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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm là ${f}'\left( x \right)=\left( 1-x \right){{e}^{-x}},\forall x\in \mathbb{R}$ và $f\left( 2 \right)=\dfrac{2}{{{e}^{2}}}.$ Biết $F\left( x \right)$ là một nguyên hàm của $f\left( x \right)$ thoả mãn $F\left( 0 \right)=3+\dfrac{2}{e},$ khi đó $F\left( 1 \right)$ bằng
A. $4.$
B. $2.$
C. $1.$
D. $3.$
A. $4.$
B. $2.$
C. $1.$
D. $3.$
${f}'\left( x \right)=\left( 1-x \right){{e}^{-x}}\Rightarrow f\left( x \right)=\int{\left( 1-x \right){{e}^{-x}}dx=-}\int{\left( 1-x \right)d\left( {{e}^{-x}} \right)}$
$=-\left( 1-x \right){{e}^{-x}}-\int{{{e}^{-x}}dx=}-\left( 1-x \right){{e}^{-x}}+{{e}^{-x}}+C=x{{e}^{-x}}+C$
$f\left( 2 \right)=2.{{e}^{-2}}+C=\dfrac{2}{{{e}^{2}}}+C=\dfrac{2}{{{e}^{2}}}\Rightarrow C=0.$
$\Rightarrow f\left( x \right)=x.{{e}^{-x}}$
$F\left( x \right)=\int{x.{{e}^{-x}}dx=-\int{xd\left( {{e}^{-x}} \right)}=-x.{{e}^{-x}}+\int{{{e}^{-x}}dx=}}-x.{{e}^{-x}}-{{e}^{-x}}+C$
$F\left( 0 \right)=-1+C=3+\dfrac{2}{e}\Rightarrow C=4+\dfrac{2}{e}$
$F\left( x \right)=-x.{{e}^{-x}}-{{e}^{-x}}+4+\dfrac{2}{e}$
Vậy $F\left( 1 \right)=-1.{{e}^{-1}}-{{e}^{-1}}+4+\dfrac{2}{e}=-\dfrac{1}{e}-\dfrac{1}{e}+4+\dfrac{2}{e}=4.$
$=-\left( 1-x \right){{e}^{-x}}-\int{{{e}^{-x}}dx=}-\left( 1-x \right){{e}^{-x}}+{{e}^{-x}}+C=x{{e}^{-x}}+C$
$f\left( 2 \right)=2.{{e}^{-2}}+C=\dfrac{2}{{{e}^{2}}}+C=\dfrac{2}{{{e}^{2}}}\Rightarrow C=0.$
$\Rightarrow f\left( x \right)=x.{{e}^{-x}}$
$F\left( x \right)=\int{x.{{e}^{-x}}dx=-\int{xd\left( {{e}^{-x}} \right)}=-x.{{e}^{-x}}+\int{{{e}^{-x}}dx=}}-x.{{e}^{-x}}-{{e}^{-x}}+C$
$F\left( 0 \right)=-1+C=3+\dfrac{2}{e}\Rightarrow C=4+\dfrac{2}{e}$
$F\left( x \right)=-x.{{e}^{-x}}-{{e}^{-x}}+4+\dfrac{2}{e}$
Vậy $F\left( 1 \right)=-1.{{e}^{-1}}-{{e}^{-1}}+4+\dfrac{2}{e}=-\dfrac{1}{e}-\dfrac{1}{e}+4+\dfrac{2}{e}=4.$
Đáp án A.