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4PM1/02R Jun 2025, Question 11: Definite Integrals and Areas

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 12 marks

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The question, in short

Curve C is y = 3e^(x/3) with point A(2, p) on it; line n is the normal at A meeting the x-axis at B. Find p exactly, then the area of shaded region R.

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

Part (a) · 1 mark

Find the exact value of p for point A(2, p) on y = 3e^(x/3).

  1. 1p = 3e^(2/3)

Answerp = 3e^(2/3)

Part (b) · 11 marks

Using algebraic integration, find the area of R bounded by C, normal n, the x-axis and y-axis, giving it as (W/V)e² + We^(2/3) − W.

  1. 1dy/dx = e^(x/3); at x = 2 the gradient is e^(2/3)
  2. 2Normal gradient = −e^(−2/3)
  3. 3n: y − 3e^(2/3) = −e^(−2/3)(x − 2)
  4. 4y = 0 gives x = 2 + 3e^(4/3), so B = (2 + 3e^(4/3), 0)
  5. 5Triangle under n: (1/2)(3e^(2/3))(3e^(4/3)) = (9/2)e²
  6. 6∫ from 0 to 2 of 3e^(x/3) dx = [9e^(x/3)] from 0 to 2 = 9e^(2/3) − 9
  7. 7Area R = (9/2)e² + 9e^(2/3) − 9

AnswerArea = (9/2)e² + 9e^(2/3) − 9

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