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Câu hỏi: Cho hàm số $f(x)$ liên tục trên $\left[0 ; \dfrac{\pi}{2}\right]$, thỏa $\int_0^{\dfrac{\pi}{2}}\left[f^2(x)-2 \sqrt{2} f(x) \sin \left(x-\dfrac{\pi}{4}\right)\right] \mathrm{d} x=\dfrac{2-\pi}{2}$. Tính tích phân $I=\int_0^{\dfrac{\pi}{2}} f(x) \mathrm{d} x$.
A. $\dfrac{\pi}{2}$.
B. $\dfrac{\pi}{4}$.
C. 1 .
D. 0 .
A. $\dfrac{\pi}{2}$.
B. $\dfrac{\pi}{4}$.
C. 1 .
D. 0 .
Ta có: $\int_0^{\dfrac{\pi}{2}} 2 \sin ^2\left(x-\dfrac{\pi}{4}\right) \mathrm{d} x=\int_0^{\dfrac{\pi}{2}}\left[1-\cos \left(2 x-\dfrac{\pi}{2}\right)\right] \mathrm{d} x=\int_0^{\dfrac{\pi}{2}}[1-\sin 2 x] \mathrm{d} x=\left.\left(x+\dfrac{1}{2} \cos 2 x\right)\right|_0 ^{\dfrac{\pi}{2}}=$ $\dfrac{\pi-2}{2}$.
Ta có: $\int_0^{\dfrac{\pi}{2}}\left[f^2(x)-2 \sqrt{2} f(x) \sin \left(x-\dfrac{\pi}{4}\right)\right] \mathrm{d} x+\int_0^{\dfrac{\pi}{2}} 2 \sin ^2\left(x-\dfrac{\pi}{4}\right) \mathrm{d} x=0$ $\Leftrightarrow \int_0^{\dfrac{\pi}{2}}\left[f(x)-\sqrt{2} \sin \left(x-\dfrac{\pi}{4}\right)\right]^2 \mathrm{~d} x=0, \forall x \in[0 ; 4] \Rightarrow f(x)=\sqrt{2} \sin \left(x-\dfrac{\pi}{4}\right)$.
Do đó: $I=\int_0^{\dfrac{\pi}{2}} f(x) \mathrm{d} x=\int_0^{\dfrac{\pi}{2}} \sqrt{2} \sin \left(x-\dfrac{\pi}{4}\right) \mathrm{d} x=-\left.\sqrt{2} \cos \left(x-\dfrac{\pi}{4}\right)\right|_0 ^{\dfrac{\pi}{2}}=-\sqrt{2} \cdot \dfrac{\sqrt{2}}{2}+\sqrt{2} \cdot \dfrac{\sqrt{2}}{2}=0$.
Ta có: $\int_0^{\dfrac{\pi}{2}}\left[f^2(x)-2 \sqrt{2} f(x) \sin \left(x-\dfrac{\pi}{4}\right)\right] \mathrm{d} x+\int_0^{\dfrac{\pi}{2}} 2 \sin ^2\left(x-\dfrac{\pi}{4}\right) \mathrm{d} x=0$ $\Leftrightarrow \int_0^{\dfrac{\pi}{2}}\left[f(x)-\sqrt{2} \sin \left(x-\dfrac{\pi}{4}\right)\right]^2 \mathrm{~d} x=0, \forall x \in[0 ; 4] \Rightarrow f(x)=\sqrt{2} \sin \left(x-\dfrac{\pi}{4}\right)$.
Do đó: $I=\int_0^{\dfrac{\pi}{2}} f(x) \mathrm{d} x=\int_0^{\dfrac{\pi}{2}} \sqrt{2} \sin \left(x-\dfrac{\pi}{4}\right) \mathrm{d} x=-\left.\sqrt{2} \cos \left(x-\dfrac{\pi}{4}\right)\right|_0 ^{\dfrac{\pi}{2}}=-\sqrt{2} \cdot \dfrac{\sqrt{2}}{2}+\sqrt{2} \cdot \dfrac{\sqrt{2}}{2}=0$.
Đáp án D.