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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 0 ; 1 \right]$ thỏa mãn $f\left( 0 \right)=0$ và $f\left( x \right)+{f}'\left( x \right)=x-2, \forall x\in \left[ 0 ; 1 \right]$. Giá trị của $\int\limits_{0}^{1}{f\left( x \right)}$ bằng
A. $\dfrac{-5e-3}{2e}$.
B. $\dfrac{3-2e}{5}$.
C. $\dfrac{e-6}{2e}$.
D. $-\dfrac{5}{2}$.
A. $\dfrac{-5e-3}{2e}$.
B. $\dfrac{3-2e}{5}$.
C. $\dfrac{e-6}{2e}$.
D. $-\dfrac{5}{2}$.
$f\left( x \right)+{f}'\left( x \right)=x-2, \forall x\in \left[ 0 ; 1 \right]$ $\Leftrightarrow {{e}^{x}}f\left( x \right)+{{e}^{x}}{f}'\left( x \right)=\left( x-2 \right){{e}^{x}}, \forall x\in \left[ 0 ; 1 \right]$ $\Leftrightarrow {{\left( {{e}^{x}}f\left( x \right) \right)}^{\prime }}=\left( x-2 \right){{e}^{x}}, \forall x\in \left[ 0 ; 1 \right]$.
Suy ra: ${{e}^{x}}f\left( x \right)=\int{\left( x-2 \right){{e}^{x}}}\text{d}x=\left( x-2 \right){{e}^{x}}-\int{{{e}^{x}}}\text{d}x=\left( x-3 \right){{e}^{x}}+C$.
$f\left( 0 \right)=0$. Suy ra: $C=3$. Do đó: $f\left( x \right)=x-3+3{{e}^{-x}}$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( x-3+3{{e}^{-x}} \right)\text{d}x}=\left. \left( \dfrac{{{x}^{2}}}{2}-3x-3{{e}^{-x}} \right) \right|_{0}^{1}=\dfrac{e-6}{2e}$.
Suy ra: ${{e}^{x}}f\left( x \right)=\int{\left( x-2 \right){{e}^{x}}}\text{d}x=\left( x-2 \right){{e}^{x}}-\int{{{e}^{x}}}\text{d}x=\left( x-3 \right){{e}^{x}}+C$.
$f\left( 0 \right)=0$. Suy ra: $C=3$. Do đó: $f\left( x \right)=x-3+3{{e}^{-x}}$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( x-3+3{{e}^{-x}} \right)\text{d}x}=\left. \left( \dfrac{{{x}^{2}}}{2}-3x-3{{e}^{-x}} \right) \right|_{0}^{1}=\dfrac{e-6}{2e}$.
Đáp án C.