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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0;1 \right]$ thoả mãn $6{{x}^{2}}.f\left( {{x}^{3}} \right)+4f\left( 1-x \right)=3\sqrt{1-{{x}^{2}}}$. Giá trị của $\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$ là
A. $\dfrac{\pi }{8}$.
B. $\dfrac{\pi }{16}$.
C. $\dfrac{\pi }{4}$.
D. $\dfrac{\pi }{20}$.
A. $\dfrac{\pi }{8}$.
B. $\dfrac{\pi }{16}$.
C. $\dfrac{\pi }{4}$.
D. $\dfrac{\pi }{20}$.
Ta có $6{{x}^{2}}.f\left( {{x}^{3}} \right)+4f\left( 1-x \right)=3\sqrt{1-{{x}^{2}}}\Rightarrow \int\limits_{0}^{1}{6{{x}^{2}}.f\left( {{x}^{3}} \right)\text{d}x}+\int\limits_{0}^{1}{4f\left( 1-x \right)\text{d}x}=\int\limits_{0}^{1}{3\sqrt{1-{{x}^{2}}}\text{d}x}$.
Mặt khác $\int\limits_{0}^{1}{6{{x}^{2}}.f\left( {{x}^{3}} \right)\text{d}x}+\int\limits_{0}^{1}{4f\left( 1-x \right)\text{d}x}=2\int\limits_{0}^{1}{f\left( {{x}^{3}} \right)\text{d}{{x}^{3}}}-4\int\limits_{0}^{1}{f\left( 1-x \right)\text{d}\left( 1-x \right)}$
$=6\int\limits_{0}^{1}{f\left( t \right)\text{dt}}=6\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$
Đặt $x=\sin t\Rightarrow \text{d}x=\cos t\text{dt}$
Vậy $\int\limits_{0}^{1}{3\sqrt{1-{{x}^{2}}}\text{d}x}=3\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{1-{{\sin }^{2}}t} \cos t\text{dt}}=3\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}t\text{dt}}=\dfrac{3}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos 2t \right)\text{dt}}$
$=\dfrac{3}{2}\left( t+\dfrac{1}{2}\sin 2t \right)\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.=\dfrac{3\pi }{4}$.
Khi đó $6\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{3\pi }{4}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{\pi }{8}$.
Mặt khác $\int\limits_{0}^{1}{6{{x}^{2}}.f\left( {{x}^{3}} \right)\text{d}x}+\int\limits_{0}^{1}{4f\left( 1-x \right)\text{d}x}=2\int\limits_{0}^{1}{f\left( {{x}^{3}} \right)\text{d}{{x}^{3}}}-4\int\limits_{0}^{1}{f\left( 1-x \right)\text{d}\left( 1-x \right)}$
$=6\int\limits_{0}^{1}{f\left( t \right)\text{dt}}=6\int\limits_{0}^{1}{f\left( x \right)\text{d}x}$
Đặt $x=\sin t\Rightarrow \text{d}x=\cos t\text{dt}$
Vậy $\int\limits_{0}^{1}{3\sqrt{1-{{x}^{2}}}\text{d}x}=3\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{1-{{\sin }^{2}}t} \cos t\text{dt}}=3\int\limits_{0}^{\dfrac{\pi }{2}}{{{\cos }^{2}}t\text{dt}}=\dfrac{3}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 1+\cos 2t \right)\text{dt}}$
$=\dfrac{3}{2}\left( t+\dfrac{1}{2}\sin 2t \right)\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.=\dfrac{3\pi }{4}$.
Khi đó $6\int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{3\pi }{4}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)\text{d}x}=\dfrac{\pi }{8}$.
Đáp án A.
