A) \[{{e}^{1000}}-1\]
B) \[\frac{{{e}^{1000}}-1}{e-1}\]
C) \[1000(e-1)\]
D) \[\frac{e-1}{1000}\]
Correct Answer: C
Solution :
We know, \[{{e}^{x-[x]}}\] is periodic function with period 1. \[\therefore \,\,I=\int_{0}^{1000}{{{e}^{x-[x]}}dx}\]? \[\int\limits_{0}^{1000}{{{e}^{x-[x]}}}\,dx=\int\limits_{0}^{1000}{{{e}^{\{x\}}}dx}\] \[=\int\limits_{0}^{1}{{{e}^{x}}}\,dx\int\limits_{1}^{2}{{{e}^{x}}dx}+\int\limits_{2}^{3}{{{e}^{x}}dx+......+\int\limits_{999}^{1000}{{{e}^{x}}dx}}\] \[=\int\limits_{0}^{1}{{{e}^{x}}}\,dx\int\limits_{0}^{1}{{{e}^{x}}dx}+\int\limits_{0}^{1}{{{e}^{x}}dx+...1000\,terms}\] \[=1000\int\limits_{0}^{1}{{{e}^{x}}dx}=1000[{{e}^{x}}]_{0}^{1}=1000[{{e}^{1}}-{{e}^{0}}]\] \[=1000\,(e-1).\]
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