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6                                                                     L. Giugiuc, A.D. Pˆırvuceanu



                                                         Z  x
                                                       2     f(t)dt
                For an arbitrary x ∈ (a, b), let k x :=   a         .
                                                      (x − a)(x − b)
                From Cauchy’s Mean Value Theorem on the intervals [a, x] and [x, b] we get that ∃u x ∈ (a, x)
            and v x ∈ (x, b) such that :

                                                                                 1
                       F(x)(2u x − a − b) = (x − a)(x − b)f(u x ) ⇐⇒ f(u x ) = (2u x − a − b)k x
                                                                                 2
                                                                                 1
                        F(x)(2v x − a − b) = (x − a)(x − b)f(v x ) ⇐⇒ f(v x ) = (2v x − a − b)k x
                                                                                 2

                From Lagrange’s Mean Value Theorem on [u x , v x ] we get that ∃c x ∈ (u x , v x ) ⊂ (a, b) such
            that
                          f(v x ) − f(u x )                       k x  2v x−a−b  −  2u x−a−b
                                                          0
                                             0
                                         = f (c x ) ⇐⇒ f (c x ) =        2         2     = k x ,
                              v x − u x                                  v x − u x
            which is the desired conclusion.
                References

                [1] Gazeta Matematic˘a, No. 11/2019.
                [2] Gazeta Matematic˘a, No. 5/2020.
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