Cho hàm số \(f(x)\) có đạo hàm liên tục trên đoạn...

A

- Hướng dẫn giải

Phương pháp giải:

Sử dụng công thức tích phân từng phần: \(\int\limits_{a}^{b}{udv}=\left. uv \right|_{a}^{b}-\int\limits_{a}^{b}{vdu}\).

Giải chi tiết:

Ta có:

\(\begin{align}  \int\limits_{0}^{1}{(x+1){{e}^{x}}f(x)dx}=\int\limits_{0}^{1}{f(x)d\left( x{{e}^{x}} \right)}=\left. x{{e}^{x}}f(x) \right|_{0}^{1}-\int\limits_{0}^{1}{x{{e}^{x}}f'(x)dx}=\frac{{{e}^{2}}-1}{4} \\  \Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}f'(x)dx}=\left. x{{e}^{x}}f(x) \right|_{0}^{1}-\frac{{{e}^{2}}-1}{4}=e\,f(1)-0-\frac{{{e}^{2}}-1}{4}=e.0-\frac{{{e}^{2}}-1}{4}=-\frac{{{e}^{2}}-1}{4} \\ \end{align}\)

Xét \(\int\limits_{0}^{1}{{{\left[ f'(x)+kx{{e}^{x}} \right]}^{2}}dx=}\int\limits_{0}^{1}{{{\left[ f'(x) \right]}^{2}}dx}+2k\int\limits_{0}^{1}{x{{e}^{x}}f'(x)dx+{{k}^{2}}\int\limits_{0}^{1}{{{x}^{2}}{{e}^{2x}}dx}}=\frac{{{e}^{2}}-1}{4}+2k.\left( -\frac{{{e}^{2}}-1}{4} \right)+{{k}^{2}}.I\)

 \(\begin{align}I=\int\limits_{0}^{1}{{{x}^{2}}{{e}^{2x}}dx}=\frac{1}{2}\int\limits_{0}^{1}{{{x}^{2}}d{{e}^{2x}}=\left. \frac{1}{2}{{x}^{2}}{{e}^{2x}} \right|_{0}^{1}-\frac{1}{2}}\int\limits_{0}^{1}{{{e}^{2x}}d{{x}^{2}}}=\frac{1}{2}{{e}^{2}}-\int\limits_{0}^{1}{x{{e}^{2x}}dx=}\frac{1}{2}{{e}^{2}}-\frac{1}{2}\int\limits_{0}^{1}{xd{{e}^{2x}}} \\  =\frac{1}{2}{{e}^{2}}-\left. \frac{1}{2}x{{e}^{2x}} \right|_{0}^{1}+\frac{1}{2}\int\limits_{0}^{1}{{{e}^{2x}}dx}=\frac{1}{2}{{e}^{2}}-\frac{1}{2}{{e}^{2}}+\frac{1}{4}\left. {{e}^{2x}} \right|_{0}^{1}=\frac{1}{4}{{e}^{2}}-\frac{1}{4} \\ \end{align}\)

\(\begin{align}  \Rightarrow \int\limits_{0}^{1}{{{\left[ f'(x)+kx{{e}^{x}} \right]}^{2}}dx}=\frac{{{e}^{2}}-1}{4}+2k.\left( -\frac{{{e}^{2}}-1}{4} \right)+{{k}^{2}}.\left( \frac{1}{4}{{e}^{2}}-\frac{1}{4} \right)=\frac{{{e}^{2}}-1}{4}.\left[ {{k}^{2}}-2k+1 \right]=\frac{{{e}^{2}}-1}{4}.{{(k-1)}^{2}}=0 \\  \Rightarrow k=1 \\ \end{align}\)

Khi đó, \(\int\limits_{0}^{1}{{{\left[ f'(x)+x{{e}^{x}} \right]}^{2}}dx}=0\Rightarrow f'(x)+x{{e}^{x}}=0\Leftrightarrow f'(x)=-x{{e}^{x}}\Rightarrow f(x)=-\int{x{{e}^{x}}dx}=-(x-1){{e}^{x}}+C\)                                                             

Mà \(f(1)=0\Rightarrow 0+C=0\Rightarrow C=0\Rightarrow f(x)=-(x-1){{e}^{x}}\)

\(\Rightarrow \int\limits_{0}^{1}{f(x)dx=}\int\limits_{0}^{1}{\left[ -(x+1){{e}^{x}} \right]dx}=-\int\limits_{0}^{1}{(x-1)d{{e}^{x}}}=-(x-1)\left. {{e}^{x}} \right|_{0}^{1}+\left. {{e}^{x}} \right|_{0}^{1}=\left. \left( -x{{e}^{x}}+2{{e}^{x}} \right) \right|_{0}^{1}=\left( -e+2e \right)-2=e-2\)

Chọn: A