If Ik=∫−2kπ2kπ |sin⁡x|[sin⁡x]dx,∀k∈N, where [.] denotes the greatest integer function, then ∑k=110 Ik is equal to

If Ik=22|sinx|[sinx]dx,kN, where [.] denotes the greatest integer function, then k=110Ik is equal to

  1. A

    -110

  2. B

    -440

  3. C

    -330

  4. D

    -220

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    Solution:

    Ik=20|sinx|[sinx]dx+02|sinx|[sinx]dx=20|sinx|[sinx]dx+02|sinx|[sinx]dx=02|sinx|([sinx]+[sinx])dx=2k0π|sinx|([sinx]+[sinx])dx=2k0πsinx(01)dx=2k[cosx]0π=4k k=110Ik=4k=110k=410112=220sinx<0,x(0,π)

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