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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 $f\left( 1 \right)=2, \int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}=8$ và $\int\limits_{0}^{1}{{{x}^{3}}.f\left( x \right)dx=10}$. Tích phân $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng
A. $-\dfrac{2}{285}$
B. $\dfrac{194}{95}$
C. $\dfrac{116}{57}$
D. $\dfrac{584}{285}$
A. $-\dfrac{2}{285}$
B. $\dfrac{194}{95}$
C. $\dfrac{116}{57}$
D. $\dfrac{584}{285}$
Tính: $I=\int\limits_{0}^{1}{{{x}^{3}}.f\left( x \right)dx}$
Đặt: $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv={{x}^{3}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)dx \\
& v=\dfrac{1}{4}{{x}^{4}} \\
\end{aligned} \right.$
Ta có: $I=\dfrac{1}{4}{{x}^{4}}.f\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.\left| -\dfrac{1}{4} \right.\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx}=\dfrac{1}{2}-\dfrac{1}{4}\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx} $ (vì $ f\left( 1 \right)=2$)
Theo giả thiết: $\int\limits_{0}^{1}{{{x}^{3}}.f\left( x \right)dx}=10\Rightarrow \int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx}=-38$
$\Leftrightarrow 8.\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx}=-38.8\Leftrightarrow 8.\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx}=-38.\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}$
$\Leftrightarrow \int\limits_{0}^{1}{\left( 8{{x}^{4}}{f}'\left( x \right)+38{{\left[ {f}'\left( x \right) \right]}^{2}} \right)dx}=0\Leftrightarrow \int\limits_{0}^{1}{{f}'\left( x \right).\left[ 8{{x}^{4}}+38{f}'\left( x \right) \right]dx}=0$
$\Rightarrow 8{{x}^{4}}+38f\left( x \right)=0\Leftrightarrow {f}'\left( x \right)=-\dfrac{4}{19}{{x}^{4}}\Rightarrow f\left( x \right)=-\dfrac{4}{95}{{x}^{5}}+C$
Với $f\left( 1 \right)=2\Rightarrow C=\dfrac{194}{195}$. Khi đó: $f\left( x \right)=-\dfrac{4}{95}{{x}^{5}}+\dfrac{194}{95}$
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( -\dfrac{4}{95}{{x}^{5}}+\dfrac{194}{95} \right)dx}=\left( -\dfrac{2}{285}{{x}^{6}}+\dfrac{194}{95} \right)x\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=\dfrac{116}{57}$
Đặt: $\left\{ \begin{aligned}
& u=f\left( x \right) \\
& dv={{x}^{3}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du={f}'\left( x \right)dx \\
& v=\dfrac{1}{4}{{x}^{4}} \\
\end{aligned} \right.$
Ta có: $I=\dfrac{1}{4}{{x}^{4}}.f\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.\left| -\dfrac{1}{4} \right.\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx}=\dfrac{1}{2}-\dfrac{1}{4}\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx} $ (vì $ f\left( 1 \right)=2$)
Theo giả thiết: $\int\limits_{0}^{1}{{{x}^{3}}.f\left( x \right)dx}=10\Rightarrow \int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx}=-38$
$\Leftrightarrow 8.\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx}=-38.8\Leftrightarrow 8.\int\limits_{0}^{1}{{{x}^{4}}{f}'\left( x \right)dx}=-38.\int\limits_{0}^{1}{{{\left[ {f}'\left( x \right) \right]}^{2}}dx}$
$\Leftrightarrow \int\limits_{0}^{1}{\left( 8{{x}^{4}}{f}'\left( x \right)+38{{\left[ {f}'\left( x \right) \right]}^{2}} \right)dx}=0\Leftrightarrow \int\limits_{0}^{1}{{f}'\left( x \right).\left[ 8{{x}^{4}}+38{f}'\left( x \right) \right]dx}=0$
$\Rightarrow 8{{x}^{4}}+38f\left( x \right)=0\Leftrightarrow {f}'\left( x \right)=-\dfrac{4}{19}{{x}^{4}}\Rightarrow f\left( x \right)=-\dfrac{4}{95}{{x}^{5}}+C$
Với $f\left( 1 \right)=2\Rightarrow C=\dfrac{194}{195}$. Khi đó: $f\left( x \right)=-\dfrac{4}{95}{{x}^{5}}+\dfrac{194}{95}$
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{\left( -\dfrac{4}{95}{{x}^{5}}+\dfrac{194}{95} \right)dx}=\left( -\dfrac{2}{285}{{x}^{6}}+\dfrac{194}{95} \right)x\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=\dfrac{116}{57}$
Đáp án C.