4PM1/02 Jun 2021, Question 11: Volumes of Revolution
Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 2 · Calculator · 8 marks
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The question, in short
Given region R between y = eˣ, y = 4e^(−x), x = a and the axes, rotated about the x-axis, the volume equals k − 8πe^(−4). Use integration to find a and k.
Summarised in our own words. Use the official paper for the exact wording and the diagram.
Solution · 8 marks
- 1Curves meet where eˣ = 4e^(−x) ⇒ e^(2x) = 4 ⇒ x = ln 2
- 2V = π∫ from 0 to ln 2 of e^(2x) dx + π∫ from ln 2 to a of 16e^(−2x) dx
- 3= π[e^(2x)/2] from 0 to ln 2 + π[−8e^(−2x)] from ln 2 to a
- 4= π(2 − 1/2) + π(−8e^(−2a) + 2) = 7π/2 − 8πe^(−2a)
- 5Compare with k − 8πe^(−4): a = 2, k = 7π/2
Answera = 2, k = 7π/2
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