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JEE Main202126 Feb 2021Morning ShiftMathematicsDefinite IntegrationActual

The value of ∑ n = 1 100 ∫ n - 1 n e x - x d x , where x is the greatest integer ≤ x , is:

Options

  1. A100 e - 1
  2. B100 e
  3. C100 1 - e
  4. D100 1 + e

Correct answer

A. 100 e - 1

Step-by-step solution

∑ n = 1 100 ∫ n - 1 n e x d x , period of x = 1 ∑ n = 1 100 ∫ 0 1 e x d x = ∑ n = 1 100 ∫ 0 1 e x d x ∑ n = 1 100 e - 1 = 100 e - 1

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