∫04 |x−1|dx is equal to

04|x1|dx is equal to

  1. A

    52

  2. B

    32

  3. C

    12

  4. D

    5

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    Solution:

    Let 04|x1|dx

    It can be seen that, (x1)0 when 0x1 and 

    (x1)0 when 1x4

     I=01|x1|dx+14|x1|dx=01(1x)dx+abf(x)dx=acf(x)dx+cbf(x)dx=xx2201+x22x14=1120+4224121=12+4+12=5

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