Theoretical Mathematics & Applications, vol.5, no.1, 2015, 1-14
ISSN: 1792-9687 (print), 1792-9709 (online)
Scienpress Ltd, 2015
On some mixed integral inequalities
and its applications
S.D. Kendre
1
and S.G. Latpate2
Abstract
In this paper, we establish some mixed integral and integro-differential
inequalities which can be used as handy tools to study properties of solutions of a certain mixed integral and differential equations.
Mathematics Subject Classification: 45A05; 45B05; 45D05
Keywords: Integral inequality; Integral equations; Explicit bound
1
Introduction
In the theory of differential, integral and integro-differential equations one
often has to deal with certain differential and integro-differential inequalities.
In the last few years with the development of the theory of nonlinear differential and integral equations, many authors have established several integral
1
Department of Mathematics, University of Pune, Pune-411007, India.
E-mail:sdkendre@yahoo.com, ksubhash@math.unipune.ac.in
2
Department of Mathematics, Nowrosjee Wadia College of Arts and Science,
Pune-411001, India. E-mail: sglatpate@gmail.com
Article Info: Received : August 20, 2014. Revised : September 30, 2014.
Published online : January 10, 2015.
2
On some mixed integral inequalities
and integro-differential inequalities, see [1, 2, 4, 6, 7, 8, 9, 10, 11, 12]. These
inequalities play an important role in the study of some properties of differential, integral and integro-differential equations. Existence of solutions of a
certain mixed integral and integro-differential equations were studied in [3, 13]
by M. B. Dhakne and H. L. Tidke.
In this paper, we establish mixed integral and integro-differential inequalities which provide an explicit bound on unknown function. In particular we
extend the result established by B. G. Pachpatte in [12]. Some applications
are also given to convey the importance of our results.
2
Preliminary
Before proceeding to the statement of our main result, we state some important integral inequalities that will be used in further discussion.
Lemma 2.1 (Fangcui Jiang and Fanwei Meng [5]). Assume that a ≥
0, p ≥ q ≥ 0, and p 6= 0, then
q
q q−p
p − q pq
ap ≤ k p a +
k , for any k > 0.
p
p
(2.1)
Theorem 2.2 (Pachpatte [12]). Let u(t), a(t), b(t), c(t) ∈ C(I = [α, β], R+ ), a(t)
be continuously differentiable on I, a0 (t) ≥ 0 and
Z
Z
t
u(t) ≤ a(t) +
α
Z
µZ
β
If p =
c(s)u(s)ds,
t ∈ I.
(2.2)
α
¶
s
c(s) exp
β
b(s)u(s)ds +
b(σ)dσ ds < 1, then
α
α
µZ
t
u(t) ≤ M exp
¶
Z
t
b(s)ds +
α
µZ
0
s
a (s) exp
α
¶
b(σ)dσ ds,
t ∈ I,
(2.3)
α
where
·
µZ s
¶
¸
Z β
Z s
1
0
M=
a(α) +
c(s)
a (τ ) exp
b(σ)dσ dτ ds ,
1−p
α
α
τ
t ∈ I. (2.4)
3
S.D. Kendre and S.G. Latpate
3
Main Results
In this section, we state and prove mixed nonlinear integral inequalities to
obtain explicit bound on solutions of a certain mixed integral equations.
Theorem 3.1. Let u(t), f (t), g(t), c(t), c0 (t) ∈ C(I = [α, β], R+ ) and p ≥
q ≥ 0, p 6= 0 be constants. If
Z t
Z β
p
q
u (t) ≤ c(t) +
f (s)u (s)ds +
g(s)up (s)ds
Z
β
and Q =
α
µZ
s
g(s) exp
α
α
¶
n1 f (σ)dσ ds < 1, then
α
¡R s
¢ ¢
0
[c
(τ
)
+
n
f
(τ
)]
exp
n
f
(σ)dσ
dτ ds
2
1
α
α
τ
up (t) ≤
1−Q
¶
¶ Z t
µZ t
µZ t
0
n1 f (σ)dσ ds,
mf (s)ds +
[c (s) + n2 f (s)] exp
× exp
c(α) +
Rβ
g(s)
¡R s
α
s
α
(3.1)
q q−p
p − q pq
where k > 0, n1 = k p and n2 =
k .
p
p
Proof. Define a function z(t) by
Z t
Z
q
z(t) = c(t) +
f (s)u (s)ds +
α
β
g(s)up (s)ds,
α
1
p
then u(t) ≤ z (t),
Z
β
z(α) = c(α) +
g(s)up (s)ds
(3.2)
α
and
q
z 0 (t) = c0 (t) + f (t)uq (t) ≤ c0 (t) + f (t)z p (t).
(3.3)
¿From Lemma 2.1 and (3.3), we have
z 0 (t) ≤ c0 (t) + n1 f (t)z(t) + n2 f (t),
or, equivalently,
0

µ
¶
Z t
z(t)
0


³ R
´ ≤ [c (t) + n2 f (t)] exp −n1
f (s)ds .
t
α
exp n1 α f (s)ds
(3.4)
4
On some mixed integral inequalities
Setting t = s; in (3.4) and integrating with respect to s from α to t, we have
µZ t
¶ Z t
µZ t
¶
0
z(t) ≤ z(α) exp
n1 f (s)ds +
[c (s) + n2 f (s)] exp
n1 f (σ)dσ ds.
α
α
s
(3.5)
As up (t) ≤ z(t) from equation (3.5), we have
µZ t
¶ Z t
µZ t
¶
p
0
u (t) ≤ z(α) exp
n1 f (s)ds +
[c (s) + n2 f (s)] exp
n1 f (σ)dσ ds.
α
α
s
(3.6)
Now from (3.2) and (3.6), we have
Z
µZ
β
¶
s
n1 f (σ)dσ ds
z(α) ≤ c(α) + z(α)
g(s) exp
α
α
¶ ¶
µZ s
µZ s
Z β
0
n1 f (σ)dσ dτ ds,
[c (τ ) + n2 f (τ )] exp
+
g(s)
τ
α
α
or, equivalently,
¶ ¶
µ
µZ s
Z β
n1 f (σ)dσ ds
z(α) 1 −
g(s) exp
α
α
¶ ¶
µZ s
µZ s
Z β
0
n1 f (σ)dσ dτ ds,
[c (τ ) + n2 f (τ )] exp
g(s)
≤ c(α) +
z(α) ≤
c(α) +
Rβ
α
τ
α
α
g(s)
¡R s
α
[c0 (τ ) + n2 f (τ )] exp
1−Q
¡R s
τ
¢ ¢
n1 f (σ)dσ dτ ds
.
(3.7)
Using inequality (3.7) in (3.6), we obtain the desired inequality (3.1). This
completes the proof.
Remark 1. It is interesting to note that when p = q = 1 the Theorem 3.1
reduces to the inequality stated in Theorem 2.2 and when p = 1 and g = 0 it
reduces to well known Gronwall-Bellman inequality.
Theorem 3.2. Let u(t), f (t), g(t), h(t) ∈ C(I, R+ ) and c ≥ 0 be a constant.
If
Z
p
t
u (t) ≤ c +
α
·
Z s
Z
q
q
h(s) u (s) +
f (σ)u (σ)dσ +
α
β
¸
p
g(σ)u (σ)dσ ds, for t ∈ I,
α
5
S.D. Kendre and S.G. Latpate
then
µZ
¶
t
p
u (t) ≤ c exp
Z
n1 A(σ)dσ +
α
where
·
Z
t
A(t) = h(t) 1 +
α
µZ
t
n2 B(s) exp
n1 A(σ)dσ ,
α
1
f (σ)dσ +
n1
Z
¶
t
(3.8)
s
¸
β
g(σ)dσ ,
·
¸
Z t
B(t) = h(t) 1 +
f (σ)dσ ,
α
α
and p, q, n1 , n2 are as same defined in Theorem 3.1.
Proof. Denoting a function z(t) by
·
Z t
Z s
Z
q
q
z(t) = c +
h(s) u (s) +
f (σ)u (σ)dσ +
α
α
¸
β
p
g(σ)u (σ)dσ ds,
α
1
we have u(t) ≤ z p (t), z(α) = c, z(t) is a nondecreasing and
·
¸
Z t
Z β
0
q
q
p
z (t) = h(t) u (t) +
f (σ)u (σ)dσ +
g(σ)u (σ)dσ
α
α
·
¸
Z t
Z β
q
q
p
p
≤ h(t) z (t) +
f (σ)z (σ)dσ +
g(σ)z(σ)dσ
α
α
½·
¸
¸
¾
·Z β
Z t
q
≤ h(t) 1 +
f (σ)dσ z p (t) +
g(σ)dσ z(t) .
α
(3.9)
α
An application of Lemma 2.1 to (3.9), we have
¸
¾
½·
¸
·Z β
Z t
0
g(σ)dσ z(t)
z (t) ≤ h(t) 1 +
f (σ)dσ [n1 z(t) + n2 ] +
α
α
½ ·
¸
·
¸¾
Z t
Z β
Z t
1
= h(t) n1 1 +
f (σ)dσ +
g(σ)dσ z(t) + n2 1 +
f (σ)dσ
n1 α
α
α
= n1 A(t)z(t) + n2 B(t),
or, equivalently,


exp
³R
t
0
z(t)
n A(σ)dσ
α 1
µ Z t
¶

´ ≤ n2 B(t) exp −
n1 A(σ)dσ .
(3.10)
α
By integrating (3.10), we get
µZ t
¶ Z t
µZ t
¶
z(t) ≤ z(α) exp
n1 A(σ)dσ +
n2 B(s) exp
n1 A(σ)dσ ds.
α
α
s
(3.11)
6
On some mixed integral inequalities
As up (t) ≤ z(t) from (3.11), we obtain
µZ t
¶ Z t
µZ t
¶
p
u (t) ≤ z(α) exp
n1 A(σ)dσ +
n2 B(s) exp
n1 A(σ)dσ ds.
α
α
s
(3.12)
Using z(α) = c in (3.12), we get the desired inequality (3.8) and hence the
proof.
Remark 2. Note that when p = q = 1 the Theorem 3.2 reduces to the
inequality established by B.G. Pachpatte in ([12] page no.40).
Corollary 3.3. Let u(t), f (t), g(t), h(t) be as same defined in Theorem 3.2
and 1 ≤ c(t) be a nondecreasing. If
¸
·
Z β
Z s
Z t
p
q
q
p
g(σ)u (σ)dσ ds, for t ∈ I,
f (σ)u (σ)dσ +
h(s) u (s) +
u (t) ≤ c(t) +
α
α
α
(3.13)
then
µZ
p
p
¶
t
Z
p
n1 A(σ)dσ + c (t)
u (t) ≤ c (t) exp
α
µZ
t
¶
t
n1 A(σ)dσ ,
n2 B(s) exp
s
α
(3.14)
where p ≥ q ≥ 1, A(t), B(t) and n1 , n2 are as same defined in Theorem 3.2 and
Theorem 3.1 respectively.
Proof. Since 1 ≤ c(t) and nondecreasing, from (3.13), we have
µ
¶p
·µ
¶q
µ
¶q
µ
¶p
¸
Z t
Z s
Z β
u(t)
u(s)
u(σ)
u(σ)
≤1+
h(s)
(s) +
f (σ)
(σ)dσ +
g(σ)
(σ)dσ ds,
c(t)
c(s)
c(σ)
c(σ)
α
α
α
(3.15)
for t ∈ I. Applying Theorem 3.2 to (3.15), we get (3.14). This completes the
proof.
Theorem 3.4. Let u(t), u0 (t), f (t), g(t), c(t), c0 (t) ∈ C(I, R+ ) and u(α) = 0.
If
Z
0
p
t
[u (t)] ≤ c(t) +
Z
q
β
f (s)u (s)ds +
α
α
g(s)[u0 (s)]p ds
(3.16)
7
S.D. Kendre and S.G. Latpate
Z
µZ
β
and Q1 =
g(s) exp
α
¶
s
q
(σ − α) n1 f (σ)dσ ds < 1, then
α
¡ Rs
¢ ¢
0
q
q
[c
(τ
)
+
(τ
−
α)
n
f
(τ
)]
exp
n
(σ
−
α)
f
(σ)dσ
dτ ds
2
1
α
α
τ
[u0 (t)]p ≤
1 − Q1
µ Z t
¶
× exp n1
(s − α)q f (s)ds
α
µ Z t
¶
Z t
0
q
q
+
[c (s) + (s − α) n2 f (s)] exp n1
(s − α) f (s)ds ds,
(3.17)
Rβ
c(α) +
g(s)
¡R s
α
s
p, q, n1 , n2 are as same defined in Theorem 3.1.
Proof. Define a function z(t) by
Z
Z t
q
f (s)u (s)ds +
z(t) = c(t) +
then u(t) ≤
Rt
α
β
g(s)[u0 (s)]p ds,
α
α
1
p
z (s), z(t) is a nondecreasing,
Z β
g(s)[u0 (s)]p (s)ds
z(α) = c(α) +
(3.18)
α
and
q
z 0 (t) ≤ c0 (t) + (t − α)q f (t)z p (t), .
(3.19)
Applying Lemma 2.1 to (3.19), we have
z 0 (t) ≤ c0 (t) + (t − α)q f (t)[n1 z(t) + n2 ]
= c0 (t) + n1 (t − α)q f (t)z(t) + n2 (t − α)q f (t)
= c0 (t) + n1 (t − α)q f (t)z(t) + n2 (t − α)q f (t),
or, equivalently,
0

µ
¶
Z t
z(t)
0
q
q

³ R
´  ≤ [c (t) + n2 (t − α) f (t)] exp −n1
(s − α) f (s)ds .
t
q
α
exp n1 α (s − α) f (s)ds
(3.20)
By integrating (3.20), we get
³
exp n1
Rt
α
Z
z(t)
(s −
α)q f (s)ds
t
´ ≤ z(α) +
[c0 (s) + n2 (s − α)q f (s)]
α
µ
Z
s
¶
q
(s − α) f (s)ds ds,
× exp −n1
α
8
On some mixed integral inequalities
µZ
¶
t
Z
q
t
[c0 (s) + n2 (s − α)q f (s)]
α
µZ t
¶
q
× exp
n1 (σ − α) f (σ)dσ ds.
n1 (s − α) f (s)ds +
z(t) ≤ z(α) exp
α
s
(3.21)
As [u0 (t)]p ≤ z(t) from (3.21), we have
µZ t
¶ Z t
0
p
q
[u (t)] ≤ z(α) exp
n1 (s − α) f (s)ds +
[c0 (s) + n2 (s − α)q f (s)]
α
α
µZ t
¶
q
exp
n1 (σ − α) f (σ)dσ ds.
s
(3.22)
Now from (3.18) and (3.22), we have
¶
µZ s
Z β
q
n1 (σ − α) f (σ)dσ ds
z(α) ≤ c(α) + z(α)
g(s) exp
α
α
¶ ¶
µZ s
µZ s
Z β
q
0
q
n1 (σ − α) f (σ)dσ dσ ds,
[c (τ ) + n2 (τ − α) f (τ )] exp
+
g(s)
τ
α
α
¶ ¶
µ
µZ s
Z β
q
n1 (σ − α) f (σ)dσ ds ≤ c(α)
z(α) 1 −
g(s) exp
α
α
¶ ¶
µZ s
µZ s
Z β
q
0
q
n1 (σ − α) f (σ)dσ dτ ds,
[c (τ ) + n2 (τ − α) f (τ )] exp
+
g(s)
τ
α
α
¡R s
¡R s
¢ ¢
Rβ
c(α) + α g(s) α [c0 (τ ) + n2 (τ − α)q f (τ )] exp τ n1 (σ − α)q f (σ)dσ dτ ds
z(α) ≤
.
1 − Q1
(3.23)
The required inequality (3.17) follows, from inequalities (3.22) and (3.23). This
completes the proof.
Theorem 3.5. Let u(t), f (t), g(t), h(t) ∈ C(I, R+ ) and c ≥ 0 be a constant.
If
Z
0
p
·
t
[u (t)] ≤ c +
Z
q
h(s) u (s) +
α
then
p
q
t
[u (t)] ≤ c exp
0
p
g(σ)[u (σ)] dσ ds, for t ∈ I,
α
¶
Z
µZ
t
n1 A1 (σ)dσ +
α
¸
β
f (σ)u (σ)dσ +
α
µZ
0
Z
s
t
n2 B1 (s) exp
α
¶
n1 A1 (σ)dσ ,
s
(3.24)
9
S.D. Kendre and S.G. Latpate
where
·
Z
t
1
A1 (t) = h(t) [t − α] +
[σ − α] f (σ)dσ +
n1
α
q
Z
·
Z
q
g(σ)dσ ,
α
¸
t
B1 (t) = h(t) [t − α] +
¸
β
q
q
[σ − α] f (σ)dσ
α
and p, q, n1 , n2 are as same defined in Theorem 3.1.
Proof. Denoting a function z(t) by
Z
·
t
z(t) = c +
Z
q
s
h(s) u (s) +
α
Z
q
f (σ)u (σ)dσ +
α
¸
β
p
g(σ)[u(σ)] dσ ds,
α
1
we have u0 (t) ≤ z p (t), z(α) = c, z(t) is a nondecreasing and
·
¸
Z t
Z β
q
q
0
p
z (t) = h(t) u (t) +
f (σ)u (σ)dσ +
g(σ)[u (σ)] dσ
α
α
·
¸
Z t
Z β
q
q
q p
q
p
≤ h(t) [t − α] z (t) +
[σ − α] f (σ)z (σ)dσ +
g(σ)z(σ)dσ
α
α
½·
¸
¸
¾
·Z β
Z t
q
q
q
≤ h(t) [t − α] +
[σ − α] f (σ)dσ z p (t) +
g(σ)dσ z(t) .
0
α
α
(3.25)
An application of Lemma 2.1 to (3.25), we have
½·
0
Z
q
z (t) ≤ h(t)
¸
t
·Z
q
β
¸
¾
g(σ)dσ z(t)
[t − α] +
[σ − α] f (σ)dσ [n1 z(t) + n2 ] +
α
α
½ ·
¸
Z t
Z β
1
q
q
= h(t) n1 [σ − α] +
[σ − α] f (σ)dσ +
g(σ)dσ z(t)
n1 α
α
·
¸¾
Z t
q
q
+n2 [t − α] +
[σ − α] f (σ)dσ
α
= n1 A1 (t)z(t) + n2 B1 (t),
or, equivalently,


exp
³R
t
α
0
z(t)
n1 A1 (σ)dσ
µ Z t
¶

´ ≤ n2 B1 (t) exp −
n1 A1 (σ)dσ .
α
(3.26)
10
On some mixed integral inequalities
By integrating (3.26), we get
µZ t
¶ Z t
µZ t
¶
z(t) ≤ z(α) exp
n1 A1 (σ)dσ +
n2 B1 (s) exp
n1 A1 (σ)dσ ds.
α
α
s
(3.27)
As [u0 (t)]p ≤ z(t) from (3.27), we obtain
µZ t
¶ Z t
µZ t
¶
0
p
[u (t)] ≤ z(α) exp
n1 A1 (σ)dσ +
n2 B1 (s) exp
n1 A1 (σ)dσ ds.
α
α
s
(3.28)
Using z(α) = c in (3.28), we get the desired inequality (3.24) and hence the
proof.
4
Applications
One of the main motivations for the study of different type inequalities
given in the previous sections is to apply them as tools in the study of various classes of integral equations. In the following section we give application
of some theorems of previous sections. In fact we discuss the boundedness
behavior of solutions of a nonlinear mixed integral equations.
Example 1. Consider the following general mixed nonlinear integral equation
Z t
Z β
p
q
y (t) = x(t) +
F (s, y (s))ds +
G(s, y p (s))ds,
(4.1)
α
α
for t ∈ I, where p ≥ q ≥ 0, p 6= 0, y(t) is unknown function, x ∈ C(I, Rn ), F, G ∈
C(I × Rn , Rn ),
I = [α, β], Rn is n dimensional Euclidean space with norm |.|.
Suppose that the functions x, y, F, G in equation (4.1) satisfy the following
conditions :
|x(t)| ≤ c(t),
q
(4.2)
q
|F (s, y )| ≤ g(s)|y| ,
p
p
|G(s, y )| ≤ f (s)|y| ,
(4.3)
(4.4)
11
S.D. Kendre and S.G. Latpate
where p ≥ q ≥ 0, p 6= 0, c, f, g are as same defined in Theorem 3.1. If y(t), t ∈ I
is a solution of equation (4.1) and Q < 1, then
¡R s
¡R s
¢ ¢
Rβ
c(α) + α g(s) α [c0 (τ ) + n2 f (τ )] exp τ n1 f (σ)dσ dτ ds
p
|y(t)| ≤
1−Q
µZ t
¶ Z t
µZ t
¶
0
× exp
mf (s)ds +
[c (s) + n2 f (s)] exp
n1 f (σ)dσ ds,
α
α
s
(4.5)
where Q, n1 , n2 are as same defined in Theorem 3.1.
¿From equation (4.1)-(4.4), we obtain
Z t
Z β
p
q
|y(t)| ≤ |x(t)| +
f (s)|y(s)| ds +
g(s)|y(s)|p ds.
α
(4.6)
α
An application of Theorem 3.1, we get (4.6). This show that solution y(t) is
bounded.
Example 2. Consider the following general mixed nonlinear integrodifferential equation
Z β
Z t
q
0
p
G (s, [y 0 (s)]p ) ds,
(4.7)
F (s, y (s)) ds +
[y (t)] = x(t) +
α
α
for t ∈ I, where p ≥ q ≥ 0, p 6= 0, y(t) is unknown function, x ∈ C(I, Rn ), F, G ∈
C(I × Rn , Rn ),
I = [α, β], Rn is n dimensional Euclidean space with norm |.|.
Suppose that the functions x, y, F, G in equation (4.7) satisfy the following
conditions
|x(t)| ≤ c(t),
(4.8)
p
|F (t, s, y p )| ≤ g(t, s) |y 0 | ,
(4.9)
|G(t, s, y q )| ≤ f (t, s)|y|q ,
(4.10)
where p ≥ q ≥ 0, p 6= 0, c, f, g are as same defined in Theorem 3.4. If y(t), t ∈ I
is a solution of equation (4.7) and Q< 1, then
¡R s 0
¡ Rs
¢ ¢
Rβ
q
q
c(α)
+
g(s)
[c
(τ
)
+
(τ
−
α)
n
f
(τ
)]
exp
n
(σ
−
α)
f
(σ)dσ
dτ ds
2
1
α
α
τ
|y(t)|p ≤
1 − Q1
µ Z t
¶
× exp n1
(s − α)q f (s)ds
α
µ Z t
¶
Z t
0
q
q
+
[c (s) + (s − α) n2 f (s)] exp n1
(s − α) f (s)ds ds,
(4.11)
α
s
12
On some mixed integral inequalities
where Q1 , n1 , n2 are as same defined in Theorem 3.4. From equation (4.7)(4.10), we obtain
Z t
Z β
0
p
q
|y (t)| ≤ c(t) +
f (s)|y(s)| ds +
g(s)|y 0 (s)|p ds.
(4.12)
α
α
Applying Theorem 3.4 to (4.12), we get (4.11).
Example 3. We calculate the explicit bound on the solution of the following
nonlinear integral equation of the form:
Z t
Z 7
1
3
u(s)ds +
su3 (s)ds
(4.13)
u (t) = 6 +
0
0 1+s
where u(t) are defined as in Theorem 3.1 and we assume that every solution
u(t) of (4.13) exists on R+ .
q−p
q
−1
kp =
By Theorem 3.1, we have p = 3, q = 2, n1 = pq k p = 23 k 3 , n2 = p−q
p
1 23
1
k , α = 0, β = 1, k > 0, f (s) = 1+s
, g(s) = s and
3
¶
µZ s
Z β
n1 f (σ)dσ ds
Q=
g(s) exp
α
α
µ
¶
Z 1
Z s
2 1
1
√
=
s exp
dσ ds
3 3k 0 σ+1
0
´√
³ √
2
√ +2
3
3 3 3 k + 23 3 k
k
³
´
=
< 1, for any k > 0.
√
2 9k 2/3 + 9 3 k + 2
Thus all the conditions of the Theorem 3.1 are satisfied, hence we obtain
¶ 
µ


R s 2 √31k
Z
Z
 1 1  2/3 s exp τ 3(σ+1) dσ

k
 ds + 6
u(t) ≤ 
s
dτ



τ +1
0 3
0
³
×
exp
1−





=




R1
0
k 3
2 √
1
3 3k
³
s exp
Rt
1
0 s+1
2 √
1
3 3k
!
2
√
3 +2
2 3 k −3
´
ds
Rs
1
0 σ+1
√
3
!
k−2
√
3
4(9k2/3 +9 k+2)
3 3
1−
´
dσ ds

t
+
k 2/3 exp
0
2
√

+ 6 (t + 1) 3 3 k
2
√
√
3 +2
3
k+2 3 k
2(9k2/3 +9
Z
!
√
3
√
3
k+2)
k
³
2 √
1
3 3k
Rt
1
τ s+1
3(τ + 1)
´
ds
 31
dτ 
 31
2
1 ³
√
+ k (t + 1) 3 3 k
2


´

−1 
 .



S.D. Kendre and S.G. Latpate
13
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14
On some mixed integral inequalities
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