Companion of Ostrowski-Type Inequality based on 5-step quadratic kernel and

Companion of Ostrowski-Type Inequality based on 5-step quadratic kernel and

Applications

Ather QAYYUM

, Muhammad SHOAIB

, Ibrahima FAYE

1

Department of fundamental and applied sciences, Universiti Teknologi Petronas, Bandar Seri Iskandar, Tronoh,

Malaysia. email: atherqayyum@gmail.com

2

Department of Mathematics, University of Hail, PO BOX 2440, Kingdom of Saudi Arabia.

May 9, 2015

Abstract

The purpose of this paper is to establish an improved version of companion of ostrowski’s type integral inequality. The inequalities are obtained by using newly developed special type of quadratic ker- nel having …ve steps. The introduction of this new Kernel is expected to improve approximation errors for various quadrature rules. Ap- plications for composite quadrature rule and Cumulative Distributive Functions are considered.

AMS Numbers: 26D15; 26D20; 26D99

Keywords: Ostrowski inequality, Numerical Integration, Composite quadra- ture rule, cumulative distributive function

1 Introduction

The …eld of inequalities appears in most of the domains of Mathematics. The importance of inequalities has increased during the past few decades and it is

now studied as a separate branch of Mathematics. Up till now, a large num- ber of research papers and books have been written on inequalities and their applications (see for instance [13]-[16]). In many practical problems, it is im- portant to bound one quantity by another quantity. The classical inequalities like Ostrowski are very useful for this purpose. Ostrowski type inequalities have immediate applications in numerical integration, optimization theory, statistics, and integral operator theory.

In 1938, Ostrowski [12] discovered the following useful integral inequality.

Theorem 1 Let f : [a; b]!R be continuous on [a; b] and di¤erentiable on (a; b); whose derivative f⁰ : (a; b)!R is bounded on (a; b); i.e.

kf⁰k₁ = sup

t2[a;b]jf⁰(t)j<1 then

f(x) 1 b a

Zb a

f(t)dt 2 41

4 + x ^a+b₂ b a

!23

5(b a)kf⁰k₁: (1)

We mention another inequality called Grüss inequality [10] which is stated as the integral inequality that establishes a connection between the integral of the product of two functions and the product of the integrals, which is given below.

Theorem 2 Let f; g : [a; b] ! R be integrable functions such that ' f(x) and g(x) ; for some constants '; ; ; and x 2[a; b].

Then

1 b a

Zb a

f(x)g(x)dx 1 b a

Zb a

f(x)dx: 1 b a

Zb a

g(x)dx (2) 1

4( ')( ):

Dragomir and Wang [5] combined Ostrowski and Grüss inequality to give a new inequality which they named Ostrowski-Grüss type inequality.

Theorem 3 Letf: [a; b]!Rsatisfy the Lipschitz condition i.e.,jf(t) f(s)j Mjt sj: Then for all x2 a;^a+b₂ , we have

f(x) +f(a+b x) 2

1 b a

Zb a

f(t)dt 2 41

8 + 2 x ^3a+b₄ b a

!23

5(b a)M:

(3) In (3), the point x= ^3a+b₄ yields the following trapezoid type inequality.

f(^3a+b₄ ) +f(^a+3b₄ ) 2

1 b a

Zb a

f(t)dt b a

8 M: (4)

The constant ¹₈ is sharp

In [3], Barnett et.al proved some Ostrowski and generalized trapezoid inequality. Dragomir [4], Liu [7] established some companions of ostrowski type integral inequalities. Alomari [1] proved the following inequality:

Let f: [a; b] ! R be a di¤erentiable mapping on (a; b). If f ⁰ 2 L¹[a; b]

and f ⁰(t) , for all t 2[a; b];then the inequality f(x) +f(a+b x)

2

1 b a

Zb a

f(t)dt 1

8(b a) ( ): (5) Recently, Liu [8] and Liu et.al [9] also proved some ostrowski type inequalities.

In all references mentioned, authors proved their results by using kernels with two or three steps.

Suppose, we take a 5-step kernel given below:

Lemma 1 Let us de…ne the kernel K(x; t) as:

K(x; t) = 8>

>>

><

>>

>>

:

t a; t 2 a;^a+x₂ ;

t ^3a+b₄ ; t2 ^a+x2 ; x ; t ^a+b₂ ; t2(x; a+b x]; t ^a+3b₄ ; t 2 a+b x;^{a+2b x}₂ ;

t b; t 2 ^{a+2b x}₂ ; b ;

(6)

for all x2 a;^a+b₂ ; then the following identity holds:

1 b a

Zb a

P(x; t)f ⁰(t)dt (7)

= 1

4 f(x) +f(a+b x) +f(a+x

2 ) +f(a+ 2b x

2 ) 1

b a Zb

a

f(t)dt:

By using above identity, we can prove the following theorem.

Theorem 4 Let f : [a; b]!R be di¤erentiable on (a; b). Iff ⁰ 2L¹[a; b]

and f ⁰(t) , for allt 2[a; b]; then the inequality 1

4 f(x) +f(a+b x) +f(a+x

2 ) +f(a+ 2b x

2 ) 1

b a Zb

a

f(t)dt 1

16(b a) ( ): (8)

holds for all x2 a;^a+b₂ :

In this paper we introduce a new 5-step quadratic kernel that further generalize (8) and various previous results [6]-[2] and [1]-[9]. With the help of this special type 5-step quadratic kernel, one can obtain di¤erent type of useful and interesting results. We will describe new results by using Grüss inequality, Cauchy inequality and Diaz-Metcalf inequality. At the end, some obtained inequalities will then be applied for quadrature formula and for cumulative distributive function.

Before we establish our main results, we need to state the following lemma.

Lemma 2 Let f : [a; b]!R be such that f ⁰ is absolutely continuous on [a; b]: Let us de…ne the kernel P(x; t) as:

P(x; t) = 8>

>>

>>

<

>>

>

1

2(t a)²; t2 a;^a+x₂ ;

1

2 t ^3a+b₄ ²; t 2 ^a+x2 ; x ;

1

2 t ^a+b₂ ²; t2(x; a+b x];

1 a+3b 2 a+2b x

(9)

for all x2 a;^a+b₂ ; the following identity holds:

1 b a

Zb a

P(x; t)f ⁰⁰dt (10)

= 1

b a Zb

a

f(t)dt 1

4f f(x) +f(a+b x) +f(a+x

2 ) +f(a+ 2b x

2 )

+ x 5a+ 3b

8 (f ⁰(x) f ⁰(a+b x)) +1

2 x 3a+b

4 f ⁰ a+x

2 f ⁰ a+ 2b x

2 g

2 Main Results

Now we will present our results by imposing three di¤erent conditions on f⁰⁰ and f ⁰⁰⁰.

Case. A: When f ⁰⁰ 2L¹[a; b] :

Theorem 5 Let f : [a; b]!R be di¤erentiable on (a; b),f ⁰ is absolutely continuous on[a; b]and f ⁰⁰(t) ,8t 2[a; b];then for allx2 a;^a+b₂ ; we have

j1

4f f(x) +f(a+b x) +f(a+x

2 ) +f(a+ 2b x

2 ) (11)

+ x 5a+ 3b

8 [f ⁰(x) f ⁰(a+b x)] + 1

2 x 3a+b 4 f ⁰ a+x

2 f ⁰ a+ 2b x

2 g f ⁰(b) f ⁰(a) (b a)²

"

1

24(x a)³ +1

3 x 3a+b 4

3 3

8 x a+b 2

3# 1 b a

Zb a

f(t)dtj (b a) (S );

and j1

4f f(x) +f(a+b x) +f(a+x

2 ) +f(a+ 2b x

2 ) (12)

+ x 5a+ 3b

8 [f ⁰(x) f ⁰(a+b x)] + 1

2 x 3a+b 4 f ⁰ a+x

2 f ⁰ a+ 2b x

2 g f ⁰(b) f ⁰(a) (b a)²

"

1

24(x a)³ +1

3 x 3a+b 4

3 3

8 x a+b 2

3# 1 b a

Zb a

f(t)dtj (b a) ( S);

where

S = f ⁰(b) f ⁰(a) b a and

= 1

96maxf a² 13ab+ 15ax 4b²+ 21bx 18x²

; 14a²+ 5ab 33ax b² 3bx+ 18x² ; a² + 11ab 9ax+ 8b² 27bx+ 18x²

; 10a² 7ab b²+ 27ax+ 9bx 18x² g: proof From (9) and using

1 b a

Zb a

f ⁰⁰(t)dt = f ⁰(b) f ⁰(a)

b a : (13)

and 1 b a

Zb a

P(x; t)dt= 1 b a

"

1

24(x a)³+1

3 x 3a+b 4

3 3

8 x a+b 2

3# (14)

we get 1 b a

Zb a

P(x; t)f⁰⁰(t)dt 1 (b a)²

Zb a

P(x; t)dt Zb

a

f ⁰⁰(t)dt (15)

= 1

b a Zb

a

f(t)dt 1

4f f(x) +f(a+b x) +f(a+x

2 ) +f(a+ 2b x

2 )

+ x 5a+ 3b

8 [f ⁰(x) f⁰(a+b x)] + 1

2 x 3a+b 4 f ⁰ a+x

2 f ⁰ a+ 2b x

2 g

f⁰(b) f⁰(a) (b a)²

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3! :

We suppose that R_n(x) = 1

b a Zb

a

P(x; t)f ⁰⁰(t)dt 1 (b a)²

Zb a

P(x; t)dt Zb

a

f ⁰⁰(t)dt: (16) If C 2R is an arbitrary constant, then we have

R_n(x) = 1 b a

Zb a

(f⁰⁰(t) C) 2

4P(x; t) 1 b a

Zb a

P(x; s)ds 3

5dt: (17) Furthermore, we have

jR_n(x)j 1

b amax

t2[a;b] P(x; t) 1 b a

Zb a

P(x; s)ds Zb a

jf⁰⁰(t) Cjdt (18) Now

max P(x; t) 1 b a

Zb a

P(x; s)ds (19)

= max 8<

:

1 2

x a 2

2 (x)

b a ; ¹₂ x ^3a+b₄ ² _{b a}^(x) ;

1

2 x ^a+b₂ ² _{b a}^(x) ; ¹₈ x ^a+b₂ ² _{b a}^(x) ; _{b a}^(x) 9=

;:

where

(x) = 1

24(x a)³+1

3 x 3a+b 4

3 3

8 x a+b 2

3

: or

1

96maxf a² 13ab+ 15ax 4b² + 21bx 18x² ; (20) 14a²+ 5ab 33ax b² 3bx+ 18x² ;

a²+ 11ab 9ax+ 8b² 27bx+ 18x² ; 10a² 7ab b²+ 27ax+ 9bx 18x² );

13a²+ 13a(b 3x) + 4b² 21bx+ 30x² g= We also have

Zb a

jf ⁰⁰(t) jdt = (S ) (b a): (21) Zb

a

jf ⁰⁰(t) jdt = ( S) (b a): (22) Therefore, we can obtain (11) and (12) by using (13) to (22) taking C = and C = in (18), respectively.

Corollary 1 If we put x=a; in (11) and (12), we get 1

2(f(a) +f(b)) 1

6(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dt

(a) (b a) (S ); (23)

1

2(f(a) +f(b)) 1

6(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dt

(a) (b a) ( S): (24)

Corollary 2 If we put x= ^a+b₂ ; in (11) and (12), we get j1

2f a+b

2 + 1

4 f(3a+b

4 ) +f(a+ 3b 4 ) +b a

32 f ⁰ a+ 3b

4 f ⁰ 3a+b 4 1

96(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dtj a+b

2 (b a) (S ); (25)

j1

2f a+b

2 + 1

4 f(3a+b

4 ) +f(a+ 3b 4 ) + 1

32(b a) f ⁰ a+ 3b

4 f ⁰ 3a+b 4 1

96(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dtj a+b

2 (b a) ( S): (26)

Corollary 3 If we put x= ^3a+b₄ ; in (11) and (12), we get j1

4 f 3a+b

4 +f a+ 3b

4 +f(7a+b

8 ) +f(a+ 7b

8 ) (27) b a

32 f ⁰ a+ 3b

4 f ⁰ 3a+b 4 5

768(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dtj 3a+b

4 (b a) (S );

j1

4 f 3a+b

4 +f a+ 3b

4 +f(7a+b

8 ) +f(a+ 7b

8 ) (28) 1

32(b a) f ⁰ a+ 3b

4 f ⁰ 3a+b 4

5

768(b a) (f ⁰(b) f ⁰(a)) 1

b a Zb a

f(t)dtj 3a+b

4 (b a) ( S):

Case. B: Whenf ⁰⁰2L²[a; b]

Theorem 6 Let f : [a;b]!R be an absolutely continuous function in(a; b) with f ⁰⁰2L²[a;b]. Then, we have

j1

4f f(x) +f(a+b x) +f(a+x

2 ) +f(a+ 2b x

2 ) + x 5a+ 3b

8 (29)

[f ⁰(x) f ⁰(a+b x)] + 1

2 x 3a+b

4 f ⁰ a+x

2 f ⁰ a+ 2b x

2 g

f ⁰(b) f ⁰(a) (b a)²

1

24(x a)³+¹₃ x ^3a+b₄ ³

3

8 x ^a+b₂ ³

! 1

b a Zb a

f(t)dtj p (f ⁰⁰)

b a f 1

320 (x a)⁵+ 0:1 x 3a+b 4

5 33

320 x a+b 2

5

1 (b a)²

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3!2

g¹² for all x2 a;^a+b₂ , where

(f ⁰⁰) =kf ⁰⁰k²2

(f(b) f(a))²

b a =kf ⁰⁰k²2 S²(b a); (30) where S is de…ned in Theorem 4.

proof LetR_n(x)is de…ned as in (16). If we choose C = _{b a}¹ Rb a

f ⁰⁰(s)ds in (17) and use the Cauchy inequality and (37), then we get

jR_n(x)j 1 b a

Zb a

f ⁰⁰(t) 1 b a

Zb a

f ⁰⁰(s)ds P(x; t) 1 b a

Zb a

P(x; s)ds dt

1 b a

2 64 Zb

a

0

@f ⁰⁰(t) 1 b a

Zb a

f⁰⁰(s)ds 1 A

2

dt 3 75

1 2

2 4 Zb

a

P(x; t) 1

b aP(x; s)ds

2

dt 3 5

1 2

p (f⁰⁰) b a

2 64

3 10 ³(x a)⁵ +₁₀¹ x ^3a+b₄ ⁵ 0:1 x ^a+b₂ ⁵

1 (b a)²

h1

24(x a)³+ ¹₃ x ^3a+b₄ ^{3 3}₈ x ^a+b₂ ³i2

3 75

1 2

:

Corollary 4 If we put x=a; in (29), we get 1

2(f(a) +f(b)) 1

6(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dt p (31)

10 ³ (f ⁰⁰) (b a)f3 (b a) 2:7g¹² :

Corollary 5 If we put x= ^a+b₂ ; in (29), we get j1

2f a+b

2 + 1

4 f(3a+b

4 ) +f(a+ 3b

4 ) (32)

+ 1

32(b a) f ⁰ a+ 3b

4 f ⁰ 3a+b 4 1

96(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dtj p10 ⁴ (f ⁰⁰) (b a)f1:9 (b a) 1g¹²

Corollary 6 If we put x= ^3a+b₄ ; in (29), we get j1

4 f 3a+b

4 +f a+ 3b

4 +f(7a+b

8 ) +f(a+ 7b

8 ) (33) 1

32(b a) f ⁰ a+ 3b

4 f ⁰ 3a+b 4 5

768(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dtj p10 ⁴ (f ⁰⁰) (b a)f(b a) 0:4g¹² :

Case. C: Whenf ⁰⁰⁰ 2L²[a; b]

Theorem 7 Let f : [a; b]!Rthree times di¤erentiable function in (a; b): If f ⁰⁰⁰ 2L²[a; b] then for all x2 a;^a+b₂ ; we have

j1

4f f(x) +f(a+b x) +f(a+x

2 ) +f(a+ 2b x

2 ) (34)

+ x 5a+ 3b

8 [f ⁰(x) f ⁰(a+b x)] + 1

2 x 3a+b 4 f ⁰ a+x

2 f ⁰ a+ 2b x

2 g f ⁰(b) f ⁰(a) (b a)²

"

1 ₃ 1 3a+b ³ 3 a+b ³#

1 Zb

b a

kf ⁰⁰⁰k2f 1

320(x a)⁵+ 1

10 x 3a+b 4

5 33

320 x a+b 2

5

1 (b a)²

1

24(x a)³+1

3 x 3a+b 4

3 3

8 x a+b 2

3!2

g¹²

proof LetR_n(x)be de…ned by (16), from (15) R_n(x) = _{b a}¹

Rb a

f(t)dt ¹₄f f(x) +f(a+b x) +f(^a+x₂ ) +f(^{a+2b x}₂ ) + x ^5a+3b₈ [f ⁰(x) f ⁰(a+b x)] + ¹₂ x ^3a+b₄ f ^{0 a+x}₂ f ^{0 a+2b x}₂ g

(35)

f ⁰(b) f ⁰(a) (b a)² [ 1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3

]:

If we choose C = f ^{00 a+b}₂ ; (17) and use the Cauchy Inequality, then we get

jR_n(x)j (36)

1 b a

Zb a

f ⁰⁰(t) f ⁰⁰ a+b

2 P(x; t) 1 b a

Zb a

P(x; s)ds dt

1 b a

2 4 Zb

a

f ⁰⁰(t) f⁰⁰ a+b 2

2

dt 3 5

1 2

2 64 Zb

a

0

@P(x; t) 1 b a

Zb a

P(x; s)ds 1 A

2

dt 3 75

1 2

:

We can use the Diaz-Metcalf inequality [11] or [17], to get Zb

a

f⁰⁰(t) f ⁰⁰ a+b 2

2

dt (b a)²

2 kf ⁰⁰⁰k²2:

Zb a

0

@P(x; t) 1 b a

Zb a

P(x; s)ds 1 A

2

dt (37)

= Zb a

P(x; t)²dt 1 (b a)²

"

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3#2

= 3 10 ³(x a)⁵ 0:1 x a+b 2

5

+ 1

10 x 3a+b 4

5

1 (b a)²

"

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3#2

:

Therefore, using the above relations, we obtain (34).

Corollary 7 If we put x=a; in (34), we get 1

2(f(a) +f(b)) 1

6(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dt (38) (b a)³

kf ⁰⁰⁰k2

p10 ³f3 (b a) 2:7g¹²

Corollary 8 If we put x= ^a+b₂ ; in (34), we get j1

2f a+b

2 + 1

4 f(3a+b

4 ) +f(a+ 3b

4 ) (39)

+ 1

32(b a) f ⁰ a+ 3b

4 f ⁰ 3a+b 4 1

96(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dtj (b a)³

kf ⁰⁰⁰k2

p10 ³ 3 (b a) 10 ¹

1 2

Corollary 9 If we put x= ^3a+b₄ ; in (34), we get j1

4 f 3a+b

4 +f a+ 3b

4 +f(7a+b

8 ) +f(a+ 7b

8 ) (40) b a

32 f ⁰ a+ 3b

4 f ⁰ 3a+b 4 5

768(b a) (f ⁰(b) f ⁰(a)) 1 b a

Zb a

f(t)dtj (b a)³

kf ⁰⁰⁰k2

p10 ⁴f(b a) 0:4g¹²

3 An application to Composite Quadrature Rules

Let I_n : a = x₀ < x₁ < x₂ < :::: < x_{n 1} < x_n = b be a division of the interval [a; b]; _i 2 [x_i; x_i+1] (i= 0;1; ::::; n 1) ; a sequence of intermediate points h_i =x_i+1 x_i (i= 0;1; :::::n 1). We have the following quadrature formula:

Consider the perturbed composite quadrature rules

A((f; ; I_n)) =

n 1

X

i=0

h_i 2 66 66 4

1 4

f(^3xi+x₄ ⁱ⁺¹) +f(^xi+3x₄ⁱ⁺¹) +f(^7xi+x₈ ⁱ⁺¹) +f(^xi+7x₈ ⁱ⁺¹)

hi

32 f ⁰(^3xi+x₄ ⁱ⁺¹) f ⁰(^xi+3x₄ ⁱ⁺¹)

5hi

768(f ⁰(x_i+1) f ⁰(x_i))

3 77 77

5 (41)

Theorem 8 Let f : [a;b] ! R be such that f ⁰ is absolutely continuous function. If f ⁰⁰ 2 L¹[a; b] and f ⁰⁰(t) for all x 2 a;^a+b₂ ; then we have the following quadrature formula:

Zb a

f(t)dt=A_n((f; ; I_n)) +R¹_n(f; ; I_n); (42) Zb

a

f(t)dt=A_n((f; ; I_n)) +R²_n(f; ; I_n); (43)

where Q_n((f; ; I_n)) is de…ned above and remainder satis…es the estimation R¹_n(f; ; I_n) (S )

n 1

X

i=0

h²_i 3x_i+x_i+1

4 ; (44)

and

R²_n(f; ; I_n) ( S)

n 1

X

i=0

h²_i 3x_i+x_i+1

4 ; (45)

for all _i 2[x_i; x_i+1]; where h_i :=x_i+1 x_i; (i= 1;2; ::::n 1):

proof Apply (27) and (28)on the interval[x_i; x_i+1], _i 2[x_i; x_i+1]; where h_i :=x_i+1 x_i (i= 1;2; ::::n 1); to get

R(f; ; I_n) =

xZi+1

xi

f(t)dt

n 1

X

i=0

h_i 2 66 4

1 4

f(^3xi+x₄ⁱ⁺¹) +f(^xi+3x₄ ⁱ⁺¹) +f(^7xi+x₈ⁱ⁺¹) +f(^xi+7x₈ⁱ⁺¹)

hi

32 f ⁰(^3xi+x₈ⁱ⁺¹) f ⁰(^xi+3x₈ⁱ⁺¹)

5hi

768(f ⁰(x_i+1) f⁰(x_i))

3 77 5

(S )

n 1

X

i=0

h²_i 3x_i+x_i+1

4 :

and

R(f; ; I_n) =

xZi+1

xi

f(t)dt

1 4

n 1

X

i=0

h_i 2 66 4

1 4

f(^3xi+x₄ⁱ⁺¹) +f(^xi+3x₄ ⁱ⁺¹) +f(^7xi+x₈ⁱ⁺¹) +f(^xi+7x₈ ⁱ⁺¹)

hi

32 f ⁰(^3xi+x₈ⁱ⁺¹) f⁰(^xi+3x₈ⁱ⁺¹)

5hi

768(f ⁰(x_i+1) f ⁰(x_i))

3 77 5

( S)

n 1

X

i=0

h²_i 3x_i+x_i+1

4 :

for (i= 1;2; ::::n 1)

Theorem 9 Letf : [a;b]!R be an absolutely continuous mapping on [a; b]

with f ⁰⁰2L²[a; b]. Then we have for all x2 a;^a+b₂ Zb

a

f(t)dt=A_n(f; I_n) +R_n(f; I_n); (46)

and remainder satis…es the estimation jR(f; I_n)j p

10 ⁴ (f ⁰⁰)

n 1

X

i=0

h²_i fh_i 0:4g¹² (47) proof Applying (33) to the interval [x_i; x_i+1], then we get

j

xi+1

Z

xi

f(t)dt

1

4h_if f ^3xi+x₄ⁱ⁺¹ +f ^xi+3x₄ ⁱ⁺¹ +f(^7xi+x₈ⁱ⁺¹) +f(^xi+7x₈ ⁱ⁺¹)

hi

32 f ^{0 xi+7x}₈ⁱ⁺¹ f ^{0 7xi+x}₈ⁱ⁺¹ g 768⁵ (h_i) (f ⁰(x_i+1) f ⁰(x_i))j p10 ⁴ (f ⁰⁰)h²_i h_i 4 10 ¹

1 2

for i= 0;1; ::::; n 1:

Now summing overifrom0ton 1;using the triangle inequality, we get (47).

Theorem 10 Let f : [a;b] !R be a three times continuously di¤erentiable function in (a; b) with f ⁰⁰⁰ 2L²[a; b]. Then we have

Zb a

f(x)dx=A(f; I_n) +R(f; I_n); (48)

and the remainder satis…es the estimation jR(f; In)j 10 ⁴kf ⁰⁰⁰k2

n 1

X

i=0

h⁶_i f4hi 1:1g: (49)

proof Applying (40) to the interval [x_i; x_i+1], then we get

j

xi+1

Z

xi

f(t)dt h_i 4

f(x_i) +f ^3xi+x₄ⁱ⁺¹ +f ^xi+3x₄ⁱ⁺¹ +f ^7xi+x₈ ⁱ⁺¹ +f ^xi+7x₈ ⁱ⁺¹ + 1

32h_i f⁰ x_i+ 3x_i+1

4 f ⁰ 3x_i+x_i+1 4 + 5

768h_iff ⁰(x_i+1) f ⁰(x_i)g j 10 ⁴kf⁰⁰⁰k2h⁶_i f4h_i 1:1g; for i= 0;1; ::::; n 1:

Now summing overifrom0ton 1;using the triangle inequality, we get (49).

4 An Application to Cumulative Distribution Function

LetX be a random variable taking values in the …nite interval[a; b]with the probability density function f : [a; b] ! [0;1] and cumulative distributive function

F (x) = Pr (X x) = Z x

a

f(t)dt: (50)

F (b) = Pr (X b) = Zb

a

f(u)du= 1: (51)

Theorem 11 With the assumptions of Theorem 2.1, we have the following

inequality which holds jb E(X)

b a

1

4f F (x) +F (a+b x) +F(a+x

2 ) +F(a+ 2b x

2 )

+ x 5a+ 3b

8 [f(x) f (a+b x)] (52)

+1

2 x 3a+b

4 f a+x

2 f a+ 2b x

2 g

f (b) f (a) (b a)²

"

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3# j (b a) (S );

jb E(X) b a

1

4f F (x) +F (a+b x) +F(a+x

2 ) +F(a+ 2b x

2 )

+ x 5a+ 3b

8 [f(x) f (a+b x)] (53)

+1

2 x 3a+b

4 f a+x

2 f a+ 2b x

2 g

f (b) f (a) (b a)²

"

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3# j (b a) ( S);

for all x2 a;^a+b₂ . Where E(X) is the expectation of X.

proof In the proof of Theorem 2.1, letf =F and using the fact that E(X) =

Z b a

tdF(t) =b Z b

a

F(t)dt: (54)

Further details are left to the interested readers.

Corollary 10 Under the assumptions of Theorem 4.1, if we put x= ^3a+b₄ in

(52) and (53) then we get jb E(X)

b a

1

4f F(3a+b

4 ) +F(a+ 3b

4 ) +F(7a+b

8 ) +F(a+ 7b 8 ) b a

8 f a+ 3b

4 f 3a+b

4 g (55)

6:5 10 ³(b a) (f(b) f (a))j (b a) (S );

jb E(X) b a

1

4f F(3a+b

4 ) +F(a+ 3b

4 ) +F(7a+b

8 ) +F(a+ 7b 8 ) b a

8 f a+ 3b

4 f 3a+b

4 g (56)

6:5 10 ³(b a) (f(b) f (a))j (b a) ( ):

Theorem 12 With the assumptions of Theorem 2.9, we have the following inequality which holds

jb E(X) b a

1

4f F (x) +F (a+b x) +F(a+x

2 ) +F(a+ 2b x

2 )

+ x 5a+ 3b

8 (f (a+b x) f(x)) (57)

+1

2 x 3a+b

4 f a+ 2b x

2 f a+x

2 g

f (b) f (a) (b a)²

"

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3# j b a

kf ⁰⁰k2f 1

320(x a)⁵+ 1

10 x 3a+b 4

5 33

320 x a+b 2

5

1 (b a)²

"

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3#2

g for all x2 a;^a+b₂ , where E(X) is the expectation of X.

Corollary 11 Under the assumptions of Theorem 4.3, if we put x= ^3a+b₄ in (57), then we get

jb E(X) b a

1

4f F(^3a+b₄ ) +F(^a+3b₄ )

+F(^7a+b₈ ) +F(^a+7b₈ ) (58) b a

8 f a+ 3b

4 f 3a+b

4 g

5

768 (b a) (f(b) f (a))j 10 ⁴(b a)³

kf ⁰⁰k2 (b a) 4:24 10 ¹

1 2

Theorem 13 With the assumptions of Theorem 4.3, we have the following inequality which holds

jb E(X) b a

1

4f F (x) +F(a+b x) +F(a+x

2 ) +F(a+ 2b x

2 )

+ x 5a+ 3b

8 [f(x) f (a+b x)] (59)

+1

2 x 3a+b

4 f a+x

2 f a+ 2b x

2 g

f (b) f (a) (b a)²

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3! j p (f ⁰⁰)

b a f3:1 10 ³(x a)⁵+ 1

10 x 3a+b 4

5

0:1 x a+b 2

5

1 (b a)²

"

1

24(x a)³+ 1

3 x 3a+b 4

3 3

8 x a+b 2

3#2

g¹² proof Applying (29) and using the same conditions that we used in The- orem 4.1, we get the required inequality (59).

Corollary 12 Under the assumptions of Theorem 4.5, if we put x= ^3a+b₄ in (59), then we get

jb E(X) b a

1

4f F(3a+b

4 ) +F(a+ 3b

4 ) +F(7a+b

8 ) +F(a+ 7b 8 ) 1

8(b a) f a+ 3b

4 f 3a+b

4 g (60)

1:3 10 ²(b a) (f(b) f (a))j

p10 ⁴ (f ⁰⁰) (b a) (b a) 4:24 10 ¹

1 2 :

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