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

  • question_answer If the ordinate \[x=a\] divides the area bounded by the curve \[y=\left( 1+\frac{8}{{{x}^{2}}} \right)\,,\] \[x-\]axis and the ordinates \[x=2,\] \[x=4\] into two equal parts, then \[a=\] [IIT 1983]

    A)            8    

    B)            \[2\sqrt{2}\]

    C)            2    

    D)            \[\sqrt{2}\]

    Correct Answer: B

    Solution :

               Let the ordinate at \[x=a\]divide the area into two equal parts                    Area of \[AMNB\]\[=\int_{2}^{4}{\left( 1+\frac{8}{{{x}^{2}}} \right)\text{ }dx=\left[ x-\frac{8}{x} \right]}_{2}^{4}=4\]                    Area of \[ACDM=\int_{2}^{a}{\left( 1+\frac{8}{{{x}^{2}}} \right)}dx=2\]                    On solving, we get \[a=\pm 2\sqrt{2}\];Since \[a>0\]Þ \[a=2\sqrt{2}\].


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