The value of ∫limits_0^1|sin 2πx| is equal to

Since $|sin 2π\,x |$ is periodic with period 1/2,

$I = \int\limits_0^1|sin 2 π\,x | dx = \int\limits_0^{1/2}sin2π\,x dx$

$= 2[-\frac{cos2π\,x}{2π}]_0^{1/2} = \frac{2}{\pi}$.

Hence (B) is the correct answer.