第16题怎么做
f(x)
=ax ; 0<x<2
=b-(1/4)x ; 2≤x<4
=0 ; elsewhere
case 1: 0<x<2
F(x) =∫(0->x) at dt = (1/2)ax^2
case 2: 2≤x<4
F(x)
= (1/2)a(2)^2 +∫(0->x) [b-(1/4)t ] dt
=2a + bx -(1/8)x^2
lim(x->4-) F(x) = 1
lim(x->4-) [2a + bx -(1/8)x^2 ] = 1
2a +4b - 2 =1
2a +4b =3 (1)
lim(x->2-) F(x) = lim(x->2+) F(x)
lim(x->2-) (1/2)ax^2 = lim(x->2+) [2a + bx -(1/8)x^2]
2a = 2a+2b - 1/2
b= 1/4
from (1)
2a +4b =3
2a +1 =3
a=1
ie
(a,b) = (1, 1/4)
分布函数
F(x)
=0 ; x≤0
=(1/2)x^2 ; 0<x<2
=2 + (1/4)x -(1/8)x^2 ; 2≤x<4
=1 ; x≥4