1.

int_(0)^(x)(2^(t))/(2^([t]))dt, where [.] denotes the greatest integer function, and x epsilonR^(+) is equal to

Answer»

`1/(1n2)([x]+2^({x})-1)`
`1/(1n2)([x]+2^({x}))`
`1/(1n2)([x]-2^({x}))`
`1/(1n2)([x]+2^({x})+1)`

Solution :Let `nlexltn+1`, where `N epsilonI`
`I=int_(0)^(x)(2^(t))/(2^([t]))DT=int_(0)^(n)2^({t})dt+int_(n)^(x)2^({t})dt`
`=n int_(0)^(1)2^({t})dt+int_(n)^(x)2^({t})dt`
`=n int_(0)^(1)2^(t)dt+int_(n)^(x)2^(t-n)dt`
`=n(2^(t))/(In2)|_(0)^(1)+1/(2^(n)) (2^(t))/(In2)|_(n)^(x)`
`=n/(In2)(2-1)+1/(2^(n)In2)(2^(x)-2^(n))`
`=n/(In2)+1/(In2)(2^(x-n)-1)`
`=([x]+2^({x})-1)/(In2)`


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