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Prove that int0^x[t]dt=([x]([x]-1))/2+[...

Prove that `int_0^x[t]dt=([x]([x]-1))/2+[x](x-[x]),` where [.] denotes the greatest integer function.

Text Solution

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Let `x=n+fAAn epsilon` and `0leflt1`
`:.[x]=n`
`int_(0)^(x)[t]dt=int_(0)^(1)[t]dt+int_(1)^(2)[t]dt+int_(2)^(3)[t]dt+………..+int_(n)^(n+f)[t]dt`
`=0+1int_(1)^(2)dt+2int_(2)^(3)dt+………….+n int_(n)^(n+f)dt`
`=(2-1)+2(3-2)+……..+n(n+f-n)`
`=1+2+3..........+(n-1)+nf`
`=((n-1)n)/2+nf`
`=([x]([x]-1))/2+[x](x-[x])` [from equation 1]
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