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Câu hỏi: Cho hàm số $f\left( x \right)$ liên tục trên $\mathbb{R}$ và thỏa mãn $f(x)+f(-x)=2\cos 2x, \forall x\in \mathbb{R}$. Khi đó $\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}$ bằng
A. $-2$.
B. $4$.
C. $2$.
D. $0$.
A. $-2$.
B. $4$.
C. $2$.
D. $0$.
Với $f(x)+f(-x)=2\cos 2x, \forall x\in \mathbb{R}$ $\Rightarrow \int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{\left( f(x)+f(-x) \right)\text{d}x}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{2\cos 2x\text{d}x}\Leftrightarrow \int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}+\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( -x \right)\text{d}x}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{2\cos 2x\text{d}x}$ (*)
Tính $I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( -x \right)\text{d}x}$
Đặt $t=-x\Rightarrow \text{d}t=-\text{d}x\Rightarrow \text{d}x=-\text{d}t$.
Đổi cận: $x=\dfrac{\pi }{2}\Rightarrow t=-\dfrac{\pi }{2}$ ; $x=-\dfrac{\pi }{2}\Rightarrow t=\dfrac{\pi }{2}$.
Khi đó $I=-\int\limits_{\dfrac{\pi }{2}}^{-\dfrac{\pi }{2}}{f\left( t \right)\text{d}t}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( t \right)\text{d}t}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}$.
Từ (*), ta được: $2\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{2\cos 2x\text{d}x}=\left. \sin 2x \right|_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}=0$ $\Rightarrow \int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=0$.
Tính $I=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( -x \right)\text{d}x}$
Đặt $t=-x\Rightarrow \text{d}t=-\text{d}x\Rightarrow \text{d}x=-\text{d}t$.
Đổi cận: $x=\dfrac{\pi }{2}\Rightarrow t=-\dfrac{\pi }{2}$ ; $x=-\dfrac{\pi }{2}\Rightarrow t=\dfrac{\pi }{2}$.
Khi đó $I=-\int\limits_{\dfrac{\pi }{2}}^{-\dfrac{\pi }{2}}{f\left( t \right)\text{d}t}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( t \right)\text{d}t}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}$.
Từ (*), ta được: $2\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=\int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{2\cos 2x\text{d}x}=\left. \sin 2x \right|_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}=0$ $\Rightarrow \int\limits_{-\dfrac{\pi }{2}}^{\dfrac{\pi }{2}}{f\left( x \right)\text{d}x}=0$.
Đáp án D.