2. Solved Papers
  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2002

  • question_answer
    \[\int_{-2}^{2}{|1-x{{|}^{2}}}dx\]is equal to:

    A)  4     

    B)                                         2

    C)         \[-2\]                                   

    D)         0

    E)         1

    Correct Answer: A

    Solution :

      \[\int_{-2}^{2}{|1-x{{|}^{2}}}dx=2\int_{0}^{2}{|1-{{x}^{2}}|}\,dx\] \[=2[\int_{0}^{1}{{{(1-x)}^{2}}x+\int_{1}^{2}{({{x}^{2}}-1)}dx}]\] \[=2\left[ \left( x-\frac{{{x}^{3}}}{3} \right)_{0}^{1}+\left( \frac{{{x}^{3}}}{3}-x \right)_{1}^{2} \right]\] \[=2\left[ \frac{2}{3}+\frac{2}{3}+\frac{2}{3} \right]=4\] 


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