∫02π (sin⁡x+|sin⁡x|)dx is equal to

02π(sinx+|sinx|)dx is equal to

  1. A

    0

  2. B

    4

  3. C

    8

  4. D

    1

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    Solution:

    02π(sinx+|sinx|)dx=0π(sinx+sinx)dx+π2π(sinxsinx)dx=0π2sinxdx+π2π0dx=2[cosx]0π+0=2(cosπcos0)=2(11)=4

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