The value of `:.int_(0)^([x]) (2^(x))/(2^([x]))dx` is

Options

A. log 2
B. `([x])/(log2)`
C. `(1)/(2)([x])/(log 2)`
D. none of these

Answer

B

Explanation

We have,
`:.underset(0)overset([x])int(2^(x))/(2^([x]))dxunderset(0)overset([x])int 2^(x-[x])dx`
Since, `2^(x-[x])` is a periodic function with period one unit.
`:.underset(0)overset([x])int 2^(x-[x])dx=[x]underset(0)overset(1)int 2x^(x-[x])=[x]underset(0)overset(2)int 2^(x)dx`
`rArr underset(0)overset([x])int 2^(x-[x])dx=([x])/(log 2 )[ 2^(x)]_(0)^(1)=([x])/(log 2) (2-1)=([x])/( log 2)`