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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 0;1 \right]$ thỏa mãn $f\left( 1 \right)=0$, $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}=7$ và $\int\limits_{0}^{1}{{{x}^{2}}f\left( x \right)dx}=\dfrac{1}{3}.$ Tích phân $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{7}{5}.$
B. 1.
C. $\dfrac{7}{4}.$
D. 4.
A. $\dfrac{7}{5}.$
B. 1.
C. $\dfrac{7}{4}.$
D. 4.
Tính: $\int\limits_{0}^{1}{{{x}^{2}}f\left( x \right)}dx.$ Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv={{x}^{2}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)dx \\
& v=\dfrac{{{x}^{3}}}{3} \\
\end{aligned} \right.$
Ta có: $\int\limits_{0}^{1}{{{x}^{2}}f\left( x \right)dx}=\dfrac{{{x}^{3}}f\left( x \right)}{3}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx}$
$=\dfrac{1.f\left( 1 \right)-0.f\left( 1 \right)}{3}-\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx}=-\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx.}$
Mà $\int\limits_{0}^{1}{{{x}^{2}}f\left( x \right)dx}=\dfrac{1}{3}\Rightarrow -\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}}.f{{\left( x \right)}^{\prime }}dx=\dfrac{1}{3}\Rightarrow \int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx=-1.}$
Ta có $\int\limits_{0}^{1}{{{\left[ f{{\left( x \right)}^{\prime }} \right]}^{2}}dx=7\left( 1 \right),}$ $\int\limits_{0}^{1}{{{x}^{6}}.dx=\dfrac{{{x}^{7}}}{7}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{1}{7}}\Rightarrow \int\limits_{0}^{1}{49{{x}^{6}}.dx}=\dfrac{1}{7}.49=7$ (2).
$\int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx=-1}\Rightarrow \int\limits_{0}^{1}{14{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx}=-14\left( 3 \right).$ Cộng hai vế (1) (2) và (3) suy ra
${{\int\limits_{0}^{1}{\left[ f{{\left( x \right)}^{\prime }} \right]}}^{2}}dx+\int\limits_{0}^{1}{49{{x}^{6}}.dx}+\int\limits_{0}^{1}{14{{x}^{3}}.f{{\left( x \right)}^{\prime }}}dx=7+7-14=0.$
$\Rightarrow \int\limits_{0}^{1}{\left\{ {{\left[ f'\left( x \right) \right]}^{2}}+14{{x}^{3}}{f}'\left( x \right)+49{{x}^{6}} \right\}}dx=0\Rightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+7{{x}^{3}} \right]}^{2}}dx=0.}$
Do ${{\left[ {f}'\left( x \right)+7{{x}^{3}} \right]}^{2}}\ge 0\Rightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+7{{x}^{3}} \right]}^{2}}}dx\ge 0$. Mà $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+7{{x}^{3}} \right]}^{2}}dx=0}$
$\Rightarrow {f}'\left( x \right)=-7{{x}^{3}}$ suy ra $\int\limits_{0}^{1}{{f}'\left( x \right)dx=\int\limits_{0}^{1}{\left( -7{{x}^{3}} \right)dx}\Rightarrow f\left( x \right)=-\dfrac{7{{x}^{4}}}{4}+C}$
Mà $f\left( 1 \right)=0\Rightarrow -\dfrac{7}{4}+C=0\Rightarrow C=\dfrac{7}{4}.$ Do đó $f\left( x \right)=-\dfrac{7{{x}^{4}}}{4}+\dfrac{7}{4}.$
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx=}\int\limits_{0}^{1}{\left( -\dfrac{7{{x}^{4}}}{4}+\dfrac{7}{4} \right)dx}=\left( -\dfrac{7{{x}^{5}}}{20}+\dfrac{7}{4}x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{7}{5}.$
& u=f\left( x \right) \\
& dv={{x}^{2}} \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)dx \\
& v=\dfrac{{{x}^{3}}}{3} \\
\end{aligned} \right.$
Ta có: $\int\limits_{0}^{1}{{{x}^{2}}f\left( x \right)dx}=\dfrac{{{x}^{3}}f\left( x \right)}{3}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.-\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx}$
$=\dfrac{1.f\left( 1 \right)-0.f\left( 1 \right)}{3}-\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx}=-\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx.}$
Mà $\int\limits_{0}^{1}{{{x}^{2}}f\left( x \right)dx}=\dfrac{1}{3}\Rightarrow -\dfrac{1}{3}\int\limits_{0}^{1}{{{x}^{3}}}.f{{\left( x \right)}^{\prime }}dx=\dfrac{1}{3}\Rightarrow \int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx=-1.}$
Ta có $\int\limits_{0}^{1}{{{\left[ f{{\left( x \right)}^{\prime }} \right]}^{2}}dx=7\left( 1 \right),}$ $\int\limits_{0}^{1}{{{x}^{6}}.dx=\dfrac{{{x}^{7}}}{7}\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{1}{7}}\Rightarrow \int\limits_{0}^{1}{49{{x}^{6}}.dx}=\dfrac{1}{7}.49=7$ (2).
$\int\limits_{0}^{1}{{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx=-1}\Rightarrow \int\limits_{0}^{1}{14{{x}^{3}}.f{{\left( x \right)}^{\prime }}dx}=-14\left( 3 \right).$ Cộng hai vế (1) (2) và (3) suy ra
${{\int\limits_{0}^{1}{\left[ f{{\left( x \right)}^{\prime }} \right]}}^{2}}dx+\int\limits_{0}^{1}{49{{x}^{6}}.dx}+\int\limits_{0}^{1}{14{{x}^{3}}.f{{\left( x \right)}^{\prime }}}dx=7+7-14=0.$
$\Rightarrow \int\limits_{0}^{1}{\left\{ {{\left[ f'\left( x \right) \right]}^{2}}+14{{x}^{3}}{f}'\left( x \right)+49{{x}^{6}} \right\}}dx=0\Rightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+7{{x}^{3}} \right]}^{2}}dx=0.}$
Do ${{\left[ {f}'\left( x \right)+7{{x}^{3}} \right]}^{2}}\ge 0\Rightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+7{{x}^{3}} \right]}^{2}}}dx\ge 0$. Mà $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+7{{x}^{3}} \right]}^{2}}dx=0}$
$\Rightarrow {f}'\left( x \right)=-7{{x}^{3}}$ suy ra $\int\limits_{0}^{1}{{f}'\left( x \right)dx=\int\limits_{0}^{1}{\left( -7{{x}^{3}} \right)dx}\Rightarrow f\left( x \right)=-\dfrac{7{{x}^{4}}}{4}+C}$
Mà $f\left( 1 \right)=0\Rightarrow -\dfrac{7}{4}+C=0\Rightarrow C=\dfrac{7}{4}.$ Do đó $f\left( x \right)=-\dfrac{7{{x}^{4}}}{4}+\dfrac{7}{4}.$
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx=}\int\limits_{0}^{1}{\left( -\dfrac{7{{x}^{4}}}{4}+\dfrac{7}{4} \right)dx}=\left( -\dfrac{7{{x}^{5}}}{20}+\dfrac{7}{4}x \right)\left| \begin{aligned}
& 1 \\
& 0 \\
\end{aligned} \right.=\dfrac{7}{5}.$
Đáp án A.