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4PM1/01 Jun 2020, Question 7: Definite Integrals and Areas

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 13 marks

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The question, in short

For f(x) = x² − 9x + 14, complete the square, find the minimum and where it occurs, intersect the curve with y = x + 5, and integrate to find the enclosed area.

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Part (a) · 2 marks

Rewrite f(x) in the form (x + a)² + b and state the constants a and b.

  1. 1x² − 9x + 14 = (x − 9/2)² − 81/4 + 14
  2. 2= (x − 9/2)² − 25/4
  3. 3a = −9/2, b = −25/4

Answera = −9/2, b = −25/4

Part (b) · 2 marks

Using your completed square, give the minimum value of f(x) and the x value where it occurs.

  1. 1(x − 9/2)² ≥ 0, so the minimum value is −25/4
  2. 2It occurs when x − 9/2 = 0, i.e. x = 9/2

Answerminimum f(x) = −25/4, at x = 9/2

Part (c) · 4 marks

Solve algebraically for the coordinates where the curve y = f(x) meets the line y = x + 5.

  1. 1x + 5 = x² − 9x + 14
  2. 2x² − 10x + 9 = 0
  3. 3(x − 9)(x − 1) = 0 ⇒ x = 1 or 9
  4. 4y = 6 or 14: (1, 6) and (9, 14)

Answer(1, 6) and (9, 14)

Part (d) · 5 marks

Use algebraic integration to calculate the exact area of the finite region between the curve and the line.

  1. 1Area = ∫ from 1 to 9 of [(x + 5) − (x² − 9x + 14)] dx
  2. 2= ∫ from 1 to 9 of (−x² + 10x − 9) dx
  3. 3= [−x³/3 + 5x² − 9x] from 1 to 9
  4. 4= (−243 + 405 − 81) − (−1/3 + 5 − 9)
  5. 5= 81 + 13/3 = 256/3

Answer256/3 (= 85⅓)

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