】令ϕ (x) = x2 f (x),由积分中值定理得2(1) 2 2 ( ) 2 2 ( ) 110f = ∫ x2 f x dx = ⋅ c f c ⋅ ,其中]2c∈[0,1 ,即12 f (1) = c2 f (c),于是有ϕ (c) =ϕ (1),由罗尔定理,存在ξ ∈(c,1) ⊂ (0,1),使得ϕ ′(ξ ) = 0.而ϕ ′(x) = x2 f ′(x) + 2xf (x),所以ξ 2 f ′(ξ ) + 2ξf (ξ ) = 0,注意到ξ ≠ 0,所以有ξf ′(ξ ) + 2 f (ξ ) = 0.