∫ [0,1] {∫ [x^2,x] kdy} dx
= k∫ [0,1] {(1/2)x^2|[上限x,下限x^2]} dx
= ∫ [0,1] (x-x^2)dx
= k (1/2 – 1/3) = k/6 = 1 -- 》k=6

f(x) = ∫[x^2,x] 6 dy = 6(x-x^2), 0
P(X>(1/2)) = 1- P(X<(1/2)) = 1-∫ [0,(1/2)]6(x-x^2) = 1/2