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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\left[ -1;1 \right]$ thỏa mãn $f\left( x \right)-1=\int\limits_{-1}^{1}{\left( x+{{e}^{t}} \right)f\left( t \right)dt.}$ Tích phân $I=\int\limits_{-1}^{1}{{{e}^{x}}f\left( x \right)dx}$ nằng:
A. $I=\dfrac{e+3}{-{{e}^{2}}+e-3}$
B. $I=\dfrac{e+3}{{{e}^{2}}-e+3}$
C. $I=\dfrac{{{e}^{2}}+3}{-{{e}^{2}}+e-3}$
D. $I=\dfrac{-2e}{{{e}^{2}}-e+3}$
A. $I=\dfrac{e+3}{-{{e}^{2}}+e-3}$
B. $I=\dfrac{e+3}{{{e}^{2}}-e+3}$
C. $I=\dfrac{{{e}^{2}}+3}{-{{e}^{2}}+e-3}$
D. $I=\dfrac{-2e}{{{e}^{2}}-e+3}$
Cách giải:
Ta có:
$f\left( x \right)-1=\int\limits_{-1}^{1}{\left( x+{{e}^{t}} \right)f\left( t \right)dt}\Leftrightarrow f\left( x \right)-1=x\int\limits_{-1}^{1}{f\left( t \right)dt}+\int\limits_{-1}^{1}{{{e}^{t}}f\left( t \right)dt}\left( * \right)$
Giả sử $\int\limits_{-1}^{1}{f\left( t \right)dt}=a,\int\limits_{-1}^{1}{{{e}^{t}}f\left( t \right)dt}=b\Rightarrow f\left( x \right)-1=xa+b\Leftrightarrow f\left( x \right)=ax+b+1.$
Thay vào (*) ta có:
$ax+b=x\int\limits_{-1}^{1}{\left( at+b+1 \right)dt+}\int\limits_{-1}^{1}{{{e}^{t}}\left( at+b+1 \right)dt}$
$\Leftrightarrow ax+b=x\left( \dfrac{a{{t}^{2}}}{2}+bt+t \right)\left| \begin{aligned}
& 1 \\
& -1 \\
\end{aligned} \right.+\int\limits_{-1}^{1}{{{e}^{t}}\left( at+b+1 \right)dt}$
$\Leftrightarrow ax+b=x\left( \dfrac{a}{2}+b+1-\dfrac{a}{2}+b+1 \right)+\left( at+b+1 \right){{e}^{t}}\left| _{-1}^{1} \right.-a\int\limits_{-1}^{1}{{{e}^{t}}dt}$
$\Leftrightarrow ax+b=x\left( 2b+2 \right)+\left( a+b+1 \right)e-\left( -a+b+1 \right){{e}^{-1}}-a\left( e-{{e}^{-1}} \right)$
$\Leftrightarrow ax+b=x\left( 2b+2 \right)+\left( b+1 \right)e+\left( 2a-b-1 \right){{e}^{-1}}$
$\Rightarrow \left\{ \begin{aligned}
& a=2b+2 \\
& b=\left( b+1 \right)e+\left( 2a-b-1 \right){{e}^{-1}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2b+2 \\
& b=\left( b+1 \right)e+\left( 3b+3 \right){{e}^{-1}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a=2b+2 \\
& b=\left( e+\dfrac{3}{e} \right)b+e+\dfrac{3}{e} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{2{{e}^{2}}}{{{e}^{2}}-e-1} \\
& b=\dfrac{e+\dfrac{3}{e}}{1-e-\dfrac{3}{e}}=\dfrac{{{e}^{2}}+3}{-{{e}^{2}}+e-3} \\
\end{aligned} \right.$
Vậy $\int\limits_{-1}^{1}{{{e}^{t}}f\left( t \right)dt}=\int\limits_{-1}^{1}{{{e}^{x}}f\left( x \right)dx}=\dfrac{{{e}^{2}}+3}{-{{e}^{2}}+e-3}=I.$
Ta có:
$f\left( x \right)-1=\int\limits_{-1}^{1}{\left( x+{{e}^{t}} \right)f\left( t \right)dt}\Leftrightarrow f\left( x \right)-1=x\int\limits_{-1}^{1}{f\left( t \right)dt}+\int\limits_{-1}^{1}{{{e}^{t}}f\left( t \right)dt}\left( * \right)$
Giả sử $\int\limits_{-1}^{1}{f\left( t \right)dt}=a,\int\limits_{-1}^{1}{{{e}^{t}}f\left( t \right)dt}=b\Rightarrow f\left( x \right)-1=xa+b\Leftrightarrow f\left( x \right)=ax+b+1.$
Thay vào (*) ta có:
$ax+b=x\int\limits_{-1}^{1}{\left( at+b+1 \right)dt+}\int\limits_{-1}^{1}{{{e}^{t}}\left( at+b+1 \right)dt}$
$\Leftrightarrow ax+b=x\left( \dfrac{a{{t}^{2}}}{2}+bt+t \right)\left| \begin{aligned}
& 1 \\
& -1 \\
\end{aligned} \right.+\int\limits_{-1}^{1}{{{e}^{t}}\left( at+b+1 \right)dt}$
$\Leftrightarrow ax+b=x\left( \dfrac{a}{2}+b+1-\dfrac{a}{2}+b+1 \right)+\left( at+b+1 \right){{e}^{t}}\left| _{-1}^{1} \right.-a\int\limits_{-1}^{1}{{{e}^{t}}dt}$
$\Leftrightarrow ax+b=x\left( 2b+2 \right)+\left( a+b+1 \right)e-\left( -a+b+1 \right){{e}^{-1}}-a\left( e-{{e}^{-1}} \right)$
$\Leftrightarrow ax+b=x\left( 2b+2 \right)+\left( b+1 \right)e+\left( 2a-b-1 \right){{e}^{-1}}$
$\Rightarrow \left\{ \begin{aligned}
& a=2b+2 \\
& b=\left( b+1 \right)e+\left( 2a-b-1 \right){{e}^{-1}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=2b+2 \\
& b=\left( b+1 \right)e+\left( 3b+3 \right){{e}^{-1}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& a=2b+2 \\
& b=\left( e+\dfrac{3}{e} \right)b+e+\dfrac{3}{e} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{2{{e}^{2}}}{{{e}^{2}}-e-1} \\
& b=\dfrac{e+\dfrac{3}{e}}{1-e-\dfrac{3}{e}}=\dfrac{{{e}^{2}}+3}{-{{e}^{2}}+e-3} \\
\end{aligned} \right.$
Vậy $\int\limits_{-1}^{1}{{{e}^{t}}f\left( t \right)dt}=\int\limits_{-1}^{1}{{{e}^{x}}f\left( x \right)dx}=\dfrac{{{e}^{2}}+3}{-{{e}^{2}}+e-3}=I.$
Đáp án C.