Lemmara logoLemmaraWorked solution

4PM1/02R Jun 2022, Question 8: Definite Integrals and Areas

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2R · Calculator · 11 marks

اقرأ هذا الحل مشروحاً بالعربي ←

The question, in short

Two curves, y = e^(3x) − 1 and y = 9 − 9e^(−3x), cross at O and A; find the x value at A and the exact shaded area between them.

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

Part (a) · 5 marks

Derive the quadratic in e^(3x) satisfied by A's x coordinate, then deduce that this coordinate equals (1/3) ln 9.

  1. 1e^(3x) − 1 = 9 − 9e^(−3x)
  2. 2Multiply by e^(3x): (e^(3x))² − e^(3x) = 9e^(3x) − 9
  3. 3(e^(3x))² − 10e^(3x) + 9 = 0
  4. 4(e^(3x) − 1)(e^(3x) − 9) = 0
  5. 5e^(3x) = 1 gives O; at A, e^(3x) = 9
  6. 6x = (1/3) ln 9

Answer(e^(3x))² − 10e^(3x) + 9 = 0, giving e^(3x) = 9 and x = (1/3) ln 9 — shown

Part (b) · 6 marks

Using calculus, calculate the exact area of the finite region enclosed by the two curves.

  1. 1Area = ∫ from 0 to (1/3) ln 9 of [(9 − 9e^(−3x)) − (e^(3x) − 1)] dx
  2. 2= ∫ (10 − 9e^(−3x) − e^(3x)) dx
  3. 3= [10x + 3e^(−3x) − e^(3x)/3] from 0 to (1/3) ln 9
  4. 4Upper: (10/3) ln 9 + 3/9 − 9/3 = (10/3) ln 9 + 1/3 − 3
  5. 5Lower: 0 + 3 − 1/3 = 8/3
  6. 6Area = (10/3) ln 9 − 8/3 − 8/3 = (10/3) ln 9 − 16/3

Answer(10/3) ln 9 − 16/3

Stuck on a step? Read the free lesson: Definite Integrals and Areas

Practise more definite integrals and areas questions

Real 4PM1 past-paper questions, marked on your working like the exam, with a worked solution for every part.

Practise free on Lemmara

More past-paper questions on definite integrals and areas

Also searched as: 4PM1 2R June 2022 · 4PM1/02R · 4pm1 jun22 2r · question 8.
Lemmara is an independent study tool, not affiliated with or endorsed by Pearson Edexcel. The worked solutions are Lemmara’s own.