Cho hàm số fx liên tục trên ℝ và thỏa mãn f(x)+f(−x)=2cos2x, ∀x∈ℝ. Khi đó ∫−π2π2fxdx bằng

Chọn D

Với $f\left(x\right)+f(-x)=2\cos 2x,\forall x\in ℝ$ $\Rightarrow \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left(f\left(x\right)+f(-x)\right)\text{d}x=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}2\cos 2x\text{d}x\iff \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)\text{d}x+\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(-x\right)\text{d}x=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}2\cos 2x\text{d}x$ (*)

Tính $I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(-x\right)\text{d}x$

Đặt $t=-x\Rightarrow \text{d}t=-\text{d}x\Rightarrow \text{d}x=-\text{d}t$.

Đổi cận: $x=\frac{\pi }{2}\Rightarrow t=-\frac{\pi }{2}$; $x=-\frac{\pi }{2}\Rightarrow t=\frac{\pi }{2}$.

Khi đó $I=-\underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}f\left(t\right)\text{d}t=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(t\right)\text{d}t=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)\text{d}x$.

Từ (*), ta được: $2\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)\text{d}x=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}2\cos 2x\text{d}x={\left.\sin 2x\right|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=0$ $\Rightarrow \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)\text{d}x=0$.