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TMUA 2021 · Paper 1 · Question 7 of 20

TMUA 2021 Paper 1 Question 7

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

TMUA 2021 · Paper 1Differentiation and integrationDefinite integrals · modulus, sign and substitution8 optionshard
The function f is such that f(0) = 0, and x f(x) > 0 for all non-zero values of x.

It is given that ∫−22 f(x) dx = 4 and ∫−22 |f(x)| dx = 8

Evaluate ∫−20 f(|x|) dx

  1. A−8
  2. B−6
  3. C−4
  4. D−2
  5. E2
  6. F4
  7. G6
  8. H8
Show the answer and worked solution
answer · G
  1. A−8
  2. B−6
  3. C−4
  4. D−2
  5. E2
  6. F4
  7. G6
  8. H8
The condition x f(x) > 0 says f has the same sign as x: negative on (−2, 0) and positive on (0, 2). Write A = ∫−20 f and B = ∫02 f. Then A + B = 4, while the modulus flips the sign on the left half, giving −A + B = 8; hence B = 6 and A = −2. In the required integral x is negative, so |x| = −x, and substituting u = −x turns ∫−20 f(−x) dx into ∫02 f(u) du = B = 6.