The value of


The value of $\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]} d x$, where $[x]$ is the greatest integer $\leq \mathrm{x}$, is

  1. $100(\mathrm{e}-1)$

  2. $100(1-\mathrm{e})$

  3. $100 \mathrm{e}$

  4. $100(1+e)$

Correct Option: 1


$\sum_{n=1}^{100} \int_{n-1}^{n} e^{\{x\}} d x$, period of $\{x\}=1$

$\sum_{n=1}^{100} \int_{0}^{1} e^{\{x\}} d x=\sum_{n=1}^{100} \int_{0}^{1} e^{x} d x$


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