Cho hàm số f (x) liên tục trên R và thoả mãn f (x) + f (−x) = 3 − 2cos x

Trả lời

Ta có: f (x) + f (−x) = 3 − 2cos x

Đặt t = −x Þ dt = − dx Û dx = − dt

Khi đó:

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

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

$=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}3-2\cos xdx={\left.\left(3x-2\sin x\right)\right|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$=3.\frac{\pi }{2}-2\sin \left(\frac{\pi }{2}\right)-3.\left(-\frac{\pi }{2}\right)+2\sin \left(-\frac{\pi }{2}\right)$

= 3p − 4

Vậy   $I=\underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}f\left(x\right)dx=\frac{3\pi }{2}-2$.