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The value of the integral ∫ 0 4 x 3 - 6 x 2 + 12 x - 4 + x - 2 cos ⁡ x - 2 d x is equal to

Options

  1. A12
  2. B16
  3. C0
  4. D10

Correct answer

B. 16

Step-by-step solution

Let, I = ∫ 0 4 x - 2 3 + x - 2 cos ⁡ x - 2 + 4 d x Let x - 2 = t ⇒ d x = d t ⇒ I = ∫ - 2 2 t 3 + t cos ⁡ t + 4 d t (As t 3 + t cos ⁡ t is an odd function) ⇒ I = 2 ∫ 0 2 4 d t = 16

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