MATH1961 · Mathematics 1a (advanced)
Integration
The second proof-heavy core. The definite integral is not ‘the antiderivative evaluated at the ends’ — that is a theorem (the FTC), and the Advanced unit makes you build the integral from its definition first. You define it with Darboux upper and lower sums: chop the interval into strips, over-estimate with the sup on each strip and under-estimate with the inf, and force the gap to zero. The key idea is the Darboux criterion — a bounded function is Riemann integrable exactly when that gap can be made arbitrarily small. Two rigour traps are planted: unbounded ⇒ not integrable (a true, provable implication) but bounded does not imply integrable (the Dirichlet counterexample). You then prove both parts of the FTC, and apply the techniques — substitution, parts, partial fractions, trig substitution — plus improper integrals and area.
What this chapter covers
- 01The Riemann integral via Darboux upper & lower sums
- 02The Darboux criterion — integrability is ‘the gap closes’
- 03The two rigour traps: unbounded ⇒ not integrable; bounded ⇏ integrable (Dirichlet)
- 04Both parts of the Fundamental Theorem of Calculus, proved
- 05Techniques: substitution, parts, partial fractions, trig substitution; improper integrals & area
Worked example: a bounded function that is not integrable (Dirichlet)
- +1Bounded. f only takes the values 0 and 1, so 0 ≤ f(x) ≤ 1 for all x: f is bounded.
- +1Upper sum. Every subinterval of any partition contains a rational, so the sup of f on each strip is 1. Hence the upper Darboux sum U(f, P) = 1 for every partition P.
- +1Lower sum. Every subinterval also contains an irrational, so the inf of f on each strip is 0. Hence the lower Darboux sum L(f, P) = 0 for every partition P.
- +1Conclude. The gap U − L = 1 for every partition, so it can never be made < ε for small ε. The Darboux criterion fails, so f is not Riemann integrable — even though it is bounded. ■
Key terms
- Darboux upper / lower sum
- For a partition of [a, b] into strips, the upper sum adds (width × sup of f on the strip) and the lower sum adds (width × inf of f on the strip). The lower sum under-estimates and the upper sum over-estimates the area, and L(f, P) ≤ U(f, P) always.
- Darboux criterion
- A bounded function f on [a, b] is Riemann integrable iff for every ε > 0 there is a partition P with U(f, P) − L(f, P) < ε — the upper and lower sums can be squeezed together. As the mesh shrinks to 0 both sums converge to the single number that is the integral.
- Riemann integrable
- A function whose Darboux upper and lower sums can be pinched arbitrarily close. Boundedness is necessary (unbounded functions are not integrable) but not sufficient (the Dirichlet function is bounded yet not integrable); every continuous function and every monotone bounded function on [a, b] is integrable.
- Fundamental Theorem of Calculus
- Two linked statements. Part I: if f is continuous, the area function F(x) = ∫ax f is differentiable with F′ = f. Part II: if F is an antiderivative of a continuous f, then ∫ab f = F(b) − F(a). Together they connect the definition (area) to computation (antiderivatives).
- Improper integral
- An integral over an unbounded interval or of an unbounded integrand, defined as a limit of proper integrals (e.g. ∫1∞ = limb→∞ ∫1b). It converges if the limit is finite and diverges otherwise.
Integration FAQ
Why build the integral from Darboux sums instead of just using antiderivatives?
Because ‘the antiderivative evaluated at the ends’ is the conclusion (the FTC), not the definition. The Advanced unit defines the integral as area trapped between upper and lower sums, then proves the FTC connects it to antiderivatives. Building from the definition is exactly what lets you handle integrability questions and the rigour traps the exam plants.
What is the difference between ‘bounded’ and ‘integrable’?
Boundedness is necessary but not sufficient. An unbounded function is never Riemann integrable (a sup or inf blows up on some strip, so the Darboux gap can never close) — a true, provable implication. But bounded does not imply integrable: the Dirichlet function is bounded yet has upper sum 1 and lower sum 0 on every partition, so its gap is stuck at 1. Keep these two statements apart.
Do I have to integrate from the definition every time?
No — you prove a couple of integrals from sums to show you can, then quote the two free theorems: every continuous function on [a, b] is integrable, and every monotone bounded function is integrable. These cover almost every integrand you meet, so the definition work is reserved for ‘show from the definition’ and counterexample questions.
What exactly does the FTC say, in two parts?
Part I: differentiating the area function gives back the integrand — for continuous f, d/dx ∫ax f = f(x). Part II: integrating using any antiderivative — ∫ab f = F(b) − F(a). MATH1961 asks you to prove both, with Part I’s proof using the picture of the area accumulating one strip at a time.
Exam move
Separate the two layers cleanly: the definition (Darboux sums, the criterion, the integrability theorems and traps) and the computation (FTC plus the techniques). For the definition, drill the Dirichlet counterexample and the ‘unbounded ⇒ not integrable’ argument until you can write each in four lines, and be able to prove a simple integral (like ∫ x dx) from sums. For computation, practise recognising which technique a given integrand needs — substitution, parts, partial fractions or a trig substitution — and learn the FTC proofs (both parts) rather than just the formula. Pair each technique with one clean worked template.
