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 \[y=f(x)\], x-axis and ordinates x = 1 and \[x=b\]is \[\frac{5}{24}\pi \], then \[f(x)\] is                    [RPET 2000]

    A)            \[3(x-1)\cos (3x+4)+\sin (3x+4)\]

    B)            \[(b-1)\sin (3x+4)+3\cos (3x+4)\]

    C)            \[(b-1)\cos (3x+4)+3\sin (3x+4)\]

    D)            None of these

    Correct Answer: A

    Solution :

                       Area bounded by the curve \[y=f(x),\,\]x-axis and the ordinates \[x=1\] and \[x=b\] is \[\int_{1}^{b}{f(x)\,dx}\]                    \ From the question \[\int_{1}^{b}{f(x)\,dx=(b-1)\sin (3b+4)}\]                    Differentiate with respect to b, we get            \[f(b)\,.\,1=3(b-1)\cos (3b+4)+\sin (3b+4)\]                    \[f(x)=3(x-1)\cos (3x+4)+\sin (3x+4)\].


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