A) 2
B) 4
C) 0
D) 3
Correct Answer: B
Solution :
\[y=\cos x\], When \[x\in \left[ 0,\frac{\pi }{2} \right],\cos x\ge 0\] When \[x\in \left[ \frac{\pi }{2},\frac{3\pi }{2} \right],\cos x\le 0\] When \[x\in \left[ \frac{3\pi }{2},2\pi \right],\cos x\ge 0\] Thus required area is given by, \[\int_{0}^{\pi /2}{\,\,ydx}=\int_{0}^{\pi /2}{\cos x\,dx+\int_{\pi /2}^{3\pi /2}{(-\cos x)dx+\int_{3\pi /2}^{2\pi }{\,\cos xdx}}}\] \[=1+2+1=4sq.\]unit.
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