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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm liên tục trên đoạn $\left[ 0;1 \right]$ và thỏa mãn $f\left( \dfrac{\pi }{2} \right)=0$. Biết $\int_{0}^{\dfrac{\pi }{2}}{{{f}^{2}}\left( x \right)dx=\pi }$ và $\int_{0}^{\dfrac{\pi }{2}}{{f}'\left( x \right)\sin 3xdx=\dfrac{-3\pi }{2}}$. Tích phân $\int_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}$ bằng.
A. $\dfrac{-2}{3}.$
B. $\dfrac{2}{3}.$
C. $\dfrac{1}{3}.$
D. $\dfrac{-5}{7}.$
A. $\dfrac{-2}{3}.$
B. $\dfrac{2}{3}.$
C. $\dfrac{1}{3}.$
D. $\dfrac{-5}{7}.$
Xét tích phân $I=\int_{0}^{\dfrac{\pi }{2}}{{f}'\left( x \right)\sin 3xdx}$
Đặt $u=\sin 3x\Rightarrow du=3cos3xdx$
$dv={f}'\left( x \right)dx\Rightarrow v=f\left( x \right)$
$I=\left. \sin 3x.f\left( x \right) \right|_{0}^{\dfrac{\pi }{2}}-\int_{0}^{\dfrac{\pi }{2}}{3f\left( x \right)\text{cos3}x\text{d}x}=-\dfrac{3\pi }{2}$
$\Leftrightarrow $ $\int_{0}^{\dfrac{\pi }{2}}{3f\left( x \right)\text{cos3}x\text{d}x}=\dfrac{3\pi }{2}\Leftrightarrow \int_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{cos3}x\text{d}x}=\dfrac{\pi }{2}$.
Xét $\int\limits_{0}^{1}{{{\left[ f\left( x \right) \right]}^{2}}\text{d}x}-\int\limits_{0}^{1}{4cos3x.f\left( x \right)\text{d}x}+\int\limits_{0}^{\dfrac{\pi }{2}}{{{\left( 2cos3x \right)}^{2}}\text{d}x}=\pi -2\pi +\pi =0$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f\left( x \right)-2cos3x \right]}^{2}}}dx=0$ $\Leftrightarrow f\left( x \right)=2cos3x$
Vậy $\int_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=\int_{0}^{\dfrac{\pi }{2}}{2cos3xdx}=\left. 2\dfrac{\sin 3x}{3} \right|_{0}^{\dfrac{\pi }{2}}=-\dfrac{2}{3}$.
Đặt $u=\sin 3x\Rightarrow du=3cos3xdx$
$dv={f}'\left( x \right)dx\Rightarrow v=f\left( x \right)$
$I=\left. \sin 3x.f\left( x \right) \right|_{0}^{\dfrac{\pi }{2}}-\int_{0}^{\dfrac{\pi }{2}}{3f\left( x \right)\text{cos3}x\text{d}x}=-\dfrac{3\pi }{2}$
$\Leftrightarrow $ $\int_{0}^{\dfrac{\pi }{2}}{3f\left( x \right)\text{cos3}x\text{d}x}=\dfrac{3\pi }{2}\Leftrightarrow \int_{0}^{\dfrac{\pi }{2}}{f\left( x \right)\text{cos3}x\text{d}x}=\dfrac{\pi }{2}$.
Xét $\int\limits_{0}^{1}{{{\left[ f\left( x \right) \right]}^{2}}\text{d}x}-\int\limits_{0}^{1}{4cos3x.f\left( x \right)\text{d}x}+\int\limits_{0}^{\dfrac{\pi }{2}}{{{\left( 2cos3x \right)}^{2}}\text{d}x}=\pi -2\pi +\pi =0$
$\Leftrightarrow \int\limits_{0}^{1}{{{\left[ f\left( x \right)-2cos3x \right]}^{2}}}dx=0$ $\Leftrightarrow f\left( x \right)=2cos3x$
Vậy $\int_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}=\int_{0}^{\dfrac{\pi }{2}}{2cos3xdx}=\left. 2\dfrac{\sin 3x}{3} \right|_{0}^{\dfrac{\pi }{2}}=-\dfrac{2}{3}$.
Đáp án A.