Mathematics Advanced • Year 12 • Module 6 • Lesson 1
Introduction to Integration
Practise HSC-style writing on antiderivatives, boundary conditions and the meaning of the constant of integration.
1. Short-answer questions
1.1 Find ∫ (4x³ − 3x² + 2x − 1) dx. Show all working. 2 marks Band 3
1.2 If dy/dx = 3x² − 4x + 2 and the curve passes through (1, 5), find the equation of the curve. 3 marks Band 3-4
1.3 A particle moves in a straight line with acceleration a(t) = 6t m/s², starting from rest at the origin.
(a) Find its velocity v(t).
(b) Find its displacement s(t).
(c) State the value of s(2) and v(2). 4 marks Band 4
2. Extended response
2.1 A particle moves in a straight line. At time t = 0 it is at the origin with velocity v(0) = 5 m/s. Its acceleration is a(t) = 6 − 2t m/s² for t ≥ 0.
(a) Find v(t) and s(t).
(b) Find the time at which the particle is instantaneously at rest, and its displacement at that time.
(c) Determine whether the particle ever returns to the origin for t > 0. Justify with full working.
(d) Explain in two sentences what the constant of integration represented at each antidifferentiation step. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — antidifferentiates a(t) to v(t) = 6t − t² + C₁ and uses v(0) = 5 to get C₁ = 5.
• 1 mark — antidifferentiates v(t) to s(t) = 3t² − t³/3 + 5t + C₂ and uses s(0) = 0 to get C₂ = 0.
Part (b) — 2 marks
• 1 mark — sets v(t) = 0, i.e. −t² + 6t + 5 = 0, solves via the quadratic formula to obtain t = 3 + √14 ≈ 6.74 s (rejecting the negative root).
• 1 mark — substitutes into s(t) and evaluates: s(3 + √14) ≈ 49.4 m (accept 2 s.f. or exact form).
Part (c) — 2 marks
• 1 mark — sets s(t) = 0 and obtains a cubic whose roots are t = 0 and the roots of −t²/3 + 3t + 5 = 0, i.e. t² − 9t − 15 = 0, so t = (9 + √141)/2 ≈ 10.4 s.
• 1 mark — states a clear conclusion: yes, the particle returns to the origin at t ≈ 10.4 s, with the supporting working shown.
Part (d) — 1 mark
• 1 mark — explains that C₁ is the initial velocity (fixed by v(0) = 5) and C₂ is the initial position (fixed by s(0) = 0).
Your response:
Stuck on (c)? Look for a positive root larger than your "at rest" time — the particle decelerates, reverses, and only returns after passing rest.How did this worksheet feel?
What I'll revisit before next class:
1.1 — ∫(4x³ − 3x² + 2x − 1) dx (2 marks)
Sample response. ∫4x³ dx = x⁴; ∫(−3x²) dx = −x³; ∫2x dx = x²; ∫(−1) dx = −x. Combining: x⁴ − x³ + x² − x + C.
Marking notes. 1 mark — correct term-by-term integration of all four terms. 1 mark — combined answer with a single + C. Missing + C: deduct 0.5. Wrong sign on the −x³ term is the most common error.
1.2 — Curve through (1, 5) with dy/dx = 3x² − 4x + 2 (3 marks)
Sample response. y = ∫(3x² − 4x + 2) dx = x³ − 2x² + 2x + C. Substituting (1, 5): 5 = 1 − 2 + 2 + C, so C = 4. y = x³ − 2x² + 2x + 4.
Marking notes. 1 mark — correct general antiderivative including + C. 1 mark — correct substitution of the point to obtain C. 1 mark — final equation stated. A response with the antiderivative but no use of the point caps at 1.
1.3 — a(t) = 6t, rest, origin (4 marks)
(a) v(t) = ∫6t dt = 3t² + C₁; v(0) = 0 ⇒ C₁ = 0; so v(t) = 3t².
(b) s(t) = ∫3t² dt = t³ + C₂; s(0) = 0 ⇒ C₂ = 0; so s(t) = t³.
(c) v(2) = 12 m/s; s(2) = 8 m.
Marking notes. (a) 1 mark — correct v(t) including the use of v(0) = 0 to set C₁ = 0. (b) 1 mark — correct s(t) including s(0) = 0 ⇒ C₂ = 0. (c) 2 marks — 1 each for the correct v(2) and s(2). Common error: stating v(t) = 3t² + C without resolving C ("from rest" was given).
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a).
v(t) = ∫(6 − 2t) dt = 6t − t² + C₁.
Using v(0) = 5: 5 = 0 − 0 + C₁ ⇒ C₁ = 5, so v(t) = 6t − t² + 5. [1 mark — v(t) with C₁ fixed by initial velocity.]
s(t) = ∫(6t − t² + 5) dt = 3t² − t³/3 + 5t + C₂.
Using s(0) = 0: 0 = C₂, so s(t) = 3t² − t³/3 + 5t. [1 mark — s(t) with C₂ fixed by initial position.]
Part (b). Set v(t) = 0:
6t − t² + 5 = 0 ⇔ t² − 6t − 5 = 0 ⇔ t = (6 ± √(36 + 20))/2 = (6 ± √56)/2 = 3 ± √14.
Reject the negative root, giving t = 3 + √14 ≈ 6.74 s. [1 mark — solves v = 0 correctly and rejects negative root.]
At t = 3 + √14: s ≈ 3(6.74)² − (6.74)³/3 + 5(6.74) ≈ 136.3 − 102.0 + 33.7 ≈ 68.0 m. [1 mark — correct substitution and evaluation; accept 67-68 m.]
Part (c). Set s(t) = 0:
3t² − t³/3 + 5t = 0 ⇔ t(3t − t²/3 + 5) = 0 ⇔ t = 0 or −t²/3 + 3t + 5 = 0.
Multiplying the second factor by −3: t² − 9t − 15 = 0 ⇒ t = (9 ± √(81 + 60))/2 = (9 ± √141)/2.
The positive root is t = (9 + √141)/2 ≈ 10.44 s. [1 mark — sets up and solves s(t) = 0 correctly.]
Conclusion: yes, the particle returns to the origin at t ≈ 10.44 s — after it stops at t ≈ 6.74 s, the (now negative) acceleration drives it back through the start. [1 mark — clear conclusion with supporting working.]
Part (d). The first constant C₁ is the initial velocity v(0) — antidifferentiating acceleration gives velocity only up to an additive constant, and v(0) = 5 fixes it. The second constant C₂ is the initial position s(0) — antidifferentiating velocity gives position only up to an additive constant, and s(0) = 0 fixes it. [1 mark — both interpretations stated clearly.]
Total: 7/7.
Band descriptors for marker.
Band 3: Correct v(t) only (without resolving C₁), or attempts s(t) but with arithmetic errors; little or no progress on (b)(c). ≈ 2-3 marks.
Band 4: Completes (a) with both constants resolved; attempts (b) but stops at a quadratic without solving cleanly; no progress on (c) or (d). ≈ 3-4 marks.
Band 5: Completes (a) and (b) correctly; sets up but does not complete (c); attempts (d) loosely. ≈ 5 marks.
Band 6: Full working in all four parts, surds left exact where appropriate, clear conclusion in (c), and (d) interprets both C₁ and C₂ explicitly. 7/7.