question_answer

Area bounded by the curve, \[y=x\,\sin \,x\] and \[x-\]axis between \[x=0\] and \[x=2\pi \] is

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 units