Name: Dhyey Shukla

1. Subject: Complex Variable And Numerical Method (2141905)
Chapter: Numerical Integration
Department Mechanical Engineering
Name of Subject Teacher
Mrs. Pragna Lad

2. Team Members
Name Enrollment Number
Kiran Patel 170990119012
Vinay Patel 170990119014
Dhyey Shukla 170990119016

3. Trapezoidal Rule
 In Numerical analysis, the Trapezoidal rule is a technique for
approximating the definite integral.
 In general, trapezoidal rule is less accurate when compared with
Simpson's rules.

4.  From the General formula, we can obtain different integration formulae by
putting different values of i.e.
 Put in general formula and neglecting second and higher differences,

5. Similarly, for the next interval
And so on for the last interval
Combining all these three Equations (1), (2) & (3)

6.  This Final Equation obtained is known as Trapezoidal rule.

7. Q-1. State Trapezoidal Rule and calculate the
value of Integral
Ans.

8.  Divide into six equal parts, then the values of for
each points of subdivision are :
By Trapezoidal Rule
4.0 4.2 4.4 4.6 4.8 5.0 5.2
1.38629 1.43508 1.48160 1.52605 1.56861 1.60943 1.64865

10. Q-2. Evaluate take
Ans.
By Trapezoidal Rule
0 0.2 0.4 0.6 0.8 1
1 0.9615 0.8621 0.7352 0.6097 0.5

12. Simpson’s 1/3 Rule

     
0
0
0 1 3 5 1 2 4 2( ) 4 ... 2 ...
3
x nh
n n n
x
h
f x y y y y y y y y y

             
( 4 2 )
3
h
X O E  
where ,
X = extreme terms
O = odd terms
E = even terms

13. example 1
X 10 11 12 13 14 15 16
Y 1.02 0.94 0.89 0.79 0.71 0.62 0.55
=1/3 ((1.02+0.55)+4(0.94+0.79+0.62)+2(0.89+0.71))
=4.7233
16
10
ydx      0 1 3 5 1 2 4 24 ... 2 ...
3
n n n
h
y y y y y y y y y             
16
10
ydx
find by simpson’s 1/3 rule.

15. Example 2
A river is 80 meter wide . the depth ‘d’ in meter at a distance x
meters from one bank is given by the following table . Calculate
the area of cross section of the river using Simpson’s 1/3 rule .
80
0
A ydx       0 1 3 5 1 2 4 24 ... 2 ...
3
n n n
h
y y y y y y y y y             
X 00 10 20 30 40 50 60 70 80
Y 0 4 7 9 12 15 14 8 7
= 10/3 ((0+7) + 4(4+9+15+8) + ( 7+12+14) )
= 723.33 meter square
80
0
A ydx 
a = 0
b = 80
c = 10

16. 

17. Gaussian Quadrature
Formula

19. One Point Gaussian Quadrature Formula
 −1
1
𝑓 𝑥 𝑑𝑥 = 2𝑓 0

20. Two Point Gaussian Quadrature Formula
 −1
1
𝑓 𝑥 𝑑𝑥 = 𝑓
−1
3
+ 𝑓(
1
3
)

21. Three Point Gaussian Quadrature Formula
 −1
1
𝑓 𝑥 𝑑𝑥 =
5
9
𝑓(−
3
5
) +
5
9
𝑓(
3
5
) +
8
9
𝑓(0)

22. Example-1
Evaluate −1
1 𝑑𝑥
1+𝑥2 by one-Point, Two Point, and three point
Gaussian formulae.
Solution:-
𝑓 𝑥 =
1
1+𝑥2
By Gaussian One-Point Formula,
 −1
1
𝑑𝑥
1+𝑥2 = 2𝑓(0)
= 2(
1
1+0
)
= 2

23. By the two point formulae,
 −1
1 𝑑𝑥
1+𝑥2 = 𝑓(
−1
3
) + 𝑓(
1
3
)
=
1
1+
1
3
+
1
1+
1
3
= 1.5
By the three point Gaussian Formulae,
 −1
1 𝑑𝑥
1+𝑥2 =
5
9
𝑓(−
3
5
) +
8
9
𝑓 0 +
5
9
𝑓(
3
5
)
=
5
9
1
1+
3
5
+
1
1+0
+
5
9
1
1+
3
5
=1.5833

24. Example-2
 Evaluate 0
1
𝑒−𝑥2 𝑑𝑥 by using the Gaussian Quadrature formula with n=3
 Solution:-
 Let 𝑥 =
𝑏−𝑎
2
𝑡 +
𝑏+𝑎
2
 Here a=0, b=1
 𝑥 =
1
2
𝑡 +
1
2
=
1
2
𝑡 + 1
 𝑑𝑥 =
1
2
𝑑𝑡
 When x=0, t=-1
 When x=1, t=1

25.  0
1
𝑒 −𝑥2 𝑑𝑥 =
1
2 −1
1
𝑒
−1
4
(𝑡 + 1)2dt
𝑓 𝑥 = 𝑒
−1
4
(𝑡+1)2
By Three Point Gaussian Quadrature formula,
 0
1
𝑒−𝑥2
𝑑𝑥 =
1
2 −1
1
𝑒
−1
4
(𝑡+1)2 𝑑𝑡
=
1
2
[
5
9
𝑓(−
3
5
) +
8
9
𝑓 0 +
5
9
𝑓(
3
5
)]
=
1
2
[
5
9
𝑒
−1
4 −
3
5
+ 1
2
+
8
9
𝑒−
1
4
𝑡+1 2+
5
9 𝑒
−
1
4
3
5
+1
2
]
=0.746815

26. Questions and Suggestions are
Welcome
Thank You