1. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Numerical Integration

2. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Objectives
• The student should be able to
– Understand the need for numerical integration
– Derive the trapezoidal rule using linear
interpolation
– Apply the trapezoidal rule
– Derive Simpson’s rule using parabolic
interpolation
– Apply Simpson’s rule

3. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Need for Numerical Integration!
( )
6
11
01
2
1
3
1
23
1
1
0
231
0
2
=−





++=






++=++= ∫ x
xx
dxxxI
( ) 11
0
1
0
1 −−−
−=−== ∫ eedxeI xx
∫
−
=
1
0
2
dxeI x

4. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Interpolation!
• If we have a function that needs to be
integrated between two points
• We may use an approximate form of the
function to integrate!
• Polynomials are always integrable
• Why don’t we use a polynomial to
approximate the function, then evaluate
the integral

5. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Example
• To perform the
definite integration of
the function between
(x0 & x1), we may
interpolate the
function between the
two points as a line.
( ) ( )0
01
01
0 xx
xx
yy
yxf −
−
−
+≈

6. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Example
• Performing the integration on the approximate
function:
( ) ( )∫∫ 





−
−
−
+≈=
1
0
1
0
0
01
01
0
x
x
x
x
dxxx
xx
yy
ydxxfI
1
0
0
2
01
01
0
2
x
x
xx
x
xx
yy
xyI 













−
−
−
+≈

7. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Example
• Performing the integration on the approximate
function:
















−
−
−
+−
















−
−
−
+≈ 00
2
0
01
01
0010
2
1
01
01
10
22
xx
x
xx
yy
xyxx
x
xx
yy
xyI
( ) ( )
2
01
01
yy
xxI
+
−≈
• Which is equivalent to the area of the trapezium!

8. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
The Trapezoidal Rule
( ) ( )
2
01
01
yy
xxI
+
−≈
( ) ( )
( ) ( )
2
2
12
12
01
01
yy
xx
yy
xxI
+
−+
+
−≈
Integrating from x0 to x2:

9. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
General Trapezoidal Rule
• For all the points equally separated
(xi+1-xi=h)
• We may write the equation of the previous slide:
( ) ( ) ( ) ( )
( )321
23
23
12
12
2
2
22
yyy
h
yy
xx
yy
xxI
++=
+
−+
+
−≈

10. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
In general






++≈ ∑
−
=
n
n
i
i yyy
h
I
1
1
0 2
2
Where n is the number if intervals and h=total interval/n

11. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Example
• Integrate
• Using the trapezoidal
rule
• Use 2 points and
compare with the
result using 3 points
∫=
1
0
2
dxxI

12. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Solution
• Using 2 points (n=1),
h=(1-0)/(1)=1
• Substituting:
( )21
2
1
yyI +≈
( ) 5.010
2
1
=+≈I

13. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Solution
• Using 3 points (n=2),
h=(1-0)/(2)=0.5
• Substituting:
( )321 2
2
5.0
yyyI ++≈
( ) 375.0125.0*20
2
5.0
=++≈I

14. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Quadratic Interpolation
• If we get to interpolate a quadratic equation
between every neighboring 3 points, we may use
Newton’s interpolation formula:
( ) ( ) ( )( )103021 xxxxbxxbbxf −−+−+≈
( ) ( ) ( )( )1010
2
3021 xxxxxxbxxbbxf ++−+−+≈

15. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Integrating
( ) ( ) ( )( )1010
2
3021 xxxxxxbxxbbxf ++−+−+≈
( ) ( ) ( )( )∫∫ ++−+−+≈
2
0
2
0
1010
2
3021
x
x
x
x
dxxxxxxxbxxbbdxxf
( ) ( )
2
0
2
0
10
2
10
3
30
2
21
232
x
x
x
x
xxx
x
xx
x
bxx
x
bxbdxxf 











++−+





−+≈∫

16. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
After substitutions and
manipulation!
( ) [ ]210 4
3
2
0
yyy
h
dxxf
x
x
++≈∫

17. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
For 4-Intervals
( ) [ ]23210 424
3
4
0
yyyyy
h
dxxf
x
x
++++≈∫

18. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
In General: Simpson’s Rule
( ) 





+++≈ ∑∑∫
−
=
−
=
n
n
i
i
n
i
i
x
x
yyyy
h
dxxf
n 2
,..4,2
1
,..3,1
0 24
30
NOTE: the number of intervals HAS TO BE even

19. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Example
• Integrate
• Using the Simpson
rule
• Use 3 points
∫=
1
0
2
dxxI

20. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Solution
• Using 3 points (n=2),
h=(1-0)/(2)=0.5
• Substituting:
• Which is the exact
solution!
( )210 4
3
5.0
yyyI ++≈
( )
3
1
125.0*40
3
5.0
=++≈I

21. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Homework #7
• Chapter 21, pp. 610-612, numbers:
21.1, 21.3, 21.5, 21.25, 21.28.
• Due date: Week 8-12 May 2005