JEE Main & Advanced AIEEE Paper (Held On 11 May 2011)

  • question_answer
    Let [.] denote the greatest integer function then the value of \[\int\limits_{0}^{1.5}{x[{{x}^{2}}]dx}\]is :     AIEEE  Solved  Paper (Held On 11 May  2011)

    A)  \[0\]                  

    B)  \[\frac{3}{2}\]

    C) \[\frac{3}{4}\]

    D) \[\frac{5}{4}\]

    Correct Answer: C

    Solution :

                    \[\int\limits_{0}^{1}{x[{{x}^{2}}]dx}+\int\limits_{1}^{\sqrt{2}}{x[{{x}^{2}}]dx}+\int\limits_{\sqrt{2}}^{1.5}{x}[{{x}^{2}}]dx\] \[\int\limits_{0}^{1}{x.0dx}+\int\limits_{1}^{\sqrt{2}}{xdx}+\int\limits_{\sqrt{2}}^{1.5}{2xdx}\] \[0+\left[ \frac{{{x}^{2}}}{2} \right]_{1}^{\sqrt{2}}+\left[ {{x}^{2}} \right]_{\sqrt{2}}^{1.5}\] \[\frac{1}{2}(2-1)+(2.25-2)\] \[\frac{1}{2}+.25\]

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