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∫2x-[x] .dx
limits from 0 to [x]
To solve this divide the integrals into o to 1 ,1 1to 2 ..... upto [x]-1 to [x]
so you get integrations of 2^(x-0) from 0 to 1 +2^(x-1) from 1 to 2 .... upto 2^(x-[x]+1) from [x-1]to [x]
each integral evaluates to 1/ln2 so we have [x] terms in our integral sum expression ..
therefore you get ans=[x]/ln2
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