A) \[2k\] sq. unit
B) 0
C) \[\frac{{{k}^{2}}}{2}\] sq. unit
D) \[k\] sq. unit
Correct Answer: A
Solution :
Required area \[=k\int_{\pi }^{2\pi }{\sin x\,\,dx=k}[-\cos x]_{\pi }^{2\pi }=-2k\] Hence, area = 2k sq. unit.
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