Riemann integrability
Objective. Prove integrability and estimate integral error.
- 1. Learn
- 2. Worked example
- 3. Practice
- 4. Feedback
- 5. Continue
Learn: the key idea
Riemann integrability asks whether the upper and lower Riemann sums of a bounded function on [a, b] converge to the same value as the partition mesh shrinks, equivalently whether upper sum minus lower sum can be made arbitrarily small. A function f is Riemann integrable on [a, b] if for every ε > 0 there exists a partition P with U(f, P) − L(f, P) < ε. Every continuous function on [a, b] is Riemann integrable, and so is any bounded function with only finitely many discontinuities. For f(x) = x on [0, 1] with n equal subintervals, U − L = 1/n, which tends to 0, confirming integrability directly from the definition.
Worked example
Show f(x) = x is Riemann integrable on [0, 1] by bounding U(f, P) − L(f, P) for n equal subintervals.
- 1Partition [0, 1] into n equal subintervals each of width 1/n.
- 2Since f is increasing, on each subinterval the sup minus inf is 1/n.
- 3So U(f, P) − L(f, P) = n × (1/n) × (1/n) = 1/n.
- 4As n → ∞, 1/n → 0, satisfying the integrability criterion for every ε > 0.
Answer: U(f,P) − L(f,P) = 1/n → 0, so f is integrable
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Practice
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Start practisingCommon mistake. Students assume integrability requires continuity everywhere; the function equal to 0 except at one point where it equals 1 is still Riemann integrable with integral 0, since a single discontinuity cannot stop U − L shrinking to zero.
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