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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 1 ;2 \right]$ thỏa mãn $\int\limits_{1}^{2}{{{\left( x-1 \right)}^{2}}f\left( x \right)\text{d}x}=-\dfrac{1}{3}$, $f\left( 2 \right)=0$ và $\int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}=7$. Tính tích phân $I=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}$.
A. $I=\dfrac{7}{5}$.
B. $I=-\dfrac{7}{5}$.
C. $I=-\dfrac{7}{20}$.
D. $I=\dfrac{7}{20}$.
A. $I=\dfrac{7}{5}$.
B. $I=-\dfrac{7}{5}$.
C. $I=-\dfrac{7}{20}$.
D. $I=\dfrac{7}{20}$.
Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& \text{d}v={{\left( x-1 \right)}^{2}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={f}'\left( x \right)\text{d}x \\
& v=\dfrac{{{\left( x-1 \right)}^{3}}}{3} \\
\end{aligned} \right.$.
Khi đó, $\int\limits_{1}^{2}{{{\left( x-1 \right)}^{2}}f\left( x \right)\text{d}x}=\left. \dfrac{{{\left( x-1 \right)}^{3}}f\left( x \right)}{3} \right|_{1}^{2}-\dfrac{1}{3}\int\limits_{1}^{2}{{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}$
$\Rightarrow -\dfrac{1}{3}=-\dfrac{1}{3}\int\limits_{1}^{2}{{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}$
$\Rightarrow \int\limits_{1}^{2}{{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}=1$.
Ta lại có: $\left\{ \begin{aligned}
& \int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}=7 \\
& \int\limits_{1}^{2}{14{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}=14 \\
& \int\limits_{1}^{2}{49{{\left( x-1 \right)}^{6}}\text{d}x}=\left. 7{{\left( x-1 \right)}^{7}} \right|_{1}^{2}=7 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}-\int\limits_{1}^{2}{14{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}+\int\limits_{1}^{2}{49{{\left( x-1 \right)}^{6}}\text{d}x}=0$
$\Rightarrow \int\limits_{1}^{2}{{{\left[ {f}'\left( x \right)-7{{\left( x-1 \right)}^{3}} \right]}^{2}}\text{d}x}=0$ $\left( 1 \right)$, mà $\int\limits_{1}^{2}{{{\left[ {f}'\left( x \right)-7{{\left( x-1 \right)}^{3}} \right]}^{2}}\text{d}x}\ge 0$.
nên $\left( 1 \right)\Rightarrow {f}'\left( x \right)-7{{\left( x-1 \right)}^{3}}=0\Rightarrow {f}'\left( x \right)=7{{\left( x-1 \right)}^{3}}\Rightarrow f\left( x \right)=\dfrac{7{{\left( x-1 \right)}^{4}}}{4}+C$.
Mà $f\left( 2 \right)=0\Leftrightarrow \dfrac{7}{4}+C=0\Leftrightarrow C=-\dfrac{7}{4}\Rightarrow f\left( x \right)=\dfrac{7}{4}\left[ {{\left( x-1 \right)}^{4}}-1 \right]$.
$\Rightarrow I=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}=\dfrac{7}{4}\int\limits_{1}^{2}{\left[ {{\left( x-1 \right)}^{4}}-1 \right]\text{d}x}=\left. \dfrac{7}{4}\left[ \dfrac{{{\left( x-1 \right)}^{5}}}{5}-x \right] \right|_{1}^{2}=-\dfrac{7}{5}$.
Vậy $I=-\dfrac{7}{5}$.
& u=f\left( x \right) \\
& \text{d}v={{\left( x-1 \right)}^{2}}\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \text{d}u={f}'\left( x \right)\text{d}x \\
& v=\dfrac{{{\left( x-1 \right)}^{3}}}{3} \\
\end{aligned} \right.$.
Khi đó, $\int\limits_{1}^{2}{{{\left( x-1 \right)}^{2}}f\left( x \right)\text{d}x}=\left. \dfrac{{{\left( x-1 \right)}^{3}}f\left( x \right)}{3} \right|_{1}^{2}-\dfrac{1}{3}\int\limits_{1}^{2}{{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}$
$\Rightarrow -\dfrac{1}{3}=-\dfrac{1}{3}\int\limits_{1}^{2}{{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}$
$\Rightarrow \int\limits_{1}^{2}{{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}=1$.
Ta lại có: $\left\{ \begin{aligned}
& \int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}=7 \\
& \int\limits_{1}^{2}{14{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}=14 \\
& \int\limits_{1}^{2}{49{{\left( x-1 \right)}^{6}}\text{d}x}=\left. 7{{\left( x-1 \right)}^{7}} \right|_{1}^{2}=7 \\
\end{aligned} \right.$
$\Rightarrow \int\limits_{1}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}\text{d}x}-\int\limits_{1}^{2}{14{{\left( x-1 \right)}^{3}}{f}'\left( x \right)\text{d}x}+\int\limits_{1}^{2}{49{{\left( x-1 \right)}^{6}}\text{d}x}=0$
$\Rightarrow \int\limits_{1}^{2}{{{\left[ {f}'\left( x \right)-7{{\left( x-1 \right)}^{3}} \right]}^{2}}\text{d}x}=0$ $\left( 1 \right)$, mà $\int\limits_{1}^{2}{{{\left[ {f}'\left( x \right)-7{{\left( x-1 \right)}^{3}} \right]}^{2}}\text{d}x}\ge 0$.
nên $\left( 1 \right)\Rightarrow {f}'\left( x \right)-7{{\left( x-1 \right)}^{3}}=0\Rightarrow {f}'\left( x \right)=7{{\left( x-1 \right)}^{3}}\Rightarrow f\left( x \right)=\dfrac{7{{\left( x-1 \right)}^{4}}}{4}+C$.
Mà $f\left( 2 \right)=0\Leftrightarrow \dfrac{7}{4}+C=0\Leftrightarrow C=-\dfrac{7}{4}\Rightarrow f\left( x \right)=\dfrac{7}{4}\left[ {{\left( x-1 \right)}^{4}}-1 \right]$.
$\Rightarrow I=\int\limits_{1}^{2}{f\left( x \right)\text{d}x}=\dfrac{7}{4}\int\limits_{1}^{2}{\left[ {{\left( x-1 \right)}^{4}}-1 \right]\text{d}x}=\left. \dfrac{7}{4}\left[ \dfrac{{{\left( x-1 \right)}^{5}}}{5}-x \right] \right|_{1}^{2}=-\dfrac{7}{5}$.
Vậy $I=-\dfrac{7}{5}$.
Đáp án B.