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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 0;1 \right]$ và thỏa mãn ${{f}^{2}}\left( x \right)-xf\left( x \right)f'\left( x \right)=2x+4 \forall x\in \left[ 0;1 \right]$. Biết $f\left( 1 \right)=3$, tích phân $I=\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}$ bằng
A. 13
B. $\dfrac{19}{3}$
C. $\dfrac{13}{3}$
D. 19
A. 13
B. $\dfrac{19}{3}$
C. $\dfrac{13}{3}$
D. 19
Ta có: ${{f}^{2}}\left( x \right)=xf\left( x \right)f'\left( x \right)+2x+4$
$\Rightarrow I=\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx=\int\limits_{0}^{1}{\left[ xf\left( x \right)f'\left( x \right)+2x+4 \right]dx=\int\limits_{0}^{1}{xf\left( x \right)f'\left( x \right)dx+\int\limits_{0}^{1}{\left( 2x+4 \right)dx=A+5\left( * \right)}}}}$
Tính $A=\int\limits_{0}^{1}{xf\left( x \right)f'\left( x \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=xf\left( x \right) \\
& dv=f'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\left( f\left( x \right)+xf'\left( x \right) \right)dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\Rightarrow A=x{{f}^{2}}\left. \left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)\left( f\left( x \right)+xf'\left( x \right) \right)dx=9-\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}-\int\limits_{0}^{1}{xf\left( x \right)f'\left( x \right)dx}}$
$\Rightarrow A=\dfrac{9-I}{2}\left( 2* \right)$. Thay (2*) vào (*), ta được: $I=\dfrac{9-I}{2}+5\Leftrightarrow I=\dfrac{19}{3}$
$\Rightarrow I=\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx=\int\limits_{0}^{1}{\left[ xf\left( x \right)f'\left( x \right)+2x+4 \right]dx=\int\limits_{0}^{1}{xf\left( x \right)f'\left( x \right)dx+\int\limits_{0}^{1}{\left( 2x+4 \right)dx=A+5\left( * \right)}}}}$
Tính $A=\int\limits_{0}^{1}{xf\left( x \right)f'\left( x \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=xf\left( x \right) \\
& dv=f'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\left( f\left( x \right)+xf'\left( x \right) \right)dx \\
& v=f\left( x \right) \\
\end{aligned} \right.$
$\Rightarrow A=x{{f}^{2}}\left. \left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{f\left( x \right)\left( f\left( x \right)+xf'\left( x \right) \right)dx=9-\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}-\int\limits_{0}^{1}{xf\left( x \right)f'\left( x \right)dx}}$
$\Rightarrow A=\dfrac{9-I}{2}\left( 2* \right)$. Thay (2*) vào (*), ta được: $I=\dfrac{9-I}{2}+5\Leftrightarrow I=\dfrac{19}{3}$
Đáp án C.