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 curve \[y=4x-{{x}^{2}}\] and the \[x-\]axis, is                                 [MP PET 1999, 2003]

    A)            \[\frac{30}{7}\] sq. unit   

    B)            \[\frac{31}{7}\] sq. unit

    C)            \[\frac{32}{3}\] sq. unit   

    D)            \[\frac{34}{3}\] sq. unit

    Correct Answer: C

    Solution :

               We have \[y=4x-{{x}^{2}}\] and \[y=0\];\[\therefore \] \[x=0\], \[4\]                    Required area \[=\int_{0}^{4}{(4x-{{x}^{2}})dx=\left[ \frac{4{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3} \right]_{0}^{4}}\]                      \[=32-\frac{64}{3}=\frac{32}{3}\]sq. unit.

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