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TMUA 2023 · Paper 1 · Question 3 of 20

TMUA 2023 Paper 1 Question 3

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

TMUA 2023 · Paper 1Differentiation and integrationDefinite integrals · additivity over intervals6 options
For any integer n ≥ 0, ∫nn+1 f(x) dx = n + 1 Evaluate ∫03 f(x) dx + ∫13 f(x) dx + ∫23 f(x) dx + ∫43 f(x) dx + ∫53 f(x) dx
  1. A−2
  2. B0
  3. C1
  4. D4
  5. E18
  6. F27
Show the answer and worked solution
answer · C
  1. A−2
  2. B0
  3. C1
  4. D4
  5. E18
  6. F27
Write In = ∫nn+1 f(x) dx = n+1, so I0 = 1, I1 = 2, I2 = 3, I3 = 4, I4 = 5. Then ∫03 = I0 + I1 + I2 = 6, ∫13 = I1 + I2 = 5 and ∫23 = I2 = 3. The last two run backwards, so they change sign: ∫43 = −I3 = −4 and ∫53 = −(I3 + I4) = −9. The total is 6 + 5 + 3 − 4 − 9 = 1.