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

  • question_answer The area enclosed between the curves \[y={{x}^{3}}\]and \[y=\sqrt{x}\] is, (in square units)                                         [Karnataka CET 2004]

    A)            \[\frac{5}{3}\]                      

    B)            \[\frac{5}{4}\]

    C)            \[\frac{5}{12}\]                   

    D)            \[\frac{12}{5}\]

    Correct Answer: C

    Solution :

               Given curves are, \[y={{x}^{3}}\] and \[y=\sqrt{x}\]                    On solving, we get \[x=0,x=1\]                    Therefore, required area =\[\int_{0}^{1}{({{x}^{3}}-\sqrt{x})}\,dx\]             \[=\left[ \frac{{{x}^{4}}}{4}-\frac{2x\sqrt{x}}{3} \right]_{0}^{1}=\left[ \frac{1}{4}-\frac{2}{3} \right]=\frac{5}{12}\],(Area can?t be negative).


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