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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên đoạn $\left[ 0;4 \right]$ và thỏa mãn điều kiện $4xf\left( {{x}^{2}} \right)+6f\left( 2x \right)=\sqrt{4-{{x}^{2}}}$. Tính tích phân $\int\limits_{0}^{4}{f\left( x \right)dx}$.
A. $I=\dfrac{\pi }{5}.$
B. $I=\dfrac{\pi }{2}.$
C. $I=\dfrac{\pi }{20}.$
D. $I=\dfrac{\pi }{10}.$
A. $I=\dfrac{\pi }{5}.$
B. $I=\dfrac{\pi }{2}.$
C. $I=\dfrac{\pi }{20}.$
D. $I=\dfrac{\pi }{10}.$
Ta có: $4xf\left( {{x}^{2}} \right)+6f\left( 2x \right)=\sqrt{4-{{x}^{2}}}\Rightarrow \int\limits_{0}^{2}{\left[ 4xf\left( {{x}^{2}} \right)+6f\left( 2x \right) \right]dx}=\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}$.
$\Leftrightarrow 4{{I}_{1}}+6{{I}_{2}}=I$.
Trong đó: ${{I}_{1}}=\int\limits_{0}^{2}{xf\left( {{x}^{2}} \right)dx}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( {{x}^{2}} \right)d\left( {{x}^{2}} \right)}=\dfrac{1}{2}\int\limits_{0}^{4}{f\left( x \right)dx}.$
$\begin{aligned}
& {{I}_{2}}=\int\limits_{0}^{2}{f\left( 2x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( 2x \right)d\left( 2x \right)}=\dfrac{1}{2}\int\limits_{0}^{4}{f\left( x \right)dx}. \\
& I=\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}=2\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{4-4{{\sin }^{2}}\left( t \right)}\cos \left( t \right)dt}=4\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}\left( t \right)dt}=2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos \left( 2t \right) \right)dt}=\left[ 2t+\sin \left( 2t \right) \right]\left| \begin{aligned}
& ^{\dfrac{\pi }{2}} \\
& _{0} \\
\end{aligned} \right.=\pi \\
\end{aligned}$
Khi đó ta có hệ: $\left\{ \begin{aligned}
& {{I}_{1}}={{I}_{2}} \\
& 4{{I}_{1}}+6{{I}_{2}}=\pi \\
\end{aligned} \right.\Leftrightarrow {{I}_{1}}={{I}_{2}}=\dfrac{\pi }{10}\Leftrightarrow \dfrac{1}{2}\int\limits_{0}^{4}{f\left( x \right)dx}=\dfrac{\pi }{10} $ hay $ \int\limits_{0}^{4}{f\left( x \right)dx}=\dfrac{\pi }{5}$.
$\Leftrightarrow 4{{I}_{1}}+6{{I}_{2}}=I$.
Trong đó: ${{I}_{1}}=\int\limits_{0}^{2}{xf\left( {{x}^{2}} \right)dx}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( {{x}^{2}} \right)d\left( {{x}^{2}} \right)}=\dfrac{1}{2}\int\limits_{0}^{4}{f\left( x \right)dx}.$
$\begin{aligned}
& {{I}_{2}}=\int\limits_{0}^{2}{f\left( 2x \right)dx}=\dfrac{1}{2}\int\limits_{0}^{2}{f\left( 2x \right)d\left( 2x \right)}=\dfrac{1}{2}\int\limits_{0}^{4}{f\left( x \right)dx}. \\
& I=\int\limits_{0}^{2}{\sqrt{4-{{x}^{2}}}dx}=2\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{4-4{{\sin }^{2}}\left( t \right)}\cos \left( t \right)dt}=4\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}\left( t \right)dt}=2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos \left( 2t \right) \right)dt}=\left[ 2t+\sin \left( 2t \right) \right]\left| \begin{aligned}
& ^{\dfrac{\pi }{2}} \\
& _{0} \\
\end{aligned} \right.=\pi \\
\end{aligned}$
Khi đó ta có hệ: $\left\{ \begin{aligned}
& {{I}_{1}}={{I}_{2}} \\
& 4{{I}_{1}}+6{{I}_{2}}=\pi \\
\end{aligned} \right.\Leftrightarrow {{I}_{1}}={{I}_{2}}=\dfrac{\pi }{10}\Leftrightarrow \dfrac{1}{2}\int\limits_{0}^{4}{f\left( x \right)dx}=\dfrac{\pi }{10} $ hay $ \int\limits_{0}^{4}{f\left( x \right)dx}=\dfrac{\pi }{5}$.
Đáp án A.