`int_0^x(2^t)/(2^([t]))dt ,w h e r e[dot]` denotes the greatest integer function and `x in R^+` , is equal to

Options

A. `1/(1n2)([x]+2^({x})-1)`
B. `1/(1n2)([x]+2^({x}))`
C. `1/(1n2)([x]-2^({x}))`
D. `1/(1n2)([x]+2^({x})+1)`

Answer

A

Explanation

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)`