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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm, liên tục trên đoạn $\left[ 0;1 \right]$ và thỏa mãn các điều kiện $f\left( 1 \right)=0$ và $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}=\int\limits_{0}^{1}{\left( x+1 \right){{e}^{x}}f\left( x \right)dx}=\dfrac{{{e}^{x}}-1}{4}$. Tính tích phân $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{e-1}{2}$.
B. $\dfrac{{{e}^{2}}}{4}$.
C. $\dfrac{e}{2}$.
D. $e-2$.
A. $\dfrac{e-1}{2}$.
B. $\dfrac{{{e}^{2}}}{4}$.
C. $\dfrac{e}{2}$.
D. $e-2$.
Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv=\left( x+1 \right){{e}^{x}}dx \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right) \\
& v=\int{\left( x+1 \right){{e}^{x}}dx}=x.{{e}^{x}} \\
\end{aligned} \right.$.
Suy ra $\int\limits_{0}^{1}{\left( x+1 \right){{e}^{x}}f\left( x \right)dx}=\left. x{{e}^{x}}.f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{x{{e}^{x}}.{f}'\left( x \right)dx}\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}.{f}'\left( x \right)dx}=\dfrac{1-{{e}^{2}}}{4}$.
Chọn k sao cho
$\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+k.x{{e}^{x}} \right]}^{2}}dx}=0\Leftrightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}+2k.\int\limits_{0}^{1}{x{{e}^{x}}.{f}'\left( x \right)dx}+{{k}^{2}}.\int\limits_{0}^{1}{{{x}^{2}}{{e}^{2x}}dx}=0$
$\Leftrightarrow \dfrac{{{e}^{2}}-1}{4}-2k.\dfrac{{{e}^{2}}-1}{4}+{{k}^{2}}.\dfrac{{{e}^{2}}-1}{4}=0\Leftrightarrow {{\left( k-1 \right)}^{2}}=0\Leftrightarrow k=1\Rightarrow {f}'\left( x \right)=-x{{e}^{x}}$.
Do đó $f\left( x \right)=\int{{f}'\left( x \right)dx}=-\int{x{{e}^{x}}dx}=-\left( x-1 \right){{e}^{x}}+C$ mà $f\left( 1 \right)=0\Rightarrow C=0$.
Vậy $f\left( x \right)=-\left( x-1 \right){{e}^{x}}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( 1-x \right){{e}^{x}}dx}=e-2$.
& u=f\left( x \right) \\
& dv=\left( x+1 \right){{e}^{x}}dx \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right) \\
& v=\int{\left( x+1 \right){{e}^{x}}dx}=x.{{e}^{x}} \\
\end{aligned} \right.$.
Suy ra $\int\limits_{0}^{1}{\left( x+1 \right){{e}^{x}}f\left( x \right)dx}=\left. x{{e}^{x}}.f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{x{{e}^{x}}.{f}'\left( x \right)dx}\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}.{f}'\left( x \right)dx}=\dfrac{1-{{e}^{2}}}{4}$.
Chọn k sao cho
$\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+k.x{{e}^{x}} \right]}^{2}}dx}=0\Leftrightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}+2k.\int\limits_{0}^{1}{x{{e}^{x}}.{f}'\left( x \right)dx}+{{k}^{2}}.\int\limits_{0}^{1}{{{x}^{2}}{{e}^{2x}}dx}=0$
$\Leftrightarrow \dfrac{{{e}^{2}}-1}{4}-2k.\dfrac{{{e}^{2}}-1}{4}+{{k}^{2}}.\dfrac{{{e}^{2}}-1}{4}=0\Leftrightarrow {{\left( k-1 \right)}^{2}}=0\Leftrightarrow k=1\Rightarrow {f}'\left( x \right)=-x{{e}^{x}}$.
Do đó $f\left( x \right)=\int{{f}'\left( x \right)dx}=-\int{x{{e}^{x}}dx}=-\left( x-1 \right){{e}^{x}}+C$ mà $f\left( 1 \right)=0\Rightarrow C=0$.
Vậy $f\left( x \right)=-\left( x-1 \right){{e}^{x}}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( 1-x \right){{e}^{x}}dx}=e-2$.
Đáp án D.