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 the curve \[y=\sin x\] between \[x=0\] and \[x=2\pi \] is

    A)            2 sq. unit                                

    B)            4 sq. unit

    C)            8 sq. unit                                

    D)            None of these

    Correct Answer: B

    Solution :

                       We have \[y=\sin x\]
    \[x\] 0 \[\pi /6\] \[\pi /2\] \[\pi \] \[3\pi /2\] \[2\pi \]
    \[y\] 0 0.5 1 0 ?1 0
                  Join these points with a free hand to obtain a rough sketch                               Required area = (area of \[OAB\]) + (area of\[BCD)\]                                          = \[\int_{0}^{\pi }{\,y\,dx+\int_{\pi }^{2\pi }{(-y)\,dx}}\],                                                    (\[\because \] Area \[BCD\] is below \[x-\]axis)                                                                          = \[\int_{0}^{\pi }{\sin x\,dx-\int_{\pi }^{2\pi }{\sin x\,dx=4}}\] sq. unit.

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