• # 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

$\therefore$ 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