Mathematics Advanced • Year 11 • Module 3 • Lesson 15
Module 3 Review
Build mixed fluency across the Module 3 toolkit: differentiation rules, sign analysis, optimisation, definite integrals, areas, volumes, and the trapezoidal rule.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 Name the differentiation rule and state when you use it:
(a) ____________ rule — for differentiating a function of a function (composite).
(b) ____________ rule — for differentiating a product of two functions.
(c) ____________ rule — for differentiating a quotient u/v.
Q1.2 Complete each formula:
(a) ∫ xⁿ dx = ____________ + C (for n ≠ −1).
(b) ∫ from a to b of f(x) dx = ____________ − ____________ , where F′(x) = f(x).
(c) Volume of revolution about x-axis: V = ____________ ∫ from a to b of [____]² dx.
Q1.3 True or false (circle one):
(a) On a closed interval the maximum value of f may occur at an endpoint, not at a stationary point. T / F
(b) The trapezoidal rule is exact for any quadratic. T / F
(c) "Signed area" and "total area" are always the same number. T / F
2. Worked example — find the maximum of f(x) = x³ − 6x² + 9x + 2 on [0, 4]
Follow each line of algebra. Every step has a reason on the right.
Step 1 — Differentiate.
f′(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3)
Reason: power rule, then factorise to find stationary points.
Step 2 — Stationary points.
f′(x) = 0 at x = 1 and x = 3 — both in [0, 4]
Reason: candidates for extreme values.
Step 3 — Evaluate at all candidates AND both endpoints.
f(0) = 2; f(1) = 1 − 6 + 9 + 2 = 6; f(3) = 27 − 54 + 27 + 2 = 2; f(4) = 64 − 96 + 36 + 2 = 6
Reason: on a closed interval, the extreme value can be at an endpoint.
Step 4 — Compare.
Largest value is 6, occurring at x = 1 and x = 4.
Reason: maximum is the largest of all candidate values.
Conclusion. Maximum value of f on [0, 4] is 6.
3. Faded example — fill in the missing steps
Differentiate y = x²·sin(3x) using the product rule (with chain rule inside). 3 marks
Step 1 — Identify u and v:
u = ____________ , v = ____________ .
Step 2 — Differentiate each (use chain rule for v):
u′ = ____________ ; v′ = ____________ × ____________ = ____________ .
Step 3 — Apply product rule y′ = u′v + uv′:
y′ = ____________ + ____________ = ____________ .
4. Graduated practice — mixed Module 3 problems
Identify which tool is needed (differentiation rule, integration, area, volume, trapezoidal rule), then apply it.
Foundation — single-step problems (4 questions)
| Q | Task | Answer |
|---|---|---|
| 4.1 1 | Differentiate y = (2x + 1)⁵. | |
| 4.2 1 | Find f″(x) for f(x) = x⁴ − 3x² + x. | |
| 4.3 1 | Evaluate ∫₁³ (2x + 1) dx. | |
| 4.4 1 | Find the antiderivative of 3x² + 2. |
Standard — multi-step problems (6 questions)
Show enough working that the marker can follow your reasoning.
4.5 Find the stationary points of f(x) = x³ − 3x² and classify each as a local maximum, local minimum, or neither (use the first-derivative sign-change test). 3 marks
4.6 Differentiate y = x² / (x + 1) using the quotient rule. 2 marks
4.7 Find the area enclosed by y = x² and y = 2x. 3 marks
4.8 The region bounded by y = √x, x = 1, x = 4 and the x-axis is rotated about the x-axis. Find the volume of the solid. 3 marks
4.9 Use the trapezoidal rule with 4 strips to approximate ∫₀² x² dx, giving the answer to 2 dp. 2 marks
4.10 A rectangle has perimeter 48 cm. Find the dimensions that maximise the area, and state the maximum area. 3 marks
Extension — multi-tool synthesis (2 questions)
4.11 For f(x) = x³ − 12x: (a) find all intervals on which f is increasing; (b) find the area enclosed between the curve and the x-axis on the interval [0, 2√3]. 4 marks
4.12 The region bounded by y = x and y = x² is rotated about the x-axis. (a) Find the intersection points. (b) Use the washer method to compute the volume of revolution. 3 marks
5. Self-check the easy 3
Tick the first three once you have checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Differentiation rules
(a) Chain. (b) Product. (c) Quotient.
Q1.2 — Formulas
(a) ∫ xⁿ dx = x^(n+1)/(n+1) + C. (b) F(b) − F(a). (c) V = π ∫ from a to b of [f(x)]² dx.
Q1.3 — True/false
(a) T. (b) F — the trapezoidal rule is exact for linear functions, not for quadratics (it overestimates concave-up curves). (c) F — signed area allows cancellation between regions above and below the axis.
Q3 — Faded example: y = x²·sin(3x)
u = x², v = sin(3x). u′ = 2x; v′ = cos(3x) × 3 = 3 cos(3x). y′ = 2x sin(3x) + x² · 3 cos(3x) = 2x sin(3x) + 3x² cos(3x).
Q4.1 — y = (2x + 1)⁵
Chain: y′ = 5(2x + 1)⁴ · 2 = 10(2x + 1)⁴.
Q4.2 — f″ for f(x) = x⁴ − 3x² + x
f′(x) = 4x³ − 6x + 1; f″(x) = 12x² − 6.
Q4.3 — ∫₁³ (2x + 1) dx
[x² + x]₁³ = (9 + 3) − (1 + 1) = 10.
Q4.4 — Antiderivative of 3x² + 2
x³ + 2x + C.
Q4.5 — Stationary points of f(x) = x³ − 3x²
f′(x) = 3x² − 6x = 3x(x − 2) = 0 at x = 0 and x = 2. Sign analysis: f′(−1) = 9 (+); f′(1) = −3 (−); f′(3) = 9 (+). At x = 0, f′ goes + to −, so local maximum (f(0) = 0). At x = 2, f′ goes − to +, so local minimum (f(2) = 8 − 12 = −4).
Q4.6 — Derivative of y = x²/(x + 1)
Quotient: u = x², v = x + 1; u′ = 2x, v′ = 1. y′ = [2x(x + 1) − x²(1)] / (x + 1)² = [2x² + 2x − x²] / (x + 1)² = (x² + 2x) / (x + 1)² = x(x + 2)/(x + 1)².
Q4.7 — Area enclosed by y = x² and y = 2x
Intersections: x² = 2x ⇒ x(x − 2) = 0 ⇒ x = 0, 2. Top is y = 2x. Area = ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3 sq units.
Q4.8 — Volume from rotating y = √x on [1, 4]
V = π ∫₁⁴ x dx = π [x²/2]₁⁴ = π(8 − 1/2) = 15π/2 cubic units.
Q4.9 — Trapezoidal rule on ∫₀² x² dx with n = 4
h = 0.5; y at 0, 0.5, 1, 1.5, 2 = 0, 0.25, 1, 2.25, 4. ≈ (0.5/2)·[0 + 2(0.25 + 1 + 2.25) + 4] = 0.25·[0 + 7 + 4] = 2.75.
Q4.10 — Rectangle perimeter 48, maximise area
Let length l and width w with 2l + 2w = 48, so l = 24 − w. Area A = w(24 − w) = 24w − w². A′(w) = 24 − 2w = 0 ⇒ w = 12; A″(w) = −2 < 0 confirms max. So l = w = 12 cm (a square), max area = 144 cm².
Q4.11 — f(x) = x³ − 12x
(a) f′ = 3x² − 12 = 3(x − 2)(x + 2). f′ = 0 at x = ±2. Sign: f′(−3) = 15 (+); f′(0) = −12 (−); f′(3) = 15 (+). Increasing on (−∞, −2) ∪ (2, ∞).
(b) On [0, 2√3] the curve passes from f(0) = 0 down through a minimum (f(2) = 8 − 24 = −16) back up to f(2√3) = (2√3)³ − 12(2√3) = 24√3 − 24√3 = 0. So the curve is below the axis on (0, 2√3). Signed integral = ∫₀^{2√3} (x³ − 12x) dx = [x⁴/4 − 6x²]₀^{2√3} = (144/4 − 6·12) − 0 = 36 − 72 = −36. Area = |−36| = 36 sq units.
Q4.12 — Region between y = x and y = x² rotated about x-axis
(a) Intersections: x = x² ⇒ x = 0, 1. (b) Outer R = x, inner r = x². V = π ∫₀¹ (x² − x⁴) dx = π [x³/3 − x⁵/5]₀¹ = π(1/3 − 1/5) = 2π/15 cubic units.