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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm trên đoạn $\left[ 1;e \right]$ và thỏa mãn $f\left( 1 \right)=0$ ; $\left[ {f}'\left( x \right)-1 \right]x=f\left( x \right),\forall x\in \left[ 1;e \right]$. Tích phân $\int\limits_{1}^{e}{f\left( x \right)\text{d}x}$ bằng
A. $\dfrac{{{e}^{2}}-1}{4}$.
B. $\dfrac{{{e}^{2}}+1}{2}$.
C. $\dfrac{{{e}^{2}}+1}{4}$.
D. $\dfrac{{{e}^{2}}-1}{2}$.
A. $\dfrac{{{e}^{2}}-1}{4}$.
B. $\dfrac{{{e}^{2}}+1}{2}$.
C. $\dfrac{{{e}^{2}}+1}{4}$.
D. $\dfrac{{{e}^{2}}-1}{2}$.
$\left[ {f}'\left( x \right)-1 \right]x=f\left( x \right)\Leftrightarrow {f}'\left( x \right)x-f\left( x \right)=x\Rightarrow \dfrac{1}{x}{f}'\left( x \right)+\dfrac{-1}{{{x}^{2}}}f\left( x \right)=\dfrac{1}{x}$
$\Leftrightarrow {{\left[ \dfrac{1}{x}f\left( x \right) \right]}^{\prime }}=\dfrac{1}{x}\Rightarrow \dfrac{1}{x}f\left( x \right)=\ln x+C$ do $x\in \left[ 1;e \right]$, mà $f\left( 1 \right)=0$ $\Rightarrow f\left( x \right)=x\ln x$.
$\int\limits_{1}^{e}{f\left( x \right)\text{d}x}=\int\limits_{1}^{e}{x\ln x\text{d}x}=\left. \dfrac{{{x}^{2}}}{2}\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\dfrac{x}{2}\text{d}x}=\dfrac{{{e}^{2}}}{2}-\left( \dfrac{{{e}^{2}}}{4}-\dfrac{1}{4} \right)=\dfrac{{{e}^{2}}+1}{4}$.
$\Leftrightarrow {{\left[ \dfrac{1}{x}f\left( x \right) \right]}^{\prime }}=\dfrac{1}{x}\Rightarrow \dfrac{1}{x}f\left( x \right)=\ln x+C$ do $x\in \left[ 1;e \right]$, mà $f\left( 1 \right)=0$ $\Rightarrow f\left( x \right)=x\ln x$.
$\int\limits_{1}^{e}{f\left( x \right)\text{d}x}=\int\limits_{1}^{e}{x\ln x\text{d}x}=\left. \dfrac{{{x}^{2}}}{2}\ln x \right|_{1}^{e}-\int\limits_{1}^{e}{\dfrac{x}{2}\text{d}x}=\dfrac{{{e}^{2}}}{2}-\left( \dfrac{{{e}^{2}}}{4}-\dfrac{1}{4} \right)=\dfrac{{{e}^{2}}+1}{4}$.
Đáp án C.