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TMUA 2019 · Paper 1 · Question 16 of 20

TMUA 2019 Paper 1 Question 16

Differentiation and integration — Definite integrals · substitution and additivity. Try it first; the answer and a full worked solution are below.

TMUA 2019 · Paper 1Differentiation and integrationDefinite integrals · substitution and additivity8 options
Given that 2∫01 f(x) dx + 5∫12 f(x) dx = 14 and ∫01 f(x+1) dx = 6 find the value of ∫02 f(x) dx
  1. A−8
  2. B−4
  3. C−2
  4. D2
  5. E4
  6. F295
  7. G325
  8. H14
Show the answer and worked solution
answer · C
  1. A−8
  2. B−4
  3. C−2
  4. D2
  5. E4
  6. F295
  7. G325
  8. H14
The second condition is the key: substituting u = x + 1 shifts the limits, so ∫01 f(x+1) dx = ∫12 f(u) du = 6. Writing A = ∫01 f and B = ∫12 f = 6, the first condition gives 2A + 30 = 14, so A = −8. Then ∫02 f = A + B = −8 + 6 = −2.