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0606/13 May/Jun 2025, Question 8: Definite Integrals and Areas

Cambridge IGCSE Additional Mathematics 0606 · Paper 13 (Non-calculator) · Non-calculator · 8 marks

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The question, in short

For the curve y = 3/(x + 1) − (x − 2)/x, points A and B have x = 1 and x = 2; find their y-values and the shaded area between chord AB and the curve.

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

Part (a) · 1 mark

Work out the y-value on the curve at x = 1 and at x = 2.

  1. 1At x = 1: y = 3/2 − (−1)/1 = 5/2
  2. 2At x = 2: y = 3/3 − 0/2 = 1

AnswerA: y = 5/2, B: y = 1

Part (b) · 7 marks

Determine the area trapped between the straight line AB and the curve, and express it as a/4 − ln(b/2) with integers a and b.

  1. 1Area of trapezium under AB = (1/2)(5/2 + 1) × 1 = 7/4
  2. 2y = 3/(x + 1) − 1 + 2/x
  3. 3∫ from 1 to 2 of (3/(x + 1) − 1 + 2/x) dx = [3ln(x + 1) − x + 2ln x] from 1 to 2
  4. 4= (3ln 3 − 2 + 2ln 2) − (3ln 2 − 1) = ln(27/2) − 1
  5. 5Shaded area = 7/4 − (ln(27/2) − 1) = 11/4 − ln(27/2)
  6. 6So a = 11, b = 27

AnswerArea = 11/4 − ln(27/2) shown, a = 11, b = 27

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Also searched as: 0606/13/M/J/25 · 0606_s25_qp_13 · 0606 s25 13 · question 8.
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