If `[dot]a n d\ {dot}` denote respectively the greatest integer and fractional part functional respectively, evaluate the following integral: `int_0^1 2^(x-[x])dx`

Explanation

`I=\int_{0}^{1} 2^{x-|x|} d x`


`0 \leq x<1`


`I=\int_{0}^{1} 2 x d x`


`I=\frac{(2^{x})}{\ln 2} _{0}^{1}=\frac{1}{\ln 2}(2^{1}-2^{\circ})`


`I=\frac{1}{\ln 2}(2-1)`


`I=\frac{1}{\ln 2}`