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JEE Main202126 Aug 2021Morning ShiftMathematicsDefinite IntegrationActual

The value of lim n → ∞ 1 n ∑ r = 0 2 n - 1 n 2 n 2 + 4 r 2 is:

Options

  1. A1 2 tan - 1 2
  2. Btan - 1 4
  3. C1 2 tan - 1 4
  4. D1 4 tan - 1 ( 4 )

Correct answer

C. 1 2 tan - 1 4

Step-by-step solution

Let I = lim n → ∞ 1 n ∑ r = 0 2 n - 1 n 2 n 2 + 4 r 2 I = lim n → ∞ 1 n ∑ r = 0 2 n - 1 n 2 n 2 + 4 r 2 I = lim n → ∞ 1 n ∑ r = 0 2 n - 1 1 1 + 4 r n 2 Let r n = x , lower limit x = 0 when r = 0 and upper limit x = 2 when r = 2 n - 1 ⇒ I = ∫ 0 2 1 1 + 4 x 2 d x ⇒ I = 1 4 ∫ 0 2 1 1 2 2 + x 2 d x ⇒ I = 1 4 × 2 tan - 1 2 x 0 2 = 1 2 tan - 1 ( 4 )

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