A) 1
B) 2
C) 0
D) 4
E) \[2\pi \]
Correct Answer: D
Solution :
Given curves,\[y=sin\text{ }x,\text{ }x=0\]and\[x=2\pi \] Area of\[OAB=2\]Area of OA (\[\because \]by symmetry) \[=2.\int_{x=0}^{\pi }{y\,dx}\] \[=2\int_{0}^{\pi }{\sin x\,dx}\] \[=-2[\cos x]_{0}^{\pi }\] \[=-2\,(\cos \pi -\cos 0)\] \[=-2(-1-1)=4\]You need to login to perform this action.
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