The value of $\int\limits_0^1|sin\,2πx|$ is equal to

0 2/π 1/π 2

2/π

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.