Ta có

$\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\left| \sin 2x \right|}dx=\int_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\left| \sin 2x \right|dx}+\int_{\frac{\pi }{2}}^{\frac{3\pi }{4}}{\left| \sin 2x \right|dx}$

$=\left. -\frac{\cos 2x}{2} \right|_{\frac{\pi }{4}}^{\frac{\pi }{2}}+\left. \frac{\cos 2x}{2} \right|_{\frac{\pi }{2}}^{\frac{3\pi }{4}}=\frac{1}{2}+\frac{1}{2}=1$

Vậy $\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}{\left| \sin 2x \right|}dx=1$. Ta chọn phương án B.