Question

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

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

Answer 1

Answer: (D)

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