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Câu hỏi: . Cho hàm số f(x) liên tục và luôn dương trên Biết rằng $f\left( 4 \right)=2,$ $\int\limits_{0}^{1}{\dfrac{xdx}{f\left( 4x \right)}=1.}$ Tính tích phân $I=\int\limits_{0}^{4}{\dfrac{{{x}^{2}}{f}'\left( x \right)dx}{{{f}^{2}}\left( x \right)}.}$
A. $I=12.$
B. $I=16.$
C. $I=6.$
D. $I=24.$
A. $I=12.$
B. $I=16.$
C. $I=6.$
D. $I=24.$
Đặt $t=4x\Rightarrow dt=4dx$, đổi cận ta được $\int\limits_{0}^{1}{\dfrac{xdx}{f\left( 4x \right)}=\int\limits_{0}^{4}{\dfrac{t.\dfrac{dt}{4}}{4f\left( t \right)}}=1\Leftrightarrow \int\limits_{0}^{4}{\dfrac{tdt}{f\left( t \right)}=16}}$
Do đó $\int\limits_{0}^{4}{\dfrac{xdx}{f\left( x \right)}}=16$, đặt $\left\{ \begin{aligned}
& u=\dfrac{1}{f\left( x \right)} \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)} \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
Suy ra $\int\limits_{0}^{4}{\dfrac{xdx}{f\left( x \right)}}=\left. \dfrac{{{x}^{2}}}{2f\left( x \right)} \right|_{0}^{4}+\dfrac{1}{2}\int\limits_{0}^{4}{\dfrac{{{x}^{2}}{f}'\left( x \right)dx}{{{f}^{2}}\left( x \right)}}\Leftrightarrow 16=\dfrac{16}{2f\left( 4 \right)}+\dfrac{1}{2}I\Leftrightarrow I=24$.
Do đó $\int\limits_{0}^{4}{\dfrac{xdx}{f\left( x \right)}}=16$, đặt $\left\{ \begin{aligned}
& u=\dfrac{1}{f\left( x \right)} \\
& dv=xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=\dfrac{-{f}'\left( x \right)}{{{f}^{2}}\left( x \right)} \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$
Suy ra $\int\limits_{0}^{4}{\dfrac{xdx}{f\left( x \right)}}=\left. \dfrac{{{x}^{2}}}{2f\left( x \right)} \right|_{0}^{4}+\dfrac{1}{2}\int\limits_{0}^{4}{\dfrac{{{x}^{2}}{f}'\left( x \right)dx}{{{f}^{2}}\left( x \right)}}\Leftrightarrow 16=\dfrac{16}{2f\left( 4 \right)}+\dfrac{1}{2}I\Leftrightarrow I=24$.
Đáp án D.