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 curve \[y=k\sin x\]between \[x=\pi \] and \[x=2\pi ,\] is

    A)            \[2k\] sq. unit                      

    B)            0

    C)            \[\frac{{{k}^{2}}}{2}\] sq. unit                                          

    D)            \[k\] sq. unit

    Correct Answer: A

    Solution :

                       Required area \[=k\int_{\pi }^{2\pi }{\sin x\,\,dx=k}[-\cos x]_{\pi }^{2\pi }=-2k\]                    Hence, area = 2k sq. unit.

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