Riemann Integration & the FTC
Core definitions, criteria, and theorems behind the Riemann integral and its link to antiderivatives.
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Questions Covered in This Set
12 cards to master
What is a partition of [a,b]?
A finite set of points a = x₀ < x₁ < ⋯ < xₙ = b dividing [a,b] into subintervals of widths Δxᵢ = xᵢ − xᵢ₋₁.
Define the lower and upper Darboux sums.
L(f,P) = Σ mᵢΔxᵢ and U(f,P) = Σ MᵢΔxᵢ, where mᵢ = inf f and Mᵢ = sup f on the i-th subinterval.
Why use inf/sup rather than min/max in Darboux sums?
Because a bounded function need not attain its extreme values on a subinterval; inf and sup always exist for bounded f.
When is f Riemann integrable (Darboux definition)?
When the lower integral sup_P L(f,P) equals the upper integral inf_P U(f,P); the common value is ∫ₐᵇ f.
State the Riemann criterion for integrability.
f is integrable iff for every ε > 0 there exists a partition P with U(f,P) − L(f,P) < ε.
What happens to L and U when a partition is refined?
Refining (adding points) raises the lower sum and lowers the upper sum, squeezing them toward each other.
Prove-sketch: why is every continuous function on [a,b] integrable?
Continuity on a compact interval is uniform, so choose δ with |f(x)−f(y)| < ε/(b−a) for |x−y| < δ; any partition of mesh < δ gives Mᵢ − mᵢ ≤ ε/(b−a), so U − L ≤ ε.
Why is every monotone function on [a,b] integrable?
With a uniform partition the differences telescope: U − L = (b−a)/n · |f(b) − f(a)| → 0.
Give a bounded non-integrable function and why it fails.
Dirichlet's function (1 on rationals, 0 on irrationals): every partition gives L = 0 and U = 1, so the lower and upper integrals differ. This motivates the Lebesgue integral.
State the Mean Value Theorem for Integrals.
If f is continuous on [a,b], there is c ∈ (a,b) with ∫ₐᵇ f = f(c)(b−a); proved by applying the IVT to the average value, which lies between min f and max f.
What does FTC Part I say?
If f is continuous on [a,b] and F(x) = ∫ₐˣ f(t)dt, then F is differentiable with F′(x) = f(x): differentiation undoes integration.
Compute ∫₀¹ x² dx from Riemann sums.
With xᵢ = i/n, U(f,Pₙ) = (1/n³)·n(n+1)(2n+1)/6 → 1/3, and U − L = 1/n → 0, so the integral is 1/3.
