Question

Integral ${\int }_0^1\left|\sin 2\pi x\right|dx$ is equal to

• A. $0$
• B. $-\frac{1}{\pi }$
• C. $\frac{1}{\pi }$
• D. $\frac{2}{\pi }$

Answer 1

The correct option is D $\frac{2}{\pi }$

Given, ${\int }_0^1\left|\sin 2\pi x\right|dx$

$={\int }_0^{1/2}\left|\sin 2\pi x\right|dx+{\int }_{1/2}^1\left|\sin 2\pi x\right|dx={\int }_0^{1/2}\sin 2\pi xdx-{\int }_{1/2}^1\sin 2\pi xdx=\frac{1}{2\pi }({[-\cos 2\pi x]}_0^{1/2}-{[-\cos 2\pi x]}_{1/2}^1)$

$=\frac{2}{\pi }$