Google
Câu hỏi: Xét hàm số $f\left( x \right)$ liên tục trên đoạn $\left[ 0;1 \right]$ và thoả mãn điều kiện $4xf\left( {{x}^{2}} \right)+3f\left( 1-x \right)=\sqrt{1-{{x}^{2}}},\forall x\in \left[ 0;1 \right].$ Tích phân $I=\int\limits_{0}^{1}{f\left( x \right)}\text{d}x$ bằng
A. $I=\dfrac{\pi }{4}.$
B. $I=\dfrac{\pi }{6}.$
C. $I=\dfrac{\pi }{16}.$
D. $I=\dfrac{\pi }{20}.$
A. $I=\dfrac{\pi }{4}.$
B. $I=\dfrac{\pi }{6}.$
C. $I=\dfrac{\pi }{16}.$
D. $I=\dfrac{\pi }{20}.$
$4xf\left( {{x}^{2}} \right)+3f\left( 1-x \right)=\sqrt{1-{{x}^{2}}},\forall x\in \left[ 0;1 \right]$ nên
$\begin{aligned}
& \int\limits_{0}^{1}{4xf\left( {{x}^{2}} \right)}\text{d}x=2\int\limits_{0}^{1}{f\left( {{x}^{2}} \right)}\text{d}\left( {{x}^{2}} \right)=2I \\
& \int\limits_{0}^{1}{3f\left( 1-x \right)}\text{d}x=-3\int\limits_{0}^{1}{f\left( 1-x \right)}\text{d}\left( 1-x \right)=3\int\limits_{0}^{1}{f\left( t \right)\text{d}t}=3I,\left( t=1-x \right) \\
\end{aligned}$
Xét $J=\int\limits_{0}^{1}{\sqrt{1-{{x}^{2}}}}\text{d}x$
Đặt $x=\sin t\Rightarrow \text{d}x=2\cos t\text{d}t$
Khi đó, $J=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{1-{{\left( \sin t \right)}^{2}}}\text{cos}}t\text{d}t=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1+\text{cos2}t}{2}}\text{d}t=\left( \dfrac{1}{2}t+\dfrac{1}{4}\sin 2t \right)\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.=\dfrac{\pi }{4}.$
Do đó $5I=\dfrac{\pi }{4}\Rightarrow I=\dfrac{\pi }{20}.$
$\int\limits_{0}^{1}{4xf\left( {{x}^{2}} \right)}\text{d}x+\int\limits_{0}^{1}{3f\left( 1-x \right)}\text{d}x=\int\limits_{0}^{1}{\sqrt{1-{{x}^{2}}}}\text{d}x$
Mà$\begin{aligned}
& \int\limits_{0}^{1}{4xf\left( {{x}^{2}} \right)}\text{d}x=2\int\limits_{0}^{1}{f\left( {{x}^{2}} \right)}\text{d}\left( {{x}^{2}} \right)=2I \\
& \int\limits_{0}^{1}{3f\left( 1-x \right)}\text{d}x=-3\int\limits_{0}^{1}{f\left( 1-x \right)}\text{d}\left( 1-x \right)=3\int\limits_{0}^{1}{f\left( t \right)\text{d}t}=3I,\left( t=1-x \right) \\
\end{aligned}$
Xét $J=\int\limits_{0}^{1}{\sqrt{1-{{x}^{2}}}}\text{d}x$
Đặt $x=\sin t\Rightarrow \text{d}x=2\cos t\text{d}t$
Khi đó, $J=\int\limits_{0}^{\dfrac{\pi }{2}}{\sqrt{1-{{\left( \sin t \right)}^{2}}}\text{cos}}t\text{d}t=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1+\text{cos2}t}{2}}\text{d}t=\left( \dfrac{1}{2}t+\dfrac{1}{4}\sin 2t \right)\left| \begin{aligned}
& \dfrac{\pi }{2} \\
& 0 \\
\end{aligned} \right.=\dfrac{\pi }{4}.$
Do đó $5I=\dfrac{\pi }{4}\Rightarrow I=\dfrac{\pi }{20}.$
Đáp án D.