可证明F(x)在[a, b]连续.而F(a) = -∫{a,b}1/f(t)dt < 0, F(b) = ∫(a,b)f(t)dt > 0.于是F(x)在[a,b]中有零点.对a ≤ x1 < x2 ≤ b, 有F(x2)-F(x1) = ∫(x1,x2)f(t)dt+∫(x1,x2)1/f(t)dt > 0.即F(x)在[a, b]为严格增函数, 故[a,b]中零点唯一.
F(a), F(b)的值是否应换一下
