Mathematics Advanced • Year 12 • Module 6 • Lesson 6

Areas Between Curves

Practise HSC-style writing on areas between curves — including a multi-part extended response with marking criteria.

Master · Past-Paper Style

1. Short-answer questions

1.1 Find the area of the region enclosed by the curves y = x² and y = 2x + 3. Show all working. 3 marks Band 3

1.2 The region between y = x³ − x and the x-axis from x = −1 to x = 1 has total area A. Explain, with a sketch or test point, why the integral must be split, then compute A. 3 marks Band 3-4

1.3 A company's marginal revenue function is r(x) = 30 − 0.2x and its marginal cost function is c(x) = 10 + 0.05x dollars per unit, for 0 ≤ x ≤ 100.
(a) Find the production level x* at which marginal revenue equals marginal cost.
(b) Compute the area between the two curves on [0, x*] and interpret what this number represents in business terms. 4 marks Band 4

Stuck on 1.3(b)? The area between marginal revenue and marginal cost represents accumulated net contribution to profit as units are produced.

2. Extended response

2.1 Two curves are given by f(x) = x² and g(x) = 4x − x².
(a) Find all points of intersection of y = f(x) and y = g(x).
(b) Sketch both curves on the same axes for −1 ≤ x ≤ 3, marking the intersections and shading the enclosed region.
(c) Find the area of the region enclosed between the two curves, showing every step (intersection step, top/bottom test, integral, antiderivative, evaluation).
(d) Without re-integrating, explain how the area of the region would change if every y-value were doubled (i.e. the new curves are 2f(x) and 2g(x)). Justify with reference to the integral formula. 7 marks Band 5-6

Explicit marking criteria

Part (a) — 1 mark. Solves f(x) = g(x): x² = 4x − x² ⇒ 2x² − 4x = 0 ⇒ x = 0 or x = 2.

Part (b) — 1 mark. Sketch shows both parabolas with correct orientation (f upward, g downward), correct intersections labelled (0, 0) and (2, 4), and shaded region.

Part (c) — 4 marks.

• 1 mark — explicit top/bottom test on (0, 2): at x = 1, g(1) = 3, f(1) = 1, so g is on top.

• 1 mark — sets up A = ∫₀² ((4x − x²) − x²) dx = ∫₀² (4x − 2x²) dx.

• 1 mark — antiderivative [2x² − (2/3)x³]₀².

• 1 mark — evaluation: (8 − 16/3) − 0 = 8/3 square units, with positive answer flagged correct.

Part (d) — 1 mark. Argues that ∫ (2g − 2f) dx = 2 ∫ (g − f) dx by the constant-multiple rule, so the new area is exactly 2 × 8/3 = 16/3 square units — area scales linearly with vertical scaling.

Your response:

Stuck on (d)? Use the rule ∫ k · h(x) dx = k · ∫ h(x) dx.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — Area between y = x² and y = 2x + 3 (3 marks)

Sample response. Intersections: x² = 2x + 3 ⇒ x² − 2x − 3 = (x − 3)(x + 1) = 0, so x = −1, 3. Test at x = 0: line gives 3, parabola gives 0, so the line is above the parabola on (−1, 3). A = ∫₋₁³ ((2x + 3) − x²) dx = [x² + 3x − x³/3]₋₁³ = (9 + 9 − 9) − (1 − 3 + 1/3) = 9 − (−5/3) = 32/3 square units.

Marking notes. 1 mark — correct intersections. 1 mark — correct top/bottom identification with brief justification. 1 mark — correct evaluation. A response that omits the top/bottom test scores 2/3.

1.2 — Total area for y = x³ − x (3 marks)

Sample response. Roots at x = −1, 0, 1. Test points: at x = −0.5 ⇒ y = −0.125 + 0.5 = 0.375 > 0; at x = 0.5 ⇒ y = 0.125 − 0.5 = −0.375 < 0. The curve is above the x-axis on [−1, 0] and below on [0, 1], so the integrand changes sign and the single integral ∫₋₁¹ (x³ − x) dx = 0 by symmetry, which would misrepresent the geometric area. Split: A = ∫₋₁⁰ (x³ − x) dx + ∫₀¹ −(x³ − x) dx = [x&sup4;/4 − x²/2]₋₁⁰ + [−x&sup4;/4 + x²/2]₀¹ = (0 − (−1/4)) + (1/4) = 1/2 square units.

Marking notes. 1 mark — explicit sign test and statement that the integral must be split. 1 mark — correct split integrals set up with the correct sign on the [0, 1] piece. 1 mark — correct numerical answer 1/2. Common error: just computing ∫₋₁¹ (x³ − x) dx = 0 and claiming area = 0.

1.3 — Marginal revenue/cost (4 marks)

(a) Sample response. 30 − 0.2x = 10 + 0.05x ⇒ 20 = 0.25x ⇒ x* = 80 units.

(b) Sample response. On (0, 80), marginal revenue > marginal cost (test x = 0: 30 > 10). A = ∫₀⁸⁰ ((30 − 0.2x) − (10 + 0.05x)) dx = ∫₀⁸⁰ (20 − 0.25x) dx = [20x − 0.125x²]₀⁸⁰ = 1600 − 800 = $800.
Interpretation: this is the total net contribution to profit generated by producing the first 80 units — each unit's marginal revenue exceeds its marginal cost, and the cumulative profit added is the area between the curves up to the point where they meet.

Marking notes. (a) 1 mark — correct x*. (b) 1 mark — top/bottom identified; 1 mark — correct evaluated area; 1 mark — interpretation references "total profit" or "net contribution" generated by producing 0 to 80 units, not just "the area".

2.1 — Extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

(a) Solve x² = 4x − x² ⇒ 2x² − 4x = 0 ⇒ 2x(x − 2) = 0, so x = 0 and x = 2. [1 mark — intersections.]

(b) Sketch: y = x² is an upward parabola through (0, 0) with vertex at origin; y = 4x − x² = −(x − 2)² + 4 is a downward parabola with vertex (2, 4) and roots at x = 0 and x = 4. The two parabolas cross at (0, 0) and (2, 4); the enclosed region lies on 0 ≤ x ≤ 2 with the downward parabola on top and is shaded. [1 mark — correct sketch with intersections labelled and region shaded.]

(c) Top/bottom test. At x = 1: g(1) = 4 − 1 = 3, f(1) = 1. Since g > f on (0, 2), g is on top. [1 mark — test.]

Set up: A = ∫₀² ((4x − x²) − x²) dx = ∫₀² (4x − 2x²) dx. [1 mark — setup.]

Antiderivative: [2x² − (2/3)x³]₀². [1 mark — antiderivative.]

Evaluation: (2 · 4 − (2/3) · 8) − 0 = 8 − 16/3 = 24/3 − 16/3 = 8/3 square units. [1 mark — value, positive.]

(d) If each curve is scaled to 2f(x) and 2g(x), the integrand becomes 2(g(x) − f(x)). By the constant-multiple rule, ∫₀² 2(g − f) dx = 2 ∫₀² (g − f) dx = 2 · 8/3 = 16/3 square units. So vertical scaling by factor 2 exactly doubles the area, because the horizontal extent is unchanged but every strip's height doubles. [1 mark — references constant-multiple rule and correct factor of 2.]

Total: 7/7.

Band descriptors for marker.

Band 3: Finds intersections and writes a partial setup but does not test which curve is on top, or evaluates with sign error. ≈ 2-3 marks.

Band 4: Complete part (c) up to evaluation with one minor slip; sketch present but unlabelled; no (d). ≈ 4-5 marks.

Band 5: Full parts (a)-(c) with correct value; attempts (d) by recomputing instead of using the linearity rule, or argues informally. ≈ 6 marks.

Band 6: Full proof in all parts, explicitly invokes the constant-multiple rule in (d), labels the sketch, and writes positive area with units. 7/7.