The value of the sum $\sum _{n=1}^{100} {\int }_{n-1}^n e^{x-\left[x\right]}dx$, where $\left[.\right]$ denotes the greatest integer function, is

Let $I=\sum _{n=1}^{100} {\int }_{n-1}^n e^{x-\left[x\right]}dx$. Then,

$\begin{array}{l}I={\int }_0^{100} e^{x-\left[x\right]}dx\\ \Rightarrow I=100{\int }_0^1 e^{x-\left[x\right]}dx\left[\because e^{x-\left[x\right]}\text{ is periodic with period }:\right.\\ \Rightarrow l=100{\int }_0^1 e^{x}dx=100{\left[e^x\right]}_0^1=100(e-1)\end{array}$