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

  • question_answer The area between the curve \[y={{\sin }^{2}}x,\] \[x-\]axis and the ordinates \[x=0\] and \[x=\frac{\pi }{2}\] is                    [RPET 1996]

    A)            \[\frac{\pi }{2}\]                 

    B)            \[\frac{\pi }{4}\]

    C)            \[\frac{\pi }{8}\]                 

    D)            \[\pi \]

    Correct Answer: B

    Solution :

               Required area \[A=\int_{0}^{\pi /2}{{{\sin }^{2}}x.\,dx}=\int_{0}^{\pi /2}{\left( \frac{1-\cos 2x}{2} \right)}\,dxa\]                                                                       \[=\frac{1}{2}[x]_{0}^{\pi /2}-\frac{1}{4}[\sin 2x]_{0}^{\pi /2}=\frac{\pi }{4}\].


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