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AP EAMCET20224 Jul 2022Morning ShiftMathematicsDefinite IntegrationActual

∫ 0 1 a k x k d x =

Options

  1. Alim n → ∞ a k 1 + 2 k + 3 k … + n k n k + 1
  2. Blim n → ∞ a k + a k + ⋯ + a k n k + 1
  3. Clim n → ∞ 1 n ∑ r n k
  4. Dlim n → ∞ 1 n ∑ 2 r n k

Correct answer

A. lim n → ∞ a k 1 + 2 k + 3 k … + n k n k + 1

Step-by-step solution

∫ 0 1 a k x k d x = a k x k + 1 k + 1 0 1 = a k k + 1 Now, lim n → ∞ a k 1 + 2 k + 3 k … + n k n k + 1 = a k lim n → ∞ ∑ r = 1 n r k n · n k = a k ∫ 0 1 x k d x = a k k + 1

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