A) \[{{e}^{1000}}-1\]
B) \[\frac{{{e}^{1000}}-1}{e-1}\]
C) \[1000(e-1)\]
D) \[\frac{e-1}{1000}\]
Correct Answer: C
Solution :
\[{{e}^{x-[x]}}\] is a periodic function with period 1. \[\therefore \int_{0}^{1000}{{{e}^{x-[x]}}dx=1000\int_{0}^{1}{{{e}^{x-[x]}}dx}}\], \[[\because [x]=0,\,\,\text{if}\,\,0<x<1]\] \[=1000\text{ }\,[{{e}^{x}}]_{0}^{1}\]\[=1000\,(e-1)\].
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