Cho hàm số \(y=f(x)\) có đạo hàm liên tục trên đoạ...

A

- Hướng dẫn giải

Đặt \(\left\{ \begin{array}{l}
u = \cos \frac{{\pi x}}{2}\\
dv = f'\left( x \right)dx
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
du =  - \frac{\pi }{2}\sin \frac{{\pi x}}{2}dx\\
v = f\left( x \right)
\end{array} \right.\) 

\(\begin{array}{l}
 \Rightarrow \int_0^1 {f'\left( x \right)\cos \frac{{\pi x}}{2}dx = \cos } \frac{{\pi x}}{2}f\left( x \right)\left| \begin{array}{l}
^1\\
_0
\end{array} \right. + \frac{\pi }{2}\int_0^1 {f\left( x \right)\sin \frac{{\pi x}}{2}dx} \\
 = f\left( 1 \right).cos\frac{\pi }{2} - f\left( 0 \right)\cos 0 + \frac{\pi }{2}\int_0^1 {f\left( x \right)\sin \frac{{\pi x}}{2}dx} \\
 = \frac{\pi }{2}\int_0^1 {f\left( x \right)\sin \frac{{\pi x}}{2}dx = \frac{{3\pi }}{4} \Rightarrow \int_0^1 {f\left( x \right)\sin \frac{{\pi x}}{2}dx = \frac{3}{2}} } 
\end{array}\) 

Xét tích phân \(\int_0^1 {{{\left[ {f\left( x \right) + k\sin \frac{{\pi x}}{2}} \right]}^2}dx = 0} \) 

\(\begin{array}{l}
 \Leftrightarrow \int_0^1 {\left[ {{f^2}\left( x \right) + 2kf\left( x \right)\sin \frac{{\pi x}}{2} + {k^2}{{\sin }^2}\frac{{\pi x}}{2}} \right]dx = 0} \\
 \Leftrightarrow \int_0^1 {{f^2}\left( x \right)dx + 2k\int_0^1 {f\left( x \right)\sin \frac{{\pi x}}{2} + {k^2}\int_0^1 {{{\sin }^2}\frac{{\pi x}}{2}dx = 0} } } \\
 \Leftrightarrow \frac{9}{2} + 2k\frac{3}{2} + \frac{1}{2}{k^2} = 0 \Leftrightarrow k =  - 3
\end{array}\)

Khi đó ta có \(\int_0^1 {{{\left[ {f\left( x \right) - 3\sin \frac{{\pi x}}{2}} \right]}^2}dx = 0 \Leftrightarrow f\left( x \right) - 3\sin \frac{{\pi x}}{2} = 0 \Leftrightarrow f\left( x \right) = 3\sin \frac{{\pi x}}{2}} \) 

Vậy \(\int_0^1 {f\left( x \right)dx = 3\int_0^1 {\sin \frac{{\pi x}}{2}dx =  - 3\frac{{\cos \frac{{\pi x}}{2}}}{{\frac{\pi }{2}}}\left| \begin{array}{l}
^1\\
\\
_0
\end{array} \right.} }  = \frac{{ - 6}}{\pi }\cos \frac{{\pi x}}{2}\left| \begin{array}{l}
^1\\
_0
\end{array} \right. =  - \frac{6}{\pi }\left( {\cos \frac{\pi }{2} - \cos 0} \right) = \frac{6}{\pi }\)