A) \[2\pi \]sq units
B) \[3\pi \]sq units
C) \[4\pi \]sq units
D) \[5\pi \]sq units
Correct Answer: C
Solution :
\[\therefore \] Required area \[=\int_{0}^{\pi }{x\,\sin \,x\,dx\,+|}\int_{\pi }^{2\pi }{(x\,\sin \,x)\,dx|}\] \[=[-x\,cos\,x+\,\int{\cos \,x\,dx]_{0}^{\pi }}\] \[+\,\left| [-x\,\cos \,x+\,\int{\cos \,xdx\,]_{\pi }^{2\pi }} \right|\] \[=[-x\,\cos \,x+\sin \,x]_{0}^{\pi }+|[-x\,\cos \,x+\sin \,x]_{\pi }^{2\pi }|\] \[=[\pi +0-(0)]+|[-2\pi +0-\pi +0]|\] \[=\pi +3\pi \] \[=4\pi \] sq unitsYou need to login to perform this action.
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