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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 $f\left( 0 \right)=0$. Biết $\int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}=\dfrac{9}{2}$ và $\int\limits_{0}^{1}{f'\left( x \right)\cos \dfrac{\pi x}{2}dx}=\dfrac{3\pi }{4}$. Tích phân $\int\limits_{0}^{1}{f\left( x \right)dx}$ bằng.
A. $\dfrac{6}{\pi }.$
B. $\dfrac{2}{\pi }.$
C. $\dfrac{4}{\pi }.$
D. $\dfrac{1}{\pi }.$
A. $\dfrac{6}{\pi }.$
B. $\dfrac{2}{\pi }.$
C. $\dfrac{4}{\pi }.$
D. $\dfrac{1}{\pi }.$
Đặt $\left\{ \begin{aligned}
& u=\cos \dfrac{\pi x}{2} \\
& dv=f'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=-\dfrac{\pi }{2}\sin \dfrac{\pi x}{2}dx \\
& v=f\left( x \right) \\
\end{aligned} \right..$
Suy ra $\begin{aligned}
& \ \ \ \int\limits_{0}^{1}{f'\left( x \right)\cos \dfrac{\pi x}{2}dx}=\cos \dfrac{\pi x}{2}f\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.+\dfrac{\pi }{2}\int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}dx} \\
& =f\left( 1 \right).\cos \dfrac{\pi }{2}-f\left( 0 \right).\cos 0+\dfrac{\pi }{2}\int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}dx} \\
& =\dfrac{\pi }{2}\int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}dx}=\dfrac{3\pi }{4}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}dx}=\dfrac{3}{2}. \\
\end{aligned}$
Xét tích phân $\int\limits_{0}^{1}{{{\left[ f\left( x \right)+k\sin \dfrac{\pi x}{2} \right]}^{2}}dx}=0$
$\begin{aligned}
& \Leftrightarrow \int\limits_{0}^{1}{\left[ {{f}^{2}}\left( x \right)+2kf\left( x \right)\sin \dfrac{\pi x}{2}+{{k}^{2}}{{\sin }^{2}}\dfrac{\pi x}{2} \right]dx}=0 \\
& \Leftrightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}+2k\int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}}+{{k}^{2}}\int\limits_{0}^{1}{{{\sin }^{2}}\dfrac{\pi x}{2}dx}=0\Leftrightarrow \dfrac{9}{2}+2k\dfrac{3}{2}+\dfrac{1}{2}{{k}^{2}}=0\Leftrightarrow k=-3. \\
\end{aligned}$
Khi đó ta có: $\int\limits_{0}^{1}{{{\left[ f\left( x \right)-3\sin \dfrac{\pi x}{2} \right]}^{2}}dx}=0\Leftrightarrow f\left( x \right)-3\sin \dfrac{\pi x}{2}=0\Leftrightarrow f\left( x \right)=3\sin \dfrac{\pi x}{2}$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=3\int\limits_{0}^{1}{\sin \dfrac{\pi x}{2}dx}=-3.\dfrac{\cos \dfrac{\pi x}{2}}{\dfrac{\pi }{2}}\left| \begin{aligned}
& ^{1} \\
& \\
& _{0} \\
\end{aligned} \right.=\dfrac{-6}{\pi }\cos \dfrac{\pi x}{2}\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=-\dfrac{6}{\pi }\left( \cos \dfrac{\pi }{2}-\cos 0 \right)=\dfrac{6}{\pi }.$
Chú ý:
Sử dụng phương pháp từng phần đối với tích phân $\int\limits_{0}^{1}{f'\left( x \right)\cos \dfrac{\pi x}{2}dx}=\dfrac{3\pi }{4}$.
Xét $\int\limits_{0}^{1}{{{\left[ f\left( x \right)+k\sin \dfrac{\pi x}{2} \right]}^{2}}dx}=0$, tìm k, từ đó suy ra $f\left( x \right)=-k\sin \dfrac{\pi x}{2}$.
$\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{-k\sin \dfrac{\pi x}{2}dx}$.
& u=\cos \dfrac{\pi x}{2} \\
& dv=f'\left( x \right)dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=-\dfrac{\pi }{2}\sin \dfrac{\pi x}{2}dx \\
& v=f\left( x \right) \\
\end{aligned} \right..$
Suy ra $\begin{aligned}
& \ \ \ \int\limits_{0}^{1}{f'\left( x \right)\cos \dfrac{\pi x}{2}dx}=\cos \dfrac{\pi x}{2}f\left( x \right)\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.+\dfrac{\pi }{2}\int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}dx} \\
& =f\left( 1 \right).\cos \dfrac{\pi }{2}-f\left( 0 \right).\cos 0+\dfrac{\pi }{2}\int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}dx} \\
& =\dfrac{\pi }{2}\int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}dx}=\dfrac{3\pi }{4}\Rightarrow \int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}dx}=\dfrac{3}{2}. \\
\end{aligned}$
Xét tích phân $\int\limits_{0}^{1}{{{\left[ f\left( x \right)+k\sin \dfrac{\pi x}{2} \right]}^{2}}dx}=0$
$\begin{aligned}
& \Leftrightarrow \int\limits_{0}^{1}{\left[ {{f}^{2}}\left( x \right)+2kf\left( x \right)\sin \dfrac{\pi x}{2}+{{k}^{2}}{{\sin }^{2}}\dfrac{\pi x}{2} \right]dx}=0 \\
& \Leftrightarrow \int\limits_{0}^{1}{{{f}^{2}}\left( x \right)dx}+2k\int\limits_{0}^{1}{f\left( x \right)\sin \dfrac{\pi x}{2}}+{{k}^{2}}\int\limits_{0}^{1}{{{\sin }^{2}}\dfrac{\pi x}{2}dx}=0\Leftrightarrow \dfrac{9}{2}+2k\dfrac{3}{2}+\dfrac{1}{2}{{k}^{2}}=0\Leftrightarrow k=-3. \\
\end{aligned}$
Khi đó ta có: $\int\limits_{0}^{1}{{{\left[ f\left( x \right)-3\sin \dfrac{\pi x}{2} \right]}^{2}}dx}=0\Leftrightarrow f\left( x \right)-3\sin \dfrac{\pi x}{2}=0\Leftrightarrow f\left( x \right)=3\sin \dfrac{\pi x}{2}$.
Vậy $\int\limits_{0}^{1}{f\left( x \right)dx}=3\int\limits_{0}^{1}{\sin \dfrac{\pi x}{2}dx}=-3.\dfrac{\cos \dfrac{\pi x}{2}}{\dfrac{\pi }{2}}\left| \begin{aligned}
& ^{1} \\
& \\
& _{0} \\
\end{aligned} \right.=\dfrac{-6}{\pi }\cos \dfrac{\pi x}{2}\left| \begin{aligned}
& ^{1} \\
& _{0} \\
\end{aligned} \right.=-\dfrac{6}{\pi }\left( \cos \dfrac{\pi }{2}-\cos 0 \right)=\dfrac{6}{\pi }.$
Chú ý:
Sử dụng phương pháp từng phần đối với tích phân $\int\limits_{0}^{1}{f'\left( x \right)\cos \dfrac{\pi x}{2}dx}=\dfrac{3\pi }{4}$.
Xét $\int\limits_{0}^{1}{{{\left[ f\left( x \right)+k\sin \dfrac{\pi x}{2} \right]}^{2}}dx}=0$, tìm k, từ đó suy ra $f\left( x \right)=-k\sin \dfrac{\pi x}{2}$.
$\int\limits_{0}^{1}{f\left( x \right)dx}=\int\limits_{0}^{1}{-k\sin \dfrac{\pi x}{2}dx}$.
Đáp án A.