Let f (x) be second-order differentiable on [a, b], and f ″ (x) ＜ 0. It is proved that: ∫ BAF (x) DX ≤ (B-A) f (a + B2)

Let f (x) be second-order differentiable on [a, b], and f ″ (x) ＜ 0. It is proved that: ∫ BAF (x) DX ≤ (B-A) f (a + B2)

Let f (x) expand f (x) at t, and we obtain f (x) = f (x) = f (x) = f (x) = f (x) = f (T) (T) (f (x) = f (T) (T) (f (x) \8704; x, t ∈ [a, b] Let f (x) be expanded at t t, and we obtain f (x) = f (x) = f (t (T) + F (T) (f (x) = f (T) + (t (T) (t) in t (T) let let f (x) expand f (x) at t (T) in t (t (x) t (x) let let t = a + B 2, let let let let let let let t = a + B 2 = a + B 2 = a + B 2, let let let let let let let let let let let let let let let let let let let let let let let let let let let let let let f (x (x (x (x (x (t = a + B 2, let let let let let let BAF ′ (a + B2) (x − a + B2) DX = (b −a)f(a+b2)+f′(a+b2)[12(x−a+b2)2]|ba =(b−a)f(a+b2)．