Because f(x) is continuous on [0, 1], F(x) is also continuous on [0, 1].

Because f (t) >; 0, so

F(0)=-∫(0， 1) 1/f(t)dt & lt； 0

F( 1)=∫(0， 1)f(t)dt & gt； 0

Therefore, according to the intermediate value theorem of continuous functions, there exists a ∈(0, 1), which makes F(a)=0.

Because f' (x) = F(x)+1f (x) > 0, f (x) strictly monotonically increases on [0, 1].

So x=a is the only zero point of the equation F(x)

That is, there is a unique a∈(0, 1) so that ∫ (0, a) f (t) dt = ∫ (a,1)/f (t) dt.

(2) Let fn (x) = ∫ (1/n, x) f (t) dt-∫ (x, 1) 1/f (t) dt for any natural number n.

Like the problem (1), it can be proved that there is a unique xn∈( 1/n, 1)? (0, 1), so ∫ (1n, xn) f (t) dt = ∫ (xn, 1) 1/f (t) dt.

Suppose there are different natural numbers n 1 and n2, so that the zeros of Fn 1(x) and Fn2(x) are both xn', then

∫( 1/n 1，xn')f(t)dt=∫(xn '， 1) 1/f(t)dt =∫( 1/N2，xn')f(t)dt

∫( 1/n 1， 1/n2)f(t)dt=0

Because f (t) >; 0, so ∫ (1/n 1, 1/N2) f (t) dt > 0, which contradicts the above conclusion.

Therefore, the zero point xn of Fn(x) changes with the change of n, combining the existence and uniqueness conclusion of xn and the conclusion of the problem (1)

It is known that finding the mapping of function zero g: {fn (x), f (x)}->; {xn, a} is a one-to-one mapping.

Because lim (n->; ∞)Fn(x)=F(x)

So lim (n->; ∞)xn

= lim(n->； ∞)g[Fn(x)]

=g[F(x)]

=a