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 geometric
insight
– Apply the trapezoidal rule
– 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
Area under the graph!
• Definite integrations always result in the
area under the graph (in x-y plane)
• Are we capable of evaluating an
approximate value for the area?
5. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Example
• To perform the
definite integration of
the function between
(x0 & x1), we may
assume that the area
is equal to that of the
trapezium:
( ) ( )01
01
2
1
0
xx
yy
dxxf
x
x
−
+
≈∫
7. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
The Trapezoidal Rule
( ) ( )
( ) ( )
2
2
12
12
01
01
yy
xx
yy
xxI
+
−+
+
−≈
Integrating from x0 to x2:
( ) ( ) ( ) ( )
2
212112101001 yxxyxxyxxyxx
I
−+−+−+−
≈
8. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
The Trapezoidal Rule
( ) ( ) hxxxx =−=− 1201
If the points are equidistant
2
2110 hyhyhyhy
I
+++
≈
( )210 2
2
yyy
h
I ++≈
9. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Dividing the whole interval into “n”
subintervals
++≈ ∑
−
=
n
n
i
i yyy
h
I
1
1
0 2
2
10. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
The Algorithm
• To integrate f(x) from a to b, determine the
number of intervals “n”
• Calculate the interval length h=(b-a)/n
• Evaluate the function at the points yi=f(xi)
where xi=x0+i*h
• Evaluate the integral by performing the
summation
++≈ ∑
−
=
n
n
i
i yyy
h
I
1
1
0 2
2
12. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Example
• Integrate
• Using the trapezoidal
rule
• Use 2,3,&4 points and
compare the results
∫=
1
0
2
dxxI
13. 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
X Y
0 0
1 1
2 points, 1 interval
14. 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
X Y
0 0
0.5 0.25
1 1
3 points, 2 interval
15. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Solution
• Using 4 points (n=3),
h=(1-0)/(3)=0.333
• Substituting:
( )4321 22
2
333.0
yyyyI +++≈
( ) 3519.01444.0*2111.0*20
2
333.0
=+++≈I
X Y
0 0
0.33 0.111
0.667 0.444
1 1
4 points, 3 interval
17. 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
18. 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 −
−
−
+≈
19. 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
−
−
−
+≈
20. 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!
21. 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:
22. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Simpson’s Rule
Using a parabola to join three
adjacent points!
23. 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 ++−+−+≈
24. 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
++−+
−+≈∫
25. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
After substitutions and
manipulation!
( ) [ ]210 4
3
2
0
yyy
h
dxxf
x
x
++≈∫
26. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Working with three points!
( ) [ ]210 4
3
2
0
yyy
h
dxxf
x
x
++≈∫
27. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
For 4-Intervals
( ) [ ]432210 44
3
4
0
yyyyyy
h
dxxf
x
x
+++++≈∫
28. 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
29. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Example
• Integrate
• Using the Simpson
rule
• Use 3 points
∫=
1
0
2
dxxI
30. 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
31. ENEM602 Spring 2007
Dr. Eng. Mohammad Tawfik
Homework #7
• Chapter 21, p. 610, numbers:
21.5, 21.6, 21.10, 21.11.