$\underset{0}{\overset{1}{\int }}\left|\sin 2\pi x\right|dx=$

$I=\underset{0}{\overset{1}{\int }}\left|\sin 2\pi x\right|dx$

$if \sin 2\pi x=0$

$\Rightarrow 2\pi x=0,\pi ,2\pi$

$\Rightarrow x=0,\frac{1}{2},1$

now we have to split the integral at 1/2

$I=\underset{0}{\overset{1/2}{\int }}\left|\sin 2\pi x\right|+\underset{1/2}{\overset{1}{\int }}\left|\sin 2\pi x\right|dx$

$I=\underset{0}{\overset{1/2}{\int }}\sin 2\pi xdx+\underset{1/2}{\overset{1}{\int }}(-\sin 2\pi x)dx$

$={\left(\frac{{\displaystyle -\cos 2\pi x}}{{\displaystyle 2\pi }}\right)}_0^{1/2}+{\left(\frac{{\displaystyle \cos 2\pi x}}{{\displaystyle 2\pi }}\right)}_{1/2}^1$

$=\frac{1}{2\pi }\left[1-(-1)\right]+\frac{1}{2\pi }[\cos 2\pi -\cos \pi ]$

$=\frac{1}{\pi }+\frac{1}{2\pi }[1-(-1)]=\frac{2}{\pi }$