Cho hàm số f (x) liên tục trên ℝ và thoả mãn f (x) + f (−x) = 2cos 2x, với mọi x thuộc R

Trả lời

Ta có: f (x) + f (−x) = 2cos 2x

Đặt t = −x Þ dt = − dx Û dx = − dt

Khi đó:

$I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=-\underset{\frac{\pi }{2}}{\overset{-\frac{\pi }{2}}{\int }}f\left(-t\right)dt=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(-t\right)dt=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(-x\right)dx$

$\Rightarrow 2I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx+\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(-x\right)dx=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\left[f\left(x\right)+f\left(-x\right)\right]dx$

$=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}2\cos 2xdx=2\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\cos 2xdx$

$={\left.\sin 2x\right|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=\sin \left(\pi \right)-\sin \left(-\pi \right)$ = 0 

Vậy  $I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=0$.