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TMUA 2019 · Paper 1 · Question 10 of 20

TMUA 2019 Paper 1 Question 10

Differentiation and integration — Definite integral of a modulus · splitting the range. Try it first; the answer and a full worked solution are below.

TMUA 2019 · Paper 1Differentiation and integrationDefinite integral of a modulus · splitting the range6 options
Evaluate ∫−13 |x|(1−x) dx
  1. A173
  2. B−173
  3. C163
  4. D−163
  5. E113
  6. F−113
Show the answer and worked solution
answer · F
  1. A173
  2. B−173
  3. C163
  4. D−163
  5. E113
  6. F−113
The modulus changes definition at x = 0, so split there. On [−1,  0], |x| = −x and the integrand is x2 − x, giving [x33 − x22]−10 = 0 − (−13 − 12) = 56. On [0,  3], |x| = x and the integrand is x − x2, giving [x22 − x33]03 = 92 − 9 = −92. The total is 56 − 92 = −113.