.3.55 walpole

Given the joint density function

f (x,y) = 6-x-y/ 8 , 0<x<2, 2<y<4

0, elsewhere,

find P (1<Y<3|X=1).

i have got g(x)=(3-x)/4 and also f(y|x) = (6-x-y) / (2(3-x))

i substitute x=1 into (6-x-y) / (2(3-x)) and i got (5-y) / 2

so i integrate (5-y) to dy from 1 to 3. Should be 2 to 3

but it is not right.

did i miss something ?