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1.9.27 Integration by substitution (reverse chain rule) (A-level only)

1.9.27 Integration by substitution (reverse chain rule) (A-level only)

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Question 24
i.

Find, in simplest form,

∫3(2x−5)4 dx \int \frac{3}{(2x - 5)^4} \, \mathrm{d}x ∫(2x−5)43​dx
[3]
ii.

A population of bacteria grows at a rate R(t)=82t+1\displaystyle R(t) = \frac{8}{2t + 1}R(t)=2t+18​ million per hour, where t t\,t is the time in hours, t≥0t \ge 0t≥0.

Show, by algebraic integration, that the total growth in the population between t=1t = 1t=1 and t=3t = 3t=3 is ln⁡b \ln b\,lnb million, where b b\,b is a rational constant to be found.

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Markscheme

1.9.27 Integration by substitution (reverse chain rule) (A-level only) Questions

  1. A Level
  2. /Maths
  3. /1.9.27 Integration by substitution (reverse chain rule) (A-level only)

36 exam-style questions on OCR (MEI) A Level Maths 1.9.27 Integration by substitution (reverse chain rule) (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.

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