Riemann Integration: Definition, Criteria & Examples

Riemann integration defines the integral of a bounded function f on [a, b] through upper and lower sums: chop the interval into subintervals, on each piece take the largest and smallest values of f, and add up the corresponding rectangles. The function is integrable when the gap between the biggest possible lower sum and the smallest possible upper sum shrinks to zero.

This construction matters because it is the precise meaning behind "area under the curve". It also tells us exactly which functions can be integrated this way — and the answer is surprising: some badly discontinuous functions are integrable, while some innocent-looking ones are not.

Key Concepts

• Partition: a finite set P = {x₀, x₁, …, xn} with a = x₀ < x₁ < … < xn = b, cutting [a, b] into subintervals of width Δxi = xi − xi−1.
• Upper and lower sums: with Mi = sup f on [xi−1, xi] and mi = inf f there, U(P, f) = Σ MiΔxi and L(P, f) = Σ miΔxi.
• Integrability: f is Riemann integrable when supP L(P, f) = infP U(P, f); the common value is the integral ∫ab f.
• ε-criterion: f is integrable if and only if for every ε > 0 there is a partition with U(P, f) − L(P, f) < ε.
• Continuous implies integrable: every continuous function on [a, b] is Riemann integrable, proved via uniform continuity.

Formulas

• Darboux sums: L(P, f) = Σ miΔxi ≤ ∫ab f ≤ Σ MiΔxi = U(P, f) for every partition P.
• Refinement: adding points to a partition can only raise L(P, f) and lower U(P, f), squeezing the integral from both sides.
• Linearity: ∫ (αf + βg) = α∫f + β∫g, and ∫ab f + ∫bc f = ∫ac f, whenever the integrals exist.
• Estimate: if m ≤ f(x) ≤ M on [a, b], then m(b − a) ≤ ∫ab f ≤ M(b − a).

Worked Example

Prove that f(x) = x is Riemann integrable on [0, 1] and find its integral using the ε-criterion.

1. Take the regular partition Pn dividing [0, 1] into n equal subintervals of width 1/n.
2. Since f is increasing, on the i-th subinterval [(i−1)/n, i/n] we have mi = (i−1)/n and Mi = i/n.
3. Upper sum: U = Σi=1n (i/n)(1/n) = (1/n²)(1 + 2 + … + n) = (1/n²)·n(n+1)/2 = 1/2 + 1/(2n).
4. Lower sum: L = Σ ((i−1)/n)(1/n) = (1/n²)(0 + 1 + … + (n−1)) = 1/2 − 1/(2n).
5. The gap U − L = 1/n, which is < ε whenever n > 1/ε. Hence f is integrable, and both sums tend to 1/2, so ∫01 x dx = 1/2.

Common Misconceptions

• "A function must be continuous to be integrable." The Thomae function (1/q at rationals p/q, 0 at irrationals) is discontinuous at every rational yet Riemann integrable with integral 0.
• "The integral is always an area." Only for non-negative f; below the x-axis the integral counts area negatively.
• "Upper sums always decrease to the integral." Only along refinements; an arbitrary sequence of partitions need not squeeze anything.

Real-World Applications

• Defining work, distance and total charge as integrals of force, velocity and current in physics and engineering.
• Justifying numerical integration rules (trapezoid, Simpson) by comparing them with upper and lower sums.
• Probability theory: expected values of continuous random variables are Riemann (or Lebesgue) integrals of density functions.

Riemann integration builds the integral from below and above with lower and upper sums, and declares the function integrable when the two squeezes meet. The ε-criterion turns this into a practical test, as the computation for f(x) = x shows: regular partitions make U − L = 1/n, which vanishes. Continuity is sufficient for integrability but not necessary, and the whole theory rests on partitions getting finer, never on guessing antiderivatives.