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Câu hỏi: Giả sử hàm số $f\left( x \right)$ có đạo hàm cấp 2 trên $\mathbb{R}$ thỏa mãn $f\left( 1 \right)={f}'\left( 1 \right)=1$ và $f\left( 1-x \right)+{{x}^{2}}.{{f}'}'\left( x \right)=2x$ với mọi $x\in \mathbb{R}$. Tính tích phân $I=\int\limits_{0}^{1}{x{f}'\left( x \right)}\text{d}x$.
A. $I=\dfrac{1}{3}$.
B. $I=\dfrac{2}{3}$.
C. $I=1$.
D. $I=2$.
A. $I=\dfrac{1}{3}$.
B. $I=\dfrac{2}{3}$.
C. $I=1$.
D. $I=2$.
Đặt $\left\{ \begin{aligned}
& u={f}'\left( x \right) \\
& dv=x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={{f}'}'\left( x \right)\text{d}x \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$.
Suy ra $I=\int\limits_{0}^{1}{x{f}'\left( x \right)}\text{d}x=\dfrac{{{x}^{2}}}{2}{f}'\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}{{f}'}'\left( x \right)}\text{d}x=\dfrac{1}{2}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}{{f}'}'\left( x \right)}\text{d}x$.
Do $f\left( 1-x \right)+{{x}^{2}}.{{f}'}'\left( x \right)=2x\Rightarrow \dfrac{{{x}^{2}}}{2}.{{f}'}'\left( x \right)=x-\dfrac{1}{2}f\left( 1-x \right)$.
Vậy $I=\dfrac{1}{2}-\int\limits_{0}^{1}{\left[ x-\dfrac{1}{2}f\left( 1-x \right) \right]}\text{d}x=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( 1-x \right)}\text{d}x$.
Đặt $t=1-x$ suy ra $I=-\dfrac{1}{2}\int\limits_{1}^{0}{f\left( t \right)}\text{d}t=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( t \right)}\text{d}t=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( x \right)}\text{d}x$.
Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv=\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={f}'\left( x \right)\text{d}x \\
& v=x \\
\end{aligned} \right.$
Suy ra$I=\dfrac{1}{2}\left[ xf\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{x{f}'\left( x \right)dx} \right]\Leftrightarrow I=\dfrac{1}{2}\left( 1-I \right)\Leftrightarrow I=\dfrac{1}{3}$.
& u={f}'\left( x \right) \\
& dv=x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={{f}'}'\left( x \right)\text{d}x \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$.
Suy ra $I=\int\limits_{0}^{1}{x{f}'\left( x \right)}\text{d}x=\dfrac{{{x}^{2}}}{2}{f}'\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}{{f}'}'\left( x \right)}\text{d}x=\dfrac{1}{2}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}{{f}'}'\left( x \right)}\text{d}x$.
Do $f\left( 1-x \right)+{{x}^{2}}.{{f}'}'\left( x \right)=2x\Rightarrow \dfrac{{{x}^{2}}}{2}.{{f}'}'\left( x \right)=x-\dfrac{1}{2}f\left( 1-x \right)$.
Vậy $I=\dfrac{1}{2}-\int\limits_{0}^{1}{\left[ x-\dfrac{1}{2}f\left( 1-x \right) \right]}\text{d}x=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( 1-x \right)}\text{d}x$.
Đặt $t=1-x$ suy ra $I=-\dfrac{1}{2}\int\limits_{1}^{0}{f\left( t \right)}\text{d}t=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( t \right)}\text{d}t=\dfrac{1}{2}\int\limits_{0}^{1}{f\left( x \right)}\text{d}x$.
Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv=\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={f}'\left( x \right)\text{d}x \\
& v=x \\
\end{aligned} \right.$
Suy ra$I=\dfrac{1}{2}\left[ xf\left( x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\int\limits_{0}^{1}{x{f}'\left( x \right)dx} \right]\Leftrightarrow I=\dfrac{1}{2}\left( 1-I \right)\Leftrightarrow I=\dfrac{1}{3}$.
Đáp án A.