0606/11 May/Jun 2025, Question 9: Definite Integrals and Areas
Cambridge IGCSE Additional Mathematics 0606 · Paper 11 (Non-calculator) · Non-calculator · 10 marks
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The question, in short
Find the area enclosed by the curve y = √(4x + 1), its tangent at the point A where x = 2, and the x-axis.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Solution · 10 marks
- 1At x = 2, y = √9 = 3
- 2dy/dx = (1/2)(4x + 1)^(−1/2) × 4 = 2/√(4x + 1); at x = 2, gradient = 2/3
- 3Tangent: y − 3 = (2/3)(x − 2); meets x-axis at x = −5/2
- 4Triangle area = (1/2) × (5/2 + 2) × 3 = 27/4
- 5Curve meets x-axis at x = −1/4
- 6∫ from −1/4 to 2 of √(4x + 1) dx = [(4x + 1)^(3/2)/6] from −1/4 to 2 = 27/6 − 0 = 9/2
- 7Shaded area = 27/4 − 9/2 = 9/4
Answer9/4
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Also searched as: 0606/11/M/J/25 · 0606_s25_qp_11 · 0606 s25 11 · question 9.
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