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4PM1/02 Nov 2023, Question 5: Definite Integrals and Areas

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 15 marks

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The question, in short

Given curve S: y = px² + qx + r through A(−2, 0), B(6, 0), P(4, −6), find its equation, the normal l at P, and the shaded area between S and l.

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

Part (a) · 3 marks

Use the three given points on S to determine p, q and r, showing the equation becomes y = x²/2 − 2x − 6.

  1. 1Roots −2 and 6: y = Q(x − 6)(x + 2)
  2. 2At (4, −6): −6 = Q(−2)(6), so Q = 1/2
  3. 3y = (1/2)(x² − 4x − 12) = x²/2 − 2x − 6

Answery = x²/2 − 2x − 6 shown

Part (b) · 5 marks

Find the gradient of S at P, then the perpendicular gradient, and show the normal l has equation 2y + x + 8 = 0.

  1. 1dy/dx = x − 2
  2. 2At x = 4: dy/dx = 2
  3. 3Normal gradient = −1/2
  4. 4y + 6 = −(1/2)(x − 4)
  5. 5y = −x/2 − 4, so 2y + x + 8 = 0

Answer2y + x + 8 = 0 shown

Part (c) · 7 marks

Set up and evaluate an algebraic integral for the region bounded by S and l, giving the exact shaded area.

  1. 1x²/2 − 2x − 6 = −x/2 − 4, so x²/2 − (3/2)x − 2 = 0
  2. 2(x − 4)(x + 1) = 0, x = −1, 4
  3. 3Area = ∫ from −1 to 4 of [(−x/2 − 4) − (x²/2 − 2x − 6)] dx
  4. 4= ∫ from −1 to 4 of (−x²/2 + (3/2)x + 2) dx
  5. 5= [−x³/6 + (3/4)x² + 2x] from −1 to 4 = 125/12

Answer125/12

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