∫−22 |[x]|dx is equal to

22|[x]|dx is equal to

  1. A

    1

  2. B

    2

  3. C

    3

  4. D

    4

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    Solution:

    22|[x]|dx=21|[x]|dx+10|[x]|dx+01|[x]|dx+12|[x]|dx=212dx+101dx+010dx+121dx=2[x]21+[x]10+0+[x]12=2(1+2)+(0+1)+(21)=2+1+1=4

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