The value of the integral `rArr int_(1)^(e^(6)) [ (log x)/(3)] dx` , where [*] denotes the greates t integeral function,l is

Options

A. 0
B. `e^(6-e^(3)`
C. `e^(6)+e^(3)`
D. `e^(3)-e^(6)`

Answer

B

Explanation

We have,
`0 lt (log_(e)x)/(3) lt 1"when" 1 lt x lt e^(3)`
and `1 le (log_(e)x)/(3) lt 2"when" e^(3) lt x lt e^(6)`
`:. [(log _(e) x)/(3)] = {{:(0,"when" 1 lt x lt e^(3)),(1,"when" e^(3) lt x lt e^(6)):}`
`rArr underset(1)overset(e^(6))int [(log _(e) x)/(3)] = dxunderset(e)overset(e^(3))int 0 dx +underset(e^(3))overset(e^(6))int 1 dx =e^(6)-e^(3)`