Cho hàm số \(f\left( x \right)\) xác định trên \(\left[ {0;\frac{\pi }{2}} \right]\) thỏa mãn \(\int\limits_0^{\frac{\pi }{2}} {\left[ {{f^2}\left( x \right) – 2\sqrt 2 f\left( x \right)\sin \left( {x – \frac{\pi }{4}} \right)} \right]{\mathop{\rm d}\nolimits} x} = \frac{{2 – \pi }}{2}\). Tích phân \(\int\limits_0^{\frac{\pi }{2}} {f\left( x \right){\mathop{\rm d}\nolimits} x} \) bằng
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Lời giải:
Báo saiTa có: \(\int\limits_0^{\frac{\pi }{2}} {2{{\sin }^2}\left( {x – \frac{\pi }{4}} \right){\mathop{\rm d}\nolimits} x} = \int\limits_0^{\frac{\pi }{2}} {\left[ {1 – \cos \left( {2x – \frac{\pi }{2}} \right)} \right]{\mathop{\rm d}\nolimits} x} = \int\limits_0^{\frac{\pi }{2}} {\left( {1 – \sin 2x} \right){\mathop{\rm d}\nolimits} x} \)
\( = \left. {\left( {x + \frac{1}{2}\cos 2x} \right)} \right|_0^{\frac{\pi }{2}} = \frac{{\pi – 2}}{2}\).
Do đó: \(\int\limits_0^{\frac{\pi }{2}} {\left[ {{f^2}\left( x \right) – 2\sqrt 2 f\left( x \right)\sin \left( {x – \frac{\pi }{4}} \right)} \right]{\mathop{\rm d}\nolimits} x} + \int\limits_0^{\frac{\pi }{2}} {2{{\sin }^2}\left( {x – \frac{\pi }{4}} \right){\mathop{\rm d}\nolimits} x} = \frac{{2 – \pi }}{2} + \frac{{\pi – 2}}{2} = 0\)
\( \Leftrightarrow \int\limits_0^{\frac{\pi }{2}} {\left[ {{f^2}\left( x \right) – 2\sqrt 2 f\left( x \right)\sin \left( {x – \frac{\pi }{4}} \right) + 2{{\sin }^2}\left( {x – \frac{\pi }{4}} \right)} \right]{\mathop{\rm d}\nolimits} x} = 0\)
\( \Leftrightarrow \int\limits_0^{\frac{\pi }{2}} {{{\left[ {f\left( x \right) – \sqrt 2 \sin \left( {x – \frac{\pi }{4}} \right)} \right]}^2}{\mathop{\rm d}\nolimits} x} = 0\)
Suy ra \(f\left( x \right) – \sqrt 2 \sin \left( {x – \frac{\pi }{4}} \right) = 0\), hay \(f\left( x \right) = \sqrt 2 \sin \left( {x – \frac{\pi }{4}} \right)\).
Bởi vậy: \(\int\limits_0^{\frac{\pi }{2}} {f\left( x \right){\mathop{\rm d}\nolimits} x} = \int\limits_0^{\frac{\pi }{2}} {\sqrt 2 \sin \left( {x – \frac{\pi }{4}} \right){\mathop{\rm d}\nolimits} x} = \left. { – \sqrt 2 \cos \left( {x – \frac{\pi }{4}} \right)} \right|_0^{\frac{\pi }{2}} = 0\)
