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The value of \sum_{n=1}^{100}\int_{n-1}^{n} e^{x-[x]}dx, where [x] is the greatest integer \leq x, is :
 
Option: 1 100(e-1)
Option: 2 100e
Option: 3 100 (1-e)  
Option: 4 100 (1+e)

Answers (1)

best_answer

\\\sum_{n=1}^{100} \int_{n-1}^{n} e^{\{x\}} d x, \text { period of }\{x\}=1 \\ \sum_{n=1}^{100} \int_{0}^{1} e^{\{x\}} d x=\sum_{n=1}^{100} \int_{0}^{1} e^{x} d x \\ \sum_{n=1}^{100}(e-1)=100(e-1)

Note that:

\\x-[x]=\{x\},\;\\\text{where [x] is the greatest integer }\leq x\;\;\text{and }\{x\}\;\text{fractional part of }x.

 

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himanshu.meshram

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