JEE Main & Advanced Mathematics Definite Integration Question Bank Area Bounded by Region, Volume and Surface Area of Solids of Revolution

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


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