$\sum\limits_{n=1}^{100} \int\limits_{n-1}^n e^{x-[x]} d x=$

$\frac{e^{100}-1}{e-1}$ $\frac{e-1}{100}$ $100(e-1)$ $\frac{e^{100}-1}{100}$

$100(e-1)$

Since the function x – [x] is periodic with period 1, therefore $\sum\limits_{n=1}^{100} \int\limits_{n-1}^n e^{x-[x]} d x=100 \int\limits_0^1 e^{x-[x]} d x$ $=100 \int\limits_0^1 e^x d x=100(e-1)$ Hence (3) is the correct answer.