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Prove ∫ [t] dt =[x]([x]-1)/2 +[x](x-[x]) where x belongs to set of real
0
positve numbers and [.] denotes greatest integer function.
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1 Answers
62theking..
the solution is not very difficult...
see first take the integral from
0-1 [t]=0
1-2 [t]=1
so on from
N-N+1 [t] = N
...
...
[x-1] to [x] [t] = [x-1]
then fro the last part [x] to x [t]=[x]
this will nail the problem..
the last integral will give the second term... [x].(x-[x])
the others will give 1.[0]+1.[1]+............ 1.[x-1]
hence ur sum :)