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Câu hỏi: Cho hàm số $f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0;2 \right]$ thỏa mãn $f\left( 2 \right)=1$, $\int\limits_{0}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}}dx=\dfrac{2}{7}$ và $\int\limits_{0}^{2}{{{x}^{2}}.f\left( x \right)}dx=\dfrac{40}{21}$. Tính tích phân $I=\int\limits_{0}^{2}{f\left( x \right)dx}$.
A. $I=21$.
B. $I=\dfrac{6}{5}$.
C. $I=\dfrac{84}{3}$.
D. $I=\dfrac{8}{5}$.
A. $I=21$.
B. $I=\dfrac{6}{5}$.
C. $I=\dfrac{84}{3}$.
D. $I=\dfrac{8}{5}$.
Đặt $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv={{x}^{2}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)dx \\
& v=\dfrac{{{x}^{3}}}{3} \\
\end{aligned} \right. $, khi đó $ \int\limits_{0}^{2}{{{x}^{2}}.f\left( x \right)dx}=\dfrac{{{x}^{3}}}{3}.f\left( x \right)\left| \begin{aligned}
& ^{2} \\
& _{0} \\
\end{aligned} \right.-\int\limits_{0}^{2}{\dfrac{{{x}^{3}}}{3}{f}'\left( x \right)dx}$
Suy ra $\dfrac{40}{21}=\dfrac{8}{3}f\left( 2 \right)-\int\limits_{0}^{2}{\dfrac{{{x}^{3}}}{3}{f}'\left( x \right)dx}\Rightarrow \int\limits_{0}^{2}{{{x}^{3}}{f}'\left( x \right)dx}=\dfrac{16}{7}$
Ta chọn k sao cho: $\int\limits_{0}^{2}{{{\left[ {f}'\left( x \right)+k{{x}^{3}} \right]}^{2}}dx}=\int\limits_{0}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx+2k\int\limits_{0}^{2}{{f}'\left( x \right){{x}^{3}}dx}+{{k}^{2}}\int\limits_{0}^{2}{{{x}^{6}}dx}=0}$
$=\dfrac{2}{7}+\dfrac{32}{7}k+\dfrac{128{{k}^{2}}}{7}=0\Rightarrow k=\dfrac{-1}{8}\Rightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)-\dfrac{1}{8}{{x}^{3}} \right]}^{2}}dx=0}$
$\Rightarrow {f}'\left( x \right)=\dfrac{{{x}^{3}}}{8}\Rightarrow f\left( x \right)=\dfrac{{{x}^{4}}}{32}+C$
Do $f\left( 2 \right)=1\Rightarrow C=\dfrac{1}{2}\Rightarrow f\left( x \right)=\dfrac{{{x}^{4}}}{32}+\dfrac{1}{2}\Rightarrow \int\limits_{0}^{2}{f\left( x \right)dx}=\dfrac{6}{5}$
& u=f\left( x \right) \\
& dv={{x}^{2}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)dx \\
& v=\dfrac{{{x}^{3}}}{3} \\
\end{aligned} \right. $, khi đó $ \int\limits_{0}^{2}{{{x}^{2}}.f\left( x \right)dx}=\dfrac{{{x}^{3}}}{3}.f\left( x \right)\left| \begin{aligned}
& ^{2} \\
& _{0} \\
\end{aligned} \right.-\int\limits_{0}^{2}{\dfrac{{{x}^{3}}}{3}{f}'\left( x \right)dx}$
Suy ra $\dfrac{40}{21}=\dfrac{8}{3}f\left( 2 \right)-\int\limits_{0}^{2}{\dfrac{{{x}^{3}}}{3}{f}'\left( x \right)dx}\Rightarrow \int\limits_{0}^{2}{{{x}^{3}}{f}'\left( x \right)dx}=\dfrac{16}{7}$
Ta chọn k sao cho: $\int\limits_{0}^{2}{{{\left[ {f}'\left( x \right)+k{{x}^{3}} \right]}^{2}}dx}=\int\limits_{0}^{2}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx+2k\int\limits_{0}^{2}{{f}'\left( x \right){{x}^{3}}dx}+{{k}^{2}}\int\limits_{0}^{2}{{{x}^{6}}dx}=0}$
$=\dfrac{2}{7}+\dfrac{32}{7}k+\dfrac{128{{k}^{2}}}{7}=0\Rightarrow k=\dfrac{-1}{8}\Rightarrow \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right)-\dfrac{1}{8}{{x}^{3}} \right]}^{2}}dx=0}$
$\Rightarrow {f}'\left( x \right)=\dfrac{{{x}^{3}}}{8}\Rightarrow f\left( x \right)=\dfrac{{{x}^{4}}}{32}+C$
Do $f\left( 2 \right)=1\Rightarrow C=\dfrac{1}{2}\Rightarrow f\left( x \right)=\dfrac{{{x}^{4}}}{32}+\dfrac{1}{2}\Rightarrow \int\limits_{0}^{2}{f\left( x \right)dx}=\dfrac{6}{5}$
Đáp án B.