A) 1 sq unit
B) 2 sq units
C) 4 sq units
D) 8 sq units
Correct Answer: C
Solution :
The graph of \[y=\sin x\]can be drawn as required area = Area of OABO + Area BCDB \[=\int\limits_{0}^{x}{|\sin \,x|dx+\int\limits_{\pi }^{2\pi }{|\sin \,x|}dx}\] \[=\int\limits_{0}^{x}{\sin x\,dx+\int\limits_{\pi }^{2\pi }{(-\sin \,x)dx}}\] (\[\because \,sin\ge 0\,\]for \[x\in [x,\pi ]\] and \[\sin x\le 0\]for \[x\in [{{\pi }_{1}}2\pi ]\]) \[=[-\cos x]_{0}^{\pi }+[\cos x]_{\pi }^{2\pi }\] \[=-\cos \pi +\cos 0+cos2\pi -\cos \pi \] \[=-(-1)+1+1-(-1)\] \[\text{=}\,\text{4sq}\,\text{units}\]
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