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

  • question_answer Let y be the function which passes through (1, 2) having slope \[(2x+1)\]. The area bounded between the curve and   x-axis is   [DCE 2005]

    A)            6 sq. unit                                

    B)            5/6 sq. unit

    C)            1/6 sq. unit                           

    D)            None of these

    Correct Answer: C

    Solution :

               \[\frac{dy}{dx}=2x+1\] Þ \[y={{x}^{2}}+x+c\]                    Þ \[y={{x}^{2}}+x\],     [\[\because \] c = 0 by putting x = 1, y = 2)                    Þ \[{{\left( x+\frac{1}{2} \right)}^{2}}=y+\frac{1}{4}\], which is a equation of parabola, whose vertices is, \[V\left( \frac{-1}{2},\,\frac{-1}{4} \right)\] \[\therefore \] Required area \[=\left. \left| \int_{-1}^{0}{({{x}^{2}}+x)\ dx} \right. \right|\] \[=\left( \frac{{{x}^{3}}}{3}+\frac{{{x}^{2}}}{2} \right)_{-1}^{0}\]                                                                       \[\left. =\left| \frac{-1}{3}+\frac{1}{2} \right. \right|=\frac{1}{6}\] sq. unit.

You need to login to perform this action.
You will be redirected in 3 sec spinner