Question

Evaluate: ${\int }_0^1|\sin 2\pi x|dx$

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

Answer 1

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

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

Since , $\sin x$ is negative in the interval $(\pi ,2\pi )$, hence integration becomes:-

$={\int }_0^{1/2}\sin \left(2\pi x\right)dx-{\int }_{1/2}^{1}sin\left(2\pi x\right)dx$

$=\frac{1}{2\pi }{[-\cos 2\pi x]}_0^{1/2}-\frac{1}{2\pi }{[-\cos 2\pi x]}_{1/2}^1$

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