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Câu hỏi: Cho hàm số $y=f(x)$ có đạo hàm liên tục trên đoạn $\left[ 1;2 \right]$ và $f(1)+f(2)=0$. Biết $\int_1^2(f(x))^2 \mathrm{~d} x=\dfrac{1}{2}, \int_1^2 f^{\prime}(x) \cos (\pi x) \mathrm{d} x=\dfrac{\pi}{2}$. Tinh $\int_1^2 f(x) \mathrm{d} x$.
A. $\pi$.
B. $\dfrac{1}{\pi}$.
C. $\dfrac{2}{\pi}$.
D. $\dfrac{-2}{\pi}$.
A. $\pi$.
B. $\dfrac{1}{\pi}$.
C. $\dfrac{2}{\pi}$.
D. $\dfrac{-2}{\pi}$.
Đặt $I=\int_{1}^{2}{{{f}^{\prime }}}(x)\cos (\pi x)\text{d}x=\dfrac{\pi }{2}$.
Đặt $\left\{ \begin{aligned}
& u=\cos (\pi x) \\
& dv={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=-\pi \sin \pi x.\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.\Rightarrow I=\left. f\left( x \right)\cos (\pi x) \right|_{1}^{2}+\pi \int\limits_{1}^{2}{f\left( x \right)}\sin \pi x.\text{d}x$
$\Rightarrow I=f\left( 2 \right)\cos 2\pi -f\left( 1 \right)\cos \pi +\pi \int\limits_{1}^{2}{f\left( x \right)}\sin \pi x.\text{d}x\Rightarrow \int\limits_{1}^{2}{f\left( x \right)}\sin \pi x.\text{d}x=\dfrac{1}{2}.$
Ta có $\int\limits_{1}^{2}{{{\sin }^{2}}\pi x.\text{d}x}=\dfrac{1}{2}\int\limits_{1}^{2}{\left( 1-\cos 2\pi x \right)\text{d}x}=\dfrac{1}{2}\left. \left( x-\dfrac{\sin 2\pi x}{2\pi } \right) \right|_{1}^{2}=\dfrac{1}{2}.$
Do đó $\int_{1}^{2}{(f(}x){{)}^{2}}~\text{d}x-2\int_{1}^{2}{f}(x)\sin (\pi x)\text{d}x+\int\limits_{1}^{2}{{{\left( \sin \pi x \right)}^{2}}\text{d}x=0}$
$\int\limits_{1}^{2}{\left[ {{f}^{2}}\left( x \right)-2f\left( x \right)\sin \pi x+{{\left( \sin \pi x \right)}^{2}} \right]}=0\Leftrightarrow {{\int\limits_{1}^{2}{\left[ f\left( x \right)-\sin \pi x \right]}}^{2}}\text{d}x=0$ $\Rightarrow f\left( x \right)=\sin \pi x$.
Do đó $\int_{1}^{2}{f}(x)\text{d}x=\int\limits_{1}^{2}{\sin \pi x}.\text{d}x=\dfrac{-1}{\pi }\left. \cos \pi x \right|_{1}^{2}=\dfrac{-2}{\pi }.$
Đặt $\left\{ \begin{aligned}
& u=\cos (\pi x) \\
& dv={f}'\left( x \right)\text{d}x \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=-\pi \sin \pi x.\text{d}x \\
& v=f\left( x \right) \\
\end{aligned} \right.\Rightarrow I=\left. f\left( x \right)\cos (\pi x) \right|_{1}^{2}+\pi \int\limits_{1}^{2}{f\left( x \right)}\sin \pi x.\text{d}x$
$\Rightarrow I=f\left( 2 \right)\cos 2\pi -f\left( 1 \right)\cos \pi +\pi \int\limits_{1}^{2}{f\left( x \right)}\sin \pi x.\text{d}x\Rightarrow \int\limits_{1}^{2}{f\left( x \right)}\sin \pi x.\text{d}x=\dfrac{1}{2}.$
Ta có $\int\limits_{1}^{2}{{{\sin }^{2}}\pi x.\text{d}x}=\dfrac{1}{2}\int\limits_{1}^{2}{\left( 1-\cos 2\pi x \right)\text{d}x}=\dfrac{1}{2}\left. \left( x-\dfrac{\sin 2\pi x}{2\pi } \right) \right|_{1}^{2}=\dfrac{1}{2}.$
Do đó $\int_{1}^{2}{(f(}x){{)}^{2}}~\text{d}x-2\int_{1}^{2}{f}(x)\sin (\pi x)\text{d}x+\int\limits_{1}^{2}{{{\left( \sin \pi x \right)}^{2}}\text{d}x=0}$
$\int\limits_{1}^{2}{\left[ {{f}^{2}}\left( x \right)-2f\left( x \right)\sin \pi x+{{\left( \sin \pi x \right)}^{2}} \right]}=0\Leftrightarrow {{\int\limits_{1}^{2}{\left[ f\left( x \right)-\sin \pi x \right]}}^{2}}\text{d}x=0$ $\Rightarrow f\left( x \right)=\sin \pi x$.
Do đó $\int_{1}^{2}{f}(x)\text{d}x=\int\limits_{1}^{2}{\sin \pi x}.\text{d}x=\dfrac{-1}{\pi }\left. \cos \pi x \right|_{1}^{2}=\dfrac{-2}{\pi }.$
Đáp án D.