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Hello majumder
Let us assume x = I + f where I is integral part and f is fractional part ..
intg ( 0 to I ) [x].dx + intg ( I to I+f) [x].dx
I(I-1)/2 + I ( I+f - I)
= I(I-1)/2 + I . f ans
Note : Intg ( o to n ) [x] .dx where n is integer = n(n-1)/2 ..so i have used it .
Also intg ( I to I+f) [x] .dx means x varies from I to I + f ...therefore [x] = I
----= intg ( I to I+f) .I.dx = I ( I+f - I ) = I.f
Is it clear or not !
Yagya
askiitians_expert
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