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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0;1 \right]$ thỏa mãn $f\left( 0 \right)=1$, $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}=\dfrac{1}{30}$, $\int\limits_{0}^{1}{\left( 2x-1 \right)f\left( x \right)dx}=-\dfrac{1}{30}$. Tích phân $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $\dfrac{11}{12}$.
B. $\dfrac{11}{4}$.
C. $\dfrac{1}{30}$.
D. $\dfrac{11}{30}$.
A. $\dfrac{11}{12}$.
B. $\dfrac{11}{4}$.
C. $\dfrac{1}{30}$.
D. $\dfrac{11}{30}$.
Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv=\left( 2x-1 \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=f\left( x \right)dx \\
& v={{x}^{2}}-x \\
\end{aligned} \right.$
$\Rightarrow -\dfrac{1}{30}=\int\limits_{0}^{1}{\left( 2x-1 \right)f\left( x \right)dx}=\left( {{x}^{2}}-x \right)f\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right){f}'\left( x \right)dx}=-\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right){f}'\left( x \right)dx}$
$\Rightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}-x \right){f}'\left( x \right)dx}=\dfrac{1}{30}$.
Ta có: $\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}dx}=\int\limits_{0}^{1}{\left( {{x}^{4}}-2{{x}^{3}}+{{x}^{2}} \right)dx}=\left( \dfrac{{{x}^{5}}}{5}-\dfrac{{{x}^{4}}}{2}+\dfrac{{{x}^{3}}}{3} \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=\dfrac{1}{30}$.
Do đó, $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)-\left( {{x}^{2}}-x \right) \right]}^{2}}dx}=\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}-2\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)f\left( x \right)dx}+\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}dx}=0$
$\Rightarrow {f}'\left( x \right)={{x}^{2}}-x\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+C$, mà $f\left( 0 \right)=1$ nên $C=1\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+1$
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( \dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+1 \right)dx}=\left( \dfrac{{{x}^{4}}}{12}-\dfrac{{{x}^{3}}}{6}+x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=\dfrac{11}{12}$
& u=f\left( x \right) \\
& dv=\left( 2x-1 \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=f\left( x \right)dx \\
& v={{x}^{2}}-x \\
\end{aligned} \right.$
$\Rightarrow -\dfrac{1}{30}=\int\limits_{0}^{1}{\left( 2x-1 \right)f\left( x \right)dx}=\left( {{x}^{2}}-x \right)f\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.-\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right){f}'\left( x \right)dx}=-\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right){f}'\left( x \right)dx}$
$\Rightarrow \int\limits_{0}^{1}{\left( {{x}^{2}}-x \right){f}'\left( x \right)dx}=\dfrac{1}{30}$.
Ta có: $\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}dx}=\int\limits_{0}^{1}{\left( {{x}^{4}}-2{{x}^{3}}+{{x}^{2}} \right)dx}=\left( \dfrac{{{x}^{5}}}{5}-\dfrac{{{x}^{4}}}{2}+\dfrac{{{x}^{3}}}{3} \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=\dfrac{1}{30}$.
Do đó, $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)-\left( {{x}^{2}}-x \right) \right]}^{2}}dx}=\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}-2\int\limits_{0}^{1}{\left( {{x}^{2}}-x \right)f\left( x \right)dx}+\int\limits_{0}^{1}{{{\left( {{x}^{2}}-x \right)}^{2}}dx}=0$
$\Rightarrow {f}'\left( x \right)={{x}^{2}}-x\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+C$, mà $f\left( 0 \right)=1$ nên $C=1\Rightarrow f\left( x \right)=\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+1$
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( \dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{2}}}{2}+1 \right)dx}=\left( \dfrac{{{x}^{4}}}{12}-\dfrac{{{x}^{3}}}{6}+x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=\dfrac{11}{12}$
Đáp án A.