Google
Câu hỏi: Cho hàm số $f(x)$ có đạo hàm $f^{\prime}(x)$ liên tục trên đoạn $[0 ; 1]$ thỏa $f(1)=0, \int_0^1\left(f^{\prime}(x)\right)^2 \mathrm{dx}=\dfrac{\pi^2}{8}$ và $\int_0^1 \cos \left(\dfrac{\pi}{2} x\right) f(x) \mathrm{d} x=\dfrac{1}{2}$. Tính $\int_0^1 f(x) \mathrm{d} x$.
A. $\dfrac{1}{\pi}$.
B. $\dfrac{2}{\pi}$.
C. $\dfrac{\pi}{2}$.
D. $\pi$.
A. $\dfrac{1}{\pi}$.
B. $\dfrac{2}{\pi}$.
C. $\dfrac{\pi}{2}$.
D. $\pi$.
Đặt $\left\{\begin{array}{l}u=f(x) \\ \mathrm{d} v=\cos \dfrac{\pi x}{2} \mathrm{~d} x\end{array} \Rightarrow\left\{\begin{array}{l}\mathrm{d} u=f^{\prime}(x) \mathrm{d} x \\ v=\dfrac{2}{\pi} \sin \dfrac{\pi x}{2}\end{array}\right.\right.$
Do đó $\int_0^1 \cos \left(\dfrac{\pi}{2} x\right) f(x) \mathrm{d} x=\dfrac{1}{2}$
$
\left.\Leftrightarrow \dfrac{2}{\pi} \sin \dfrac{\pi x}{2} f(x)\right|_0 ^1-\dfrac{2}{\pi} \int_0^1 \sin \left(\dfrac{\pi}{2} x\right) f^{\prime}(x) \mathrm{d} x=\dfrac{1}{2} \Leftrightarrow \int_0^1 \sin \left(\dfrac{\pi}{2} x\right) f^{\prime}(x) \mathrm{d} x=-\dfrac{\pi}{4} .
$
Lại có: $\int_0^1 \sin ^2\left(\dfrac{\pi}{2} x\right) \mathrm{d} x=\dfrac{1}{2}$
$
\begin{gathered}
\Rightarrow I=\int_0^1\left(-\dfrac{2}{\pi} \cdot f^{\prime}(x)\right)^2 \mathrm{~d} x-2\left(-\dfrac{2}{\pi}\right) \int_0^1 \sin \left(\dfrac{\pi}{2} x\right) f^{\prime}(x) \mathrm{d} x+\int_0^1 \sin ^2\left(\dfrac{\pi}{2} x\right) \mathrm{d} x \\
=\int_0^1\left(-\dfrac{2}{\pi} f^{\prime}(x)-\sin \left(\dfrac{\pi}{2} x\right)\right)^2 \mathrm{~d} x=\dfrac{4}{\pi^2} \dfrac{\pi^2}{8}-\dfrac{2}{\pi} \cdot \dfrac{\pi}{2}+\dfrac{1}{2}=0
\end{gathered}
$
Vì $\left(-\dfrac{2}{\pi} f^{\prime}(x)-\sin \left(\dfrac{\pi}{2} x\right)\right)^2 \geq 0$ trên đoạn $[0 ; 1]$ nên
$
\int_0^1\left(-\dfrac{2}{\pi} f^{\prime}(x)-\sin \left(\dfrac{\pi}{2} x\right)\right)^2 \mathrm{~d} x=0 \Leftrightarrow-\dfrac{2}{\pi} f^{\prime}(x)=\sin \left(\dfrac{\pi}{2} x\right) \Leftrightarrow f^{\prime}(x)=-\dfrac{\pi}{2} \sin \left(\dfrac{\pi}{2} x\right) \text {. }
$
Suy ra $f(x)=\cos \left(\dfrac{\pi}{2} x\right)+C$ mà $f(1)=0$ do đó $f(x)=\cos \left(\dfrac{\pi}{2} x\right)$.
Vậy $\int_0^1 f(x) \mathrm{d} x=\int_0^1 \cos \left(\dfrac{\pi}{2} x\right) \mathrm{d} x=\dfrac{2}{\pi}$.
Do đó $\int_0^1 \cos \left(\dfrac{\pi}{2} x\right) f(x) \mathrm{d} x=\dfrac{1}{2}$
$
\left.\Leftrightarrow \dfrac{2}{\pi} \sin \dfrac{\pi x}{2} f(x)\right|_0 ^1-\dfrac{2}{\pi} \int_0^1 \sin \left(\dfrac{\pi}{2} x\right) f^{\prime}(x) \mathrm{d} x=\dfrac{1}{2} \Leftrightarrow \int_0^1 \sin \left(\dfrac{\pi}{2} x\right) f^{\prime}(x) \mathrm{d} x=-\dfrac{\pi}{4} .
$
Lại có: $\int_0^1 \sin ^2\left(\dfrac{\pi}{2} x\right) \mathrm{d} x=\dfrac{1}{2}$
$
\begin{gathered}
\Rightarrow I=\int_0^1\left(-\dfrac{2}{\pi} \cdot f^{\prime}(x)\right)^2 \mathrm{~d} x-2\left(-\dfrac{2}{\pi}\right) \int_0^1 \sin \left(\dfrac{\pi}{2} x\right) f^{\prime}(x) \mathrm{d} x+\int_0^1 \sin ^2\left(\dfrac{\pi}{2} x\right) \mathrm{d} x \\
=\int_0^1\left(-\dfrac{2}{\pi} f^{\prime}(x)-\sin \left(\dfrac{\pi}{2} x\right)\right)^2 \mathrm{~d} x=\dfrac{4}{\pi^2} \dfrac{\pi^2}{8}-\dfrac{2}{\pi} \cdot \dfrac{\pi}{2}+\dfrac{1}{2}=0
\end{gathered}
$
Vì $\left(-\dfrac{2}{\pi} f^{\prime}(x)-\sin \left(\dfrac{\pi}{2} x\right)\right)^2 \geq 0$ trên đoạn $[0 ; 1]$ nên
$
\int_0^1\left(-\dfrac{2}{\pi} f^{\prime}(x)-\sin \left(\dfrac{\pi}{2} x\right)\right)^2 \mathrm{~d} x=0 \Leftrightarrow-\dfrac{2}{\pi} f^{\prime}(x)=\sin \left(\dfrac{\pi}{2} x\right) \Leftrightarrow f^{\prime}(x)=-\dfrac{\pi}{2} \sin \left(\dfrac{\pi}{2} x\right) \text {. }
$
Suy ra $f(x)=\cos \left(\dfrac{\pi}{2} x\right)+C$ mà $f(1)=0$ do đó $f(x)=\cos \left(\dfrac{\pi}{2} x\right)$.
Vậy $\int_0^1 f(x) \mathrm{d} x=\int_0^1 \cos \left(\dfrac{\pi}{2} x\right) \mathrm{d} x=\dfrac{2}{\pi}$.
Đáp án B.