Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 2, Article 29, 2002
A NOTE ON THE PERTURBED TRAPEZOID INEQUALITY
XIAO-LIANG CHENG AND JIE SUN
D EPARTMENT OF M ATHEMATICS
Z HEJIANG U NIVERSITY,
X IXI C AMPUS , Z HEJIANG 310028,
T HE P EOPLE ’ S R EPUBLIC OF C HINA .
xlcheng@mail.hz.zj.cn
Received 21 May, 2001; accepted 01 February, 2002.
Communicated by N.S. Barnett
A BSTRACT. In this paper, we utilize a variant of the Grüss inequality to obtain some new perturbed trapezoid inequalities. We improve the error bound of the trapezoid rule in numerical
integration in some recent known results. Also we give a new Iyengar’s type inequality involving a second order bounded derivative for the perturbed trapezoid inequality.
Key words and phrases: Grüss inequality, Perturbed trapezoid inequality, Sharp bounds.
2000 Mathematics Subject Classification. 26D15, 26D10.
In the literature [2], [4] – [8], [11], [12] on numerical integration, the following estimation is
well known as the trapezoid inequality:
Z b
1
≤ 1 M2 (b − a)3 ,
(1.1)
f
(x)
dx
−
(b
−
a)(f
(a)
+
f
(b))
12
2
a
where the mapping f : [a, b] → R is supposed to be twice differentiable on the interval (a, b),
with the second derivative bounded on (a, b) by M2 = supx∈(a,b) |f 00 (x)| < +∞. In [5], the
authors derived the error bounds for the trapezoid inequality (1.1) by different norm of mapping
f . In [2, 7, 11], the authors obtained the trapezoid inequality by the difference of sup and inf
bound of the first derivative, that is,
Z b
1
1
f (x) dx − (b − a)(f (a) + f (b)) ≤ (Γ1 − γ1 )(b − a)2 ,
2
8
a
where Γ1 = supx∈(a,b) f 0 (x) < +∞ and γ1 = inf x∈(a,b) f 0 (x) > −∞.
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
046-01
2
X IAO - LIANG C HENG AND J IE S UN
For the perturbed trapezoid inequality, S. Dragomir et al. [5] obtained the following inequality
by an application of the Grüss inequality:
Z b
1
1
2
0
0
(1.2) f (x) dx − (b − a)(f (a) + f (b)) + (b − a) (f (b) − f (a))
2
12
a
1
≤ (Γ2 − γ2 )(b − a)3 ,
32
where f is supposed to be twice differentiable on the interval (a, b), with the second derivative
bounded on (a, b) by Γ2 = supx∈(a,b) f 00 (x) < +∞ and γ2 = inf x∈(a,b) f 00 (x) > −∞. The
1
constant 32
is smaller than 6√1 5 given in [11] and 181√3 given in [2].
1
In this note we first improve the constant 32
in the inequality (1.2) to the best possible one
1√
of 36 3 . Then we give two new perturbed trapezoid inequalities for high-order differentiable
mappings. We need the following variant of the Grüss inequality:
Theorem 1.1. Let h, g : [a, b] → R be two integrable functions such that φ ≤ g(x) ≤ Φ for
some constants φ, Φ for all x ∈ [a, b], then
Z b
Z b
Z b
1
1
(1.3) h(x)g(x) dx −
h(x)
dx
g(x)
dx
b−a a
(b − a)2 a
a
Z b Z b
1
1
h(x) −
dx (Φ − φ).
≤
h(y)dy
2
b−a a
a
Proof. We write the left hand of inequality (1.3) as
Z b
Z b
Z b
Z b
Z b
1
1
h(x) dx
h(y) dy)g(x) dx.
h(x)g(x) dx −
g(x) dx =
(h(x) −
b−a a
b−a a
a
a
a
Denote
Z b
Z b
1
+
I =
max(h(x) −
h(y) dy, 0) dx
b−a a
a
and
Z b
Z b
1
−
I =
min(h(x) −
h(y) dy, 0) dx.
b−a a
a
Obviously I + + I − = 0. For φ ≤ g(x) ≤ Φ, then
Z b
Z b
1
(h(x) −
h(y) dy)g(x) dx ≤ I + Φ + I − φ
b
−
a
a
a
and
Z b
Z b
1
−
(h(x) −
h(y) dy)g(x) dx ≤ −I + φ − I − Φ
b−a a
a
and hence the obtained result (1.3) follows.
Theorem 1.2. Let f : [a, b] → R be a twice differentibale mapping on (a, b) with Γ2 =
supx∈(a,b) f 00 (x) < +∞ and γ2 = inf x∈(a,b) f 00 (x) > −∞, then we have the estimation
Z b
1
1
2
0
0
(1.4) f (x) dx − (b − a)(f (a) + f (b)) + (b − a) (f (b) − f (a))
2
12
a
1
≤ √ (Γ2 − γ2 )(b − a)3 ,
36 3
1√
where the constant 36 3 is the best one in the sense that it cannot be replaced by a smaller one.
J. Inequal. Pure and Appl. Math., 3(2) Art. 29, 2002
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A N OTE O N THE P ERTURBED T RAPEZOID I NEQUALITY
3
Proof. We choose in (1.3), h(x) = − 21 (x − a)(b − x) and g(x) = f 00 (x), we get
Z x2 Z Z b
1 b 1
1
2
dx = h(x)
−
h(y)dy
h(x)
+
(b
−
a)
dx
2 a b−a a
12
x1
1
√ (b − a)3 ,
=
36 3
√
3− 3
(b
6
where x1 = a +
− a) and x2 = a +
√
3+ 3
(b
6
− a). Thus from (1.3), we derive
Z b
1
1
2
0
0
f
(x)
dx
−
(b
−
a)(f
(a)
+
f
(b))
+
(b
−
a)
(f
(b)
−
f
(a))
2
12
a
Z b
Z b
Z b
1
1
1
00
00
= − (x − a)(b − x)f (x) dx −
− (x − a)(b − x) dx
f (x) dx
2
b−a a 2
a
a
1
≤ √ (Γ2 − γ2 )(b − a)3 .
36 3
To explain the best constant 361√3 in the inequality (1.4), we can construct the function f (x) =
Rx Ry
j(z) dz dy to attain the inequality in (1.4),
a
a
√

3
−
3


γ2 , a ≤ x < x1 = a +
(b − a),



6



√
j(x) =
3+ 3

Γ2 , x1 ≤ x < x2 = a +
(b − a),


6





γ2 , x2 ≤ x ≤ b.
The proof is complete.
Theorem 1.3. Let f : [a, b] → R be a third-order differentibale mapping on (a, b) with Γ3 =
supx∈(a,b) f 000 (x) < +∞ and γ3 = inf x∈(a,b) f 000 (x) > −∞, then we have the estimation
(1.5)
Z b
1
1
2
0
0
f
(x)
dx
−
(b
−
a)(f
(a)
+
f
(b))
+
(b
−
a)
(f
(b)
−
f
(a))
2
12
a
1
≤
(Γ3 − γ3 )(b − a)4 ,
384
where the constant
1
384
is the best one in the sense that it cannot be replaced by a smaller one.
1
(x
12
Proof. We choose in (1.3), h(x) =
1
2
Z
a
b
1
|h(x) −
b−a
Z
− a)(2x − a − b)(b − x), g(x) = f 000 (x), to get
b
Z
h(y) dy| dx =
a
a+b
2
h(x) dx =
a
1
(b − a)4 ,
384
Thus from (1.3) in Theorem 1.1,
(1.5) immediately. Finally, we
inequality
R x we
R ycan
R zderive the
construct the function f (x) = a a
j(s) ds dz dy, where j(x) = Γ3 for a ≤ x < a+b
2
and j(x) = γ3 for a+b
≤ x ≤ b, then the equality holds in (1.5).
2
J. Inequal. Pure and Appl. Math., 3(2) Art. 29, 2002
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4
X IAO - LIANG C HENG AND J IE S UN
Theorem 1.4. Let f : [a, b] → R be a fourth-order differentibale mapping on (a, b) with
M4 = supx∈(a,b) |f (4) (x)| < +∞, then
Z b
1
1
2
0
0
(1.6) f (x) dx − (b − a)(f (a) + f (b)) + (b − a) (f (b) − f (a))
2
12
a
1
≤
M4 (b − a)5 ,
720
where
1
720
is the best possible constant.
Proof. We may write the remainder of the perturbed trapezoid inequality in the kernel form
Z b
Z b
1
1
2
0
0
(1.7)
f (x) dx − (b − a)(f (a) + f (b)) + (b − a) (f (b) − f (a)) =
f (4) (x)k4 (x) dx,
2
12
a
a
where k4 (x) =
(1.8)
1
(x
24
− a)2 (b − x)2 . Then we get
Z 1
Z b
1
1
x2 (1 − x)2 dx =
.
|k4 (x)| dx =
24 0
720
a
Then (1.7) – (1.8) imply (1.6). The equality holds for f (x) = x4 , a ≤ x ≤ b in inequality (1.6).
Remark 1.5. We also can prove Theorem 1.2 and 1.3 in the kernel form
Z b
Z b
1
1
2
0
0
(1.9)
f (x) dx − (b − a)(f (a) + f (b)) + (b − a) (f (b) − f (a)) =
f (n) (x)kn (x) dx,
2
12
a
a
1
1
where k2 (x) = − 12 (x − a)(b − x) + 12
and k3 (x) = 12
(x − a)(2x − a − b)(b − x). By the
formula (1.7) and (1.9), and derive the perturbed trapezoid inequality for different norms as
shown in [5].
Now we present the composite perturbed trapezoid quadrature for an equidistant partitioning
of interval [a, b] into n subintervals. Applying Theorems 1.2 – 1.4, we obtain
Z b
f (x) dx = Tn (f ) + Rn (f ),
a
where
n−1 b−aX
b−a
b−a
(b − a)2 0
Tn (f ) =
f a+i
+ f a + (i + 1)
−
(f (b) − f 0 (a)),
2n i=0
n
n
12n2
and the remainder Rn (f ) satisfies the error estimate

(b − a)3


√
(Γ2 − γ2 ), if γ2 ≤ f 00 (x) ≤ Γ2 , ∀x ∈ (a, b),


2

36 3n





(b − a)4
(1.10)
|Rn (f )| ≤
(Γ3 − γ3 ), if γ3 ≤ f 000 (x) ≤ Γ3 , ∀x ∈ (a, b)

3
384n






 (b − a)5


M4 ,
if |f (4) (x)| ≤ M4 , ∀x ∈ (a, b).
720n4
Then we can use (1.10) to get different error estimates of the composite perturbed trapezoid
quadrature.
J. Inequal. Pure and Appl. Math., 3(2) Art. 29, 2002
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A N OTE O N THE P ERTURBED T RAPEZOID I NEQUALITY
5
As in [5], we may also apply the Theorems 1.2, 1.3 and 1.4 to special means. In this case we
may improve some of the bounds related to inequalities about special means as given in [5, p.
492-494].
Furthermore, we discuss the Iyengar’s type inequality for the perturbed trapezoidal quadrature rule for functions whose first and second order derivatives are bounded. In [1, 3, 9, 10] they
proved the following interesting inequality involving bounded derivatives.
If f is a differentiable function on (a, b) and |f 0 (x)| ≤ M1 , then
Z b
M1 (b − a)2 (f (b) − f (a))2
1
≤
f
(x)
dx
−
(b
−
a)(f
(a)
+
f
(b))
−
.
2
4
4M1
a
If |f 00 (x)| ≤ M2 , x ∈ [a, b] for positive constant M2 ∈ R, then
Z b
1
1
2
0
0
(b
−
a)(f
(a)
+
f
(b))
+
(b
−
a)
(f
(b)
−
f
(a))
f
(x)
dx
−
2
8
a
M2
≤
24
(b − a)3 −
|∆|
M2
3 !
,
Z b
1
1
2
0
0
f
(x)
dx
−
(b
−
a)(f
(a)
+
f
(b))
+
(b
−
a)
(f
(b)
−
f
(a))
2
8
a
∆2 (b − a)
M2
≤
(b − a)3 − 1
,
24
16M2
where
a+b
(f (b) − f (a))
0
0
+ f 0 (b),
∆1 = f 0 (a) − 2
+ f 0 (b).
(1.11)
∆ = f (a) − 2f
2
b−a
We will prove the following inequality.
Theorem 1.6. Let f : I → R, where I ⊆ R is an interval. Suppose that f is twice differentiable
◦
◦
in the interior I of I, and let a, b ∈ I with a < b. If |f 00 (x)| ≤ M2 , x ∈ [a, b] for positive
constant M2 ∈ R. Then
Z b
1
1
2
0
0
(1.12) f (x) dx − (b − a)(f (a) + f (b)) + (b − a) (f (b) − f (a))
2
8
a
s
1
|∆1 |3 (b − a)3
≤ M2 (b − a)3 −
,
24
72M2
where ∆1 is defined as (1.11).
Proof. Denote
b
Z
Jf =
a
1
1
f (x) dx − (b − a)(f (a) + f (b)) + (b − a)2 (f 0 (b) − f 0 (a)).
2
8
It is easy to see that
Z
Jf =
a
b
1
2
a+b
x−
2
2
f 00 (x) dx,
and
(1.13)
2(f (b) − f (a))
1
∆1 = f (a) −
+ f 0 (b) =
b−a
b−a
0
J. Inequal. Pure and Appl. Math., 3(2) Art. 29, 2002
Z
a
b
a+b
2 x−
2
f 00 (x) dx.
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6
X IAO - LIANG C HENG AND J IE S UN
For any |ε| ≤ 18 , we get for |f 00 (x)| ≤ M2 , x ∈ [a, b],
2
!
Z b
a
+
b
a
+
b
1
x−
+ 2ε(b − a) x −
f 00 (x) dx
Jf + ε(b − a)2 ∆1 =
2
2
2
a
≤ F (ε)M2 (b − a)3 ,
where
2
Z b 1
1
a
+
b
a
+
b
F (ε) =
x−
+ 2ε(b − a) x −
dx
(b − a)3 a 2
2
2
2
Z 1 1
1 1
=
x−
+ 2ε x −
dx.
2
2 0 2
For the case 0 ≤ ε ≤ 18 , we have
2
Z 1 1
1 1
x−
+ 2ε x −
F (ε) =
dx
2
2 0 2
(Z 1
2
!
−4ε
2
1
1
1
=
x−
+ 2ε x −
dx
2
2
2
0
2
!
Z 1
2
1
1
1
−
x−
+ 2ε x −
dx
1
2
2
2
−4ε
2
2
! )
Z 1
1
1
1
+
x−
+ 2ε x −
dx
1
2
2
2
2
=
1
32
+ ε3 .
24
3
For the case − 18 ≤ ε ≤ 0, we have similarly
F (ε) =
1
32
− ε3 .
24
3
We can prove |∆1 | ≤ 21 (b − a)M2 easily from (1.13). Thus we choose the parameter
s
|∆1 |
1
ε∗ = sign(∆1 )
,
|ε∗ | ≤ .
32(b − a)M2
8
By the above inequalities, we obtain
1
Jf ≤ F (ε∗ )M2 − ε∗ (b − a)2 ∆1 ≤ (b − a)3 M2 −
24
s
(b − a)3 |∆1 |3
.
72M2
Replacing f with −f , we have
J−f = −Jf ≤
1
(b − a)3 M2 −
24
s
(b − a)3 |∆1 |3
.
72M2
Thus we obtain bounds for |Jf | and prove the inequality (1.12).
J. Inequal. Pure and Appl. Math., 3(2) Art. 29, 2002
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A N OTE O N THE P ERTURBED T RAPEZOID I NEQUALITY
7
Remark 1.7. As |∆1 | ≤ 21 M2 (b − a), we have
s
r
|∆1 |
2 |∆1 |
≥
,
M2
b − a M2
Z b
1
1
2
0
0
f
(x)
dx
−
(b
−
a)(f
(a)
+
f
(b))
+
(b
−
a)
(f
(b)
−
f
(a))
2
8
a
M2
∆2 (b − a)
≤
(b − a)3 − 1
.
24
6M2
For the case f 0 (a) = f 0 (b) = 0, we have
Z b
M2
1
2 |f (b) − f (a)|2
3
≤
(1.14)
f
(x)
dx
−
(b
−
a)(f
(a)
+
f
(b))
(b
−
a)
−
.
2
24
3 M2 (b − a)
a
The inequality (1.14) is sharper than that stated in [9, p. 69].
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J. Inequal. Pure and Appl. Math., 3(2) Art. 29, 2002
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