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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2006

  • question_answer
                    Let\[f(x)=x-[x],\]for every real\[x,\]where\[[x]\]is the greatest integer less than or equal to\[x\]. Then,\[\int\limits_{-1}^{1}{f(x)}dx\]is:

    A)  1                            

    B)         2

    C)  3                            

    D)         0

    E)  \[\frac{1}{2}\]

    Correct Answer: A

    Solution :

    \[\therefore \] \[\int\limits_{-1}^{1}{(x-[x])dx=}\int\limits_{-1}^{0}{(x-[x])}dx\] \[+\int\limits_{0}^{1}{(x-[x])}dx\] \[=\int\limits_{-1}^{0}{(x+1)}dx+\int\limits_{0}^{1}{(x-0)}dx\] \[=\left[ \frac{{{(x+1)}^{2}}}{2} \right]_{-1}^{0}+\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{1}=\frac{1}{2}+\frac{1}{2}=1\]


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