A) \[\frac{({{\pi }^{3}}-8)}{4}\]
B) \[\frac{{{\pi }^{3}}}{4}\]
C) \[\frac{({{\pi }^{3}}-16)}{4}\]
D) \[\frac{({{\pi }^{3}}-8)}{2}\]
Correct Answer: A
Solution :
Area of the circle in first quadrant is \[\frac{\pi ({{\pi }^{2}})}{4}\]i.e., \[\frac{{{\pi }^{3}}}{4}\]. Also area bounded by curve \[y=\sin x\]and \[x\]-axis is 2 sq. unit. Hence required area is \[\frac{{{\pi }^{3}}}{4}-2=\frac{{{\pi }^{3}}-8}{4}\].You need to login to perform this action.
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