`rArrint_(0)^(oo) [(2)/(e^(x))]dx` (where [*] denotes the greatest integer function) equals

Options

A. `log_(e)2`
B. ` e^(2)`
C. 0
D. `2//e`

Answer

A

Explanation

We have,
`e^(x) ge 1 "for all" x ge-0`
`rArr 0 le (1)/(e^(x)) le1 "for all" x ge0`
`rArr 0 le (1)/(e^(x)) le 2 "for all" x ge0`
`rArr 0 le (1)/(e^(x)) le 1 "for" x gt log_(e)2`
`rArr 1 le (1)/(e^(x)) le 2 "for" 0x le log_(e)2`
`rArr underset(0)overset(oo)int[(2)/(e^(x))]dx=underset(0)overset(log_(e)2)int1 dx+underset(log_(e)2)overset(oo)int 0.dx = log_(e)2`