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.