A) 0
B) 1
C) 2
D) 4
Correct Answer: D
Solution :
\[\int_{0}^{2\pi }{|\sin x|dx=\int_{0}^{\pi }{\sin x\,dx+\int_{\pi }^{2\pi }{-\sin x\,dx}}}\] \[=[-\cos x]_{0}^{\pi }+[\cos x]_{\pi }^{2\pi }=1+1+1+1=4\].
You need to login to perform this action.
You will be redirected in
3 sec
