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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm dương, liên tục trên đoạn $\left[ 0;1 \right]$, thỏa mãn $f\left( 0 \right)=1$ và $3\int\limits_{0}^{1}{\left[ f'\left( x \right).{{f}^{2}}\left( x \right)+\dfrac{1}{9} \right]dx}=2\int\limits_{0}^{1}{\sqrt{f'\left( x \right)}.f\left( x \right)dx}$. Tính $I=\int\limits_{0}^{1}{{{f}^{3}}\left( x \right)dx}.$
A. $I=\dfrac{3}{2}.$
B. $I=\dfrac{5}{4}.$
C. $I=\dfrac{5}{6}.$
D. $I=\dfrac{7}{6}.$
A. $I=\dfrac{3}{2}.$
B. $I=\dfrac{5}{4}.$
C. $I=\dfrac{5}{6}.$
D. $I=\dfrac{7}{6}.$
Giả thiết tương đương với $3\int\limits_{0}^{1}{{{\left[ \sqrt{f'\left( x \right)}.f\left( x \right) \right]}^{2}}dx}+\dfrac{1}{3}=2\int\limits_{0}^{1}{\sqrt{f'\left( x \right)}.f\left( x \right)dx}$.
$\begin{aligned}
& \Leftrightarrow \int\limits_{0}^{1}{{{\left[ 3\sqrt{f'\left( x \right)}.f\left( x \right) \right]}^{2}}dx}-2\int\limits_{0}^{1}{3\sqrt{f'\left( x \right)}.f\left( x \right)dx}+\int\limits_{0}^{1}{dx}=0 \\
& \Leftrightarrow \int\limits_{0}^{1}{{{\left[ 3\sqrt{f'\left( x \right)}.f\left( x \right)-1 \right]}^{2}}dx=0\Leftrightarrow 3\sqrt{f'\left( x \right)}.f\left( x \right)=1,\forall x\in \left[ 0;1 \right]\Leftrightarrow 9f'\left( x \right).{{f}^{2}}\left( x \right)=1} \\
\end{aligned}$
$\Rightarrow \int{9f'\left( x \right).{{f}^{2}}\left( x \right)dx}=\int{dx}$ hay $9\int{{{f}^{2}}\left( x \right)d\left( f\left( x \right) \right)dx}=\int{dx}\Rightarrow 3{{f}^{3}}\left( x \right)=x+C$.
Do $f\left( 0 \right)=1$, nên ta có: $3{{f}^{3}}\left( 0 \right)=0+C\Leftrightarrow C=3$.
Vậy ${{f}^{3}}\left( x \right)=\dfrac{1}{3}x+1\Rightarrow \int\limits_{0}^{1}{{{f}^{3}}\left( x \right)dx}=\dfrac{7}{6}$.
$\begin{aligned}
& \Leftrightarrow \int\limits_{0}^{1}{{{\left[ 3\sqrt{f'\left( x \right)}.f\left( x \right) \right]}^{2}}dx}-2\int\limits_{0}^{1}{3\sqrt{f'\left( x \right)}.f\left( x \right)dx}+\int\limits_{0}^{1}{dx}=0 \\
& \Leftrightarrow \int\limits_{0}^{1}{{{\left[ 3\sqrt{f'\left( x \right)}.f\left( x \right)-1 \right]}^{2}}dx=0\Leftrightarrow 3\sqrt{f'\left( x \right)}.f\left( x \right)=1,\forall x\in \left[ 0;1 \right]\Leftrightarrow 9f'\left( x \right).{{f}^{2}}\left( x \right)=1} \\
\end{aligned}$
$\Rightarrow \int{9f'\left( x \right).{{f}^{2}}\left( x \right)dx}=\int{dx}$ hay $9\int{{{f}^{2}}\left( x \right)d\left( f\left( x \right) \right)dx}=\int{dx}\Rightarrow 3{{f}^{3}}\left( x \right)=x+C$.
Do $f\left( 0 \right)=1$, nên ta có: $3{{f}^{3}}\left( 0 \right)=0+C\Leftrightarrow C=3$.
Vậy ${{f}^{3}}\left( x \right)=\dfrac{1}{3}x+1\Rightarrow \int\limits_{0}^{1}{{{f}^{3}}\left( x \right)dx}=\dfrac{7}{6}$.
Đáp án D.