A) 0
B) 1
C) 2
D) 4
Correct Answer: D
$ \int_0^{2\pi }{|\sin x|dx=\int_0^{\pi }{\sin xdx+\int _{\pi }^{2\pi }{-\sin xdx}}} $
$ =[-\cos x] _0^{\pi }+[\cos x] _{\pi }^{2\pi }=1+1+1+1=4 $ .
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