A) n
B) 2n
C) ?2n
D) None of these
Correct Answer: B
Solution :
\[\int_{0}^{2n\pi }{\left( |\sin x|-\frac{1}{2}|\sin x| \right)}\ dx\]=\[\frac{1}{2}\int_{0}^{2n\pi }{\ \ \ \ |\sin x|dx}\] \[=\frac{2n}{2}\times 2\int_{0}^{\pi /2}{\sin x\ dx=2n}[-\cos x]_{0}^{\pi /2}=2n.\]
You need to login to perform this action.
You will be redirected in
3 sec
