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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0;1 \right]$ thỏa mãn $f\left( 0 \right)=1$ và ${{\left( f'\left( x \right) \right)}^{2}}+4\left( 6{{x}^{2}}-1 \right)f\left( x \right)=40{{x}^{6}}-44{{x}^{4}}+32{{x}^{2}}-4,\forall x\in \left[ 0;1 \right]$. Tích phân $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{23}{15}$
B. $-\dfrac{17}{15}$
C. $\dfrac{13}{15}$
D. $-\dfrac{7}{15}$
A. $\dfrac{23}{15}$
B. $-\dfrac{17}{15}$
C. $\dfrac{13}{15}$
D. $-\dfrac{7}{15}$
Từ (1) ta có
$\int\limits_{0}^{1}{{{\left( f'\left( x \right) \right)}^{2}}dx+}4\int\limits_{0}^{1}{\left( 6{{x}^{2}}-1 \right)dx}=\int\limits_{0}^{1}{\left( 40{{x}^{6}}-44{{x}^{4}}+32{{x}^{2}}-4 \right)dx=\dfrac{376}{105}\left( 2 \right)}$
Theo công thức tích phân từng phần ta có
$\begin{aligned}
& \int\limits_{0}^{1}{\left( 6{{x}^{1}}-1 \right)f\left( x \right)dx=\int\limits_{0}^{1}{f\left( x \right)d\left( 2{{x}^{3}}-x \right)=\left( 2{{x}^{3}}-x \right)}}f\left. \left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\left( 2{{x}^{3}}-x \right)f'\left( x \right)dx} \\
& \Leftrightarrow \int\limits_{0}^{1}{\left( 6{{x}^{2}}-1 \right)}f\left( x \right)dx=1-\int\limits_{0}^{1}{\left( 2{{x}^{3}}-x \right)}f'\left( x \right)dx \\
\end{aligned}$
Thay vào (*) ta được
$\begin{aligned}
& \int\limits_{0}^{1}{{{\left( f'\left( x \right) \right)}^{2}}dx+4\left( 1-\int\limits_{0}^{1}{\left( 2{{x}^{3}}-x \right)f'\left( x \right)dx} \right)}=\dfrac{376}{105} \\
& \Leftrightarrow \int\limits_{0}^{1}{{{\left( f'\left( x \right) \right)}^{2}}dx-4\int\limits_{0}^{1}{\left( 2{{x}^{3}}-x \right)f'\left( x \right)dx+\dfrac{44}{105}=0}} \\
& \Leftrightarrow \int\limits_{0}^{1}{{{\left( f'\left( x \right)-2\left( 2{{x}^{3}}-x \right) \right)}^{2}}dx=0\Leftrightarrow f'\left( x \right)=2\left( 2{{x}^{3}}-x \right)},\forall x\in \left[ 0;1 \right] \\
& \Rightarrow f\left( x \right)=4{{x}^{4}}-{{x}^{2}}+C \\
\end{aligned}$
Mặt khác $f\left( 1 \right)=1\Rightarrow C=1\Rightarrow f\left( x \right)={{x}^{4}}-{{x}^{2}}+1\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx=\int\limits_{0}^{1}{\left( {{x}^{4}}-{{x}^{2}}+1 \right)}}dx=\dfrac{13}{15}$
$\int\limits_{0}^{1}{{{\left( f'\left( x \right) \right)}^{2}}dx+}4\int\limits_{0}^{1}{\left( 6{{x}^{2}}-1 \right)dx}=\int\limits_{0}^{1}{\left( 40{{x}^{6}}-44{{x}^{4}}+32{{x}^{2}}-4 \right)dx=\dfrac{376}{105}\left( 2 \right)}$
Theo công thức tích phân từng phần ta có
$\begin{aligned}
& \int\limits_{0}^{1}{\left( 6{{x}^{1}}-1 \right)f\left( x \right)dx=\int\limits_{0}^{1}{f\left( x \right)d\left( 2{{x}^{3}}-x \right)=\left( 2{{x}^{3}}-x \right)}}f\left. \left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\left( 2{{x}^{3}}-x \right)f'\left( x \right)dx} \\
& \Leftrightarrow \int\limits_{0}^{1}{\left( 6{{x}^{2}}-1 \right)}f\left( x \right)dx=1-\int\limits_{0}^{1}{\left( 2{{x}^{3}}-x \right)}f'\left( x \right)dx \\
\end{aligned}$
Thay vào (*) ta được
$\begin{aligned}
& \int\limits_{0}^{1}{{{\left( f'\left( x \right) \right)}^{2}}dx+4\left( 1-\int\limits_{0}^{1}{\left( 2{{x}^{3}}-x \right)f'\left( x \right)dx} \right)}=\dfrac{376}{105} \\
& \Leftrightarrow \int\limits_{0}^{1}{{{\left( f'\left( x \right) \right)}^{2}}dx-4\int\limits_{0}^{1}{\left( 2{{x}^{3}}-x \right)f'\left( x \right)dx+\dfrac{44}{105}=0}} \\
& \Leftrightarrow \int\limits_{0}^{1}{{{\left( f'\left( x \right)-2\left( 2{{x}^{3}}-x \right) \right)}^{2}}dx=0\Leftrightarrow f'\left( x \right)=2\left( 2{{x}^{3}}-x \right)},\forall x\in \left[ 0;1 \right] \\
& \Rightarrow f\left( x \right)=4{{x}^{4}}-{{x}^{2}}+C \\
\end{aligned}$
Mặt khác $f\left( 1 \right)=1\Rightarrow C=1\Rightarrow f\left( x \right)={{x}^{4}}-{{x}^{2}}+1\Rightarrow \int\limits_{0}^{1}{f\left( x \right)dx=\int\limits_{0}^{1}{\left( {{x}^{4}}-{{x}^{2}}+1 \right)}}dx=\dfrac{13}{15}$
Đáp án C.