Ankit_Practical_File-1.pdf A detailed overview of Rizir as a brand

`Numerical iNtegratioN
Department Of Mathematics
Research Project
By
aNKit
(Roll-No:210087302014)
Under The Supervision
Dr. DileeP Kumar
goverNmeNt Post graDuate college, NoiDa
(chauDhary charaN siNgh uNiversity)
June, 2024

> INTRODUCTION
The term “numerical integration” first appears in 1915 in the publication. A course in
interpolation and Numeric Integration for the Mathematical Laboratory by David Gibb.
The beginnings of numerical integration have its roots in antiquity. A prime example of how
ancient these methods are, is the Greek quadrature of the circle by means of inscribed and
circumscribed regular polygons. This process led Archimedes to an upper bound and lower
bound for the value Pi. These methods were used widely due to the lack of formal calculus.
The method of the sum of an infinitesimal area over a finite range was unknown until the
sixteenth century when Newton formalized the concepts of what we know now know as
calculus. The earliest forms of numerical integration are similar to that of the Greek method
of inscribing regular polygons into curved functions. This process broken down was taking a
known area and overlapping it with an unknown area to approximate the area of the unknown
shape. One could improve accuracy by choosing a better fitting shape. Later methods decided
to improve upon estimation area under a curve decided to use more polygons but smaller in
area. Such as example is the use of rectangles evenly spaced under a curve to estimate the
area. Even further improvements saw the use of trapezoids instead of rectangles to better fit
the curvature of the function being analyzed. Today the best methods for numerical integration
are known as quadrature methods that have a very small error.
The process of evaluating a definite integral from a set of tabulated values of the integrand
f(x) is called numerical integration. This process when applied to a function of a single
variable, is known as quadrature.
The problem of numerical integration, like that ot numerical differentiation, is solved by
representing f(x) by an interpolation formula and then integrating it between the given limits.
In this way, we can derive quadrature formulae for approximate integration of a function
defined by a set of numerical values only.
The problem of numerical integration is started below:
Given a set of data points (𝑥0, 𝑦0), (𝑥1𝑦1), ………. , (𝑥𝑛, 𝑦𝑛) of a function y = 𝒇(𝒙), it is
required to find the value of the definite integral ∫ 𝒇(𝒙)𝒅𝒙
𝒃
𝒂
. The function 𝑓(𝑥) is replaced
By a suitable interpolating polynomial ∅(𝑥).
Then the approximate value of the definite integral is calculated using the following formula.
∫ 𝒇(𝒙)𝒅𝒙 ≅ ∫ ∅(𝒙)𝒅𝒙
𝒃
𝒂
𝒃
𝒂
Thus, different integration formulae can be derived depending on the type of the
interpolation formulae used.
A numerical integration formula is said to be of closed type, if the limits of integration
a and b are taken as interpolating points. If a and b are not taken as interpolating
points then the formula is known as open type formula.

General Quadrature Formula Based On Newton’s Forward
Interpolation.
The Newton’s forward interpolation formula for the equispaced points 𝑥𝑖, 𝑖= 0,1, . . . . . . , n,
𝑥𝑖 = 𝑥0 + 𝑖ℎ is
∅(𝑥) = 𝑦0 + 𝑢∆ 𝑦0 +
𝑢(𝑢−1)
2!
∆2
𝑦0 +
𝑢(𝑢−1)(𝑢−2)
3!
∆3
𝑦0 + . . . . . ,
Where 𝑢 =
𝑥−𝑥0
ℎ
, ℎ is the spacing.
Let the interval [𝑎 , 𝑏] be divided into 𝑛 equal subintervals such that 𝑎 = 𝑥0 <
𝑥1 < 𝑥2 < . . . . . . < 𝑥𝑛 = 𝑏. Then,
𝐼 = ∫ 𝑓(𝑥)𝑑𝑥 ≅ ∫ ∅(𝑥)𝑑𝑥
𝑥𝑛
𝑥0
𝑏
𝑎
= ∫ [𝑦0 + 𝑢∆ 𝑦0 +
𝑢(𝑢−1)
2!
∆2
𝑦0 +
𝑢(𝑢−1)(𝑢−2)
3!
∆3
𝑦0+ . . . . . ] 𝑑𝑥
𝑥𝑛
𝑥0
.
Since 𝑥 = 𝑥0 + 𝑢ℎ, 𝑑𝑥 = ℎ𝑑𝑢, when 𝑥 = 𝑥0 𝑡ℎ𝑒𝑛 𝑢 = 0 and when 𝑥 = 𝑥𝑛 then 𝑢 = 𝑛.
Thus,
𝐼 = ∫ [𝑦0
+ 𝑢∆ 𝑦0
+
𝑢2−𝑢
2!
∆2
𝑦0
+
𝑢3−3𝑢2+2𝑢
3!
∆3
𝑦0
+ .....] ℎ𝑑𝑢
𝑛
0
= ℎ[𝑦0[𝑢]0
𝑛
+
∆2 𝑦0
2!
[
𝑢3
3
−
𝑢2
2
]
0
𝑛
+
∆3 𝑦0
3!
[
𝑢4
4
− 𝑢3
+ 𝑢2
]
0
𝑛
+ ⋯ ]
= 𝑛ℎ[
𝑛
2
∆𝑦0 +
2𝑛2−3𝑛
12
∆2
𝑦0 +
𝑛3−4𝑛2+4𝑛
24
∆3
𝑦0 + ⋯ ] . . . . . ( 1 )
From this formula, one can generate different integration formulae by substituting
n = 1,2,3, . . . .
1. Trapezoidal Rule
Substituting n = 1 in equation ( 1). In this case all differences higher than the first difference
become zero. Then,
∫ 𝑓(𝑥)𝑑𝑥 = ℎ [ 𝑦0 +
1
2
∆𝑦0] = ℎ [𝑦0 +
1
2
(𝑦1 − 𝑦0)] =
ℎ
2
𝑥𝑛
𝑥0
(𝑦0 + 𝑦1) . . . (2)
The formula is known as the trapezoidal rule.

In this formula, the interval [𝑎, 𝑏] is considered as a single interval, and it gives a very rough
answer. But, if the interval [𝑎, 𝑏] is divided into several subintervals and this formula is
applied to each of these subintervals then a better approximate result may be obtained.
Error in trapezoidal rule
The error of trapezoidal rule is
𝐸 = ∫ 𝑓(𝑥)𝑑𝑥 −
𝑏
𝑎
ℎ
2
(𝑦0 + 𝑦1) . . .( 3 )
Let 𝑦 = 𝑓(𝑥) be continuous and processes continuous derivates of all orders. Also,
It is assumed that there exists a function 𝐹(𝑥) such that 𝐹′(𝑥) = 𝑓(𝑥) 𝑖𝑛 [𝑥0, 𝑥1]
Then,
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝐹′(𝑥)𝑑𝑥 = 𝐹(𝑥1) − 𝐹(𝑥0)
𝑥1
𝑥0
𝑏
𝑎
= 𝐹(𝑥0 + ℎ) − 𝐹(𝑥0) = 𝐹(𝑥0) + ℎ𝐹′(𝑥0)
+
ℎ2
2!
𝐹′′(𝑥0) +
ℎ3
3!
𝐹′′′(𝑥0) + ⋯ 𝐹(𝑥0)
= ℎ𝑓(𝑥0) +
ℎ2
2!
𝑓′(𝑥0) +
ℎ3
3!
𝑓′′(𝑥0) + ⋯
= ℎ𝑦0 +
ℎ2
2!
𝑦′0 +
ℎ3
6!
𝑦′′0 + ⋯ . . . ( 4 )
Again,
ℎ
2
(𝑦0 + 𝑦1) =
ℎ
2
[𝑦0 + 𝑦(𝑥0 + ℎ)]
=
ℎ
2
[𝑦0 + 𝑦(𝑦0) + ℎ𝑦′(𝑥0 ) +
ℎ2
2!
𝑦′′(𝑥0 ) + ⋯ ]
=
ℎ
2
[𝑦0 + 𝑦0 + ℎ𝑦′
0
+
ℎ2
2!
𝑦′′
0
+ ⋯ ] . . .( 5 )
Using (4) and (5), equation (3) becomes,
𝐸 =
ℎ
2
[𝑦0 +
ℎ2
2!
𝑦′
0
+
ℎ2
6!
𝑦′′
0
+ ⋯ ] −
ℎ
2
[2𝑦0 + ℎ𝑦′
0
+
ℎ2
2!
𝑦′′
0
+ ⋯ ]
= −
ℎ3
12!
𝑦′′
0
+ ⋯

= −
ℎ3
12!
𝑓′′(𝑥0) + ⋯ ≅ −
ℎ3
12!
𝑓′′(𝜉), ⋯ (𝟔)
Where 𝑎 = 𝑥0 < 𝜉 < 𝑥1 = 𝑏.
Equation (6) gives the error in the interval [𝑥0,𝑥1].
The total error in the composite rule is
𝐸 = −
ℎ3
12
( 𝑦′′
0
+ 𝑦′′
1
+ ⋯ + 𝑦𝑛−1
𝑛
)
If 𝑦′′(𝜉) is the largest among the n quantities 𝑦′′
0
, 𝑦′′
1
, ⋯ , 𝑦𝑛−1
𝑛
then
𝐸 ≤ −
1
12
ℎ3
𝑛𝑦′′(𝜉) = −
(𝑏−𝑎)
12
ℎ2
𝑛𝑦′′(𝜉), as 𝑛ℎ = 𝑏 − 𝑎
Note:
Trapezoidal rule: The error term shows that if the second and higher order derivatives of
𝑓(𝑥) vanish then the trapezoidal rule gives exact result of the integral. This means, the
method gives exact result when 𝑓(𝑥) is linear.
Geometrical Interpretation Of Trapezoidal Rule
In this rule, the curve 𝑦 = 𝑓(𝑥) is replaced by the line joining the points 𝐴(𝑥0 + 𝑦0) and
𝐵(𝑥1 + 𝑦1) Figure(a). Thus the area bounded by the curve 𝑦 = 𝑓(𝑥), the ordinates
𝑥 = 𝑥0, 𝑥 = 𝑥1 and the x axis is then approximately equivalent to the area of the trapezium
(ABCD) bounded by the line AB, 𝑥 = 𝑥0, 𝑥 = 𝑥1 and x axis.
Figure(a): Geometrical Interpretation Of Trapezoidal Rule
Implementation Of Trapezoidal Rule Via Coding :
To The Implement Integral We Used Walfram Mathematica Software.

Questions :- Implement the following integrals using the Trapezoidal Rule
(i) ∫ 1/𝑥𝑑𝑥
3
2
(ii)∫
𝑑𝑥
1+𝑥2
1
0
(iii)∫ 𝑐𝑜𝑠2
𝑥𝑑𝑥
0.25
−0.25
Solution 1.
f[x_] = 1/x;
a = 2;
b = 3;
exactIn1 = Integrate[f[x],{x, a, b}];
approxiIn1 = (b-a) * ((f[a] + f[b] )/2);
error1 = Abs(exactIn1 - apporxiIn1);
print["Exact value of integration is: ", N[exactIn1]]
print["Integration using Trapezoidal rule is: ", N[approxiIn1]]
print["Error in the approximation is: ", N[error1]]
Output:-
print[ Exact value of integration is: , 0.405465 ]
print [ Integration using Trapezoidal Rule is: , 0.416667 ]
print [ Error in the Approximation is: , -0.0112016 Abs ]
Solution 2.
f[x_] = 1/(1+x^2);
a = 0;

b = 1;
output:-
print [ Integration using Trapezoidal Rule is: , 0.75]
print [ Error in the Approximation is: , Abs( 0. 785396 -1. approxiIn1 ) ]
Solution 3.
f[x_] = Cos[x]^2;
a = -0.25;

b = 0.25;
output:-
print [ Error in the Approximation is: , 0. 0203171 Abs ]
1.1. Composite Trapezoidal Rule
Let the interval [𝑎, 𝑏] be divided into n equal subintervals by the points 𝑎 =
𝑥0, 𝑥1, 𝑥2, ⋯ , 𝑥𝑛 = 𝑏, where 𝑥𝑖 = 𝑥0 + 𝑖ℎ, 𝑖 = 1, 2, 3, ⋯ , 𝑛.
Applying the trapezoidal rule to each of the subintervals, one can find the
composite formula as,

∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓′(𝑥)𝑑𝑥 +
𝑥1
𝑥0
𝑏
𝑎
∫ 𝑓(𝑥)𝑑𝑥 + ⋯ + ∫ 𝑓(𝑥)𝑑𝑥
𝑥𝑛
𝑥𝑛−1
𝑥2
𝑥1
≅
ℎ
2
[𝑦0 + 𝑦1] +
ℎ
2
[𝑦1 + 𝑦2] +
ℎ
2
[𝑦2 + 𝑦3] + ⋯ +
ℎ
2
[𝑦𝑛−1 + 𝑦𝑛]
=
ℎ
2
[𝑦0 + 2( 𝑦1 + 𝑦2 + 𝑦3 + ⋯ + 𝑦𝑛−1) + 𝑦𝑛] . . . (7)
The above equation no 7 is called Composite Trapezoidal Rule
Error for Composite Trapezoidal Rule:
𝐸 ≤ −
1
12
ℎ3
𝑛𝑦′′(𝜉) = −
(𝑏−𝑎)
12
ℎ2
𝑛𝑦′′(𝜉), as 𝑛ℎ = 𝑏 − 𝑎
Geometrical Interpretation Of Composite Trapezoidal Rule:
The geometrical significance of composite trapezoidal rule is that the curve 𝑦 = 𝑓(𝑥) is
replaced by n straight lines joining the points (𝑥0, 𝑦0) and (𝑥1, 𝑦1) ; (𝑥1, 𝑦1) 𝑎𝑛𝑑 (𝑥2, 𝑦2);
⋯ , (𝑥𝑛−1, 𝑦𝑛−1) and (𝑥𝑛, 𝑦𝑛). Then the area bounded by the curve 𝑦 = 𝑓(𝑥), the lines 𝑥 =
𝑥0, , 𝑥 = 𝑥𝑛 and the x-axis is then approximately equivalent to the sum of the area of n
trapeziums (Figure b).
Implementation Of Composite Trapezoidal Rule Via Coding
Questions :- Implement the following integrals using the Composite
Trapezoidal Rule the indicated values of n.
(i)∫ 𝑥𝑙𝑜𝑔(𝑥 + 1)𝑑𝑥
0
−0.5
(ii) ∫ 𝑐𝑜𝑠2
𝑥𝑑𝑥
0.25
−0.25
, n=4
Solution 1.

ClearAll[a, b, exactIn, approxiIn, error, f, x, t, n, m]
n = 4;
f[t_] = 𝑥𝑙𝑜𝑔(𝑥 + 1)𝑑𝑥;
a = -0.5;
b = 0;
h = (b - a) / n;
x = Range[a, b, h];
m = Length[x];
exactIn = ∫ 𝑓[𝑡]𝑑𝑡
𝑏
𝑎
approxiIn = (ℎ
2
⁄ ) ∗ ( 𝑓[𝑥[𝑖]] + 𝑓[𝑥[𝑚]] + 2 ∗ ∑ 𝑓[𝑥[𝑖]]
𝑚−1
𝑖=2 )
error = Abs[exactIn - approxiIn]
Print["Exact value of integral is : ", N[exactIn]]
Print["Approximate value of integral using trapezoidal rule is : ", N[approxiIn]]
Print["Error in the approximation is ", N[error]]
Output:
Exact value of integral is : 0.0525698
Approximate value of integral using trapezoidal rule is : 0.0535485
Error in the approximation is 0.000978701

Solution 2.
ClearAll[a, b, exactIn, approxiIn, error, f, x, t, n, m]
n = 4;
f[t_] = Cos[t]^2;
a = -0.25;
b = 0.25;
h = (b - a) / n;
x = Range[a, b, h];
m = Length[x];
𝑏
𝑎
approxiIn = (ℎ
2
⁄ ) ∗ ( 𝑓[𝑥[𝑖]] + 𝑓[𝑥[𝑚]] + 2 ∗ ∑ 𝑓[𝑥[𝑖]]
𝑚−1
𝑖=2 )
Print["Approximate value of integral using trapezoidal rule is : ", N[approxiIn]]
Output:
Exact value of integral is : 0.489713
Approximate value of integral using trapezoidal rule is : 0.488463
Error in the approximation is 0.00124981

2. Simpson rule:
In this formula the interval [𝑎, 𝑏] is divided into two equal subintervals by the points
𝑥0, 𝑥1, 𝑥2, where ℎ =
𝑏−𝑎
2
, 𝑥1 = 𝑥0 + ℎ 𝑎𝑛𝑑 𝑥2 = 𝑥1 + ℎ.
This rule is obtained by putting n =2 in (1). In this case, the third and higher order
differences do not exist.
The equation (1) is simplified as,
∫ 𝑓(𝑥)𝑑𝑥 ≅ 2ℎ [ 𝑦0 + ∆𝑦0 +
1
6
∆2
𝑦0]
𝑥𝑛
𝑥0
= 2ℎ [ 𝑦0 + (𝑦1 − 𝑦0) +
1
6
(𝑦2 − 2𝑦1 + 𝑦0)]
=
ℎ
3
[𝑦0 + 4𝑦1 + 𝑦2] . . . . . (8)
The above rule is known as Simpson rule.
Error in Simpson rule
The error in this formula is
𝐸 = ∫ 𝑓(𝑥)𝑑𝑥 −
𝑥𝑛
𝑥0
ℎ
3
(𝑦0 + 4𝑦1 + 𝑦2)
Let the function f(x) be continuous in [𝑥0, 𝑥2] and possesses continuous derivatives of all
order. Also, let there exists a function F(x) in [𝑥0, 𝑥2], such that F’(x) = f(x).
Then,
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝐹′(𝑥)𝑑𝑥 = 𝐹(𝑥2) − 𝐹(𝑥0)
𝑥2
𝑥0
𝑥2
𝑥0
= 𝐹(𝑥0 + 2ℎ) − 𝐹(𝑥0)
= 𝐹(𝑥0) + 2𝐹′(𝑥0) +
(2ℎ)2
2!
𝐹′′(𝑥0) +
(2ℎ)3
3!
𝐹′′′(𝑥0) +
(2ℎ)4
4!
𝐹𝑖𝑣(𝑥0) +
(2ℎ)5
5!
𝐹𝑣(𝑥0) + ⋯ − 𝐹(𝑥0)
= 2ℎ𝑓(𝑥0) + 2ℎ𝑓′(𝑥0) +
4
3
ℎ3
𝑓′′(𝑥0) +
2
3
ℎ3
𝑓′′′(𝑥0) +
4
15
ℎ3
𝑓𝑖𝑣(𝑥0) + ⋯ . . . . . . . . (9)
Again,
=
ℎ
3
[𝑦0 + 4𝑦1 + 𝑦2] =
ℎ
3
[𝑓(𝑥0) + 4𝑓(𝑥0 + ℎ) + 𝑓(𝑥0 + 2ℎ)]

=
ℎ
3
[ 𝑓(𝑥0) + 4 { 𝑓(𝑥0) + ℎ𝑓′(𝑥0) +
ℎ2
2!
𝑓′′(𝑥0) +
ℎ3
3!
𝑓′′′(𝑥0) +
ℎ4
4!
𝑓𝑖𝑣(𝑥0) + ⋯ } + { 𝑓(𝑥0) + 2ℎ𝑓′(𝑥0) +
(2ℎ)2
2!
𝑓′′(𝑥0) +
(2ℎ)3
3!
𝑓′′′(𝑥0) +
(2ℎ)4
4!
𝑓𝑖𝑣(𝑥0) + ⋯ } ]
= 2ℎ𝑓(𝑥0) + 2ℎ2
𝑓′(𝑥0) +
4
3
ℎ3
𝑓′′(𝑥0) +
2
3
ℎ4
𝑓′′′(𝑥0) +
5
4
ℎ5
𝑓𝑖𝑣(𝑥0) +
⋯ . . . . . .. (10)
Using eq (9) and (10),equation (8) becomes,
𝐸 = (
4
15
−
5
18
) ℎ5
𝑓𝑖𝑣(𝑥0) + ⋯ ≅ −
ℎ5
90
𝑓𝑖𝑣(𝜉), ⋯ (𝟏𝟏)
Where 𝑥0 < 𝜉 < 𝑥2
This is the error in the interval [𝑥0, 𝑥2]
The total error in composite formula is
𝐸 = −
ℎ5
90
{𝑓𝑖𝑣(𝑥0) + 𝑓𝑖𝑣(𝑥2) + ⋯ + 𝑓𝑖𝑣(𝑥𝑛−2)}
= −
ℎ5
90
𝑛
2
𝑓𝑖𝑣(𝜉)
= −
𝑛ℎ5
180
𝑓𝑖𝑣(𝜉),
( 𝑤ℎ𝑒𝑟𝑒 𝑓𝑖𝑣(𝜉) 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑎𝑚𝑜𝑛𝑔 𝑓𝑖𝑣(𝑥0),
𝑓𝑖𝑣(𝑥2), ⋯ , 𝑓𝑖𝑣(𝑥𝑛−2))
= −
(𝑏−𝑎)
180
ℎ4
𝑓𝑖𝑣(𝜉), . . . . . . . . (12)
Geometrical interpretation of Simpson rule:
In Simpson’s 1/3 rule, the curve y = f(x) is replaced by the second degree parabola passing
through the points 𝐴(𝑥0, 𝑦0), 𝐵(𝑥1, 𝑦1) 𝑎𝑛𝑑 𝐶(𝑥2, 𝑦2). Therefore, the area bounded by the
curve 𝑦 = 𝑓(𝑥), the ordinates 𝑥 = 𝑥0, 𝑥 = 𝑥2 𝑎𝑛𝑑 𝑥 − 𝑎𝑥𝑖𝑠, 𝑖. 𝑒, the area of the shaded
region ABCDEA.

Implementation of Simpson rule via Coding:
Questions :- Implement the following integrals using the Simpson’s 1/3 Rule
(i) ∫ 1/𝑥𝑑𝑥
3
2
(ii) ∫
𝑑𝑥
1+𝑥2
1
0
(iii)∫ 𝑐𝑜𝑠2
𝑥𝑑𝑥
0.25
−0.25
Solution 1.
f[x_] = 1/x;
a = 2;
b = 3;
h = (b-a)/2
approxiIn1 = (h/3) * ( [f[a] + 4*f[(a = b) / 2] + f[b]);
Output:-

print [ Error in the Approximation is: , -0.177868 Abs ]
Solution 2.
f[x_] = 1/(1+x^2);
a = 0;
b = 1;
h = (b-a)/2
Output:-

print [ Error in the Approximation is: , 0.00206483Abs ]
Solution 3.
f[x_] = Cos[x]^2;
a = -0.25;
b = 0.25;
h = (b-a)/2
Output:-

print [ Error in the Approximation is: , 0. 00509549 Abs ]
Simpson Composite rule
Let the interval [𝑎, 𝑏] be divided into n (an even number ) equal subintervals by the points
𝑥0, 𝑥1, 𝑥2, ⋯ , 𝑥𝑛, where 𝑥𝑖 = 𝑥0 + 𝑖ℎ, 𝑖 = 1, 2, ⋯ , 𝑛. Then,
∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 +
𝑥1
𝑥0
∫ 𝑓(𝑥)𝑑𝑥 + ⋯ + ∫ 𝑓(𝑥)𝑑𝑥
𝑥𝑛
𝑥𝑛−2
𝑥4
𝑥2
𝑏
𝑎
=
ℎ
3
[𝑦0 + 4𝑦1 + 𝑦2] +
ℎ
3
[𝑦2 + 4𝑦3 + 𝑦4] + ⋯ +
ℎ
3
[𝑦𝑛−2 + 4𝑦𝑛−1 + 𝑦𝑛]
=
ℎ
3
[𝑦0 + 4(𝑦1 + 𝑦3 + ⋯ + 𝑦𝑛−1 ) + 2(𝑦2 + 𝑦4 +
⋯ + 𝑦𝑛−2 ) + 𝑦𝑛 ] . . . . . . . (13)
This formula is known as Simpson composite rule for numerical integration.
Error in Simpson Composite rule
𝑬 = −
(𝑏−𝑎)
180
ℎ4
𝑓𝑖𝑣
(𝜉)

Geometrical interpretation of Simpson Composite rule:
Putting 𝑛 = 2 in (1) above and taking the curve through (𝑥0, 𝑦0), (𝑥1, 𝑦1) 𝑎𝑛𝑑 (𝑥2, 𝑦2) as a
parabola (Figure C), i.e., a polynomial of the second order so that differences of order higher
than the second vanish, we get
(figure C)
Implementation of Simpson Composite rule via Coding
Questions :- Implement the following integrals using the Simpson’s 1/3
Composite Rule the indicated values of n.
(i)∫ 𝑥𝑙𝑜𝑔(𝑥 + 1)𝑑𝑥
0
−0.5
(ii) ∫ 𝑐𝑜𝑠2
𝑥𝑑𝑥
0.25
−0.25
, n=4
Solution 1.
ClearAll[a, b, exactIn, approxiIn, error, f, x, t, n, m, h]
n = 6;
f[t_] = 𝑡 ∗ 𝐿𝑜𝑔[𝑡 + 1];
a = -0.5;
b = 0;
h = (b - a) / (2*n);
x = Range[a, b, h];
m = Length[x];
𝑏
𝑎
approxiIn = (ℎ
3
⁄ ) ∗ ( 𝑓[𝑥[𝑖]] + 4 ∗ ∑ 𝑓[𝑥[2 ∗ 𝑖]]
(𝑚−1)/2
𝑖=1 + 2 ∗ ∑ 𝑓[𝑥[2 ∗
(𝑚−1)/2
𝑖=2
𝑖 − 1]] + 𝑓[𝑥[𝑚]]);
Print["Approximate value of integral using Simpson’s rule is : ", N[approxiIn]]

Output:-
print[ Exact value of integral is: , 0.0525698 ]
print [ Error in the Approximation is: , 0.294004 ]
Solution 2.
ClearAll[a, b, exactIn, approxiIn, error, f, x, t, n, m, h]
n = 4;
f[t_] = Cos[t]^2;
a = -0.25;
b = 0.25;
h = (b - a) / (2*n);
x = Range[a, b, h];
m = Length[x];
𝑏
𝑎
approxiIn = (ℎ
3
⁄ ) ∗ ( 𝑓[𝑥[𝑖]] + 4 ∗ ∑ 𝑓[𝑥[2 ∗ 𝑖]]
(𝑚−1)/2
𝑖=1 + 2 ∗ ∑ 𝑓[𝑥[2 ∗
(𝑚−1)/2
𝑖=2
𝑖 − 1]] + 𝑓[𝑥[𝑚]] );
Print["Approximate value of integral using Simpson’s rule is : ", N[approxiIn]]

Output:-
print[ Exact value of integral is: , 0.489713 ]
print [ Error in the Approximation is: , 0.136148 ]
Reasons For Numerical Integration:
There are several reasons for carrying out numerical integration.
1. The integrand f(x) may be known only at certain points, such as obtained by
sampling. Some embedded system and other computer applications may need
numerical integration for this reason.
2. A formula for the integrand may be known, but it may be difficult or impossible
to find an antiderivative that is an elementary function. An example of such an
integrand is f(x) = exp (−x2), the antiderivative of which (the error function, times
a constant) cannot be written in elementary form.
3. It may be possible to find an antiderivative symbolically, but it may be easier to
compute a numerical approximation than to compute the antiderivative. That
may be the case if the antiderivative is given as an infinite series or product, or
if its evaluation requires a special function that is not available.

Application Of Integration:
It Helps To:
➢ Locate the centroid.
➢ Find the arc length of graph
➢ Find the surface area of solid
➢ Find the volume of a solid figure
➢ Solve for the work done
➢ Solve the moment of inertia
➢ Find the area
It is also used to find:
➢ Water plane area
➢ Sectional area
➢ Submerged volume
Pros:
➢ Possible to integrate any function
➢ Multidimensional integrals are straightforward
➢ Ability to solve integrals along irregular domains in multidimensional spaces
(any shape).
➢ Numerical integration gives you an answer to some problems that analytic
techniques don’t.
Cons:
➢ There is an intrinsic error in calculation
➢ Numerical integrals are, always, computationally expensive.
Summary And References:
In numerical analysis, Numerical Integration constitutes a broad family of algorithms
for calculating the numerical value of a definite integral, and by extension, the term is
also sometimes used to describe the numerical solution of differential equations. This
article focuses on calculation of definite integrals. The term numerical quadrature
(often abbreviated to quadrature) is more or less a synonym for numerical integration,
especially as applied to one-dimensional integrals. Integration has been there since
even before the proper use of calculus. In modern day integration has led to some great
creations including the Petronas towers and the Sydney opera house.

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