The value of

$\sum_{n=1}^{100} \int_{n-1}^{n} e^{x-[x]} d x$

$=\int_{0}^{1} e^{[x\}} d x+\int_{1}^{2} e^{\{x\}} d x+\int_{2}^{3} e^{\{x\}} d x+\ldots \ldots \int_{99}^{100} e^{\{x\}} d x \quad(\because\{x\}=x-[x])$

$=\left.e^{x}\right|_{0} ^{1}+\left.e^{(x-1)}\right|_{1} ^{2}+\left.e^{(x-2)}\right|_{2} ^{3}+\ldots \ldots+\left.e^{(x-99)}\right|_{99} ^{100}$

$=(e-1)+(e-1)+(e-1)+\ldots \ldots(e-1)$

$=100(e-1)$