Mathematics Advanced • Year 11 • Module 3 • Lesson 12
Areas
Build procedural fluency in evaluating definite integrals and using them to measure areas — including regions partly below the x-axis and regions between two curves.
1. Quick recall
Answer each question in the space provided. 1 mark each
Q1.1 State the Fundamental Theorem of Calculus by completing the blank:
∫ from a to b of f(x) dx = ______________ − ______________, where F′(x) = f(x).
Q1.2 True or false (circle one):
(a) A definite integral always gives a positive number. T / F
(b) For the area between two curves on [a, b], you integrate (top − bottom). T / F
(c) If a region lies entirely above the x-axis, the definite integral equals the area. T / F
Q1.3 Find an antiderivative (with +C):
(a) ∫ 3x² dx = ____________ (b) ∫ (4x − 1) dx = ____________ (c) ∫ 1 dx = ____________
2. Worked example — ∫₁³ (2x + 1) dx
Follow each line of algebra. Every step has a reason on the right.
Step 1 — Find the antiderivative.
∫ (2x + 1) dx = x² + x (+ C, omitted for definite integrals)
Reason: power rule term-by-term.
Step 2 — Evaluate at the upper limit and subtract evaluation at the lower limit.
[x² + x]₁³ = (3² + 3) − (1² + 1) = 12 − 2 = 10
Reason: Fundamental Theorem of Calculus: F(b) − F(a).
Step 3 — Interpret.
y = 2x + 1 > 0 on [1, 3], so signed area = total area.
Reason: when the function is positive on the interval, the definite integral equals the area between the curve and the axis.
Conclusion. The area is 10 square units.
3. Faded example — fill in the missing steps
Find the area enclosed by y = x² and y = x. 3 marks
Step 1 — Intersections.
x² = x ⇒ x² − x = 0 ⇒ x(x − ____) = 0 ⇒ x = ____ or x = ____.
Step 2 — Which curve is on top?
Test x = 0.5: y = x gives 0.5; y = x² gives 0.25. Top curve is y = ____ .
Step 3 — Integrate (top − bottom).
Area = ∫₀¹ (____________) dx = [____________ − ____________]₀¹
= ____________ − ____________ = ____________ square units.
4. Graduated practice — definite integrals and areas
Show the antiderivative, then F(b) − F(a).
Foundation — clean integrals (4 questions)
| Q | Integral | Value |
|---|---|---|
| 4.1 1 | ∫₀² 3x² dx | |
| 4.2 1 | ∫₁⁴ (2x − 1) dx | |
| 4.3 1 | ∫₀³ 4 dx | |
| 4.4 1 | ∫₋₁¹ (x² + 1) dx |
Standard — areas and split integrals (6 questions)
Split at x-intercepts when the curve crosses the x-axis; take absolute values for total area.
4.5 Find the area bounded by y = x², the x-axis, x = 1 and x = 3. 2 marks
4.6 Evaluate ∫₀² (3x² + 2x) dx. 2 marks
4.7 Find the total area between y = x² − 1 and the x-axis from x = 0 to x = 2. (Hint: the curve crosses the x-axis at x = 1.) 3 marks
4.8 Find the area enclosed by y = x² and y = 2x. 2 marks
4.9 Find the area enclosed by y = 4 − x² and the x-axis. 2 marks
4.10 Find the area between y = x³ and y = x for 0 ≤ x ≤ 1. 2 marks
Extension — multi-step regions (2 questions)
4.11 Find the total area bounded by y = x² − 4x + 3, the x-axis, x = 0 and x = 4. Use the fact that y has x-intercepts at x = 1 and x = 3. 3 marks
4.12 Find the area enclosed by the curves y = x² and y = 2 − x². 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 — FTC
∫ from a to b of f(x) dx = F(b) − F(a), where F′(x) = f(x).
Q1.2 — True/false
(a) F — if the function is negative on the interval, the definite integral is negative. (b) T. (c) T.
Q1.3 — Antiderivatives
(a) x³ + C. (b) 2x² − x + C. (c) x + C.
Q3 — Faded example: area enclosed by y = x² and y = x
x² = x ⇒ x(x − 1) = 0, so x = 0 or x = 1. Top curve is y = x (since 0.5 > 0.25). Area = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6 sq units.
Q4.1 — ∫₀² 3x² dx
[x³]₀² = 8 − 0 = 8.
Q4.2 — ∫₁⁴ (2x − 1) dx
[x² − x]₁⁴ = (16 − 4) − (1 − 1) = 12.
Q4.3 — ∫₀³ 4 dx
[4x]₀³ = 12 − 0 = 12.
Q4.4 — ∫₋₁¹ (x² + 1) dx
[x³/3 + x]₋₁¹ = (1/3 + 1) − (−1/3 − 1) = 4/3 − (−4/3) = 8/3.
Q4.5 — Area under y = x² from x = 1 to x = 3
y > 0 on [1, 3], so area = ∫₁³ x² dx = [x³/3]₁³ = 9 − 1/3 = 26/3 sq units.
Q4.6 — ∫₀² (3x² + 2x) dx
[x³ + x²]₀² = (8 + 4) − 0 = 12.
Q4.7 — Total area between y = x² − 1 and the x-axis from 0 to 2
The curve crosses the x-axis at x = 1. Split: ∫₀¹ (x² − 1) dx = [x³/3 − x]₀¹ = 1/3 − 1 = −2/3 (below axis). ∫₁² (x² − 1) dx = [x³/3 − x]₁² = (8/3 − 2) − (1/3 − 1) = 2/3 − (−2/3) = 4/3 (above axis). Total area = |−2/3| + 4/3 = 2/3 + 4/3 = 2 sq units.
Q4.8 — Area enclosed by y = x² and y = 2x
x² = 2x ⇒ x(x − 2) = 0, so x = 0, 2. On (0, 2), 2x > x² (test x = 1: 2 > 1). Area = ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3 sq units.
Q4.9 — Area enclosed by y = 4 − x² and the x-axis
x-intercepts: x = ±2. y ≥ 0 between them. Area = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 16/3 − (−16/3) = 32/3 sq units.
Q4.10 — Area between y = x and y = x³ on [0, 1]
On (0, 1), x > x³ (test x = 0.5: 0.5 > 0.125). Area = ∫₀¹ (x − x³) dx = [x²/2 − x⁴/4]₀¹ = 1/2 − 1/4 = 1/4 sq unit.
Q4.11 — Total area under y = x² − 4x + 3 from 0 to 4
y = (x − 1)(x − 3), so zeros at x = 1, 3. Split into [0, 1], [1, 3], [3, 4]. 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 = 64/3 − 20 = 4/3. ∫₀¹ = 4/3 (above); ∫₁³ = 0 − 4/3 = −4/3 (below); ∫₃⁴ = 4/3 − 0 = 4/3 (above). Total area = 4/3 + 4/3 + 4/3 = 4 sq units.
Q4.12 — Area enclosed by y = x² and y = 2 − x²
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.