\(\int \limits_0^1 x{e^{1 - x}}dx\) bằng
A. 1 - e
B. e - 2
C. 1
D. - 1
Đặt \(\left\{ \begin{array}{l}
u = x\\
dv = {e^{1 - x}}dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du = dx\\
v = - {e^{1 - x}}
\end{array} \right.\)
\(\begin{array}{l}
\Rightarrow \int\limits_0^1 {x{e^{1 - x}}dx = - x\left. {{e^{1 - x}}} \right|_0^1 + \int\limits_0^1 {{e^{1 - x}}dx} } \\
= - 1 - \left. {{e^{1 - x}}} \right|_0^1 = - 1 - 1 + e = e - 2
\end{array}\)
Chọn B.
-- Mod Toán 12
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