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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên $\left[ 0;\dfrac{\pi }{2} \right]$ đồng thời thỏa mãn $\int\limits_{0}^{\dfrac{\pi }{2}}{{{f}^{2}}\left( x \right)\text{d}x}=3\pi , \int\limits_{0}^{\pi }{\left( \text{sin}x-x \right){f}'\left( \dfrac{x}{2} \right)\text{d}x}=-2\pi $ và $f\left( \dfrac{\pi }{2} \right)=4$. Giá trị của $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{sin}x\text{d}x}$ bằng
A. $\dfrac{8+5\sqrt{3}}{3}$.
B. $\dfrac{8+5\sqrt{2}}{3}$.
C. $\dfrac{8-5\sqrt{2}}{3}$.
D. $\dfrac{8-5\sqrt{3}}{3}$.
A. $\dfrac{8+5\sqrt{3}}{3}$.
B. $\dfrac{8+5\sqrt{2}}{3}$.
C. $\dfrac{8-5\sqrt{2}}{3}$.
D. $\dfrac{8-5\sqrt{3}}{3}$.
Có $I=\int\limits_{0}^{\pi }{\left( \text{sin}x-x \right){f}'\left( \dfrac{x}{2} \right)\text{d}x}=-2\pi $.
Đặt $t=\dfrac{x}{2}\Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin 2t-2t \right){f}'\left( t \right)\text{d}t}=2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin 2x-2x \right){f}'\left( x \right)\text{d}x}$.
Tích phân từng phần cho $I$, ta được:
$\dfrac{I}{2}=\left. \left( \sin 2x-2x \right)f\left( x \right) \right|_{0}^{\dfrac{\pi }{2}}-\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-2 \right)f\left( x \right)\text{d}x}$ $ =-\pi f\left( \dfrac{\pi }{2} \right)-2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \cos 2x-1 \right)f\left( x \right)\text{d}x}$
$\Leftrightarrow -\pi =-4\pi +4\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x}f\left( x \right)\text{d}x\Leftrightarrow 4\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x}f\left( x \right)\text{d}x=3\pi $
Có $\int\limits_{0}^{\dfrac{\pi }{2}}{{{f}^{2}}\left( x \right)\text{d}x}-2\int\limits_{0}^{\dfrac{\pi }{2}}{4{{\sin }^{2}}xf\left( x \right)\text{d}x}+16\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{4}}x\text{d}x=3\pi -2.3\pi +3\pi =0}$
$\begin{aligned}
& \Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{\left( f\left( x \right)-4{{\sin }^{2}}x \right)}^{2}}}\text{d}x=0 \\
& \Leftrightarrow f\left( x \right)-4{{\sin }^{2}}x=0\Leftrightarrow f\left( x \right)=4{{\sin }^{2}}x, \forall x\in \left[ 0;\dfrac{\pi }{2} \right]. \\
\end{aligned}$
Vậy $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{sin}x\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{4}}{\text{4si}{{\text{n}}^{3}}x\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 3\sin x-\sin 3x \right)\text{d}x}$ $=\left. \left( \dfrac{1}{3}\cos 3x-3\cos x \right) \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{8-5\sqrt{3}}{3}$.
Đặt $t=\dfrac{x}{2}\Rightarrow I=2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin 2t-2t \right){f}'\left( t \right)\text{d}t}=2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \sin 2x-2x \right){f}'\left( x \right)\text{d}x}$.
Tích phân từng phần cho $I$, ta được:
$\dfrac{I}{2}=\left. \left( \sin 2x-2x \right)f\left( x \right) \right|_{0}^{\dfrac{\pi }{2}}-\int\limits_{0}^{\dfrac{\pi }{2}}{\left( 2\cos 2x-2 \right)f\left( x \right)\text{d}x}$ $ =-\pi f\left( \dfrac{\pi }{2} \right)-2\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \cos 2x-1 \right)f\left( x \right)\text{d}x}$
$\Leftrightarrow -\pi =-4\pi +4\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x}f\left( x \right)\text{d}x\Leftrightarrow 4\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{2}}x}f\left( x \right)\text{d}x=3\pi $
Có $\int\limits_{0}^{\dfrac{\pi }{2}}{{{f}^{2}}\left( x \right)\text{d}x}-2\int\limits_{0}^{\dfrac{\pi }{2}}{4{{\sin }^{2}}xf\left( x \right)\text{d}x}+16\int\limits_{0}^{\dfrac{\pi }{2}}{{{\sin }^{4}}x\text{d}x=3\pi -2.3\pi +3\pi =0}$
$\begin{aligned}
& \Leftrightarrow \int\limits_{0}^{\dfrac{\pi }{2}}{{{\left( f\left( x \right)-4{{\sin }^{2}}x \right)}^{2}}}\text{d}x=0 \\
& \Leftrightarrow f\left( x \right)-4{{\sin }^{2}}x=0\Leftrightarrow f\left( x \right)=4{{\sin }^{2}}x, \forall x\in \left[ 0;\dfrac{\pi }{2} \right]. \\
\end{aligned}$
Vậy $\int\limits_{0}^{\dfrac{\pi }{4}}{f\left( x \right)\text{sin}x\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{4}}{\text{4si}{{\text{n}}^{3}}x\text{d}x}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( 3\sin x-\sin 3x \right)\text{d}x}$ $=\left. \left( \dfrac{1}{3}\cos 3x-3\cos x \right) \right|_{0}^{\dfrac{\pi }{4}}=\dfrac{8-5\sqrt{3}}{3}$.
Đáp án D.
