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4PM1/01R Jun 2023, Question 9: Definite Integrals and Areas

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1R · Calculator · 14 marks

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The question, in short

Expand a binomial with a fractional negative index, state when it converges, build a series for a rational function with parameter k, determine k from a given x³ coefficient, then integrate the series to estimate a definite integral.

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

Part (a) · 3 marks

Write (1 + 2x)^(−1/3) as a series in ascending powers of x through the x³ term, giving each coefficient as a simplified fraction.

  1. 1(1 + 2x)^(−1/3) = 1 + (−1/3)(2x) + (−1/3)(−4/3)/2! (2x)² + (−1/3)(−4/3)(−7/3)/3! (2x)³
  2. 2= 1 − (2/3)x + (2/9)(4x²) − (14/81)(8x³)
  3. 3= 1 − (2/3)x + (8/9)x² − (112/81)x³

Answer1 − (2/3)x + (8/9)x² − (112/81)x³

Part (b) · 1 mark

State the interval of x values over which the series expansion converges.

  1. 1Valid for |2x| < 1
  2. 2So −1/2 < x < 1/2

Answer−1/2 < x ≤ 1/2 (accept −1/2 < x < 1/2 or |x| < 1/2)

Part (c) · 3 marks

Express (2 + kx²)/(1 + 2x)^(1/3) as a series up to the x³ term, with coefficients involving k.

  1. 1f(x) = (2 + kx²)(1 − (2/3)x + (8/9)x² − (112/81)x³ + …)
  2. 2= 2 − (4/3)x + (16/9)x² − (224/81)x³ + kx² − (2k/3)x³ + …
  3. 3= 2 − (4/3)x + (16/9 + k)x² − (224/81 + 2k/3)x³

Answer2 − (4/3)x + (16/9 + k)x² − (224/81 + 2k/3)x³

Part (d) · 2 marks

Use the stated x³ coefficient of −8/3 to work out the exact value of k.

  1. 1−224/81 − 2k/3 = −8/3
  2. 2−2k/3 = −8/3 + 224/81 = 8/81
  3. 3k = −4/27

Answerk = −4/27

Part (e) · 5 marks

Integrate the series term by term from 0.1 to 0.2 to approximate the integral of f(x), correct to 4 decimal places.

  1. 1With k = −4/27: f(x) ≈ 2 − (4/3)x + (44/27)x² − (8/3)x³
  2. 2∫ f(x) dx = 2x − (2/3)x² + (44/81)x³ − (2/3)x⁴
  3. 3Substitute 0.2 and 0.1 and subtract
  4. 4= 0.1828

Answer0.1828

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