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Numerical Integration
1
Introduction to Numerical
Integration



Definitions
Upper and Lower Sums
Trapezoid Method (Newton-Cotes Methods)
2
Integration
Indefinite Integrals
2
x
 x dx  2  c
Indefinite Integrals of a
function are functions
that differ from each
other by a constant.
Definite Integrals
1
2 1
x
0 xdx  2
0
1

2
Definite Integrals are
numbers.
3
Fundamental Theorem of Calculus
If f is continuous on an interval [a,b],
F is antideriva tive of f (i.e., F ' (x)  f(x) )

b
a
f(x)dx  F(b)  F(a)
There is no antideriva tive for : e
b
x2
No closed form solution for :  e dx
x2
a
4
The Area Under the Curve
One interpretation of the definite integral is:
Integral = area under the curve
f(x)
Area  
a
b
a
f(x)dx
b
5
Upper and Lower Sums
The interval is divided into subintervals.
Partition P  a  x0  x1  x2  ...  xn  b
Define
mi  min  f ( x) : xi  x  xi 1
M i  max  f ( x) : xi  x  xi 1 
f(x)
n 1
Lower sum L( f , P)   mi  xi 1  xi 
i 0
n 1
Upper sum U ( f , P)   M i xi 1  xi 
i 0
x0 x1 x2 x3
a
b
6
Upper and Lower Sums
n 1
Lower sum L( f , P )   mi  xi 1  xi 
i 0
n 1
Upper sum U ( f , P )   M i  xi 1  xi 
f(x)
i 0
Estimate of the integral 
Error 
L U
2
U L
2
x0 x1 x2 x3
CISE301_Topic7
a
b
7
Example

1
0
x 2 dx
 1 2 3 
Partition : P  0, , , ,1
 4 4 4 
n  4 (four equal intervals )
1
1
9
m0  0,
m1  , m2  , m3 
16
4
16
1
1
9
M 0  , M1  , M 2  , M 3  1
16
4
16
1
xi 1  xi 
4
for i  0,1, 2, 3
0
1
4
1
2
3
4
1
8
Example
n 1
Lower sum L( f , P )   mi  xi 1  xi 
i 0
L( f , P ) 
1
1 1 9  14
0

  

4  16 4 16  64
n 1
Upper sum U ( f , P )   M i  xi 1  xi 
i 0
U ( f , P) 
11 1 9
 30



1



4 16 4 16  64
1  30 14  11
Estimate of the integral     
2  64 64  32
1  30 14  1
Error     
2  64 64  8
0
1
4
1
2
3
4
1
9
Upper and Lower Sums
• Estimates based on Upper and Lower
Sums are easy to obtain for monotonic
functions (always increasing or always
decreasing).
• For non-monotonic functions, finding
maximum and minimum of the function
can be difficult and other methods can be
more attractive.
10
Newton-Cotes Methods
In Newton-Cote Methods, the function
is approximated by a polynomial of
order n.
 Computing the integral of a polynomial
is easy.


b

b
a
a
f ( x)dx  
b
a
a

n

a
x

...

a
x
dx
0
1
n
(b 2  a 2 )
(b n1  a n1 )
f ( x)dx a0 (b  a)  a1
 ...  an
2
n 1
11
Newton-Cotes Methods

Trapezoid Method (First Order Polynomials are used)
b
a

f ( x)dx  
b
a
a0  a1x dx
Simpson 1/3 Rule (Second Order Polynomials are used)
b
a
f ( x)dx  
b
a
a
0

 a1x  a2 x dx
2
12
Trapezoid Method




Derivation-One Interval
Multiple Application Rule
Estimating the Error
Recursive Trapezoid Method
13
Trapezoid Method
b
f (b)  f (a)
f (a) 
( x  a)
ba
f(x)
I   f ( x)dx
a
f (b)  f (a )


I    f (a) 
( x  a ) dx
a
ba


b
b
f (b)  f (a ) 

  f (a)  a
x
ba

 a
2 b
f (b)  f (a ) x

ba
2
a
b
a
f (b)  f (a )
 b  a 
2
14
Trapezoid Method
Derivation-One Interval
f (b)  f (a )


I   f ( x)dx    f (a ) 
( x  a ) dx
a
a
ba


b
f (b)  f (a ) f (b)  f (a ) 
I    f (a)  a

x dx
a
ba
ba


b
b
b
2 b
f (b)  f (a ) 
f (b)  f (a ) x

  f (a)  a
x 
ba
ba
2

 a
a
f (b)  f (a ) 
f (b)  f (a ) 2

  f (a)  a
(b  a 2 )
b  a  
ba
2(b  a )


f (b)  f (a )
 b  a 
2
15
Trapezoid Method
f(x)
f (b )
f (a )
ba
 f (a)  f (b) 
Area 
2
a
b
16
Trapezoid Method
Multiple Application Rule
The interval
[a, b] is
f(x)
f ( x2 )  f ( x1 )
x2  x1 
Area 
2
partitione d into n segments
a  x0  x1  x2  ...  xn  b

b
a
f ( x)dx  sum of the areas
of the trapezoid s
x
x0
a
x1
x2
x3
b
17
Trapezoid Method
General Formula and Special Case
If the interval is divided into n segments (not necessaril y equal)
a  x0  x1  x2  ...  xn  b

n 1
1
f ( x)dx    xi 1  xi  f ( xi 1 )  f ( xi ) 
i 0 2
b
a
Special Case ( Equaliy spaced base points)
xi 1  xi  h for all i

b
a
n 1
1

f ( x)dx  h   f ( x0 )  f ( xn )   f ( xi ) 
i 1
2

18
Example
Given a tabulated
values of the velocity of
an object.
Time (s)
0.0
1.0
2.0
3.0
Velocity (m/s) 0.0
10
12
14
Obtain an estimate of
the distance traveled in
the interval [0,3].
Distance = integral of the velocity
Distance 

3
0
V (t ) dt
19
Example 1
The interval is divided
Time (s)
0.0
1.0
2.0
3.0
into 3 subinterva ls
Velocity
(m/s)
0.0
10
12
14
Base points are0,1,2,3
Trapezoid Method
h  xi 1  xi  1
1
 n 1

T  h  f ( xi )   f ( x0 )  f ( xn ) 
2
 i 1

1


Distance  1(10  12)  (0  14)  29
2


20
Error in estimating the integral
Theorem
Assumption :
f ' ' ( x) is continuous on [a,b]
Equal intervals (width  h)
Theorem :
approximat e
If Trapezoid Method is used to

b
a
f ( x)dx
then
b  a 2 ''
Error  
h f ( ) where   [a,b]
12
ba 2
Error 
h max f ' ' ( x)
x[ a ,b ]
12
CISE301_Topic7
21
Estimating the Error
For Trapezoid Method
How many equally spaced intervals are
needed to compute


0
sin( x)dx
to 5 decimal digit accuracy ?
22
Example


sin( x )dx,
0
1
find h so that error   105
2
ba 2
Error 
h max f ' ' ( x )
12
x[ a ,b]
b   ; a  0; f ' ( x )  cos( x ); f ' ' ( x )   sin( x )

1
f ' ' ( x )  1  Error  h   105
12
2
2 6
 h   105  h  0.00437
2

(b  a )

 n

 719 intervals
h
0.00437
23
Example
x
1.0
1.5
2.0
2.5
3.0
f(x)
2.1
3.2
3.4
2.8
2.7
Use Trapezoid method to compute :

3
1
f ( x)dx
n 1
1
Trapezoid T ( f , P)    xi 1  xi  f ( xi 1 )  f ( xi ) 
i 0 2
Special Case : h  xi 1  xi for all i,
1
 n 1

T ( f , P)  h  f ( xi )   f ( x0 )  f ( xn ) 
2
 i 1

24
Example
x
1.0
1.5
2.0
2.5
3.0
f(x)
2.1
3.2
3.4
2.8
2.7
3
1
n1

1
f ( x)dx  h  f ( xi )   f ( x0 )  f ( xn ) 
2
 i 1

1


 0.5  3.2  3.4  2.8  2.1  2.7 
2


 5.9
25
Recursive Trapezoid Method
Estimate based on one interval :
f(x)
h ba
ba
 f ( a )  f ( b) 
R (0,0) 
2
a
ah
26
Recursive Trapezoid Method
Estimate based on 2 intervals :
f(x)
h
ba
2
R (1,0) 
ba 
1



f
(
a

h
)

f
(
a
)

f
(
b
)

2 
2
R(1,0) 
1
R(0,0)  h f ( a  h )
2
Based on previous estimate
a
ah
a  2h
Based on new point
27
Recursive Trapezoid Method
f(x)
ba
h
4
R (2,0) 
R(2,0) 
ba
 f (a  h)  f (a  2h)  f (a  3h)
4
1

  f (a )  f (b) 
2

1
R(1,0)  h f (a  h)  f (a  3h)
2
Based on previous estimate
a
a  2h
a  4h
Based on new points
28
Recursive Trapezoid Method
Formulas
ba
 f (a)  f (b)
R (0,0) 
2


1
R (n,0)  R (n  1,0)  h   f a  (2k  1)h 
2
 k 1

2 ( n1)
ba
h n
2
29
Recursive Trapezoid Method
ba
,
2
ba
 f (a)  f (b)
2
1
1

R(1,0)  R(0,0)  h  f a  (2k  1)h 
2
 k 1

ba
h 2 ,
2
1
 2

R(2,0)  R(1,0)  h  f a  (2k  1)h 
2
 k 1

ba
h 3 ,
2
2

1
R(3,0)  R(2,0)  h  f a  (2k  1)h 
2
 k 1

h  b  a,
h
R (0,0) 
2
..................
h
ba
,
n
2
2

1
R(n,0)  R(n  1,0)  h   f a  (2k  1)h 
2
 k 1

( n 1)
30
Example on Recursive Trapezoid
Use Recursive Trapezoid method to estimate :

 /2
sin( x )dx by computing R(3,0) then estimate the error
0
n h
R(n,0)
0
(b-a)=/2
(/4)[sin(0) + sin(/2)]=0.785398
1
(b-a)/2=/4
R(0,0)/2 + (/4) sin(/4) = 0.948059
2
(b-a)/4=/8
R(1,0)/2 + (/8)[sin(/8)+sin(3/8)] = 0.987116
3
(b-a)/8=/16
R(2,0)/2 + (/16)[sin(/16)+sin(3/16)+sin(5/16)+
sin(7/16)] = 0.996785
Estimated Error = |R(3,0) – R(2,0)| = 0.009669
31
Advantages of Recursive Trapezoid
Recursive Trapezoid:
 Gives the same answer as the standard
Trapezoid method.
 Makes use of the available information to
reduce the computation time.
 Useful if the number of iterations is not
known in advance.
32