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

  • question_answer Area under the curve \[y=\sqrt{3x+4}\] between \[x=0\] and \[x=4,\] is                                               [AI CBSE 1979, 80]

    A)            \[\frac{56}{9}\] sq. unit   

    B)            \[\frac{64}{9}\] sq. unit

    C)            8 sq. unit                                

    D)            None of these

    Correct Answer: D

    Solution :

                       Area \[=\int_{0}^{4}{\sqrt{3x+4}}dx=\left| \frac{{{(3x+4)}^{3/2}}}{3.(3/2)} \right|_{0}^{4}\]                             \[=\frac{2}{9}\times 56=\frac{112}{9}sq.\]unit.


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