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

  • question_answer
    The area bounded by the x-axis, the curve \[y=f(x)\] and the lines \[x=1,\,x=b\] is equal to \[\sqrt{{{b}^{2}}+1}-\sqrt{2}\] for all b > 1, then \[f(x)\] is  \[\]                                   [MP PET 2000; AMU 2000]

    A)            \[\sqrt{x-1}\]                       

    B)            \[\sqrt{x+1}\]

    C)            \[\sqrt{{{x}^{2}}+1}\]       

    D)            \[\frac{x}{\sqrt{1+{{x}^{2}}}}\]

    Correct Answer: D

    Solution :

                       \[\int_{1}^{b}{f(x)\,dx=\sqrt{{{b}^{2}}+1}-\sqrt{2}}=\sqrt{{{b}^{2}}+1}-\sqrt{1+1}=[\sqrt{{{x}^{2}}+1}]_{1}^{b}\]            \[\therefore \] \[f(x)=\frac{d}{dx}\sqrt{{{x}^{2}}+1}\]\[=\frac{2x}{2\sqrt{{{x}^{2}}+1}}=\frac{x}{\sqrt{{{x}^{2}}+1}}\].


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