xk][x_{k-1}, x_k] for k=1,2,,nk = 1, 2, \ldots, n.

Definition 6.2: Norm of Partition

The norm (or mesh) of partition Δ\Delta is:

Δ=max{Δx1,Δx2,,Δxn}\|\Delta\| = \max\{\Delta x_1, \Delta x_2, \ldots, \Delta x_n\}

where Δxk=xkxk1\Delta x_k = x_k - x_{k-1} is the length of the kk-th subinterval.

Definition 6.3: Riemann Sum

Given partition Δ\Delta and sample points ξk[xk1,xk]\xi_k \in [x_{k-1}, x_k], the Riemann sum is:

SΔ(f,ξ)=k=1nf(ξk)Δxk=k=1nf(ξk)(xkxk1)S_\Delta(f, \xi) = \sum_{k=1}^{n} f(\xi_k) \Delta x_k = \sum_{k=1}^{n} f(\xi_k)(x_k - x_{k-1})

This represents the sum of rectangle areas with heights f(ξk)f(\xi_k) and widths Δxk\Delta x_k.

Example 6.1: Computing a Riemann Sum

Problem: Compute the Riemann sum for f(x)=x2f(x) = x^2 on [0,1][0, 1] with uniform partition n=4n = 4 and right endpoints.

Solution:

Partition: xk=k/4x_k = k/4 for k=0,1,2,3,4k = 0, 1, 2, 3, 4. So Δxk=1/4\Delta x_k = 1/4.

Right endpoints: ξk=k/4\xi_k = k/4.

S4=k=14(k4)214=164(1+4+9+16)=3064=1532S_4 = \sum_{k=1}^{4} \left(\frac{k}{4}\right)^2 \cdot \frac{1}{4} = \frac{1}{64}(1 + 4 + 9 + 16) = \frac{30}{64} = \frac{15}{32}

Note: The exact value is 01x2dx=1/30.333\int_0^1 x^2\,dx = 1/3 \approx 0.333, while 15/320.46915/32 \approx 0.469.

3. Darboux Upper and Lower Sums

Definition 6.4: Darboux Sums

For partition Δ\Delta of [a,b][a, b] and bounded function ff, define:

Mk=supx[xk1,xk]f(x),mk=infx[xk1,xk]f(x)M_k = \sup_{x \in [x_{k-1}, x_k]} f(x), \quad m_k = \inf_{x \in [x_{k-1}, x_k]} f(x)

The upper Darboux sum and lower Darboux sum are:

SΔ=k=1nMkΔxk\overline{S}_\Delta = \sum_{k=1}^{n} M_k \Delta x_k

Upper sum

SΔ=k=1nmkΔxk\underline{S}_\Delta = \sum_{k=1}^{n} m_k \Delta x_k

Lower sum

Theorem 6.1: Bounds on Riemann Sums

For any partition Δ\Delta and sample points {ξ}\{\xi\}:

SΔSΔ(f,ξ)SΔ\underline{S}_\Delta \leq S_\Delta(f, \xi) \leq \overline{S}_\Delta
Proof of Theorem 6.1:

Since mkf(ξk)Mkm_k \leq f(\xi_k) \leq M_k for each kk, multiplying by Δxk>0\Delta x_k > 0 and summing:

k=1nmkΔxkk=1nf(ξk)Δxkk=1nMkΔxk\sum_{k=1}^n m_k \Delta x_k \leq \sum_{k=1}^n f(\xi_k) \Delta x_k \leq \sum_{k=1}^n M_k \Delta x_k

4. Refinement of Partitions

Definition 6.5: Refinement

Partition Δ2\Delta_2 is a refinement of Δ1\Delta_1 (written Δ1Δ2\Delta_1 \subset \Delta_2) if every partition point of Δ1\Delta_1 is also a partition point of Δ2\Delta_2.

The common refinement of Δ1\Delta_1 and Δ2\Delta_2 is Δ=Δ1Δ2\Delta^* = \Delta_1 \cup \Delta_2.

Theorem 6.2: Refinement Lemma

If Δ2Δ1\Delta_2 \supset \Delta_1 and Δ2\Delta_2 has kk more points than Δ1\Delta_1, then:

0SΔ1SΔ2kΔ1(Mm)0 \leq \overline{S}_{\Delta_1} - \overline{S}_{\Delta_2} \leq k\|\Delta_1\|(M - m)
0SΔ2SΔ1kΔ1(Mm)0 \leq \underline{S}_{\Delta_2} - \underline{S}_{\Delta_1} \leq k\|\Delta_1\|(M - m)

where M=supfM = \sup f, m=inffm = \inf f on [a,b][a, b].

Corollary 6.1: Upper and Lower Sum Comparison

For any two partitions Δ1,Δ2\Delta_1, \Delta_2 of [a,b][a, b]:

SΔ1SΔ2\underline{S}_{\Delta_1} \leq \overline{S}_{\Delta_2}

Every lower sum is ≤ every upper sum (even for different partitions).

5. Upper and Lower Integrals

Definition 6.6: Darboux Integrals

For bounded ff on [a,b][a, b], define the upper integral and lower integral:

abf(x)dx=infΔSΔ\overline{\int_a^b} f(x)\,dx = \inf_\Delta \overline{S}_\Delta

Upper (Darboux) integral

abf(x)dx=supΔSΔ\underline{\int_a^b} f(x)\,dx = \sup_\Delta \underline{S}_\Delta

Lower (Darboux) integral

Theorem 6.3: Darboux's Theorem

For bounded ff on [a,b][a, b]: ϵ>0\forall \epsilon > 0, δ>0\exists \delta > 0 such that for any partition Δ\Delta with Δ<δ\|\Delta\| < \delta:

abfϵ<SΔ<abf+ϵ\overline{\int_a^b} f - \epsilon < \overline{S}_\Delta < \overline{\int_a^b} f + \epsilon
abfϵ<SΔ<abf+ϵ\underline{\int_a^b} f - \epsilon < \underline{S}_\Delta < \underline{\int_a^b} f + \epsilon
Remark 6.1

The upper and lower integrals always exist for bounded functions, and we always have:

abfabf\underline{\int_a^b} f \leq \overline{\int_a^b} f

6. Definition of the Riemann Integral

Definition 6.7: Riemann Integrability

Let ff be bounded on [a,b][a, b]. If there exists ARA \in \mathbb{R} such that:

ϵ>0,δ>0:Δ<δSΔ(f,ξ)A<ϵ\forall \epsilon > 0, \exists \delta > 0: \|\Delta\| < \delta \Rightarrow |S_\Delta(f, \xi) - A| < \epsilon

for any partition Δ\Delta and any sample points {ξ}\{\xi\}, then ff is Riemann integrable on [a,b][a, b].

We write fR[a,b]f \in R[a, b] and define:

abf(x)dx=A=limΔ0k=1nf(ξk)Δxk\int_a^b f(x)\,dx = A = \lim_{\|\Delta\| \to 0} \sum_{k=1}^n f(\xi_k) \Delta x_k
Theorem 6.4: Uniqueness

If fR[a,b]f \in R[a, b], then the integral value AA is unique.

Theorem 6.5: Necessary Condition

If fR[a,b]f \in R[a, b], then ff is bounded on [a,b][a, b].

Proof of Theorem 6.5:

Suppose ff is unbounded on [a,b][a, b]. For any partition Δ\Delta, ff is unbounded on at least one subinterval [xk1,xk][x_{k-1}, x_k]. By choosing sample points appropriately, we can make SΔ(f,ξ)|S_\Delta(f, \xi)| arbitrarily large, contradicting convergence to a finite AA.

7. Integrability Criteria

Theorem 6.6: Darboux Criterion

For bounded ff on [a,b][a, b]:

fR[a,b]abf=abff \in R[a, b] \Leftrightarrow \overline{\int_a^b} f = \underline{\int_a^b} f
Theorem 6.7: Oscillation Criterion

Define the oscillation on [xk1,xk][x_{k-1}, x_k] as ωk=Mkmk\omega_k = M_k - m_k. Then:

fR[a,b]ϵ>0,δ>0:Δ<δk=1nωkΔxk<ϵf \in R[a, b] \Leftrightarrow \forall \epsilon > 0, \exists \delta > 0: \|\Delta\| < \delta \Rightarrow \sum_{k=1}^n \omega_k \Delta x_k < \epsilon

Equivalently: SΔSΔ<ϵ\overline{S}_\Delta - \underline{S}_\Delta < \epsilon.

Theorem 6.8: Du Bois-Reymond Criterion

For bounded ff on [a,b][a, b]:

fR[a,b]ϵ>0,σ>0, partition Δ:f \in R[a, b] \Leftrightarrow \forall \epsilon > 0, \forall \sigma > 0, \exists \text{ partition } \Delta:
k:ωkϵΔxk<σ\sum_{k: \omega_k \geq \epsilon} \Delta x_k < \sigma

The total length of “bad” subintervals (where oscillation ≥ ε) can be made arbitrarily small.

8. Classes of Integrable Functions

Example 6.2: Continuous Functions

Theorem: If fC[a,b]f \in C[a, b], then fR[a,b]f \in R[a, b].

Proof Sketch:

Continuous on closed interval ⟹ uniformly continuous.

Given ϵ>0\epsilon > 0, choose δ\delta from uniform continuity.

For Δ<δ\|\Delta\| < \delta: ωk<ϵ/(ba)\omega_k < \epsilon/(b-a) on each subinterval, so ωkΔxk<ϵ\sum \omega_k \Delta x_k < \epsilon.

Example 6.3: Monotonic Functions

Theorem: If ff is monotonic on [a,b][a, b], then fR[a,b]f \in R[a, b].

Proof Sketch:

For increasing ff: Mk=f(xk)M_k = f(x_k), mk=f(xk1)m_k = f(x_{k-1}).

SS=k=1n[f(xk)f(xk1)]ΔxkΔ[f(b)f(a)]\overline{S} - \underline{S} = \sum_{k=1}^n [f(x_k) - f(x_{k-1})]\Delta x_k \leq \|\Delta\|[f(b) - f(a)]

This can be made <ϵ< \epsilon by choosing Δ\|\Delta\| small.

Example 6.4: Bounded Functions with Finitely Many Discontinuities

Theorem: If ff is bounded on [a,b][a, b] and has only finitely many discontinuities, then fR[a,b]f \in R[a, b].

Proof Idea:

Enclose each discontinuity in a small interval. The total contribution from these intervals can be bounded by 2Mkδ2M \cdot k\delta where kk is the number of discontinuities.

Example 6.5: The Riemann Function (Thomae's Function)

Definition:

R(x)={1qx=pq in lowest terms,q>00x irrationalR(x) = \begin{cases} \frac{1}{q} & x = \frac{p}{q} \text{ in lowest terms}, q > 0 \\ 0 & x \text{ irrational} \end{cases}

Key Facts:

  • RR is discontinuous at every rational
  • RR is continuous at every irrational
  • RR[0,1]R \in R[0, 1] with 01R(x)dx=0\int_0^1 R(x)\,dx = 0
Example 6.6: Dirichlet Function (NOT Integrable)

Definition:

D(x)=1Q(x)={1xQ0xQD(x) = \mathbf{1}_\mathbb{Q}(x) = \begin{cases} 1 & x \in \mathbb{Q} \\ 0 & x \notin \mathbb{Q} \end{cases}

Why NOT Integrable:

For any partition: Mk=1M_k = 1, mk=0m_k = 0 on every subinterval.

Thus S=ba\overline{S} = b - a, S=0\underline{S} = 0 for all partitions, so \overline{\int} \neq \underline{\int}.

9. Basic Conventions

Remark 6.2: Notation Conventions

Integration variable: The definite integral is independent of the variable name:

abf(x)dx=abf(t)dt=abf(u)du\int_a^b f(x)\,dx = \int_a^b f(t)\,dt = \int_a^b f(u)\,du

Reversed limits:

baf(x)dx=abf(x)dx\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx

Equal limits:

aaf(x)dx=0\int_a^a f(x)\,dx = 0

10. Common Mistakes

❌ Bounded ⟹ Integrable

The Dirichlet function is bounded but not integrable!

❌ Discontinuous ⟹ Not Integrable

Functions with “few” discontinuities (measure zero) can be integrable.

❌ Confusing Riemann and Darboux sums

Riemann sums use sample points; Darboux uses sup/inf.

❌ Forgetting ||Δ|| → 0 means ALL partitions

The limit must work for any sequence of partitions with shrinking norm.

11. Additional Examples

Example 6.7: Computing ∫₀¹ x² Using Definition

Problem: Compute 01x2dx\int_0^1 x^2\,dx directly from the definition using Riemann sums.

Solution:

Use uniform partition with right endpoints: xk=k/nx_k = k/n, Δxk=1/n\Delta x_k = 1/n.

Sn=k=1n(kn)21n=1n3k=1nk2S_n = \sum_{k=1}^{n} \left(\frac{k}{n}\right)^2 \cdot \frac{1}{n} = \frac{1}{n^3} \sum_{k=1}^{n} k^2

Using the formula k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}:

Sn=1n3n(n+1)(2n+1)6=(n+1)(2n+1)6n2S_n = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{(n+1)(2n+1)}{6n^2}

Taking the limit:

limnSn=limn(1+1/n)(2+1/n)6=126=13\lim_{n \to \infty} S_n = \lim_{n \to \infty} \frac{(1+1/n)(2+1/n)}{6} = \frac{1 \cdot 2}{6} = \frac{1}{3}
Example 6.8: Computing ∫₀¹ eˣ dx

Problem: Compute 01exdx\int_0^1 e^x\,dx using Riemann sums.

Solution:

With right endpoints: ξk=k/n\xi_k = k/n.

Sn=k=1nek/n1n=1nk=1n(e1/n)kS_n = \sum_{k=1}^{n} e^{k/n} \cdot \frac{1}{n} = \frac{1}{n} \sum_{k=1}^{n} (e^{1/n})^k

This is a geometric series with ratio r=e1/nr = e^{1/n}:

Sn=1ne1/n(e1)e1/n1=e1/n(e1)n(e1/n1)S_n = \frac{1}{n} \cdot \frac{e^{1/n}(e - 1)}{e^{1/n} - 1} = \frac{e^{1/n}(e-1)}{n(e^{1/n}-1)}

Since limnn(e1/n1)=1\lim_{n \to \infty} n(e^{1/n} - 1) = 1 (by L'Hôpital or Taylor):

01exdx=limnSn=e1\int_0^1 e^x\,dx = \lim_{n \to \infty} S_n = e - 1
Example 6.9: Upper and Lower Sums for Step Function

Problem: Let f(x)=xf(x) = \lfloor x \rfloor on [0,3][0, 3]. Show fR[0,3]f \in R[0, 3] and find the integral.

Solution:

f(x)=0f(x) = 0 on [0,1)[0, 1), f(x)=1f(x) = 1 on [1,2)[1, 2), f(x)=2f(x) = 2 on [2,3][2, 3].

Use partition Δ={0,1,2,3}\Delta = \{0, 1, 2, 3\}:

  • On [0,1][0, 1]: M1=1,m1=0M_1 = 1, m_1 = 0
  • On [1,2][1, 2]: M2=2,m2=1M_2 = 2, m_2 = 1
  • On [2,3][2, 3]: M3=2,m3=2M_3 = 2, m_3 = 2

SS=(10)+(21)+(22)=2\overline{S} - \underline{S} = (1-0) + (2-1) + (2-2) = 2.

With finer partitions avoiding integers: Δ={0,1ϵ,1+ϵ,2ϵ,2+ϵ,3}\Delta' = \{0, 1-\epsilon, 1+\epsilon, 2-\epsilon, 2+\epsilon, 3\}:

SS4ϵ0\overline{S}' - \underline{S}' \leq 4\epsilon \to 0

So fR[0,3]f \in R[0, 3] with 03f=01+11+21=3\int_0^3 f = 0 \cdot 1 + 1 \cdot 1 + 2 \cdot 1 = 3.

Example 6.10: Verifying Integrability via Oscillation

Problem: Show that f(x)=sin(1/x)f(x) = \sin(1/x) for x(0,1]x \in (0, 1], f(0)=0f(0) = 0 is integrable on [0,1][0, 1].

Solution:

ff is continuous on (0,1](0, 1] and bounded by 1.

For any ϵ>0\epsilon > 0, choose δ=ϵ/4\delta = \epsilon/4.

Split [0,1]=[0,δ][δ,1][0, 1] = [0, \delta] \cup [\delta, 1]:

  • On [0,δ][0, \delta]: oscillation ≤ 2, contribution to sum ≤ 2δ=ϵ/22\delta = \epsilon/2
  • On [δ,1][\delta, 1]: ff is uniformly continuous, can make oscillation sum <ϵ/2< \epsilon/2

Total ωkΔxk<ϵ\sum \omega_k \Delta x_k < \epsilon, so fR[0,1]f \in R[0, 1].

Example 6.11: Showing Riemann Sums Converge

Problem: Prove limn1nk=1n11+(k/n)2=π4\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{1}{1 + (k/n)^2} = \frac{\pi}{4}.

Solution:

Recognize this as a Riemann sum for f(x)=11+x2f(x) = \frac{1}{1+x^2} on [0,1][0, 1]:

1nk=1nf(k/n)=k=1nf(ξk)Δxk0111+x2dx\frac{1}{n}\sum_{k=1}^n f(k/n) = \sum_{k=1}^n f(\xi_k) \Delta x_k \to \int_0^1 \frac{1}{1+x^2}\,dx

Computing the integral:

0111+x2dx=arctanx01=arctan1arctan0=π4\int_0^1 \frac{1}{1+x^2}\,dx = \arctan x \Big|_0^1 = \arctan 1 - \arctan 0 = \frac{\pi}{4}

12. Practice Problems

Problem Set A: Riemann Sums

  1. 1.Compute 02x3dx\int_0^2 x^3\,dx using Riemann sums with right endpoints.
  2. 2.Show limn1nk=1nk/n=2/3\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \sqrt{k/n} = 2/3.
  3. 3.For f(x)=xf(x) = x on [0,1][0, 1], find SΔ\overline{S}_\Delta and SΔ\underline{S}_\Delta for uniform partition with nn points.
  4. 4.Evaluate limn1p+2p++npnp+1\lim_{n \to \infty} \frac{1^p + 2^p + \cdots + n^p}{n^{p+1}} using Riemann sums.

Problem Set B: Integrability

  1. 5.Prove that if ff is bounded and has countably many discontinuities, then fR[a,b]f \in R[a, b].
  2. 6.Show that f(x)=xD(x)f(x) = x \cdot D(x) (where DD is Dirichlet) is integrable on [0,1][0, 1].
  3. 7.If fR[a,b]f \in R[a, b] and gg differs from ff at finitely many points, show gR[a,b]g \in R[a, b].
  4. 8.Give an example of a function integrable on [0,1][0, 1] that is discontinuous at every rational.

Problem Set C: Darboux Sums

  1. 9.Prove: SΔ1Δ2min(SΔ1,SΔ2)\overline{S}_{\Delta_1 \cup \Delta_2} \leq \min(\overline{S}_{\Delta_1}, \overline{S}_{\Delta_2}).
  2. 10.Show that f+g(f+g)\overline{\int} f + \overline{\int} g \geq \overline{\int}(f + g) with an example where equality fails.
  3. 11.For f(x)=x2f(x) = x^2 on [0,2][0, 2], compute S\overline{S} and S\underline{S} for partition {0,1,2}\{0, 1, 2\}.
  4. 12.Prove: If fgf \leq g on [a,b][a, b], then fg\overline{\int} f \leq \overline{\int} g.
Remark 6.3: Answers to Selected Problems

Problem 1: 02x3dx=4\int_0^2 x^3\,dx = 4

Problem 4: 1p+1\frac{1}{p+1}

Problem 8: Riemann function R(x)R(x)

Problem 11: S=5,S=1\overline{S} = 5, \underline{S} = 1

13. Historical Context

Bernhard Riemann (1826-1866)

German mathematician who formalized the definition of the integral in his 1854 habilitation lecture. He introduced the concept of using arbitrary partitions and sample points, asking when the limit exists independent of these choices.

Jean-Gaston Darboux (1842-1917)

French mathematician who developed the upper and lower sum approach (1875), providing an equivalent but often more convenient criterion for integrability. This approach eliminates dependence on sample points.

Paul du Bois-Reymond (1831-1889)

German mathematician who gave a precise characterization of integrable functions in terms of their oscillation on “bad” subintervals, leading to measure-theoretic interpretations.

Henri Lebesgue (1875-1941)

French mathematician who extended integration theory (1902) to handle more functions. His measure-theoretic approach showed that Riemann integrable functions are exactly those continuous “almost everywhere.”

Remark 6.4: The Evolution of Integration

Integration concepts evolved over centuries:

  • Archimedes (c. 287-212 BC): Method of exhaustion for areas
  • Newton & Leibniz (1660s): Infinitesimal calculus, antiderivatives
  • Cauchy (1823): First rigorous definition for continuous functions
  • Riemann (1854): General definition allowing discontinuities
  • Lebesgue (1902): Measure-theoretic extension

14. Connections to Other Topics

Measure Theory

Riemann integrability is characterized by the set of discontinuities having measure zero. This connects to Lebesgue's criterion and motivates the Lebesgue integral.

Numerical Analysis

Riemann sums form the basis of numerical integration methods: left/right rectangle rules, midpoint rule, and their generalizations (trapezoidal, Simpson's rule).

Probability Theory

The integral of a PDF over an interval gives probability. The definition of expected valueE[X]=xf(x)dxE[X] = \int x f(x)\,dx relies on Riemann/Lebesgue integration.

Physics

Work, center of mass, moments of inertia—all computed as integrals. The Riemann sum interpretation gives physical meaning to these quantities.

15. Summary of Key Results

ConceptFormula/Statement
Riemann SumSΔ=k=1nf(ξk)ΔxkS_\Delta = \sum_{k=1}^n f(\xi_k)\Delta x_k
Upper SumSΔ=k=1nMkΔxk\overline{S}_\Delta = \sum_{k=1}^n M_k \Delta x_k
Lower SumSΔ=k=1nmkΔxk\underline{S}_\Delta = \sum_{k=1}^n m_k \Delta x_k
IntegrabilityfR[a,b]f=ff \in R[a,b] \Leftrightarrow \overline{\int} f = \underline{\int} f
Oscillation Criterionϵ>0,Δ:ωkΔxk<ϵ\forall \epsilon > 0, \exists \Delta: \sum \omega_k \Delta x_k < \epsilon
Continuous ⟹ IntegrablefC[a,b]fR[a,b]f \in C[a,b] \Rightarrow f \in R[a,b]
Monotone ⟹ Integrableff monotonic on [a,b]fR[a,b][a,b] \Rightarrow f \in R[a,b]

16. Advanced Topics Preview

Lebesgue's Criterion

A bounded function ff on [a,b][a, b] is Riemann integrable if and only if the set of discontinuities of ff has Lebesgue measure zero.

This explains why the Riemann function is integrable (discontinuous only at rationals, a set of measure zero) while the Dirichlet function is not (discontinuous everywhere).

Improper Integrals

When the interval is unbounded or the function is unbounded, we define improper integrals as limits:

af(x)dx=limbabf(x)dx\int_a^\infty f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx
ab1(xa)pdx=limϵ0+a+ϵb1(xa)pdx\int_a^b \frac{1}{(x-a)^p}\,dx = \lim_{\epsilon \to 0^+} \int_{a+\epsilon}^b \frac{1}{(x-a)^p}\,dx

Riemann-Stieltjes Integral

A generalization where we integrate with respect to a function gg rather than xx:

abf(x)dg(x)=limΔ0k=1nf(ξk)[g(xk)g(xk1)]\int_a^b f(x)\,dg(x) = \lim_{\|\Delta\| \to 0} \sum_{k=1}^n f(\xi_k)[g(x_k) - g(x_{k-1})]

This is useful in probability (expectations), physics (moments), and functional analysis.

Multiple Integrals

The Riemann integral extends to higher dimensions. For a function on a rectangle R=[a,b]×[c,d]R = [a,b] \times [c,d]:

Rf(x,y)dA=limΔ0i,jf(ξij,ηij)ΔxiΔyj\iint_R f(x,y)\,dA = \lim_{\|\Delta\| \to 0} \sum_{i,j} f(\xi_{ij}, \eta_{ij}) \Delta x_i \Delta y_j

Fubini's theorem connects double integrals to iterated single integrals.

17. Computational Tips

Recognizing Riemann Sums

To evaluate limn1nk=1nf(k/n)\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n f(k/n):

  • Identify this as a Riemann sum on [0,1][0, 1]
  • The limit equals 01f(x)dx\int_0^1 f(x)\,dx
  • Generalize: 1nf(a+k(ba)/n)abf(x)dx1ba\frac{1}{n}\sum f(a + k(b-a)/n) \to \int_a^b f(x)\,dx \cdot \frac{1}{b-a}

Power Sum Formulas

Useful for computing Riemann sums:

k=1nk=n(n+1)2\sum_{k=1}^n k = \frac{n(n+1)}{2}

k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}

k=1nk3=[n(n+1)2]2\sum_{k=1}^n k^3 = \left[\frac{n(n+1)}{2}\right]^2

Proving Integrability

Strategy checklist:

  1. Check if ff is continuous → done!
  2. Check if ff is monotonic → done!
  3. Check if discontinuities are “sparse”
  4. Use oscillation criterion directly

Showing Non-Integrability

To show fR[a,b]f \notin R[a,b]:

  • Find partitions where SS\overline{S} - \underline{S} is bounded away from 0
  • Show MkmkM_k - m_k is constant on every subinterval
  • Example: Dirichlet function has Mk=1,mk=0M_k = 1, m_k = 0 always

18. Study Guide

Key Definitions to Memorize

  • Partition, norm of partition
  • Riemann sum with sample points
  • Upper/lower Darboux sums
  • Upper/lower Darboux integrals
  • Oscillation ωk=Mkmk\omega_k = M_k - m_k

Key Theorems to Understand

  • Refinement decreases upper sums, increases lower sums
  • Darboux criterion: =fR[a,b]\overline{\int} = \underline{\int} \Leftrightarrow f \in R[a,b]
  • Oscillation criterion: ωkΔxk0\sum \omega_k \Delta x_k \to 0
  • Continuous functions are integrable
  • Monotonic functions are integrable

Important Examples

  • Computing xn\int x^n from definition
  • Dirichlet function (not integrable)
  • Riemann/Thomae function (integrable, discontinuous at rationals)
  • Step functions (integrable)

Common Exam Questions

  • Evaluate limits using Riemann sum interpretation
  • Prove a specific function is/isn't integrable
  • Compute upper/lower sums for a given partition
  • State and prove the oscillation criterion

19. What's Next?

In the next section, we explore the properties of definite integrals:

Linearity: (αf+βg)=αf+βg\int(\alpha f + \beta g) = \alpha\int f + \beta\int g
Interval additivity: ab=ac+cb\int_a^b = \int_a^c + \int_c^b
Comparison theorems and estimation
Absolute value and product integrability
Riemann Integral Quiz
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1
What is a partition of [a,b][a,b]?
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The Riemann sum SΔ(f,ξ)S_\Delta(f, \xi) equals:
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The Darboux upper sum SΔ\overline{S}_\Delta uses:
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For any Riemann sum, we have:
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5
If Δ2\Delta_2 is a refinement of Δ1\Delta_1, then:
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A bounded function ff is Riemann integrable on [a,b][a,b] iff:
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Which function is NOT Riemann integrable on [0,1][0,1]?
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A continuous function on [a,b][a,b] is:
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A monotonic function on [a,b][a,b] is:
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The Du Bois-Reymond criterion states fR[a,b]f \in R[a,b] iff:
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Frequently Asked Questions

What is the intuition behind Riemann sums?

Riemann sums approximate the area under a curve by dividing it into rectangles. As the partition gets finer (||Δ|| → 0), these approximations converge to the true area if the function is integrable.

Why do we need Darboux sums?

Darboux sums provide upper and lower bounds independent of sample point choice. They make proving integrability easier since we only need to show the upper and lower integrals are equal.

What's the difference between Riemann and Darboux integrals?

They are equivalent for bounded functions! A function is Riemann integrable iff it's Darboux integrable, and the integral values are the same.

Why isn't the Dirichlet function integrable?

For any partition, every subinterval contains both rationals and irrationals, so M_k = 1, m_k = 0. Thus S̄ = 1, S = 0 for all partitions.

How does the Riemann function (Thomae's function) differ?

The Riemann function R(x) = 1/q for x = p/q in lowest terms, R(x) = 0 for irrational x. It's discontinuous at rationals but integrable because rationals are 'sparse'.

What does 'norm of partition' ||Δ|| mean?

||Δ|| = max{Δx₁, Δx₂, ..., Δxₙ} is the length of the longest subinterval. As ||Δ|| → 0, the partition becomes finer.

Can a function with infinitely many discontinuities be integrable?

Yes! The key is that discontinuities must form a 'small' set (measure zero). The Riemann function has countably many discontinuities but is integrable.

What's the relationship between uniform continuity and integrability?

Uniform continuity on [a,b] implies integrability. Given ε, uniform continuity provides δ such that ||Δ|| < δ ensures S̄ - S < ε.

Key Takeaways

Riemann Sum

SΔ=f(ξk)ΔxkS_\Delta = \sum f(\xi_k)\Delta x_k

Darboux Bounds

SSΔS\underline{S} \leq S_\Delta \leq \overline{S}

Integrability

fR[a,b]=f \in R[a,b] \Leftrightarrow \overline{\int} = \underline{\int}

Key Classes

Continuous, monotonic functions are integrable

Refinement Property

Δ2Δ1SΔ2SΔ1\Delta_2 \supset \Delta_1 \Rightarrow \overline{S}_{\Delta_2} \leq \overline{S}_{\Delta_1}

Oscillation Criterion

ωkΔxk0\sum \omega_k \Delta x_k \to 0

Quick Reference: Integrability Tests

✓ Always Integrable

  • Continuous functions
  • Monotonic functions
  • Bounded + finite discontinuities

? May Be Integrable

  • Bounded + countable disc.
  • Riemann/Thomae function
  • Piecewise continuous

✗ Not Integrable

  • Dirichlet function
  • Unbounded functions
  • Dense discontinuities

Key Definitions Checklist

Partition: a=x0<x1<<xn=ba = x_0 < x_1 < \cdots < x_n = b
Norm: Δ=maxΔxk\|\Delta\| = \max \Delta x_k
Upper sum: S=MkΔxk\overline{S} = \sum M_k \Delta x_k
Lower sum: S=mkΔxk\underline{S} = \sum m_k \Delta x_k
Oscillation: ωk=Mkmk\omega_k = M_k - m_k
Integral: limΔ0SΔ\lim_{\|\Delta\| \to 0} S_\Delta

Quick Decision Guide

When Computing Riemann Sums

  • Use uniform partition xk=a+k(ba)/nx_k = a + k(b-a)/n
  • Choose convenient sample points (right, left, or midpoint)
  • Apply power sum formulas for polynomials
  • Use geometric series for exponentials

When Proving Integrability

  • Check continuity first (easiest case)
  • Check monotonicity next
  • Use oscillation criterion for tricky cases
  • Apply Du Bois-Reymond for “sparse” discontinuities