A) 4 sq units
B) 3 sq units
C) 2 sq units
D) 1 sq unit
Correct Answer: A
Solution :
Required area\[=\int_{0}^{2\pi }{\cos x\,dx}\] \[=4\int_{0}^{2\pi }{\cos x\,dx}\] \[=4[\sin x]_{0}^{x/2}\] \[=4\left[ \sin \frac{\pi }{2}-\sin 0 \right]\] \[=4\,sq\,units\]
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