Question:
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
Correct Option: 1
Solution:
$\sum_{n=1}^{100} \int_{n-1}^{n} e^{\{x\}} d x$, period of $\{x\}=1$
$\sum_{n=1}^{100} \int_{0}^{1} e^{\{x\}} d x=\sum_{n=1}^{100} \int_{0}^{1} e^{x} d x$
$\sum_{n=1}^{100}(e-1)=100(e-1)$