The value of ∫limits_0^1|sin 2 π x| d x is equal to

Since $|\sin 2 \pi x|$ is periodic with period $\frac{1}{2}$,

$I=\int\limits_0^1 |\sin 2 \pi x| dx=2 \int\limits_0^{1 / 2} \sin 2 \pi ~dx$

$=2\left[-\frac{\cos 2 \pi x}{2 \pi}\right]_0^{1 / 2}=\frac{2}{\pi}$

Hence (2) is the correct answer.