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0606/12 Feb/Mar 2025, Question 7: Definite Integrals and Areas

Cambridge IGCSE Additional Mathematics 0606 · Paper 12 (Non-calculator) · Non-calculator · 10 marks

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The question, in short

A straight line meets a reciprocal-type curve at A: find A, then integrate to get the exact area the two enclose with the y-axis.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 4 marks

Solve the curve and line equations together for A.

  1. 15/(x + 1) + 2 = 2x + 1
  2. 25 = (2x − 1)(x + 1)
  3. 32x² + x − 6 = 0
  4. 4(x + 2)(2x − 3) = 0
  5. 5x = 3/2 (A is in the first quadrant), y = 2(3/2) + 1 = 4

Answer(3/2, 4)

Part (b) · 6 marks

Integrate to get the exact area of the region next to the y-axis.

  1. 1Area = ∫ from 0 to 3/2 of (5/(x + 1) + 2) dx − area under the line
  2. 2Area under the line (trapezium) = (1/2)(1 + 4)(3/2) = 15/4
  3. 3∫ (5/(x + 1) + 2) dx = 5 ln(x + 1) + 2x
  4. 4[5 ln(x + 1) + 2x] from 0 to 3/2 = 5 ln(5/2) + 3
  5. 5Shaded area = 5 ln(5/2) + 3 − 15/4 = 5 ln(5/2) − 3/4

Answer5 ln(5/2) − 3/4

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Also searched as: 0606/12/F/M/25 · 0606_m25_qp_12 · 0606 m25 12 · question 7.
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