1exactly prajith..........ive already posted tht doubt.....
20 Answers
62yup it will be.. and i have changed that already.. tx for pointing out :)
1ok................thnx................so wont d answer be........./4???
62no you have to take
0∫1([x]-[x-1/2])dx=
0∫1/2([x]-[x-1/2])dx+1/2∫1([x]-[x-1/2])dx
= 0∫1/2(1)dx+1/2∫1(1/2)dx
= 1/2+0 =1/2
1how????????? example.....
x=2.3
[x]=2
[x-1/2]=1
so integral [x]-[x-1/2]=2-1=1
0∫11dx=1...........rite??
1wel we cant say prajith..........coz d limits are variables of x.....so anything can happen........but i think we may be rite.....
62yes exactly akand.. so the integral from [k] to [k+1] will have it as 1/2
right?
hence there are [x] such integrals .. so it will become [x]/2!
1it is either 1 or 0..................
1 if fractional part of x is lesser than 1/2 and 0 if its greater than 1/2
1Bhaiyya the in question itz [x- 1/2] and not [(x-1)/2]......so it should be [x] im i rite bhaiyya!!!!!!!!!!
62the proof comes like this..
[x] is 1 more than [x-1/2] when x has fractional part less than 1/2
otherwise it is zero
thus the first part will be [x]/2
now we have to integrate this from 0 to [x]
which will be 1/2+2/2+3/2.....+([x]-1)/2
= [x].([x]-1)/4
= [x].([x]-1)/2
