Mathematics Advanced • Year 11 • Module 3 • Lesson 10
Integration
Practise HSC-style writing on integration — including a structured "recover the function then interpret" extended response.
1. Short-answer questions
1.1 Find ∫ (4x³ − 6x² + 2) dx. 2 marks Band 3
1.2 Find ∫ (2x⁴ + 3x²)/x dx. Simplify the integrand before applying the power rule. 3 marks Band 3-4
1.3 Given f′(x) = 6x² − 4x + 1 and f(2) = 10, find f(x). 4 marks Band 4
Stuck on 1.3? Integrate first to get f(x) + C; substitute the known value to solve for C.2. Extended response
2.1 A particle moves along a straight line with acceleration (in m/s²)
a(t) = 6t − 4, t ≥ 0.
At t = 0 the particle is at the origin with velocity v(0) = 1 m/s.
(a) Find a general expression for the velocity v(t) by integrating a(t), and determine the specific v(t) using v(0).
(b) Find a general expression for the displacement s(t) by integrating v(t), and determine the specific s(t) using s(0) = 0.
(c) Hence find the displacement of the particle at t = 3 s, and the velocity at that instant.
(d) Verify your s(t) by differentiating it twice and confirming you recover a(t). 8 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — integrates a(t): v(t) = 3t² − 4t + C₁.
• 1 mark — uses v(0) = 1 to find C₁ = 1; states v(t) = 3t² − 4t + 1.
Part (b) — 2 marks
• 1 mark — integrates v(t): s(t) = t³ − 2t² + t + C₂.
• 1 mark — uses s(0) = 0 to find C₂ = 0; states s(t) = t³ − 2t² + t.
Part (c) — 2 marks
• 1 mark — s(3) = 27 − 18 + 3 = 12 m.
• 1 mark — v(3) = 27 − 12 + 1 = 16 m/s.
Part (d) — 2 marks
• 1 mark — differentiates s(t) once: s′(t) = 3t² − 4t + 1 = v(t). ✓
• 1 mark — differentiates again: s″(t) = 6t − 4 = a(t). ✓ Confirms the integration was correct.
Your response:
Stuck on (d)? Use the differentiation rules from earlier in this module; the verification should reduce back to a(t) = 6t − 4.How did this worksheet feel?
What I'll revisit before next class:
1.1 — ∫ (4x³ − 6x² + 2) dx (2 marks)
Sample response. ∫ 4x³ dx = x⁴; ∫ (−6x²) dx = −2x³; ∫ 2 dx = 2x. Combine: x⁴ − 2x³ + 2x + C.
Marking notes. 1 — correct anti-derivatives of all three terms. 1 — combined answer with +C. Missing +C automatically loses 1 mark — this is a non-negotiable HSC convention.
1.2 — ∫ (2x⁴ + 3x²)/x dx (3 marks)
Sample response. Simplify first: (2x⁴ + 3x²)/x = 2x³ + 3x.
∫ (2x³ + 3x) dx = (2x⁴)/4 + (3x²)/2 + C = x⁴/2 + 3x²/2 + C, equivalently (x⁴ + 3x²)/2 + C.
Marking notes. 1 — simplifies the integrand (most efficient route). 1 — integrates each term correctly. 1 — combines with +C in a tidy form. Trying to apply the quotient rule for integration (which doesn't exist in this syllabus) and giving up scores 0.
1.3 — f′(x) = 6x² − 4x + 1, f(2) = 10 (4 marks)
Sample response. f(x) = ∫ (6x² − 4x + 1) dx = 2x³ − 2x² + x + C.
Substitute x = 2: f(2) = 2(8) − 2(4) + 2 + C = 16 − 8 + 2 + C = 10 + C.
Setting 10 + C = 10: C = 0. So f(x) = 2x³ − 2x² + x.
Marking notes. 1.5 — correctly integrates each term. 1 — substitutes x = 2 into the antiderivative. 1 — solves for C. 0.5 — writes the specific f(x) with C substituted. A response that omits +C in the integration step cannot pick up the C = 0 mark and loses 2 marks total.
2.1 — Particle motion (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). v(t) = ∫ a(t) dt = ∫ (6t − 4) dt = 3t² − 4t + C₁. [1 mark — general v(t).]
v(0) = 0 − 0 + C₁ = 1 ⇒ C₁ = 1. Therefore v(t) = 3t² − 4t + 1 m/s. [1 mark — specific v(t).]
Part (b). s(t) = ∫ v(t) dt = ∫ (3t² − 4t + 1) dt = t³ − 2t² + t + C₂. [1 mark — general s(t).]
s(0) = 0 − 0 + 0 + C₂ = 0 ⇒ C₂ = 0. Therefore s(t) = t³ − 2t² + t m. [1 mark — specific s(t).]
Part (c). s(3) = 27 − 18 + 3 = 12 m. [1 mark — displacement.]
v(3) = 27 − 12 + 1 = 16 m/s. [1 mark — velocity.]
Part (d). Verify by differentiating s(t) twice:
s′(t) = 3t² − 4t + 1 = v(t). ✓ [1 mark — first derivative matches v.]
s″(t) = 6t − 4 = a(t). ✓ [1 mark — second derivative matches a; integration verified.]
Total: 8/8.
Band descriptors for marker.
Band 3: Integrates once with +C but does not use the initial condition; quotes only v(t) (not s(t)); cannot evaluate at t = 3. ≈ 2-3 marks.
Band 4: Integrates twice with arbitrary constants but uses initial conditions inconsistently or only on one constant; finds s(3) but not v(3); skips verification. ≈ 4-5 marks.
Band 5: All algebra correct; both initial conditions used; both s(3) and v(3) found. Verification missing or partial. ≈ 6-7 marks.
Band 6: Clear progression a → v → s with each constant of integration named and solved using the appropriate initial condition; both numerical answers at t = 3 stated with units; explicit verification by differentiating s twice back to a; precise use of "general" vs "specific" anti-derivative. 8/8.