Σlimits_n=1^100 ∫limits_n-1^n e^x-[x] d x=

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.