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Discussion Response Post to: another Definite integrals..

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Akhil (2725)

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total posts: 1451
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see takin denominator in power we get
2^(x-[x])
=2^{x}
{x} is periodic with period 1

so 2^{x} also repaets after every 1 unit
also integrating from 0 to [x]
we have integral no of such units
area of 1 unit
=integral (from 0 to 1) 2^x
=2^x/log 2 from 0 to 1
=(2-1)/log 2
=1/log 2
and there are [x] such units

so total=[x]/log 2

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