Let f (x) be differentiable on [0,1], and f (1) = 2 ∫ 0 ~ 1 / 2 XF (x) DX. It is proved that there exists ξ belonging to (0,1), such that f (ξ) + ξ f '(ξ) = 1

Let f (x) be differentiable on [0,1], and f (1) = 2 ∫ 0 ~ 1 / 2 XF (x) DX. It is proved that there exists ξ belonging to (0,1), such that f (ξ) + ξ f '(ξ) = 1

It is proved that: according to the integral mean value theorem, there exists η∈ (0,1 / 2) such that 2 ∫ [0 → 1 / 2] XF (x) DX = 2 * η f (η) * (1 / 2) = η f (η) = f (1) such that G (x) = XF (x), then G (η) = η f (η) = f (1), G (1) = f (1). Therefore, G (x) satisfies the Rolle mean value theorem condition in [η, 1], that is, there exists ξ∈ (η, 1), such that G '(ξ) = 0, and G (1) satisfies the Rolle mean value theorem