Mathematics Advanced • Year 12 • Module 6 • Lesson 2
Integrating Power Functions
Practise HSC-style writing on rewriting integrands, applying the power rule to fractional and negative powers, and identifying the n = −1 exception.
1. Short-answer questions
1.1 Find ∫ (x³ + 4x² − x)/x² dx. Show all working. 3 marks Band 3-4
1.2 Find ∫ (√x + 1/√x)² dx. Expand first, then integrate term by term. 3 marks Band 4
1.3 A curve has gradient dy/dx = 2x − 4/x² and passes through the point (2, 3). Find the equation of the curve. 4 marks Band 4
Stuck on 1.3? Rewrite 4/x² as 4x−2 before integrating, then use the point to determine C.2. Extended response
2.1 A water tank is being filled at rate dV/dt = 6√t litres per minute, starting empty at t = 0.
(a) Find V(t).
(b) Find the volume of water that enters during the first 9 minutes.
(c) Find the time at which the total volume reaches 144 litres.
(d) The owner claims "the rate is increasing, so doubling the volume takes less than double the time". Verify or refute this claim by comparing the time to fill 72 L with the time to fill 144 L, and explain in one sentence the relationship between the rate's behaviour and this comparison. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — V(t) = 4 t3/2 + C with C = 0 (justified by V(0) = 0).
Part (b) — 1 mark
• 1 mark — V(9) = 4 · 27 = 108 L.
Part (c) — 2 marks
• 1 mark — sets 4 t3/2 = 144 and isolates t3/2 = 36.
• 1 mark — solves t = 362/3 ≈ 10.9 minutes (accept exact form or 1-d.p. decimal).
Part (d) — 3 marks
• 1 mark — computes time to 72 L: t72 = 182/3 ≈ 6.87 minutes.
• 1 mark — quantitative comparison: 2 × 6.87 ≈ 13.74 > 10.9, so the time to fill 144 L is less than twice the time to fill 72 L; the claim is supported.
• 1 mark — explains: because dV/dt = 6√t is increasing in t, each subsequent litre arrives faster, so doubling the volume takes less than double the time.
Your response:
Stuck on (d)? Compute t72 and t144 separately, then test "is t144 < 2 · t72?".How did this worksheet feel?
What I'll revisit before next class:
1.1 — ∫ (x³ + 4x² − x)/x² dx (3 marks)
Sample response. (x³ + 4x² − x)/x² = x + 4 − x−1. ∫(x + 4 − 1/x) dx = x²/2 + 4x − ln|x| + C.
Marking notes. 1 mark — correct rewriting as a sum of powers (recognises the 1/x = x−1 term). 1 mark — power-rule integration of the polynomial terms x²/2 + 4x. 1 mark — correctly handles the 1/x term as −ln|x| (with absolute value) and includes + C. Common error: writing the 1/x term as x⁰/0, which is undefined — this should score 2/3 maximum.
1.2 — ∫ (√x + 1/√x)² dx (3 marks)
Sample response. (√x + 1/√x)² = x + 2 + 1/x. ∫(x + 2 + 1/x) dx = x²/2 + 2x + ln|x| + C.
Marking notes. 1 mark — correct expansion (note the cross-term: 2 · √x · 1/√x = 2). 1 mark — integration of x and the constant 2. 1 mark — correctly integrates 1/x as ln|x| + C. Forgetting the cross-term 2 is the most common loss-of-mark.
1.3 — Curve through (2, 3) with dy/dx = 2x − 4/x² (4 marks)
Sample response. Rewrite dy/dx = 2x − 4 x−2. y = ∫(2x − 4 x−2) dx = x² − 4 · x−1/(−1) + C = x² + 4/x + C. At (2, 3): 3 = 4 + 2 + C ⇒ C = −3. y = x² + 4/x − 3.
Marking notes. 1 mark — rewrites 4/x² in power form. 1 mark — correct antiderivative of −4 x−2 as +4/x (sign from the negative exponent is the common slip). 1 mark — uses the point (2, 3) to solve for C. 1 mark — final stated equation. A response with the antiderivative form but wrong sign on the 4/x term should score 2/4.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) V(t) = ∫ 6 t1/2 dt = 6 · (2/3) t3/2 + C = 4 t3/2 + C. V(0) = 0 ⇒ C = 0, so V(t) = 4 t3/2. [1 mark.]
(b) V(9) = 4 · 93/2 = 4 · 27 = 108 L. [1 mark.]
(c) Set 4 t3/2 = 144. Then t3/2 = 36. [1 mark — isolates correctly.]
Hence t = 362/3. Since 361/3 ≈ 3.302, t ≈ 3.302² ≈ 10.90 minutes. [1 mark — final time.]
(d) Time to 72 L: 4 t3/2 = 72 ⇒ t3/2 = 18 ⇒ t72 = 182/3 ≈ 6.87 minutes. [1 mark.]
Test the claim: 2 × t72 ≈ 13.74 minutes, but t144 ≈ 10.90 minutes. Since 10.90 < 13.74, the time to fill 144 L is less than twice the time to fill 72 L — the claim is supported. [1 mark — quantitative comparison and clear conclusion.]
This is because the flow rate dV/dt = 6√t is increasing in t, so each successive litre arrives faster than the previous one; consequently doubling the cumulative volume takes less than double the time. [1 mark — links the result to the monotonicity of dV/dt.]
Total: 7/7.
Band descriptors for marker.
Band 3: Completes (a) and (b) only; little or no progress on (c)(d). ≈ 2 marks.
Band 4: Completes (a)(b)(c) but struggles with the algebra of t2/3 in (c); attempts (d) without a quantitative comparison. ≈ 3-4 marks.
Band 5: All numeric parts correct; (d) gives the comparison but does not explain why the rate's behaviour drives the result. ≈ 5-6 marks.
Band 6: All parts complete, with (d) including both the numeric comparison (10.9 < 13.74) and the interpretation (increasing rate ⇒ later litres arrive faster ⇒ doubling takes less than double the time). 7/7.