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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\left[ -1 ; 1 \right]$ thoả $f\left( x \right)+2=\dfrac{3}{2}\int\limits_{-1}^{1}{\left( x+t \right)f\left( t \right)\text{d}t}, \forall x\in \left[ -1 ; 1 \right]$. Tính $I=\int\limits_{-1}^{1}{f\left( x \right)} \text{d}x$ ?
A. $I=4$.
B. $I=3$.
C. $I=2$.
D. $I=1$.
A. $I=4$.
B. $I=3$.
C. $I=2$.
D. $I=1$.
Ta có $f\left( x \right)=\dfrac{3}{2}x\int\limits_{-1}^{1}{f\left( t \right)\text{d}t}+\dfrac{3}{2}\int\limits_{-1}^{1}{t.f\left( t \right)\text{d}t}-2 , \left( * \right)$. Đặt $A=\int\limits_{-1}^{1}{f\left( t \right)\text{d}t} , B=\int\limits_{-1}^{1}{t.f\left( t \right)\text{d}t}$.
$\left( * \right)\Leftrightarrow f\left( x \right)=\dfrac{3}{2}x.A+\dfrac{3}{2}B-2 , \left( 1 \right)$
$\Rightarrow xf\left( x \right)=\dfrac{3}{2}A{{x}^{2}}+\dfrac{3}{2}Bx-2x , \left( 2 \right)$.
Lấy tích phân từ $-1$ đến $1$ của $\left( 1 \right)$ và $\left( 2 \right)$ ta được
$\left\{ \begin{aligned}
& \int\limits_{-1}^{1}{f\left( x \right)\text{d}x}=\int\limits_{-1}^{1}{\left( \dfrac{3}{2}x.A+\dfrac{3}{2}B-2 \right)\text{d}x} \\
& \int\limits_{-1}^{1}{x.f\left( x \right)\text{d}x}=\int\limits_{-1}^{1}{\left( \dfrac{3}{2}A{{x}^{2}}+\dfrac{3}{2}Bx-2x \right)\text{d}x} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& A=\left. \left( \dfrac{3A{{x}^{2}}}{4}+\dfrac{3Bx}{2}-2x \right) \right|_{-1}^{1}=3B-4 \\
& B=\left. \left( \dfrac{A{{x}^{3}}}{2}+\dfrac{3B{{x}^{2}}}{4}-{{x}^{2}} \right) \right|_{-1}^{1}=A \\
\end{aligned} \right.\Leftrightarrow A=B=2$
Vậy $I=A=\int\limits_{-1}^{1}{f\left( x \right)} \text{d}x=2$.
$\left( * \right)\Leftrightarrow f\left( x \right)=\dfrac{3}{2}x.A+\dfrac{3}{2}B-2 , \left( 1 \right)$
$\Rightarrow xf\left( x \right)=\dfrac{3}{2}A{{x}^{2}}+\dfrac{3}{2}Bx-2x , \left( 2 \right)$.
Lấy tích phân từ $-1$ đến $1$ của $\left( 1 \right)$ và $\left( 2 \right)$ ta được
$\left\{ \begin{aligned}
& \int\limits_{-1}^{1}{f\left( x \right)\text{d}x}=\int\limits_{-1}^{1}{\left( \dfrac{3}{2}x.A+\dfrac{3}{2}B-2 \right)\text{d}x} \\
& \int\limits_{-1}^{1}{x.f\left( x \right)\text{d}x}=\int\limits_{-1}^{1}{\left( \dfrac{3}{2}A{{x}^{2}}+\dfrac{3}{2}Bx-2x \right)\text{d}x} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& A=\left. \left( \dfrac{3A{{x}^{2}}}{4}+\dfrac{3Bx}{2}-2x \right) \right|_{-1}^{1}=3B-4 \\
& B=\left. \left( \dfrac{A{{x}^{3}}}{2}+\dfrac{3B{{x}^{2}}}{4}-{{x}^{2}} \right) \right|_{-1}^{1}=A \\
\end{aligned} \right.\Leftrightarrow A=B=2$
Vậy $I=A=\int\limits_{-1}^{1}{f\left( x \right)} \text{d}x=2$.
Đáp án C.