Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 8 (1997), 79{83
FURTHER GENERALIZATIONS OF INEQUALITIES FOR AN INTEGRAL QI Feng Using the Taylor's formula we prove two integral inequalities, that generalize K. S. K. Iyengar's inequality and some other integral inequalities.
Let f(x) be a dierentiable function satisfying jf 0 (x)j N for a x b. The value of the integral of f(x) over [a; b] can be estimated in a variety of ways. For example, the useful inequality of Iyengar in [1{6] states that (1)
b R f(x) dx a
(b a)(f(a) + f(b)) (b a)2N 2 4
(f(b) f(a))2 ; 4N
which reduces to [2,3,4] (2)
b R f(x) dx a
2 (b 4a) N
when f(a) = f(b) = 0: If the bounds for the derivative of f(x) are expressed in the form m f 0 (x) M, then other useful inequalities are [1, 3] (3)
(b a)2 m Rb f(x) dx (b a)f(a) (b a)2 M ; 2 2 a
(4)
mM(b a)2 + 2(b a)(Mf(a) mf(b)) + (f(a) f(b))2 Rb f(x) dx 2(M m) a 2 Mf(b)) + (f(a) f(b))2 : mM(b a) + 2(b a)(mf(a) 2(M m)
0
1991 Mathematics Subject Classi cation: Primary 26D15; Secondary 26D10
79
80
QI Feng When f(a) = f(b) = 0, the inequality (4) reduces to b R f(x) dx
(5)
a
a)2 : mM(b 2(M m)
If we set M = m = N, then (1) follows from (4); if we set f(a) = f(b) = 0 and M = m = N, then (2) follows from (5). Suppose that f(x) is 2n-times dierentiable for x 2 [a; b], and f (2n) (x) N, f (r) (a) = f (r) (b) = 0; r = 0; 1; : : :; n 1. Then we have [1, 2] b R f(x) dx
(6)
a
(n!) N (b a)2n+1: (2n)!(2n + 1)! 2
We remark that the inequalities from (1) to (5) are not included in (6). There is a related integral inequality [2]: Let f(x) be dierentiable of class C 2; jf 00 (x)j N; x 2 [a; b]: Then we have (7)
1
f(a) + f(b) + 1 + Q2 (b a) f 0 (b) f 0 (a) f(x) dx b aa 2 3 Rb
N(b24 a) (1 3Q2); 2
f(a) f(b) + f 0 (a) + f 0 (b) 2 where Q = 2 : N (b a)2 f 0 (b) f 0 (a) 2 In this article, we generalize these results, giving a short proof which is based on the Taylor's formula. Theorem. Let f(x) be a dierentiable function of C n [a; b] satisfying m f (n) (x) M . If we denote 2
(8)
Sn (u; v; w) =
n X1
( 1)k uk f (k 1) (v) + w un; k! n! k=1
@ k Sn = S (k) (u; v; w); n @uk then, when n is even, we have for any t 2 [a; b]
(9)
(10)
n +1 X
( 1)i S (i) (a; a; m) S (i) (b; b; m)ti Rb f(x) dx n+1 n+1 i! a i=0
n +1 X
( 1)i S (i) (a; a; M) S (i) (b; b; M)ti ; n+1 n+1 i! i=0
Further generalizations of inequalities for an integral
81
whereas when n is odd, we have n +1 X
( 1)i S (i) (a; a; m) S (i) (b; b; M)ti Rb f(x) dx n+1 n+1 i! a i=0
(11)
n +1 X
( 1)i S (i) (a; a; M) S (i) (b; b; m)ti ; n+1 n+1 i! i=0
t 2 [a; b]:
Proof. Let t be a parameter satisfying a < t < b, and write Rb
(12)
a
Rt
Rb
a
t
f(x) dx = f(x)dx + f(x) dx:
The Taylor's formula states that (13)
f(x) =
(14)
f(x) =
n X1
f (i) (a) (x a)i + f (n) () (x a)n; 2 (a; x); i! n! i=0 n X1
f (i) (b) (x b)i + f (n) () (x b)n; 2 (x; b): i! n! i=0
Integrating on both sides of (13) over [a; t], we obtain t
Z (n) (i) f () (x a)n dx: i+1 + f(x) dx = (if +(a) (t a) 1)! n! a i=0
Rt
(15)
n X1
a
Since m f (n) (x) M, then m (t a)n+1 f(x) dx (16) (n + 1)! a Rt
n X1
f (i) (a) (t a)i+1 M (t a)n+1 : (i + 1)! (n + 1)! i=0
Integrating on both sides of (14) over [t; b], we get (17)
Rb t
b
f (i) (b) (t b)i+1 + Z f (n) () (x b)n dx: f(x) dx = (i + 1)! n! i=0 n X1
t
When n is even, from (17), it follows that n X1 (i) Rb M (t b)n+1 : i+1 (18) (n +m 1)! (t b)n+1 f(x) dx+ (if +(b) (t b) 1)! (n + 1)! t i=0
When n is odd, the reversed inequalities of (18) hold.
82
QI Feng From (12), (16) and (18), when n is even, we have
(19)
n X1 i=0
f (i) (a) (t a)i+1 (i + 1)! Rb
f(x) dx a
n X1
f (i) (b) (t b)i+1 + m (t a)n+1 (t b)n+1 (i + 1)! (n + 1)! i=0 n X1
f (i) (a) (t a)i+1 (i + 1)! i=0
n+1 (t b)n+1; (t a) + (n M + 1)!
n X1
f (i) (b) (t b)i+1 (i + 1)! i=0
when n is odd, we obtain (20)
n X1
f (i) (a) (t a)i+1 (i + 1)! i=0
n X1
f (i) (b) (t b)i+1 (i + 1)! i=0
m (t a)n+1 M n+1 + (n + 1)! (n + 1)! (t b) n n X1 f (i) (a) X1 f (i) (b) Rb i +1 (t b)i+1 f(x) dx (i + 1)! (t a) (i + 1)! a i=0 i=0 M m n +1 n +1 + (n + 1)! (t a) (n + 1)! (t b) : Considering (8) and (9), rewriting (19) and (20), the desired inequalities (10) and (11) follow. Corollary. Let f(x) be a two times dierentiable function satisfying m f 00 (x) M . Then m(a3 b3 ) f(a) f(b) + bf 0 (b) af 0 (a) + m(b2 a2 )=2 ((21) 6 2 (a b)m + f 0 (a) f 0 (b) 0
0
f(x) dx bf(b) + af(a) + b f (b) 2 a f (a) Rb
2
2
a
3 3 + bf 0 (b) af 0 (a) + M(b2 a2 )=2 : M(a 6 b ) f(a) f(b) 2 (a b)M + f 0 (a) f 0 (b)
Proof. Taking n = 2 in inequality (10) and
0 (a) bf 0 (b) + m(a2 b2 )=2 t = f(a) f(b)(a+ afb)m ; + f 0 (a) f 0 (b) the left-hand side inequality of (21) follows. By the symmetry of inequality (10), the right-hand side inequality of (21) follows easily.
Further generalizations of inequalities for an integral Remark. If we set n = 1; t = bM
83
am + f(a) f(b) =(M m), and t = bm aM + f(a) f(b) =(m M), then (4) follows from (11), so (4) is included in (10). If we set M = m = N, then (7) follows from (21), therefore (7) is included in (10). Since the inequalities (1), (2), (4), (5) and (7) are all included in (10) and (11), inequalities (10) and (11) are general. Moreover, an uni ed proof and an uni ed representative of these inequalities are given by the theorem in this article. REFERENCES
1. D. S. Mitrinovic: Analytic Inequalities. Springer-Verlag, 1970. 2. Kuang Jichang: Applied Inequalities (in Chinese), 2nd edition. Hunan Education Press, Changsha, China, 1993. 3. QI Feng: Inequalities for an integral. Math. Gaz. 80, No 488 (1996), 376{377. 4. Xu Lizhi, Wang Xinghua: Methods of Mathematical Analysis and Selected Examples (in Chinese). Revised Edition. Higher Education Press, Beijing, China, 1984. 5. D. S. Mitrinovic, A. M. Fink: Inequalities Involving Functions and Their Integrals and Derivatives, Chapter XV. Kluwer Academic Publishers, 1991. 6. G. V. Milovanovic, J. E. Pecaric: Some considerations on Iyengar's inequality and some related applications. Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. Fiz. No 544{576 (1976), 166{170. Department of Mathematics, University of Science and Technology of China, Hefei 230026, Anhui, People's Republic of China Current E-mail address to December of 1998: qi
[email protected]
(Received January 7, 1997)