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

The value of ∑n=1100∫n-1nex-xdx, where x is the greatest integer ≤x, is:
  1. 100e-1
  2. 100e
  3. 1001-e
  4. 1001+e

Solution

∑n=1100∫n-1nexdx, period of x=1

∑n=1100∫01exdx=∑n=1100∫01exdx

∑n=1100e-1=100e-1

Asked in: JEE Main 2021 (26 Feb Shift 1)

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