B.tech ii unit-2 material beta gamma function

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About This Presentation

B.Tech -II sem E.M-II


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Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 1
Unit-II: GAMMA, BETA FUNCTION
Sr. No. Name of the Topic Page No.
1 Definition of Gamma function 2
2 Examples Based on Gamma Function 3
3 Beta function 5
4 Relation between Beta and Gamma Functions 5
5 Dirichlet’s Integral 9
6 Application to Area & Volume: Liouville’s
extension of dirichlet theorem
11
7 Reference Book 13

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 2
GAMMA, BETA FUNCTION

 The Gamma function and Beta functions belong to the category of the
special transcendental functions and are defined in terms of improper
definite integrals.

1.1 Definition of Gamma function :

The gamma function is denoted and defined by the integral
Γ�=∫�
−�
�
�−1
�� (�>0)

0


1.2 Properties of Gamma function :

1) Γ(�+1)=�Γ�
2) Γ(�+1)=�! When m is a positive integer.
3) Γ(�+??????)=(�+??????−1)(�+??????−2)………??????Γ??????, when n is a
positive integer.
4) Γ�=2∫�
−�
2
�
2�−1
�� (�>0)

0

5)
Γ�
�
�
=∫�
−��
�
�−1
�� (�>0)

0

6) Γ
1
2
=√�
7) ∫�
−�
2
��=
√??????
2

0

8) ∫�
�
(��� �)
�
��=
(−1)
�
(�+1)�+1
Γ(�+1)
1
0




2.1 Examples Based on Gamma Function:
Example 1: Evaluate ??????(−
�
�
).
Solution: We know that Γ(�+1)=�Γ�

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 3
 Γ(−
1
2
+1)=−
1
2
Γ(−
1
2
)
 Γ(
1
2
)=−
1
2
Γ(−
1
2
)
 √�=−
1
2
Γ(−
1
2
)
∴ ??????(−
�
�
)=−�√�. __________Ans.
Example 2: Evaluate ∫ √�
�
�
−√�
��

�

Solution: Let ??????=∫ �
1
4 �
−√�
��

0
__________(i)
Putting √�=� ⟹ �=�
2
so that ��=2� in (i), we get
??????=∫ �
12⁄
�
−�
2� ��

0

=2∫�
32⁄
�
−�
��

0

=2∫�
5
2
−1
�
−�
��

0

=2Γ(
5
2
)
=(2×
3
2
)Γ(
3
2
)
=(2×
3
2
×
1
2
)Γ(
1
2
)
=
3
2
√�
∴ ∫ √�
�
�
−√�
��

�
=
�
�
√� ________Ans.

Example 3: Evaluate ∫
�
�
�
�
��

�
.
Solution: Let ??????=∫
�
??????
�
??????
��

0
_______ (i)
Putting ??????
�
=�
�

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 4
⟹ �log??????=�
⟹ �=
1
log�

⟹ ��=
��
log�
in (i), we have
??????=∫(
�
log�
)
�
�
−�


0
��
log�

=
1
(log�)??????+1
∫�
−�
�
�
��

0

=
1
(log�)??????+1
∫�
(�+1)−1
�
−�
��

0

=
1
(log�)??????+1
Γ(??????+1)
∴ ∫
�
�
�
�
��

�
=
�
(??????????????????�)�+�
??????(�+�) ________ Ans.
Example 4: Prove that ∫(� ��??????�)
�
��=
�!
�
�
�
�

Solution: We know that
∫�
�
(��� �)
�
��=
(−1)
�
(�+1)�+1
Γ(�+1)
1
0
_______(i)
Now, ∫(� ����)
4
��=
1
0
∫�
4
1
0
(����)
4
��
Putting �=�=4 in (i), we get
∫�
4
1
0
(����)
4
��=
(−1)
4
(4+1)
4+1
Γ(4+1)
=
Γ5
5
5

=
4!
5
5
__________ proved.
2.2 EXERCISE:

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 5
1) Evaluate: (a) Γ(−
3
2
) (b) Γ(
7
2
) (c)Γ(0)
2) ∫�
−ℎ
2
�
2
��

0

3) ∫
��
√−��??????�
1
0

4) ∫(� ����)
3
��
1
0


3.1 BETA FUNCTION:
Definition: The Beta function denoted by �(�,�) or ??????(�,�) is defined as
??????(�,�)=∫�
�−1
(1−�)
�−1
��, (�>0,�>0)
1
0


3.2 Properties of Beta function:

1) B(m,n) = B(n,m)
2) ??????(�,�)=2∫ �??????�
2�−1
?????? ���
2�−1
?????? �??????
??????
2

0

3) ??????(�,�)=∫
�
�−1
(1+�)�+�
��

0

4) ??????(�,�)=∫
�
�−1
+�
�−1
(1+�)
�+�
��
1
0


4.1 Relation between Beta and Gamma Functions:

Relation between Beta and gamma functions is

�(�,�)=
Γm .Γn
Γ(m+n)


 Using above relation we can derive following results:

 ∫�??????�
�
?????? ���
�
?????? �??????=
1
2
�(
�+1
2
,
�+1
2
)=
Γ(
�+1
2
).(
�+1
2
)
2Γ(
�+�+2
2
)
??????
2

0

 Γ(
1
2
)=√�
 Euler’s formula:
Γ� .Γ(1−�)=
??????
sin�??????

 Duplication formula:

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 6
Γ� .Γ(�+
1
2
)=
√?????? Γ(2�)
2
2�−1



4.2 EXAMPLES:

Example 1: Evaluate ∫�
�
(�−√�)
�
��
�
�

Solution: Let √�=� ⟹ �=�
2
so that ��=2� ��
∫�
4
(1−√�)
5
��=
1
0
∫(�
2
)
4
(1−�)
5
(2� ��)
1
0

=2∫�
9
(1−�)
5
��
1
0

=2 ??????(10,6)
=2
Γ10 Γ6
Γ16

=2×
9!5!
15!

=
2×1×2×3×4×5
15×14×13×12×11×10

=
1
11×13×7×15

=
1
15015

∴ ∫�
4
(1−√�)
5
��=
1
0
1
15015
_________ Ans.
Example 2: Find the value of ??????(
�
�
).
Solution: We know that,
∫�??????�
�
?????? ���
�
?????? �??????=
Γ(
�+1
2
).(
�+1
2
)
2Γ(
�+�+2
2
)
??????
2

0

Putting �=�=0, we get ∫�??????=
??????(
�
�
) ??????(
�
�
)
2 ??????�
??????
2
0

 [??????]
0
??????2⁄
=
1
2

1
2
)
2

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 7

??????
2
=
1
2

1
2
)
2

 (Γ
1
2
)
2
=�
 Γ(
1
2
)=√� _______ Ans.

Example 3: show that ∫√���?????? �??????=
�
�
??????(
�
�
)??????(
�
�
)
�
�
�

Solution: We know that,
∫�??????�
�
?????? ���
�
?????? �??????=
Γ(
�+1
2
).(
�+1
2
)
2Γ(
�+�+2
2
)
??????
2

0

∫√���??????�??????=∫
���
12⁄
??????
�??????�
12⁄
??????
�??????
??????
2
0
??????
2
0

=∫�??????�
−12⁄
??????
??????
2
0
���
12⁄
?????? �??????

On applying formula (1), we have

∫√���??????�??????=
Γ(

1
2
+1
2
) Γ(
1
2
+1
2
)
2Γ(

1
2
+
1
2
+2
2
)
??????
2
0

=
Γ(
1
4
) Γ(
3
4
)
2 Γ(1)

=
1
2
Γ(
1
4
)Γ(
3
4
)
∴ ∫√���?????? �??????=
1
2
Γ(
1
4
)Γ(
3
4
)
??????
2
0
__________ Ans.

Example 4: Evaluate ∫(�+�)
�−�
(�−�)
�−�
��
+�
−�


Solution: Put �=2cos2??????, then ��=−2sin2?????? �?????? in

∫(1+�)
�−1
(1−�)
�−1
��
+1
−1


=∫(1+���2??????)
�−1
(1−���2??????)
�−1
(−2�??????�2??????)
0
??????
2

=∫(1+2���
2
??????−1)
�−1
(1−1+2�??????�
2
??????)
�−1
(−4�??????�?????? ���?????? �??????)
0
??????
2

=4∫2
�−1
���
2�−2
?????? .2
�−1
�??????�
2�−2
?????? .�??????�??????���?????? �??????
??????
2
0

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 8
=2
�+�
∫�??????�
2�−1
?????? ���
2�−1
??????

0
�??????
=2
�+�

Γ(
2�
2
) Γ(
2�
2
)
2Γ(
2�+2�
2
)

=2
�+�−1

Γ(�) Γ(�)
Γ(�+�)
__________Ans.

Example 5: Show that ??????(�)??????(�−�)=
�
�??????� ��
(&#3627409358;<&#3627408475;<1)
Solution: We know that
&#3627409149;(&#3627408474;,&#3627408475;)=∫
&#3627408485;
&#3627408475;−1
(1+&#3627408485;)
&#3627408474;+&#3627408475;
&#3627408465;&#3627408485;

0


Γ&#3627408474; Γ&#3627408475;
Γ(&#3627408474;+&#3627408475;)
=∫
&#3627408485;
&#3627408475;−1
(1+&#3627408485;)
&#3627408474;+&#3627408475;
&#3627408465;&#3627408485;

0


Putting &#3627408474;+&#3627408475;=1 &#3627408476;&#3627408479; &#3627408474;=1−&#3627408475;, we get


Γ(1−&#3627408475;) Γ&#3627408475;
Γ1
=∫
&#3627408485;
&#3627408475;−1
(1+&#3627408485;)
1
&#3627408465;&#3627408485;

0

Γ(1−&#3627408475;)Γ&#3627408475;=∫
&#3627408485;
&#3627408475;−1
1+&#3627408485;
&#3627408465;&#3627408485;

0
[∵ ∫
&#3627408485;
&#3627408475;−1
1+&#3627408485;
&#3627408465;&#3627408485;

0
=
??????
&#3627408480;??????&#3627408475; &#3627408475;??????
]

∴ Γ(&#3627408475;)Γ(1−&#3627408475;)=
??????
&#3627408480;??????&#3627408475; &#3627408475;??????
______ proved.

4.3 EXERCISE:
1) Evaluate ∫(1−&#3627408485;
3
)
−12⁄
&#3627408465;&#3627408485;
1
0

2) Evaluate ∫
&#3627408485;
&#3627408474;−1
+&#3627408485;
&#3627408475;−1
(1+&#3627408485;)&#3627408474;+&#3627408475;
&#3627408465;&#3627408485;
1
0

3) Evaluate ∫(
&#3627408485;
3
1−&#3627408485;
3
)
1
2
&#3627408465;&#3627408485;
1
0

4) Prove that Γ(
1
4
) Γ(
3
4
)=&#3627409163;√2
5) Show that &#3627409149;(&#3627408477;,&#3627408478;)=&#3627409149;(&#3627408477;+1,&#3627408478;)+(&#3627408477;,&#3627408478;+1)

5.1 DIRICHLET’S INTEGRAL :

If &#3627408473;,&#3627408474;,&#3627408475; are all positive, then the triple integral
∭&#3627408485;
&#3627408473;−1
&#3627408486;
&#3627408474;−1
&#3627408487;
&#3627408475;−1
&#3627408465;&#3627408485; &#3627408465;&#3627408486; &#3627408465;&#3627408487;

??????
=
Γ(l)Γ(m)Γ(n)
Γ(&#3627408473;+&#3627408474;+&#3627408475;+1)

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 9

Where V is the region &#3627408485;≥0,&#3627408486;≥0,&#3627408487;≥0 and &#3627408485;+&#3627408486;+&#3627408487;≤1.

Note:
∭ &#3627408485;
&#3627408473;−1
&#3627408486;
&#3627408474;−1
&#3627408487;
&#3627408475;−1
&#3627408465;&#3627408485; &#3627408465;&#3627408486; &#3627408465;&#3627408487;

??????
=
Γ(l)Γ(m)Γ(n)
Γ(&#3627408473;+&#3627408474;+&#3627408475;+1)

&#3627408473;+&#3627408474;+&#3627408475;


Where V is the domain, &#3627408485;≥0,&#3627408486;≥0,&#3627408487;≥0 and &#3627408485;+&#3627408486;+&#3627408487;≤ℎ

5.2 Corollary: Dirichlet’s theorem for n variables, the theorem status that

∭…∫&#3627408485;
1
&#3627408473;
1−1
&#3627408485;
2
&#3627408473;
2−1
… &#3627408485;
&#3627408475;
&#3627408473;
&#3627408475;−1
&#3627408465;&#3627408485;
1 &#3627408465;&#3627408485;
2 &#3627408465;&#3627408485;
3… &#3627408465;&#3627408485;
&#3627408475;
=
Γ&#3627408473;
1Γ&#3627408473;
2Γ&#3627408473;
3… Γ&#3627408473;
&#3627408475;
Γ(1+&#3627408473;
1+&#3627408473;
2+⋯+&#3627408473;
&#3627408475;)

&#3627408473;
1+&#3627408473;
2+⋯+&#3627408473;
&#3627408475;

Example 1: Prove that ∫
&#3627408537;
&#3627409362;
(&#3627409359;+&#3627408537;
&#3627409363;
)
(&#3627409359;+&#3627408537;
&#3627409359;&#3627409363;
)
&#3627408517;&#3627408537;=
&#3627409359;
&#3627409363;&#3627409358;&#3627409358;&#3627409363;

&#3627409358;

Solution: Let ??????=∫
&#3627408537;
&#3627409362;
(&#3627409359;+&#3627408537;
&#3627409363;
)
(&#3627409359;+&#3627408537;)
&#3627409359;&#3627409363;
&#3627408517;&#3627408537;

&#3627409358;

 ??????=∫
&#3627408485;
4
(1+&#3627408485;)
15
&#3627408465;&#3627408485;

0
+∫
&#3627408485;
9
(1+&#3627408485;)
15
&#3627408465;&#3627408485;

0

 ??????= ??????
1+??????
2 __________ (i)
Now, put &#3627408485;=
&#3627408481;
1+&#3627408481;
, when &#3627408485;=0, &#3627408481;=0; when &#3627408485;=∞,&#3627408481;=1
1+&#3627408485;=1+
&#3627408481;
1−&#3627408481;
=
1
1−&#3627408481;

 &#3627408465;&#3627408485;=
&#3627408465;&#3627408481;
(1−&#3627408481;)
2

∴ ??????
1=∫(
&#3627408481;
1−&#3627408481;
)
4
.(1−&#3627408481;)
15
.
1
(1−&#3627408481;)
2
&#3627408465;&#3627408481;
1
0

=∫&#3627408481;
4
(1−&#3627408481;)
9
&#3627408465;&#3627408481;
1
0

=&#3627409149;(5,10) _______(2)
And ??????
2=∫(
&#3627408481;
1−&#3627408481;
)
9
.(1−&#3627408481;)
15
.
1
(1−&#3627408481;)
2
&#3627408465;&#3627408481;
1
0

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 10
=∫&#3627408481;
9
(1−&#3627408481;)
4
&#3627408465;&#3627408481;
1
0

=&#3627409149;(10,5) ________(3)
∴ ??????=??????
1+??????
2
=&#3627409149;(5,10)+&#3627409149;(10,5) [Using(2) and (3)]
=&#3627409149;(5,10)+&#3627409149;(5,10) [&#3627409149;(&#3627408474;,&#3627408475;)=&#3627409149;(&#3627408475;,&#3627408474;)]
=2&#3627409149;(5,10)
=
2Γ5Γ10
Γ15

=
2×4!×9!
14!

=
2×4×3×2×1×9!
14×13×12×11×10×9!

=
1
5005
_______ Proved.
5.3 EXERCISE:
1) Find the value of ∫
&#3627408485;
3
−2&#3627408485;
4
+&#3627408485;
5
(1+&#3627408485;)
7
&#3627408465;&#3627408485;
1
0

2) Show that ∫
&#3627408485;
&#3627408474;−1
(1−&#3627408485;)
&#3627408475;−1
(&#3627408462;+&#3627408485;)
&#3627408474;+&#3627408475;
&#3627408465;&#3627408485;=
&#3627409149;(&#3627408474;,&#3627408475;)
&#3627408462;
&#3627408475;
(1+&#3627408462;)
&#3627408474;
1
0

3) &#3627409149;(&#3627408474;+1,&#3627408475;)=
&#3627408474;
&#3627408474;+&#3627408475;
&#3627409149;(&#3627408474;,&#3627408475;)





6.1 Application to Area & Volume:

 Liouville’s extension of dirichlet theorem:

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 11
∭&#3627408467;(&#3627408485;+&#3627408486;+&#3627408487;)&#3627408485;
&#3627408473;−1
&#3627408486;
&#3627408474;−1
&#3627408487;
&#3627408475;−1
&#3627408465;&#3627408485; &#3627408465;&#3627408486; &#3627408465;&#3627408487;
=
Γ(l)Γ(m)Γ(n)
Γ(l+m+n)
∫&#3627408467;(&#3627408482;) &#3627408482;
&#3627408473;+&#3627408474;+&#3627408475;−1
&#3627408465;&#3627408482;

2

1

Example1: Show that ∭
&#3627408517;&#3627408537; &#3627408517;&#3627408538; &#3627408517;&#3627408539;
(&#3627408537;+&#3627408538;+&#3627408539;+&#3627409359;)
&#3627409361;
=
&#3627409359;
&#3627409360;
&#3627408525;&#3627408528;??????&#3627409360;−
&#3627409363;
&#3627409359;&#3627409364;
, the integral being
taken throughout the volume bounded by
&#3627408537;=&#3627409358;,&#3627408538;=&#3627409358;,&#3627408539;=&#3627409358;,&#3627408537;+&#3627408538;+&#3627408539;=&#3627409359;.
Solution: By Liouville’s theorem when 0<&#3627408485;+&#3627408486;+&#3627408487;<1

&#3627408465;&#3627408485; &#3627408465;&#3627408486; &#3627408465;&#3627408487;
(&#3627408485;+&#3627408486;+&#3627408487;+1)
3
=∭
&#3627408485;
&#3627408473;−1
&#3627408486;
&#3627408474;−1
&#3627408487;
&#3627408475;−1
&#3627408465;&#3627408485; &#3627408465;&#3627408486; &#3627408465;&#3627408487;
(&#3627408485;+&#3627408486;+&#3627408487;+1)
3
(0≤&#3627408485;+&#3627408486;+&#3627408487;≤1)
=
Γ1Γ1Γ1
Γ(l+m+n)

1
(u+1)
3
u
3−1
du
1
0

=
1
2

&#3627408482;
2
(&#3627408482;+1)
3
&#3627408465;&#3627408482;
1
0

=∫[
1
&#3627408482;+1

2
(&#3627408482;+1)
2
+
1
(&#3627408482;+1)
3
]&#3627408465;&#3627408482;
1
0
(Partial fractions)
=
1
2
[log(&#3627408482;+1)+
2
&#3627408482;+1

1
2(&#3627408482;+1)
2
]
0
1

=
1
2
[&#3627408473;&#3627408476;&#3627408468;2+2(
1
2
−1)−(
1
8

1
2
)]
=
1
2
&#3627408473;&#3627408476;&#3627408468;2−
5
16

∴∭
&#3627408517;&#3627408537; &#3627408517;&#3627408538; &#3627408517;&#3627408539;
(&#3627408537;+&#3627408538;+&#3627408539;+&#3627409359;)
&#3627409361;
=
&#3627409359;
&#3627409360;
&#3627408525;&#3627408528;??????&#3627409360;−
&#3627409363;
&#3627409359;&#3627409364;
_______ Proved.
Example 2: Find the mass of an octant of the ellipsoid
&#3627408537;
&#3627409360;
&#3627408514;
&#3627409360;
+
&#3627408538;
&#3627409360;
&#3627408515;
&#3627409360;
+
&#3627408539;
&#3627409360;
&#3627408516;
&#3627409360;
=&#3627409359;,
the density at any point being &#3627409222;=&#3627408524; &#3627408537; &#3627408538; &#3627408539;.
Solution: Mass =∭&#3627409164; &#3627408465;&#3627408483;
=∭(&#3627408472; &#3627408485; &#3627408486; &#3627408487;)&#3627408465;&#3627408485; &#3627408465;&#3627408486; &#3627408465;&#3627408487;

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 12
=&#3627408472;∭(&#3627408485; &#3627408465;&#3627408485;)(&#3627408486; &#3627408465;&#3627408485;)(&#3627408487; &#3627408465;&#3627408487;) _______ (1)
Putting
&#3627408485;
2
&#3627408462;
2
=&#3627408482;,
&#3627408486;
2
&#3627408463;
2
=&#3627408483;,
&#3627408487;
2
&#3627408464;
2
=&#3627408484; and &#3627408482;+&#3627408483;+&#3627408484;=1
So that
2&#3627408485; &#3627408465;&#3627408485;
&#3627408462;
2
=&#3627408465;&#3627408482;,
2&#3627408486; &#3627408465;&#3627408486;
&#3627408463;
2
=&#3627408465;&#3627408483;,
2&#3627408487; &#3627408465;&#3627408487;
&#3627408464;
2
=&#3627408465;&#3627408484;
Mass=&#3627408472;∭(
&#3627408462;
2
&#3627408465;&#3627408482;
2
)(
&#3627408463;
2
&#3627408465;&#3627408483;
2
)(
&#3627408464;
2
&#3627408465;&#3627408484;
2
)
=
&#3627408472; &#3627408462;
2
&#3627408463;
2
&#3627408464;
2
8
∭&#3627408465;&#3627408482; &#3627408465;&#3627408483; &#3627408465;&#3627408484;, Where &#3627408482;+&#3627408483;+&#3627408484;≤1
=
&#3627408472; &#3627408462;
2
&#3627408463;
2
&#3627408464;
2
8
∭&#3627408482;
&#3627408473;−1
&#3627408483;
&#3627408473;−1
&#3627408484;
&#3627408473;−1
&#3627408465;&#3627408482; &#3627408465;&#3627408483; &#3627408465;&#3627408484;
=
&#3627408472; &#3627408462;
2
&#3627408463;
2
&#3627408464;
2
8

Γ1Γ1Γ1
Γ3+1

=
&#3627408472; &#3627408462;
2
&#3627408463;
2
&#3627408464;
2
8×6

=
&#3627408472; &#3627408462;
2
&#3627408463;
2
&#3627408464;
2
48

∴ ??????&#3627408514;&#3627408532;&#3627408532;=
&#3627408524; &#3627408514;
&#3627409360;
&#3627408515;
&#3627409360;
&#3627408516;
&#3627409360;
&#3627409362;??????
Ans.

Unit-2 GAMMA, BETA FUNCTION
RAI UNIVERSITY, AHMEDABAD 13
6.2 EXERCISE:
1) Find the value of ∭&#3627408473;&#3627408476;&#3627408468;(&#3627408485;+&#3627408486;+&#3627408487;) &#3627408465;&#3627408485; &#3627408465;&#3627408486; &#3627408465;&#3627408487; the integral extending
over all positive and zero values of &#3627408485;,&#3627408486;,&#3627408487; subject to the condition &#3627408485;+
&#3627408486;+&#3627408487;<1.
2) Evaluate ∭
√1−&#3627408485;
2
−&#3627408486;
2
−&#3627408487;
2
1+&#3627408485;
2
+&#3627408486;
2
+&#3627408487;
2
&#3627408465;&#3627408485; &#3627408465;&#3627408486; &#3627408465;&#3627408487;, integral being taken over all
positive values of &#3627408485;,&#3627408486;,&#3627408487; such that &#3627408485;
2
+&#3627408486;
2
+&#3627408487;
2
≤1.
3) Find the area and the mass contained m the first quadrant enclosed by
the curve (
&#3627408485;
&#3627408462;
)
&#3627409148;
+(
&#3627408486;
&#3627408463;
)
&#3627409149;
=1 &#3627408484;ℎ&#3627408466;&#3627408479;&#3627408466; &#3627409148;>0,&#3627409149;>0 given that density at
any point &#3627408477;(&#3627408485;&#3627408486;) is &#3627408472; √&#3627408485;&#3627408486;.


7.1 REFERENCE BOOK :
1) Introduction to Engineering Mathematics
By H. K. DASS. & Dr. RAMA VERMA
2) Higher Engineering Mathematics
By B.V.RAMANA
3) A text book of Engineering Mathematics
By N.P.BALI
4) www1.gantep.edu.tr/~olgar/C6.SP.pdf
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