The Riemann Integral — Mathematics Knowledge System

Overview

The Riemann integral is the rigorous definition, due to Bernhard Riemann, of the integral of a function on an interval — the definition that makes precise the intuitive picture used throughout introductory calculus of 'area under a curve' as a limit of sums of rectangle areas. It defines the integral by approximating the region under the graph of f with finite sums of areas of thin vertical rectangles (Riemann sums), then asking whether these sums converge to a single value as the rectangles are made arbitrarily thin (the partition is refined). For every continuous function on a closed, bounded interval, this limiting value exists and defines the integral; Riemann sums that are close to the limit also serve as numerical approximations to it.

Intuition

Take the region under a curve y=f(x) between x=a and x=b, and slice it into thin vertical strips using a partition a=x₀<x₁<...<xₙ=b. Approximate each strip by a rectangle of width (xᵢ₊₁-xᵢ) and height f(tᵢ) for some sample point tᵢ inside the strip. Adding up all these rectangle areas gives a Riemann sum — a first estimate of the true area. The Riemann integral asks: as you slice more and more finely (every strip's width shrinking toward zero), do these estimates converge to one specific number, no matter how you chose the sample points tᵢ within each strip? If they do, that common limiting value is defined to be the integral.

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Formal Definition

Definition

Let f be a bounded function on [a,b]. A tagged partition consists of points a=x₀<x₁<⋯<xₙ=b together with sample points tᵢ∈[xᵢ,xᵢ₊₁]. The Riemann sum for this tagged partition is:

\sum_{i=0}^{n-1} f(t_i)\,(x_{i+1}-x_i)

The sum of rectangle areas — width times sampled height — over the partition

Riemann sum

\forall \varepsilon > 0,\ \exists \delta > 0 \text{ such that, for any tagged partition with mesh} < \delta,\quad \left|\left(\sum_{i=0}^{n-1} f(t_i)(x_{i+1}-x_i)\right) - s\right| < \varepsilon

f is Riemann integrable with integral s if, for every tolerance ε, sufficiently fine partitions (mesh — the widest sub-interval — below some δ) force every Riemann sum within ε of s, regardless of which sample points tᵢ were chosen

Riemann integral (definition via mesh)

s = \int_a^b f(x)\,dx

The common limiting value is denoted the definite integral of f over [a,b]

Notation

Notation	Meaning
$\int_a^b f(x)\,dx$	The Riemann integral of f over [a,b]
$\\|P\\| = \max_i (x_{i+1}-x_i)$	The mesh (or norm) of a partition — the width of its widest sub-interval
$L(f,P) = \sum m_i \Delta x_i,\quad U(f,P) = \sum M_i \Delta x_i$	Lower and upper Darboux sums, using the infimum mᵢ and supremum Mᵢ of f on each sub-interval — an equivalent formulation to tagged Riemann sums
$t_i \in [x_i, x_{i+1}]$	A sample (tag) point chosen within the i-th sub-interval of the partition

Derivation

Sketch of why continuity on a closed bounded interval guarantees integrability: the key tool is uniform continuity, which lets a single δ control oscillation everywhere at once.

f \text{ continuous on } [a,b] \implies f \text{ uniformly continuous on } [a,b]

By the Heine–Cantor theorem, continuity on a compact interval upgrades to uniform continuity

\forall \varepsilon>0\ \exists \delta>0: |x-y|<\delta \implies |f(x)-f(y)| < \frac{\varepsilon}{b-a}

Uniform continuity gives one δ that bounds oscillation across the ENTIRE interval simultaneously, not just near one point

\|P\| < \delta \implies U(f,P) - L(f,P) = \sum_i (M_i-m_i)\Delta x_i < \frac{\varepsilon}{b-a}\sum_i \Delta x_i = \varepsilon

Once the partition mesh is below δ, each sub-interval's oscillation (Mᵢ-mᵢ) is small, so the total gap between upper and lower sums shrinks below ε

\sup_P L(f,P) = \inf_P U(f,P) \implies f \text{ is Riemann integrable}

The upper/lower sums can be squeezed arbitrarily close together, satisfying the Darboux criterion for integrability

Properties

Continuity implies integrability

\text{If } f \text{ is continuous on } [a,b], \text{ then } f \text{ is Riemann integrable on } [a,b].

Linearity

\int_a^b (\alpha f + \beta g)\,dx = \alpha\int_a^b f\,dx + \beta \int_a^b g\,dx

Additivity over intervals

\int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx \quad (a<c<b)

Darboux (upper/lower sum) equivalence

f \text{ is Riemann integrable} \iff \sup_P L(f,P) = \inf_P U(f,P)

Condition: The lower and upper Darboux integrals coincide — an equivalent, often more tractable, characterization avoiding the choice of sample points tᵢ.

Boundedness is necessary

\text{Riemann integrability on } [a,b] \text{ requires } f \text{ to be bounded on } [a,b].

Condition: Unbounded functions require the extension to improper integrals.

Applications

Work done by a variable force, and center-of-mass computations for continuous bodies, are defined as Riemann integrals of force or density functions over a spatial interval.

Worked Examples

Partition [0,1] into 4 equal pieces of width 1/4, with right endpoints tᵢ = 1/4, 2/4, 3/4, 4/4.
$\Delta x = \frac{1}{4}$
Sum f(tᵢ)Δx for each right endpoint.
$\frac{1}{4}\left[\left(\tfrac14\right)^2+\left(\tfrac24\right)^2+\left(\tfrac34\right)^2+\left(\tfrac44\right)^2\right] = \frac{1}{4}\cdot\frac{30}{16}$

Answer: ≈ 0.46875 (the exact integral is 1/3 ≈ 0.333; the right-sum overestimates since x² is increasing)

Practice Problems

Difficulty 5/10

Estimate ∫₀² x dx using a left-endpoint Riemann sum with n=2 equal sub-intervals.

Difficulty 6/10

Which condition on f is SUFFICIENT (but not necessary) to guarantee Riemann integrability on [a,b]?

f is bounded on [a,b]f is continuous on [a,b]f is defined at every rational point of [a,b]f is unbounded on [a,b]

Common Mistakes

Common Mistake

Believing every bounded function is Riemann integrable.

Boundedness is necessary but not sufficient. The Dirichlet function (1 on rationals, 0 on irrationals) is bounded on [0,1] but not Riemann integrable, since its upper and lower Darboux sums never agree (1 vs. 0 on every sub-interval).

Common Mistake

Thinking the choice of sample points tᵢ can affect whether the limit exists, as long as SOME choice gives a nice answer.

The definition requires convergence to the SAME value s for every possible choice of tags tᵢ, once the mesh is small enough — not just for a convenient choice like left or right endpoints. This universality over tag choices is exactly what the Darboux upper/lower sum criterion is designed to test without enumerating all choices.

Quiz

The Riemann integral is defined as the limit of:

derivativessums of areas of rectangles (Riemann sums) as their width → 0a single antiderivativea Taylor series

Which condition GUARANTEES a function is Riemann integrable on [a,b]?

it is continuous on [a,b]it is defined at 0it is unboundedit is a polynomial only

Summary

The Riemann integral defines ∫ₐᵇf dx as the common limiting value of Riemann sums Σf(tᵢ)Δxᵢ as the partition mesh → 0, for every possible choice of sample points tᵢ.
Formally: ∀ε>0 ∃δ>0 such that any tagged partition with mesh<δ gives a Riemann sum within ε of the integral s.
Every continuous function on a closed, bounded interval is Riemann integrable (via uniform continuity, Heine–Cantor theorem).
The equivalent Darboux formulation (upper sums U(f,P) and lower sums L(f,P) using sup/inf per sub-interval) avoids picking sample points and gives a clean integrability criterion: sup L = inf U.
Boundedness is necessary but not sufficient for integrability — the Dirichlet function is the classic bounded-but-not-integrable counterexample.

References

WebsiteWikipedia — Riemann integral
BookRudin, W. Principles of Mathematical Analysis, 3rd ed. Ch. 6.