Đáp án đúng: D
Giải chi tiết:Đặt \(\left\{ \begin{array}{l}u = f\left( x \right)\\dv = \left( {x + 1} \right){e^x}dx\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = f'\left( x \right)dx\\v = x{e^x}\end{array} \right.\)
Khi đó \(\frac{{{e^2} - 1}}{4} = \left. {x.{e^x}f\left( x \right)} \right|_0^1 - \int\limits_0^1 {x.{e^x}f'\left( x \right)dx} \Rightarrow \int\limits_0^1 {x.{e^x}f'\left( x \right)dx} = - \frac{{{e^2} - 1}}{4}\)
Xét
\(\begin{array}{l}\int\limits_0^1 {{{\left[ {f'\left( x \right) + kx{e^x}} \right]}^2}dx} = \int\limits_0^1 {{{\left[ {f'\left( x \right)} \right]}^2}dx} + 2k\int\limits_0^1 {x.{e^x}f'\left( x \right)dx} + {k^2}\int\limits_0^1 {{x^2}{e^{2x}}dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{e^2} - 1}}{4} - 2k\frac{{{e^2} - 1}}{4} + {k^2}\frac{{{e^2} - 1}}{4} = 0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \frac{{{e^2} - 1}}{4}{\left( {k - 1} \right)^2} = 0 \Leftrightarrow k = 1\\ \Leftrightarrow f'\left( x \right) + x{e^x} = 0\\ \Rightarrow f'\left( x \right) = - x{e^x} \Rightarrow f\left( x \right) = - \int {x{e^x}dx} = - \int\limits_{}^{} {xd\left( {{e^x}} \right)} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - \left( {x{e^x} - \int\limits_{}^{} {{e^x}dx} + C} \right) = - x{e^x} + {e^x} + C\\ \Rightarrow f\left( 1 \right) = - e + e + C = 0 \Leftrightarrow C = 0\\ \Rightarrow f\left( x \right) = - x{e^x} + {e^x} = {e^x}\left( {1 - x} \right)\end{array}\)
Do đó
\(\begin{array}{l}\int\limits_0^1 {f\left( x \right)dx} = \int\limits_0^1 {\left( {1 - x} \right){e^x}} dx = \int\limits_0^1 {\left( {1 - x} \right)d\left( {{e^x}} \right)} \\ = \left. {\left( {1 - x} \right){e^x}} \right|_0^1 + \int\limits_0^1 {{e^x}dx} = - 1 + e - 1 = e - 2\end{array}\)
Chọn D.
