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

  • question_answer For \[0\le x\le \pi ,\] the area bounded by \[y=x\] and \[y=x+\sin x,\] is                [Roorkee Qualifying 1998]

    A)            2    

    B)            4

    C)            \[2\pi \]                                  

    D)            \[4\pi \]

    Correct Answer: A

    Solution :

                       The curves \[y=x\]and \[y=x\] intersect at (0, 0) and \[(\pi ,\pi )\]. Hence area bounded by the two curves                    \[=\int\limits_{0}^{\pi }{(x+\sin x)dx-\int\limits_{0}^{\pi }{x\,dx}=\int\limits_{0}^{\pi }{\sin x\,dx}}\]                    \[=[-\cos x]_{0}^{\pi }=-\cos \pi +\cos 0=-(-1)+(1)=2\].

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