Build the definite integral rigorously from Riemann sums and Darboux sums, establishing precise conditions for when a function is integrable.
How do we compute the area under the curve from to ?
The ancient Greeks approximated areas using inscribed/circumscribed shapes. We formalize this by dividing into subintervals and summing rectangle areas.
Let be defined on . A partition of is a finite set of points:
This divides into subintervals for .
The norm (or mesh) of partition is:
where is the length of the -th subinterval.
Given partition and sample points , the Riemann sum is:
This represents the sum of rectangle areas with heights and widths .
Problem: Compute the Riemann sum for on with uniform partition and right endpoints.
Solution:
Partition: for . So .
Right endpoints: .
Note: The exact value is , while .
For partition of and bounded function , define:
The upper Darboux sum and lower Darboux sum are:
Upper sum
Lower sum
For any partition and sample points :
Since for each , multiplying by and summing:
Partition is a refinement of (written ) if every partition point of is also a partition point of .
The common refinement of and is .
If and has more points than , then:
where , on .
For any two partitions of :
Every lower sum is ≤ every upper sum (even for different partitions).
For bounded on , define the upper integral and lower integral:
Upper (Darboux) integral
Lower (Darboux) integral
For bounded on : , such that for any partition with :
The upper and lower integrals always exist for bounded functions, and we always have:
Let be bounded on . If there exists such that:
for any partition and any sample points , then is Riemann integrable on .
We write and define:
If , then the integral value is unique.
If , then is bounded on .
Suppose is unbounded on . For any partition , is unbounded on at least one subinterval . By choosing sample points appropriately, we can make arbitrarily large, contradicting convergence to a finite .
For bounded on :
Define the oscillation on as . Then:
Equivalently: .
For bounded on :
The total length of “bad” subintervals (where oscillation ≥ ε) can be made arbitrarily small.
Theorem: If , then .
Proof Sketch:
Continuous on closed interval ⟹ uniformly continuous.
Given , choose from uniform continuity.
For : on each subinterval, so .
Theorem: If is monotonic on , then .
Proof Sketch:
For increasing : , .
This can be made by choosing small.
Theorem: If is bounded on and has only finitely many discontinuities, then .
Proof Idea:
Enclose each discontinuity in a small interval. The total contribution from these intervals can be bounded by where is the number of discontinuities.
Definition:
Key Facts:
Definition:
Why NOT Integrable:
For any partition: , on every subinterval.
Thus , for all partitions, so .
Integration variable: The definite integral is independent of the variable name:
Reversed limits:
Equal limits:
The Dirichlet function is bounded but not integrable!
Functions with “few” discontinuities (measure zero) can be integrable.
Riemann sums use sample points; Darboux uses sup/inf.
The limit must work for any sequence of partitions with shrinking norm.
Problem: Compute directly from the definition using Riemann sums.
Solution:
Use uniform partition with right endpoints: , .
Using the formula :
Taking the limit:
Problem: Compute using Riemann sums.
Solution:
With right endpoints: .
This is a geometric series with ratio :
Since (by L'Hôpital or Taylor):
Problem: Let on . Show and find the integral.
Solution:
on , on , on .
Use partition :
.
With finer partitions avoiding integers: :
So with .
Problem: Show that for , is integrable on .
Solution:
is continuous on and bounded by 1.
For any , choose .
Split :
Total , so .
Problem: Prove .
Solution:
Recognize this as a Riemann sum for on :
Computing the integral:
Problem 1:
Problem 4:
Problem 8: Riemann function
Problem 11:
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.
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.
German mathematician who gave a precise characterization of integrable functions in terms of their oscillation on “bad” subintervals, leading to measure-theoretic interpretations.
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.”
Integration concepts evolved over centuries:
Riemann integrability is characterized by the set of discontinuities having measure zero. This connects to Lebesgue's criterion and motivates the Lebesgue integral.
Riemann sums form the basis of numerical integration methods: left/right rectangle rules, midpoint rule, and their generalizations (trapezoidal, Simpson's rule).
The integral of a PDF over an interval gives probability. The definition of expected value relies on Riemann/Lebesgue integration.
Work, center of mass, moments of inertia—all computed as integrals. The Riemann sum interpretation gives physical meaning to these quantities.
| Concept | Formula/Statement |
|---|---|
| Riemann Sum | |
| Upper Sum | |
| Lower Sum | |
| Integrability | |
| Oscillation Criterion | |
| Continuous ⟹ Integrable | |
| Monotone ⟹ Integrable | monotonic on |
A bounded function on is Riemann integrable if and only if the set of discontinuities of 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).
When the interval is unbounded or the function is unbounded, we define improper integrals as limits:
A generalization where we integrate with respect to a function rather than :
This is useful in probability (expectations), physics (moments), and functional analysis.
The Riemann integral extends to higher dimensions. For a function on a rectangle :
Fubini's theorem connects double integrals to iterated single integrals.
To evaluate :
Useful for computing Riemann sums:
Strategy checklist:
To show :
In the next section, we explore the properties of definite integrals:
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.
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.
They are equivalent for bounded functions! A function is Riemann integrable iff it's Darboux integrable, and the integral values are the same.
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.
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'.
||Δ|| = max{Δx₁, Δx₂, ..., Δxₙ} is the length of the longest subinterval. As ||Δ|| → 0, the partition becomes finer.
Yes! The key is that discontinuities must form a 'small' set (measure zero). The Riemann function has countably many discontinuities but is integrable.
Uniform continuity on [a,b] implies integrability. Given ε, uniform continuity provides δ such that ||Δ|| < δ ensures S̄ - S < ε.
Continuous, monotonic functions are integrable
✓ Always Integrable
? May Be Integrable
✗ Not Integrable