Mathematics Advanced • Year 11 • Module 3 • Lesson 12
Areas
HSC-style writing on definite integrals — including an extended response on a region partly above and partly below the x-axis.
1. Short-answer questions
1.1 Evaluate ∫₁⁴ (2x − 3) dx. 2 marks Band 3
1.2 Find the area enclosed by the curves y = x² and y = 2 − x². 4 marks Band 4
1.3 The region bounded by y = x² − x − 2 and the x-axis lies between the curve's x-intercepts. Find the area of this region. 3 marks Band 4
Stuck on 1.3? Factorise to find intercepts. The region is below the x-axis on the enclosed interval, so the integral is negative and the area is its absolute value.2. Extended response
2.1 Consider the curve y = x² − 4x + 3.
(a) Find the x-intercepts of the curve.
(b) Hence find the total area bounded by the curve, the x-axis, x = 0 and x = 4. (You must split the integral at every x-intercept inside [0, 4] and take absolute values where the curve dips below the axis.)
(c) Show by direct integration that ∫₀⁴ (x² − 4x + 3) dx = 4/3. Explain in one sentence the difference between this signed value and your answer to (b).
(d) On the same diagram, sketch y = x² − 4x + 3 from x = 0 to x = 4. Shade the three sub-regions (above-axis from 0 to 1, below-axis from 1 to 3, above-axis from 3 to 4) and label each with its area. 8 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correctly factorises (x − 1)(x − 3) and states intercepts x = 1 and x = 3.
Part (b) — 3 marks
• 1 mark — correctly identifies the need to split at x = 1 and x = 3 inside [0, 4].
• 1 mark — correctly computes each of ∫₀¹, ∫₁³, ∫₃⁴ using F(x) = x³/3 − 2x² + 3x.
• 1 mark — takes absolute value of negative piece and sums: total area = 4/3 + 4/3 + 4/3 = 4 sq units.
Part (c) — 2 marks
• 1 mark — correctly evaluates [x³/3 − 2x² + 3x]₀⁴ = 64/3 − 32 + 12 = 4/3.
• 1 mark — explains that the signed integral (4/3) is the net (positive minus negative) signed area, while total area (4) is the sum of absolute areas; the two differ because part of the region is below the axis.
Part (d) — 2 marks
• 1 mark — sketch shows correct parabola, x-intercepts at 1 and 3, minimum at x = 2 (y = −1), with the three sub-regions correctly distinguished.
• 1 mark — each sub-region labelled with its correct area (4/3, 4/3 and 4/3) and the below-axis region clearly identified.
Your response:
Stuck on (b)? After finding the antiderivative F(x), compute F at each of 0, 1, 3, 4 and take differences for each subinterval.How did this worksheet feel?
What I'll revisit before next class:
1.1 — ∫₁⁴ (2x − 3) dx (2 marks)
Sample response. Antiderivative: x² − 3x. Evaluate: [x² − 3x]₁⁴ = (16 − 12) − (1 − 3) = 4 − (−2) = 6.
Marking notes. 1 mark — correct antiderivative. 1 mark — correct evaluation including sign handling on F(1) = −2. Common error: writing F(1) = 1 + 3 = 4 (sign error in −3x term).
1.2 — Area enclosed by y = x² and y = 2 − x² (4 marks)
Sample response. Intersections: x² = 2 − x² ⇒ 2x² = 2 ⇒ x = ±1. Top curve is y = 2 − x² (test x = 0: 2 > 0). Area = ∫₋₁¹ ((2 − x²) − x²) dx = ∫₋₁¹ (2 − 2x²) dx = [2x − 2x³/3]₋₁¹ = (2 − 2/3) − (−2 + 2/3) = 4/3 − (−4/3) = 8/3 sq units.
Marking notes. 1 mark — correct intersections. 1 mark — correctly identifies the top curve. 1 mark — correct integrand (top − bottom) and antiderivative. 1 mark — correct evaluation and final answer. Common errors: subtracting the wrong way (giving −8/3, which a student should recognise as the absolute value and report as 8/3); forgetting the symmetric trick and integrating only over [0, 1].
1.3 — Area enclosed by y = x² − x − 2 and the x-axis (3 marks)
Sample response. Factor: y = (x − 2)(x + 1), so x-intercepts at x = −1 and x = 2. The parabola opens upward, so the enclosed region (between the intercepts) lies below the axis. Compute ∫₋₁² (x² − x − 2) dx = [x³/3 − x²/2 − 2x]₋₁² = (8/3 − 2 − 4) − (−1/3 − 1/2 + 2) = (8/3 − 6) − (−1/3 + 3/2) = (8/3 − 18/3) − (−2/6 + 9/6) = −10/3 − 7/6 = −20/6 − 7/6 = −27/6 = −9/2. Area = |−9/2| = 9/2 sq units.
Marking notes. 1 mark — correct intercepts. 1 mark — correct antiderivative and signed integral. 1 mark — takes absolute value and reports area as 9/2. Common error: leaving the answer as −9/2 (area cannot be negative).
2.1 — Extended response (8 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a). y = x² − 4x + 3 = (x − 1)(x − 3), so x-intercepts at x = 1 and x = 3. [1 mark.]
Part (b). Antiderivative F(x) = x³/3 − 2x² + 3x. F(0) = 0; F(1) = 1/3 − 2 + 3 = 4/3; F(3) = 9 − 18 + 9 = 0; F(4) = 64/3 − 32 + 12 = 4/3.
| Interval | Sign of y | Integral F(b) − F(a) | |Integral| = area |
|---|---|---|---|
| [0, 1] | positive | 4/3 − 0 = 4/3 | 4/3 |
| [1, 3] | negative | 0 − 4/3 = −4/3 | 4/3 |
| [3, 4] | positive | 4/3 − 0 = 4/3 | 4/3 |
[1 mark — splits at 1 and 3; 1 mark — correct integrals on each piece; 1 mark — takes absolute value of the negative piece and sums.] Total area = 4/3 + 4/3 + 4/3 = 4 square units.
Part (c). ∫₀⁴ (x² − 4x + 3) dx = [x³/3 − 2x² + 3x]₀⁴ = (64/3 − 32 + 12) − 0 = 64/3 − 20 = 64/3 − 60/3 = 4/3. [1 mark.] The signed integral 4/3 is the algebraic sum: the positive areas above the axis (4/3 + 4/3 = 8/3) minus the negative area below the axis (4/3), giving 8/3 − 4/3 = 4/3. Total area (part b) sums the absolute values without cancellation. The values differ because part of the region lies below the x-axis, so signed and unsigned contributions partly cancel in the signed integral but not in the total area. [1 mark.]
Part (d). Sketch description: parabola opening upward, vertex at (2, −1), x-intercepts at (1, 0) and (3, 0), y-intercept at (0, 3), passing through (4, 3).
y |
3 |• •
| \ /
| \ shaded 1 /
0 |____\(1)________(3)/_____ x
| \ shaded /
| \ −4/3 /
−1| \ ____ /
vertex (2,−1)
Shading: region from x = 0 to x = 1 lies above the axis (area = 4/3); region from x = 1 to x = 3 lies below the axis (area = 4/3, shown shaded under the axis); region from x = 3 to x = 4 lies above the axis (area = 4/3). [1 mark — correct shape and intercepts and three sub-regions distinguished; 1 mark — each region labelled with its area.]
Total: 8/8.
Band descriptors for marker.
Band 3: Finds the antiderivative and computes the signed integral 4/3 only; does not realise the curve crosses the axis inside [0, 4]. Total area incorrectly reported as 4/3. ≈ 3-4 marks.
Band 4: Splits at x-intercepts but forgets to take absolute value of the middle piece, producing 4/3 + (−4/3) + 4/3 = 4/3 (same as signed). Sketch correct but areas not labelled. ≈ 5-6 marks.
Band 5: Completes (a)-(c) correctly; sketch is qualitatively right but does not differentiate the shading per sub-region or omits the area labels. ≈ 6-7 marks.
Band 6: All four parts complete, total area is 4 sq units, signed integral is 4/3, the difference is explicitly explained, and the sketch clearly shows three sub-regions each labelled 4/3. 8/8.