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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 $f\left( 1 \right)=0$, $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}=5$ và $\int\limits_{0}^{1}{xf\left( x \right)\text{d}x}=\dfrac{1}{2}$. Tích phân $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$ bằng
A. $\dfrac{10}{9}$.
B. $\dfrac{11}{4}$.
C. $-\dfrac{10}{9}$.
D. $-\dfrac{11}{4}$.
A. $\dfrac{10}{9}$.
B. $\dfrac{11}{4}$.
C. $-\dfrac{10}{9}$.
D. $-\dfrac{11}{4}$.
Theo giả thiết, ta có $\int\limits_{0}^{1}{xf\left( x \right)\text{d}x}=\dfrac{1}{2}$. Đặt: $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& \text{d}v=x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={f}'\left( x \right)\text{d}x \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$.
Ta có $\dfrac{1}{2}=\int\limits_{0}^{1}{xf\left( x \right)\text{d}x}=\left. \dfrac{{{x}^{2}}}{2}f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}.{f}'\left( x \right)\text{d}x}$
$\Leftrightarrow 0-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}.{f}'\left( x \right)\text{d}x}=\dfrac{1}{2}$ $\Leftrightarrow -\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}.{f}'\left( x \right)\text{d}x}=\dfrac{1}{2}$ $\Leftrightarrow \int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)\text{d}x}=-1.$
Ta có: $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}=5$
$\int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)\text{d}x}=-1$ $\Leftrightarrow 10\int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)\text{d}x}=10\left( -1 \right)$ $\Leftrightarrow \int\limits_{0}^{1}{2\left( 5{{x}^{2}} \right).{f}'\left( x \right)\text{d}x}=-10.$
$\int\limits_{0}^{1}{{{\left( -5{{x}^{2}} \right)}^{2}}\text{d}x}=5$
Từ đó, ta có $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}+\int\limits_{0}^{1}{2\left( 5{{x}^{2}} \right).{f}'\left( x \right)\text{d}x}+\int\limits_{0}^{1}{{{\left( 5{{x}^{2}} \right)}^{2}}\text{d}x}=5-10+5.$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+5{{x}^{2}} \right]}^{2}}}dx=0$ $\Leftrightarrow {f}'\left( x \right)+5{{x}^{2}}=0\Leftrightarrow {f}'\left( x \right)=-5{{x}^{2}}\Leftrightarrow f\left( x \right)=-\dfrac{5}{3}{{x}^{3}}+C$
Mà $f\left( 1 \right)=0\Leftrightarrow 0=-\dfrac{5}{3}+C\Leftrightarrow C=\dfrac{5}{3}$. Khi đó: $f\left( x \right)=-\dfrac{5}{3}{{x}^{2}}+\dfrac{5}{3}$.
Vậy: $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( -\dfrac{5}{3}{{x}^{2}}+\dfrac{5}{3} \right)}\text{d}x$ $=\left. \left( -\dfrac{5{{x}^{3}}}{9}+\dfrac{5x}{3} \right) \right|_{0}^{1}$ $=\dfrac{10}{9}$.
& u=f\left( x \right) \\
& \text{d}v=x\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={f}'\left( x \right)\text{d}x \\
& v=\dfrac{{{x}^{2}}}{2} \\
\end{aligned} \right.$.
Ta có $\dfrac{1}{2}=\int\limits_{0}^{1}{xf\left( x \right)\text{d}x}=\left. \dfrac{{{x}^{2}}}{2}f\left( x \right) \right|_{0}^{1}-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}.{f}'\left( x \right)\text{d}x}$
$\Leftrightarrow 0-\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}.{f}'\left( x \right)\text{d}x}=\dfrac{1}{2}$ $\Leftrightarrow -\int\limits_{0}^{1}{\dfrac{{{x}^{2}}}{2}.{f}'\left( x \right)\text{d}x}=\dfrac{1}{2}$ $\Leftrightarrow \int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)\text{d}x}=-1.$
Ta có: $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}=5$
$\int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)\text{d}x}=-1$ $\Leftrightarrow 10\int\limits_{0}^{1}{{{x}^{2}}.{f}'\left( x \right)\text{d}x}=10\left( -1 \right)$ $\Leftrightarrow \int\limits_{0}^{1}{2\left( 5{{x}^{2}} \right).{f}'\left( x \right)\text{d}x}=-10.$
$\int\limits_{0}^{1}{{{\left( -5{{x}^{2}} \right)}^{2}}\text{d}x}=5$
Từ đó, ta có $\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}+\int\limits_{0}^{1}{2\left( 5{{x}^{2}} \right).{f}'\left( x \right)\text{d}x}+\int\limits_{0}^{1}{{{\left( 5{{x}^{2}} \right)}^{2}}\text{d}x}=5-10+5.$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)+5{{x}^{2}} \right]}^{2}}}dx=0$ $\Leftrightarrow {f}'\left( x \right)+5{{x}^{2}}=0\Leftrightarrow {f}'\left( x \right)=-5{{x}^{2}}\Leftrightarrow f\left( x \right)=-\dfrac{5}{3}{{x}^{3}}+C$
Mà $f\left( 1 \right)=0\Leftrightarrow 0=-\dfrac{5}{3}+C\Leftrightarrow C=\dfrac{5}{3}$. Khi đó: $f\left( x \right)=-\dfrac{5}{3}{{x}^{2}}+\dfrac{5}{3}$.
Vậy: $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\int\limits_{0}^{1}{\left( -\dfrac{5}{3}{{x}^{2}}+\dfrac{5}{3} \right)}\text{d}x$ $=\left. \left( -\dfrac{5{{x}^{3}}}{9}+\dfrac{5x}{3} \right) \right|_{0}^{1}$ $=\dfrac{10}{9}$.
Đáp án A.