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WMA12/01 Jan 2022, Question 9: Definite Integrals and Areas

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 10 marks

The question, in short

Curve C is y = x − x² and line l is y = mx with 0 < m < 1, meeting at O and P; regions R₁ and R₂ are shaded. Find P, the area of R₁, and m when the two areas are equal.

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

Part (a) · 2 marks

Find the coordinates of the second intersection point P of the line and the curve, giving them in terms of m.

  1. 1mx = x − x², so x(1 − m − x) = 0
  2. 2At P (x ≠ 0): x = 1 − m
  3. 3y = m(1 − m), so P is (1 − m, m(1 − m))

AnswerP(1 − m, m(1 − m))

Part (b) · 5 marks

Show that the area of the region R₁ between the curve and the line equals (1 − m)³ ÷ 6.

  1. 1Area of R₁ = ∫₀^(1−m) (x − x² − mx) dx = ∫₀^(1−m) ((1 − m)x − x²) dx
  2. 2= [(1 − m)x²/2 − x³/3] from 0 to 1 − m
  3. 3= (1 − m)(1 − m)²/2 − (1 − m)³/3
  4. 4= (1 − m)³(1/2 − 1/3) = (1 − m)³/6

AnswerShown: area of R₁ = ∫₀^(1−m) (x − x² − mx) dx = (1 − m)³/6

Part (c) · 3 marks

Given that the areas of R₁ and R₂ are equal, find the exact value of m.

  1. 1Area of R₁ + R₂ = ∫₀¹ (x − x²) dx = 1/2 − 1/3 = 1/6
  2. 2R₁ = R₂, so R₁ = 1/12
  3. 3(1 − m)³/6 = 1/12, so (1 − m)³ = 1/2
  4. 41 − m = 1/∛2, so m = 1 − 1/∛2

Answerm = 1 − 1/∛2

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Also searched as: WMA12 January 2022 · WMA12/01 · wma12 jan22 · IAS Maths Pure Mathematics 2 January 2022 · question 9.
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