Cho hàm số f (x) liên tục trên R và thoả mãn f(x) + f(-x) = căn 2 + 2cos2x

Trả lời

Ta có:  $f\left(x\right)+f\left(-x\right)=\sqrt{2+2\cos 2x}$

$=\sqrt{2+2\left(2{\cos }^{2}x-1\right)}=2\sqrt{{\cos }^{2}x}=2\left|\cos x\right|$

Đặt t = −x Þ dt = − dx Û dx = − dt

Khi đó:

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

$\Rightarrow 2I=\underset{-\frac{3\pi }{2}}{\overset{\frac{3\pi }{2}}{\int }}f\left(x\right)dx+\underset{-\frac{3\pi }{2}}{\overset{\frac{3\pi }{2}}{\int }}f\left(-x\right)dx=\underset{-\frac{3\pi }{2}}{\overset{\frac{3\pi }{2}}{\int }}\left[f\left(x\right)+f\left(-x\right)\right]dx$

$=\underset{-\frac{3\pi }{2}}{\overset{\frac{3\pi }{2}}{\int }}2\left|\cos x\right|dx=-2\underset{-\frac{3\pi }{2}}{\overset{\frac{3\pi }{2}}{\int }}\cos xdx$

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

= (−2).(−1) + 2.1 = 4

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