Đặt \(\displaystyle \left\{ \begin{array}{l}u = x\\dv = {e^{1 - x}}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = dx\\v = - {e^{1 - x}}\end{array} \right.\)
\(\displaystyle \Rightarrow \int\limits_0^1 {x{e^{1 - x}}dx} = \left. { - x{e^{1 - x}}} \right|_0^1 + \int\limits_0^1 {{e^{1 - x}}dx} \) \(\displaystyle = - 1 - \left. {{e^{1 - x}}} \right|_0^1\) \(\displaystyle = - 1 - 1 + e = e - 2\)
Chọn B.
