MATH1061 · Mathematics 1a
Integration
Integration is differentiation run backwards — and, separately, the measure of accumulation and area. The Fundamental Theorem of Calculus ties these two ideas together: it lets you compute a definite integral ∫ab f as F(b) − F(a) using any antiderivative F, and it says differentiating an integral recovers the integrand. In MATH1061 the marks are in the techniques: substitution (reverse chain rule), integration by parts (reverse product rule), partial fractions and trig integrals, then the applications — area between curves, volumes by discs/shells, and improper integrals. The exam wants the right technique chosen quickly and the working shown, including the substitution's limit change and the parts table.
What this chapter covers
- 014.1 Antiderivatives — the standard table
- 024.2 The definite integral as area — Riemann sums
- 034.3 The Fundamental Theorem of Calculus (both parts)
- 044.4 Technique I — substitution (reverse chain rule)
- 054.5 Technique II — integration by parts (reverse product rule)
- 064.6 Technique III — partial fractions
- 074.7 Technique IV — trig integrals / trig substitution
- 08Applications — areas between curves, volumes, improper integrals
Worked example: substitution then by parts
- +1(a) Choose u: let u = x² + 1, so du = 2x dx — and 2x dx already sits in the integral.
- +1(a) Rewrite & integrate: ∫ u⁴ du = u⁵/5 + C.
- +1(a) Back-substitute: = (x² + 1)⁵/5 + C.
- +1(b) Choose parts: u = x (so du = dx), dv = ex dx (so v = ex) — pick u to simplify on differentiating.
- +1(b) Apply ∫u dv = uv − ∫v du: = x ex − ∫ ex dx.
- +1(b) Finish: = x ex − ex + C = ex(x − 1) + C.
Key terms
- Antiderivative
- A function F whose derivative is the integrand: F′ = f. Indefinite integrals are antiderivatives plus an arbitrary constant C, since differentiation kills constants.
- Fundamental Theorem of Calculus
- Part 1: differentiating an integral with a variable upper limit returns the integrand. Part 2 (the evaluation rule): ∫ab f = F(b) − F(a) for any antiderivative F. This is the bridge between area and antiderivatives.
- Substitution
- Integration's reverse chain rule: set u equal to an inner function, replace dx via du, integrate in u, then back-substitute. For a definite integral, change the limits to u-values instead of back-substituting.
- Integration by parts
- Integration's reverse product rule: ∫ u dv = uv − ∫ v du. Choose u to become simpler when differentiated (LIATE is a guide) so the new integral ∫ v du is easier than the original.
- Improper integral
- A definite integral with an infinite limit or an unbounded integrand, defined as a limit of ordinary integrals. It converges if that limit is finite; the p-test settles the standard power cases.
Integration FAQ
How do I pick u for a substitution?
Choose the inner function whose derivative also appears (up to a constant) elsewhere in the integrand — typically what's inside a power, a root, an exponential or a trig function. After setting u, the whole integrand including dx must rewrite cleanly in u; if leftover x's remain, the choice was wrong. For definite integrals, convert the limits to u rather than back-substituting.
How do I choose u and dv for integration by parts?
Pick u to be the part that gets simpler when differentiated, and dv to be the part you can integrate. The LIATE order (Logs, Inverse trig, Algebraic, Trig, Exponential) is a reliable priority list for u. For ∫ x eˣ dx, x is algebraic and eˣ is exponential, so u = x, dv = eˣ dx — and the leftover integral is easier.
Substitution or by parts — how do I tell?
Substitution works when the integrand contains a function and (a constant times) its derivative — a composition with its inner derivative present. By parts works for a product of two unrelated functions where one simplifies on differentiation, like x·eˣ or x·sin x or x·ln x. If a plain substitution leaves leftover factors, reach for parts.
What does the Fundamental Theorem actually let me do?
It lets you compute a definite integral — an area — without summing rectangles: find any antiderivative F, then evaluate F(b) − F(a). It also says the two big operations of calculus are inverses, so differentiating an accumulation function returns the original integrand. Almost every definite-integral mark in the exam uses Part 2.
Exam move
Memorise the antiderivative table and the FTC evaluation rule first — they underlie everything. Then train technique-spotting under time: composition-with-inner-derivative → substitution; product that simplifies on differentiating → by parts; proper rational function → partial fractions. Always carry the +C on indefinite integrals, and for definite integrals by substitution, change the limits rather than back-substituting (it is cleaner and examiners reward it). For improper integrals, write the limit explicitly and apply the p-test. Because the work is by hand, lay out the substitution and the parts table line by line so the method marks are unambiguous.
