Simpson's Rule is a numerical integration technique used to approximate the definite integral of a function. It involves dividing the interval of integration into equal subintervals and taking weighted averages of the function at the left and right endpoints and midpoint of each subinterval. The weights are 1/3 for the endpoints and 4/3 for the midpoints. An example calculates the integral from 0 to 1 of e^-x^2 using Simpson's Rule with 10 intervals, obtaining an approximate value of 0.746825. The document outlines the mathematical formulation and algorithm for Simpson's Rule and its application to numerical integration problems.

1. Simpson’s Rule of Integration
Dr. Varun Kumar
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 1 / 9

2. Outlines
1 Introduction to Simpson’s Rule
2 Mathematical Formulation
3 Example
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 2 / 9

3. Introduction to Simpson’s Rule
Important points
⇒ Numerical integration means the numerical evaluation of integrals
J =
Z b
a
f (x)dx = F(b) − F(a), where F0
(x) = f (x) (1)
a and b are given.
f (x) is a function given analytically.
J is the area under curve of f (x) between a and b.
⇒ As per Simpson’s rule
We divide the interval of integration a 5 x 5 b into an even number of
interval.
Ex- n = 2m or sub-interval length h = b−a
n
Interval points are x0 = a, x1, x2, ......, x2m−1, x2m = b.
Including end points, total 2m + 1 points are taken for consideration.
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 3 / 9

4. Continued–
⇒ We take and find the two sub-interval and approximate f (x) in the
interval x0 5 x 5 x2 = x0 + 2h.
⇒ Applying the Lagrange’s polynomial p2(x) through (x0, f0), (x1, f1),
(x2, f2)
p2(x) =
(x − x1)(x − x2)
(x0 − x1)(x0 − x2)
f0 +
(x − x0)(x − x2)
(x1 − x0)(x1 − x2)
f1 +
(x − x0)(x − x1)
(x2 − x0)(x2 − x1)
f2 (2)
⇒ The denominator terms are 2h2
, -h2
and 2h2
, respectively.
⇒ Let s = (x−x1)
h
then
p2(x) =
1
2
s(s − 1)f0 − (s + 1)(s − 1)f1 +
1
2
(s + 1)sf2 (3)
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 4 / 9

5. Geometrical Interpretation of Simpson’s Rule
Figure: Simpson’s Rule
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 5 / 9

6. Algorithm of Simpson’s Rule
Algorithm
It computes the integral
R b
a f (x)dx
Steps
⇒ INPUT: a, b, m, f0, f1, ....., f2m
⇒ Output: Approximate value ˜
J of J.
⇒ Compute s0 = f0 + f2m
s1 = f1 + f3 + ... + f2m−1
s2 = f2 + f4 + ... + f2m−2
h = (b−a)
2m
˜
J = h
3 (s0 + 4s1 + 2s2)
⇒ Mathematical expression of numerical integration using Simpson’s
Rule
Z b
a
f (x)dx ≈
h
3
f0+4f1+2f2+4f3+...+2f2m−2+42f2m−1+f2m
(4)
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 6 / 9

7. Example–
Example
Q Evaluate the following definite integral J with 2m = 10, where
J =
Z 1
0
e−x2
dx
⇒ Solve the above integral using Simpson’s rule and estimate the error.
Ans Solution: According to question, a = 0, b = 1, 2m = 10 and
h = 1−0
10 = 0.1
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 7 / 9

8. Continued–
˜
J ≈ h
3
h
1.367879 + 4 × 3.740266 + 2 × 3.037901
i
= 0.746825
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 8 / 9

9. Integration accuracy
Solve the standard integral and check its accuracy using rectangular
and trapezoidal rule
1
R 1
0 x2dx
2
R 3
1
1
1+x dx
3
R π/2
0
cos x
sin5
x
dx
4
R 2
1 x2e2x dx
5
R 11
4
x3
1+x5 dx
6
R 1
0 x cos xdx
Note: Students can also observe the effect of n on definite integral
accuracy.
Dr. Varun Kumar (IIIT Surat) Unit 5 / Lecture-2 9 / 9