The value of $\int\limits_0^1\left(1+e^{-x}\right) d x$ is equal to :

$2-\frac{1}{e}$ $2+\frac{1}{e}$ $e+\frac{1}{e}$ none of these

$2-\frac{1}{e}$

$I=\int\limits_0^1\left(1+e^{-x}\right) d x=\left|x-e^{-x}\right|_0^1=\left(1-e^{-1}\right)-(0-1)=2-e^{-1}$ Hence (1) is the correct answer.