CUET Preparation Today
The value of $\int\limits_0^1|\sin 2 \pi x| d x$ is equal to |
0 $\frac{2}{\pi}$ $\frac{1}{\pi}$ 2 |
$\frac{2}{\pi}$ |
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. |
