A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{8}\]
D) \[\pi \]
Correct Answer: B
Solution :
Required area \[A=\int_{0}^{\pi /2}{{{\sin }^{2}}x.\,dx}=\int_{0}^{\pi /2}{\left( \frac{1-\cos 2x}{2} \right)}\,dxa\] \[=\frac{1}{2}[x]_{0}^{\pi /2}-\frac{1}{4}[\sin 2x]_{0}^{\pi /2}=\frac{\pi }{4}\].You need to login to perform this action.
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