Question

${\int }_0^{2n\pi }\{\left|\sin x\right|-\left|\frac{1}{2}\sin x\right|\}dx$ equals.

• A. $n$
• B. $2n$
• C. $-2n$
• D. None of the above

Answer 1

The correct option is B $2n$

Let $I={\int }_0^{2n\pi }\{\left|\sin x\right|-\left|\frac{1}{2}\sin x\right|\}dx$
$I=2n{\int }_0^{\pi }\{\left|\sin x\right|-\left|\frac{1}{2}\sin x\right|\}dx$
$I=2n{\int }_0^{\pi }\{\left(\sin x\right)-\frac{1}{2}\sin x\}dx[{\int }_0^{Tx}ydy=T{\int }_0^{x}ydy]$ (where T is regular period of y)
$I=2n{\int }_0^{\pi }\frac{1}{2}\sin xdx$
$I=\frac{2n}{2}{\int }_0^{\pi }\sin xdx$
$I=n{[-\cos x]}_0^{\pi }$
$I=n[-\cos \pi +\cos 0]$
$I=n[-(-1)+1]$
$I=2n$