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Let · denote the greatest integer function then the value of

015xx2dx is

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a
0
b
3/2
c
3/4
d
5/4

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detailed solution

Correct option is C

Put x2=t, so that xdx=12dtI=12∫02⋅25 [t]dt=12∫01 [t]dt+∫12 [t]dt+∫22⋅25 [t]dt=120+∫12 dt+2∫22⋅25 dt=12[1+2×.25]=34


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