Numerical Methods for Eng [ENGR 391]
[Lyes KADEM 2007]
CHAPTER VI
Numerical Integration
Topics
-
Riemann sums
Trapezoidal rule
Simpson’s rule
Richardson’s extrapolation
Gauss quadrature rule
Mathematically, integration is just finding the area under a curve from one point to another. It is
b
represented by
 f ( x)dx , where the symbol 
is an integral sign, the numbers a and b are the
a
lower and upper limits of integration, respectively, the function f is the integrand of the integral, and x
is the variable of integration. Figure 1 represents a graphical demonstration of the concept.
Why are we interested in integration: because most equations in physics are differential equations
that must be integrated to find the solution(s). Furthermore, some physical quantities can be obtained
by integration (example: displacement from velocity).
The problem is that sometimes integrating analytically some functions can easily become laborious.
For this reason, a wide variety of numerical methods have been developed to find the integral.
Figure.6.1- Integration.
I. Riemann Sums
Let f be defined on the closed interval [a, b], and let ∆ be an arbitrary partition of [a, b] such as: a = x0
< x1 < x2 < … < xn-1 <xn = b, where ∆xi is the length of the ith subinterval.
If ci is any point in the ith subinterval, then the sum
n
 f (c )x , x
i 1
i
i
Numerical Integration
i 1
 ci  xi
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∆xi
x1
x2
...
xi-1
xi
…
xn-1
xn
is called a Riemann Sum of the function f for the partition ∆ on the interval [a , b].
For a given partition ∆, the length of the longest subinterval is called the norm of the partition. It is
denoted by ||∆|| (the norm of ∆). The following limit is used to define the definite integral:
n
lim
 0
 f (c )x
i 1
i
i
I
This limit exists if and only if for any positive number ε, there exists a positive number δ such that for
every partition ∆ of [a, b] with ||∆|| < δ, it follows that
n
I   f (ci )xi  
i 1
for any choice of the numbers ci in the ith subinterval of ∆.
If the limit of a Riemann Sum of f exists, then the function f is said to be integrable over [a, b] and that
the Riemann Sums of f on [a, b] approach the number I.
n
lim
 0
 f (c )x
i 1
i
i
I,
b
Where
I   f ( x)dx
a
Example
Find the area of the region between the parabola y = x2 and the x-axis on the interval [0,
4.5]. Use Riemann’s Sum with four partitions.
2. TRAPEZOIDAL RULE
Trapezoidal rule is based on the Newton-Cotes formula that if we approximate the integrand by an nth
order polynomial, then the integral of the function is approximated by the integral of that n th order
polynomial.
 b n1  a n1 
n

, n  0
x
dx

a
 n 1 
b
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So if we want to approximate the integral
b
I   f ( x)dx
a
to find the value of the above integral, we write our function under polynomial form:
f ( x)  f n ( x)
where
f n ( x)  a 0  a1 x  ...  a n 1 x n 1  a n x n
where
f n (x) is an n th order polynomial. Trapezoidal rule assumes n  1, that is, the area under
the linear polynomial (straight line),
b
b
a
a
 f ( x)dx   f ( x)dx
1
2.1. DERIVATION OF THE TRAPEZOIDAL RULE
We have:
b
b
a
a
 f ( x)dx   f ( x)dx
1
b
  (a0  a1 x)dx
a
 b2  a2 
 .
 a0 (b  a)  a1 
 2 
But what is a0 and a1? Now if we choose, (a, f (a )) and (b, f (b)) as the two points to approximate
f (x ) by a straight line from a to b ,
f (a)  f1 (a)  a0  a1a
f (b)  f1 (b)  a0  a1b
Solving the above two equations for
f (b)  f (a )
ba
f (a)b  f (b)a
a0 
ba
a and b ,
a1 
b
Hence we get,

a
f ( x)dx 
f (a)b  f (b)a
f (b)  f (a) b 2  a 2
(b  a) 
ba
ba
2
b
 f ( x)
a
Numerical Integration
 f (a)  f (b) 
dx  (b  a) 

2

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Numerical Methods for Eng [ENGR 391]
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Example
The vertical distance covered by a rocket from t  8 to t  30 seconds is given by
30


140000


x    2000 ln 
 9.8t dt

140000  2100t 

8
a) Use single segment Trapezoidal rule to find the distance covered.
b) Find the true error, Et for part (a).
c) Find the absolute relative true error for part (a).
3.1. Multiple-segment Trapezoidal Rule:
One way to increase the accuracy of the trapezoidal rule is to increase the number of segments
between a and b. So in this procedure, we will divide [ a, b] into n equal segments and apply the
Trapezoidal rule over each segment, the sum of the results obtained for each segment is the
approximate value of the integral.
Divide (b  a ) into
segment is
n equal segments as shown in the figure below. Then the width of each
h
ba
n
The integral  can be broken into h integrals as
b
I   f ( x)dx
a
ah


a  ( n 1) h
a2h
f ( x)dx 
a

ah
f ( x)dx  ... 

f ( x)dx 
a ( n2) h
b
 f ( x)dx
a  ( n 1) h
Figure.6.2- Multiple-segment Trapezoidal rule.
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Applying Trapezoidal rule on each segment gives:
 f (a)  f (a  h) 

2
b
 f ( x)dx  (a  h)  a 
a
 f ( a  h)  f ( a  2 h) 
 (a  2h)  (a  h) 

2

 ………………

b
a
f ( x)dx 

ba 
 n1

f
(
a
)

2
 f (a  ih )  f (b)

2n 
 i 1


Example
The vertical distance covered by a rocket from t  8 to t  30 seconds is given by
30


140000


x    2000 ln 
 9.8t dt

140000  2100t 

8
a) Use two-segment Trapezoidal rule to find the distance covered.
b) Find the true error, Et for part (a).
c) Find the absolute relative true error for part (a).
3.1.1. Why increasing the number of segments
To illustrate the importance of increasing the number of segments in the Trapezoidal rule, let us
consider the following integral:
10
300 x
dx
1 ex
0
I
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The following table represents the variation in the absolute and relative error with the number of
segments used. Note that with a small number of segments, the error is very high.
Approximate Value
Et
t
1
0.681
245.91
99.724%
2
50.535
196.05
79.505%
4
170.61
75.978
30.812%
n
8
227.04
19.546
7.927%
16
241.70
4.887
1.982%
32
245.37
1.222
0.495%
64
246.28
0.305
0.124%
3.1.2. Error in Multiple-segment Trapezoidal Rule
The true error for a single segment Trapezoidal rule is given by
Et 
(b  a) 3
f " ( ), a    b
12
 
where  is some point in a, b .
What is the error, then, in the multiple-segment Trapezoidal rule? It will be simply the sum of the
errors from each segment, where the error in each segment is that of the single segment Trapezoidal
rule. The error in each segment is
E1
3

( a  h)  a 

12
f " ( 1 )
a  1  a  h
h3
f " ( 1 )
12
(a  2h)  (a  h)3 f "( )
E2 
2
12
h3

f " ( 2 )
12

a  h   2  a  2h
.
.
En

3

b  a  (n  1)h

12
f " ( n )
a  (n  1)h   n  b
h3
f " ( n )
12
Hence the total error in multiple-segment Trapezoidal rule is
n
Et   Ei
i 1
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h3 n
 f " ( i )
12 i 1
(b  a) 3 n

 f " ( i )
12n 3 i 1

n
 f " (
(b  a) 3

12n 2
i 1
i
)
n
n
 f " (
i 1
The term
i
n
)
is an approximate average value of the second derivative f " ( x), a  x  b .
Hence
n
(b  a) 3
Et 
12n 2
 f " (
i 1
i
)
n
4. SIMPSON’S 1/3RD RULE
Trapezoidal rule was based on approximating the integrand by a first order polynomial, and then
integrating the polynomial in the interval of integration. Simpson’s 1/3rd rule is an extension of
Trapezoidal rule where the integrand is approximated by a second order polynomial.
We have,
b
b
a
a
I   f ( x)dx   f 2 ( x)dx
where
f 2 ( x) is a second order polynomial.
f 2 ( x)  a0  a1 x  a 2 x 2
 a  b  a  b 

, f
 , and (b, f (b)) as the three points of the function to
 2 
 2
evaluate a 0 , a1 and a 2 .
Choose (a, f (a )),
f (a )  f 2 (a )  a 0  a1 a  a 2 a 2
ab
ab
ab
ab
f
  f2 
  a 0  a1 
  a2 

 2 
 2 
 2 
 2 
f (b)  f 2 (b)  a 0  a1b  a 2 b 2
2
Solving the above three equations for unknowns,
a 0 , a1 and a 2 gives
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ab
2
a 2 f (b)  abf (b)  4abf 
  abf (a)  b f (a )
 2 
a0 
2
a  2ab  b 2
ab
ab
af (a )  4af 
  3af (b)  3bf (a)  4bf 
  bf (b)
2 
2 


a1  
a 2  2ab  b 2


ab
2 f (a)  2 f 
  f (b) 
 2 

a2  
2
2
a  2ab  b
Then
b
I   f 2 ( x)dx
a
b


  a0  a1 x  a2 x 2 dx
a
b

x2
x3 
 a 0 x  a1
 a2 
2
3 a

b2  a2
b3  a3
 a0 (b  a)  a1
 a2
2
3
Substituting values of
b
f
2
( x)dx 
a
a 0 , a1 and a 2 gives

ba 
ab
f (a)  4 f 
  f (b)

6 
 2 

Since for Simpson’s 1/3rd Rule, the interval
h
a, b is broken into 2 segments, the segment width is
ba
2
Hence the Simpson’s 1/3rd rule is given by
b

a b
  f ( b) 
2 

h
 f ( x)dx  3  f (a)  4 f 
a
Since the above form has 1/3 in its formula, it is called Simpson’s 1/3rd Rule.
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Example
t  8 to t  30 is given by


140000


x    2000 ln 
 9.8t dt

140000  2100t 

8
The distance covered by a rocket from
30
a) Use Simpson’s 1/3rd Rule to find the approximate value of x .
b) Find the true error, E t
c) Find the absolute relative true error, t .
4.1. Multiple Segment Simpson’s 1/3rd Rule
 
Just like in multiple-segment Trapezoidal Rule, we can subdivide the interval a, b into n segments
and apply Simpson’s 1/3rd Rule over every two segments. Note that n needs to be even. Divide
interval
a, b into n equal segments, hence the segment width h  b  a .
n
b
xn
a
x0
 f ( x)dx   f ( x)dx
where
x0  a
xn  b
b
x2
x4
xn  2
xn
a
x0
x2
xn  4
xn  2
 f ( x)dx   f ( x)dx   f ( x)dx  ......   f ( x)dx   f ( x)dx
Apply Simpson’s 1/3rd Rule over each interval,
b
 f ( x)dx  ( x
a
2
 f ( x0 )  4 f ( x1 )  f ( x2 ) 
 f ( x 2 )  4 f ( x3 )  f ( x 4 ) 
 x0 ) 
 ( x4  x2 ) 

  ...
6
6



 f ( x n  4 )  4 f ( x n 3 )  f ( x n  2 ) 
 f ( xn 2 )  4 f ( xn 1 )  f ( xn ) 
 ( xn2  xn4 )
 ( xn  xn2 )


6
6




Since
x i  x i  2  2h
i  2, 4, ..., n
then
 f ( x0 )  4 f ( x1 )  f ( x2 ) 
 f ( x 2 )  4 f ( x3 )  f ( x 4 ) 
 2h 

  ...
6
6


b
 f ( x)dx  2h 
a
 f ( x n  4 )  4 f ( x n 3 )  f ( x n  2 ) 
 f ( x n 2 )  4 f ( x n 1 )  f ( xn ) 
 2h 
 2h 


6
6




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h
 f ( x0 )  4 f ( x1 )  f ( x3 )  ...  f ( xn1 )  2 f ( x2 )  f ( x4 )  ...  f ( xn2 )  f ( xn )
3


n 1
n2
h

f ( x 0 )  4  f ( xi )  2  f ( xi )  f ( x n ) 

3
i 1
i 2


i  odd
i  even
b

a


n 1
n 2
ba 
f ( x )dx 
f ( x 0 )  4  f ( xi )  2  f ( xi )  f ( x n ) 


3n
i 1
i 2
i odd
i  even


4.2. Error in Multiple Segment Simpson’s 1/3rd Rule
The true error in a single application of Simpson’s 1/3rd Rule is given by
Et  
(b  a) 5 ( 4)
f ( ), a    b
2880
In Multiple Segment Simpson’s 1/3rd Rule, the error is the sum of the errors in each application of
Simpson’s 1/3rd Rule. The error in n segment Simpson’s 1/3rd Rule is given by
( x 2  x0 ) 5 ( 4)
f ( 1 ), x0   1  x 2
2880
h 5 ( 4)

f ( 1 )
90
( x  x2 ) 5 ( 4)
E2   4
f ( 2 ), x2   2  x4
2880
h 5 ( 4)

f ( 2 )
90
E1  
:
( xn  xn2 ) 5 4  
En  
f   n , x n  2   n  x n
2880
2
2
 2
5
h ( 4)  

f   n 
90
 2
Hence, the total error in Multiple Segment Simpson’s 1/3rd Rule is
n
2
Et   Ei
i 1

5
n
2
h
f ( 4) ( i )

90 i 1
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
(b  a )
90n 5
5
n
2
f
n
2
f
f
5
( 4)
( i )
i 1
n
( i )
( 4)
i 1
The term
( i )
i 1
n
2
(b  a )

90n 4
( 4)
[Lyes KADEM 2007]
is an approximate average value of
n
Et  
f ( 4) ( x), a  x  b . Hence
(b  a ) 5 ( 4 )
f
90n 4
where
n
2
f
( 4)

f
( 4)
i 1
( i )
n
5. Richardson’s Extrapolation Formula for Trapezoidal Rule
The true error in a multiple segment Trapezoidal Rule with n segments for an integral
b
 f x dx
a
is given by
n
Et
f  i 
3

b  a 
i 1

12n 2
where for each i,
i
n
is a point somewhere in the domain
a  i  1h, a  ih, and
n
 f  
the term
i
i 1
n
can be viewed as an approximate average value of
us to say that the true error, Et can be written under the form:
Et  
Numerical Integration
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1
n2
f x  in a, b . This leads
Numerical Methods for Eng [ENGR 391]
Et 
Or
[Lyes KADEM 2007]
C
n2
where
C is an approximate constant of proportionality.
Since, we have
Et  TV  I n
where
TV = true value
I n = approximate value using n-segments.
Then, we can write,
C
 TV  I n
n2
If the number of segments is doubled from n to 2n in the Trapezoidal rule,
C
2n 2
 TV  I 2 n
The above equations can be combined to get:
TV  I 2n 
I 2n  I n
3
6. GAUSS QUADRATURE RULE
6.1. Derivation of two-point Gaussian Quadrature Rule
The two-point Gauss Quadrature Rule is an extension of the Trapezoidal Rule approximation where
the arguments of the function are not predetermined as
a and b , but as unknowns x1 and x 2 . So
in the two-point Gauss Quadrature Rule, the integral is approximated as
b
I   f ( x)dx
a
 c1 f ( x1 )  c2 f ( x2 )
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There are four unknowns
[Lyes KADEM 2007]
x1 , x 2 , c1 and c2 . These are found by assuming that the formula gives
exact results for integrating a general third order polynomial, f ( x)  a 0  a1 x  a 2 x  a3 x .
2
3
Hence
b

b


f ( x)dx   a0  a1 x  a 2 x 2  a3 x 3 dx
a
a

x2
 a 0 x  a1
 a2
2

 b2
 a0 b  a   a1 

b
x3
x4 
 a3 
3
4 a
 b3  a3 
 b4  a4 
 a2 
  a 2 
  a3 

2 
3
4




The formula gives

b
 f ( x)dx  c f ( x )  c
1
1
2
 
f ( x2 )  c1 a0  a1 x1  a2 x1  a3 x1  c2 a0  a1 x2  a2 x2  a3 x2
2
3
2
a
Equating the above equations gives
 b2  a 2 
 b3  a 3 
 b4  a 4 
  a 2 
  a3 

a0 b  a   a1 
 2 
 3 
 4 

 
 c x   a c x
 c1 a0  a1 x1  a 2 x1  a3 x1  c2 a0  a1 x2  a 2 x2  a3 x2
2
 a0 c1  c2   a1 c1 x1
3
2
Since in this equation, the constants
2
2
1 1
2
2

2
3
3

a 0 , a1 , a 2 , and a 3 are arbitrary, the coefficients of a 0 , a1 ,
b  a  c1  c2
b2  a2
 c1 x1  c 2 x 2
2
b3  a3
2
2
 c1 x1  c 2 x 2
3
4
b  a4
3
3
 c1 x1  c2 x2
4
and
ba
2
Numerical Integration

 c2 x2  a3 c1 x1  c2 x2
a2 , and a 3 are equal. This gives us the four following equations:
c1 

3
93
3

Numerical Methods for Eng [ENGR 391]
[Lyes KADEM 2007]
ba
2
 b  a  1  b  a
x1  
 
 
2
3
 2 
 b  a  1  b  a
x2  
 

2
 2  3 
c2 
Hence
ba ba
 f ( x)dx  c f x   c f x   2 f  2  
b
1
1
2
2
a
1  ba ba

 
2 
2
3
ba 1  ba

f 

 
2 
 2  3
6.2. Higher point Gaussian Quadrature Formulas
If we write the integral of the function f(x) under the following form:
b
 f ( x)dx  c
1
f ( x1 )  c2 f ( x2 )  c3 f ( x3 )
a
This is called the three-point Gauss Quadrature Rule. The coefficients
function arguments
c1 , c2 and c3 , and the
x1 , x 2 and x3 are calculated by assuming the formula gives exact expressions
for integrating a fifth order polynomial
 a

b
0
 a1 x  a2 x 2  a3 x 3  a4 x 4  a5 x 5 dx . General n-point rules would approximate the integral
a
b
 f ( x)dx  c
1
f ( x1 )  c2 f ( x2 )  . . . . . . .  cn f ( xn )
a
6.2.1. Arguments and weighing factors for n-point Gauss Quadrature Rules
Usually coefficients and arguments for n-point Gauss Quadrature Rule are tabulated. But, they are
given for integrals of the form
1
n
1
i 1
 g ( x)dx   ci g ( xi )
Numerical Integration
94
Numerical Methods for Eng [ENGR 391]
Table 1: Weighting factors
[Lyes KADEM 2007]
c and function arguments x used in Gauss Quadrature formulas
Points
2
3
4
5
6
Weighting
Function
Factors
Arguments
c1  1.000000000
x1  0.577350269
c2  1.000000000
x2  0.577350269
c1  0.555555556
x1  0.774596669
c2  0.888888889
x2  0.000000000
c3  0.555555556
x3  0.774596669
c1  0.347854845
x1  0.861136312
c2  0.652145155
x2  0.339981044
c3  0.652145155
x3  0.339981044
c4  0.347854845
x4  0.861136312
c1  0.236926885
x1  0.906179846
c2  0.478628670
x2  0.538469310
c3  0.568888889
x3  0.000000000
c4  0.478628670
x4  0.538469310
c5  0.236926885
x5  0.906179846
c1  0.171324492
x1  0.932469514
c2  0.360761573
x2  0.661209386
c3  0.467913935
x3  0.238619186
c4  0.467913935
x4  0.238619186
c5  0.360761573
x5  0.661209386
c6  0.171324492
x6  0.932469514
1
Note: if the table is given for
 g ( x)dx integrals, how can we solve
1
Any integral with limits of
Numerical Integration
b
 f ( x)dx ?
a
a, b can be converted into an integral with limits  1, 1 .
95
Let
Numerical Methods for Eng [ENGR 391]
[Lyes KADEM 2007]
x  mt  c
If x  a , then t  1
If x  b, then t 
1
such that
a  m(1)  c
b  m(1)  c
Solving these two simultaneous linear Equations (21) gives
ba
2
ba
c
2
m
Hence
ba
ba
t
2
2
ba
dx 
dt
2
x
Substituting our values of
x and dx into the integral gives us
b

1
f ( x )dx 
a
baba
ba
x
dx

2
2  2
 f 
1
Example
Use three-point Gauss Quadrature Rule to approximate the distance covered by a rocket from
to
t  30 as given by
30


140000


x    2000 ln 
 9.8t dt

140000  2100t 

8
Also, find the absolute relative true error.
Numerical Integration
96
t 8