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JEE Main20199 Jan 2019Evening ShiftMathematicsIndefinite IntegrationActual

If f x = ∫ 5 x 8 + 7 x 6 x 2 + 1 + 2 x 7 2 d x , x ≥ 0 , and f 0 = 0 , then the value of f ( 1 ) is

Options

  1. A- 1 4
  2. B1 2
  3. C1 4
  4. D- 1 2

Correct answer

C. 1 4

Step-by-step solution

We have, f x = ∫ 5 x 8 + 7 x 6 x 2 + 1 + 2 x 7 2   d x = ∫ 5 x 8 + 7 x 6 x 14 1 x 5 + 1 x 7 + 2 2 d x = ∫ 5 x 6 + 7 x 8 1 x 5 + 1 x 7 + 2 2 d x Put, 1 x 5 + 1 x 7 + 2 = t ⇒ - 5 x 6 - 7 x 8 d x = d t ∴ f x = - ∫ d t t 2 = 1 t + c , where c is the constant of integration. = x 7 x 2 + 1 + 2 x 7 + c Now,   f 0 = 0 ⇒   c = 0 ∴     f 1 = 1 1 2 + 2 + 2.1 7 = 1 4

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