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N! ) b + f (x)g(x)h(n) (σ ) a (x − y)n dy. ) b a b a n−1 g(x)h(x) I0 + ∑ Ik k=1 n−1 + f (x)g(x) L0 + ∑ Lk dx k=1 g(x)h(x) f (n) (ξ ) + h(x) f (x)g(n) (η ) + f (x)g(x)h(n) (σ ) b a (x − y)n dy dx. 60). The proof is complete. 3. 45 Taking h(x) = 1 and hence h(k) (x) = 0, k = 1, . . 60), reduces to 1 b−a b a f (x)g(x)dx − 1 2(b − a)2 b a b 1 2(b − a)2 a n−1 n−1 g(x) I0 + ∑ Ik + f (x) J0 + ∑ Jk k=1 |g(x)| f (n) ∞ dx k=1 + | f (x)| g(n) Mn (x)dx. 60). 5. 6 ˇ ¨ Discrete Inequalities of the Gruss-and Cebyˇ sev-type ˇ A number of Gr¨uss-and Cebyˇ sev-type discrete inequalities have been investigated by different researchers, see [79,144], where further references are also given.

N−1 (k) g (y) g(x) = g(y) + ∑ k=1 k! n−1 (k) h (y) h(x) = h(y) + ∑ k=1 k! 62) (x − y)k + 1 (n) g (η )(x − y)n , n! 63) (x − y)k + 1 (n) h (σ )(x − y)n , n! 64) where ξ = y + α (x − y) (0 < α < 1), η = y + β (x − y) (0 < β < 1), σ = y + γ (x − y) (0 < γ < 1). From the definitions of Ik , Jk , Lk , I0 , J0 , L0 and integration by parts (see, ), we have the relations n−1 n−1 k=1 k=1 n−1 n−1 k=1 k=1 n−1 n−1 k=1 k=1 I0 + ∑ Ik = nI0 − (b − a) ∑ Fk (x), J0 + ∑ Jk = nJ0 − (b − a) ∑ Gk (x), L0 + ∑ Lk = nL0 − (b − a) ∑ Hk (x).

1) A simple integration by parts, gives f (a) + f (b) (b − a) − 2 b b x− f (x)dx = a a a+b 2 f (x)dx. 2, we get b 1 b−a 1 b−a a b a x− a+b 2 x− a+b 1 − 2 b−a f (x)dx− b a b 1 b−a a x− a+b dx 2 y− a+b dy × 2 x− a+b dx = 0, 2 b 1 b−a f (x) − 1 b−a f (x)dx a b f (y)dy dx. a As b a we get b a x− a+b 2 a+b f (b) − f (a) dx f (x) − 2 b−a a a+b f (b) − f (a) b dx x− max f (x) − b − a 2 x∈(a,b) a b f (x)dx = x− (b − a)2 f (b) − f (a) . 1). 4) for all a, b ∈ I and x ∈ (a, b). , | f (u) − f (v)| M|u − v|, M > 0, then | f (b) − f (a) − (b − a) f (x)| = | f (c) − f (x)||b − a| M|c − x||b − a| M(b − a)2 , where c ∈ (a, b).

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