question_answer

Area bounded by \[y=x\sin x\] and \[x-\]axis between \[x=0\] and \[x=2\pi ,\] is                 [Roorkee 1981; RPET 1995]

A)            0    

B)            \[2\pi \] sq. unit

C)            \[\pi \] sq. unit                    

D)            \[4\pi \] sq. unit

Correct Answer: D

Solution :

                   Required area is \[{{A}_{1}}+{{A}_{2}}=\int_{0}^{\pi }{y\,\,dx+\left| \int_{\pi }^{2\pi }{y\,dx} \right|=4\pi \,sq.}\,unit\]