The value of ∑$$n=11000$$ ∫n−$$1n$$ ex−[x]dx is $$($$[x] is greatest integer $$function)$$

Correct option is 31000 (e – 1)

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The value of $\sum_{n = 1}^{1000} \int_{n - 1}^{n} e^{x - [x]} d x$ is ([x] is greatest integer function)

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e1000−11000

e1000−1e−1

1000 (e – 1)

e−11000

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detailed solution

Correct option is C

Since the period of the function x-[x] is 1 so∑n=11000 ∫n−1n ex−[x]dx=∫01000 ex−[x]dx=1000∫01 ex−[x]dx=1000∫01 exdx=1000(e−1).

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The value of ∑n=11000 ∫n−1n ex−[x]dx is ([x] is greatest integer function)

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The value of $\sum_{n = 1}^{1000} \int_{n - 1}^{n} e^{x - [x]} d x$ is ([x] is greatest integer function)