$\int_{o}^{x}[x]dx$

11 years ago

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