4.4: Composite Numerical
Integration
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Composite Numerical Integration: Overcoming Large Intervals with Piecewise & Simpson's, Lecture notes of Calculus
The challenges of using Newton-Cotes methods, specifically Simpson's Rule, for large integration intervals. The text suggests a piecewise approach, dividing the interval into several subintervals and applying Simpson's Rule to each one. The document also introduces the concept of Generalized Simpson's Rule and provides a summary of the method's application.
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4.4: Composite Numerical
Integration
Problem arising from large
integration interval
)]
( ) ( 4 ) ( [ 3 ) (
2
1
0
(^20)
x f x f x f h
dx x
f x x^
≈
Use Simpson’s Rule
76958 .
56
)
4
(
2 3
4
2
0
(^40)
1
=
∫
=
=
e
e
e
dx
e
f
x
Analytical Solution:
f
59819 .
53
0
4
(^40)
2
=
−
=
∫
=
e
e
dx
e
x
1714
3
.
3
|
|
2
1
=
−
=
f
f
error
Solution:
piecewise
technique (divide [0,4] into
several subinterval) e.g. [0,4]= [0,1]+ [1,2]+ [2,3]+ [3,4]and use Simpson’ rule (
i.e. low-order
Newton-Cotes
) in each subinterval.
)]
( ) ( 4 ) ( [ 3 )
(^
2
1
0
(^20)
x f x f x f h
dx x f x x^
≈
∫
2
3 2
1 2
4 0 0
(^10)
2 1
(^32)
(^43)
3
∫
∫
∫
∫
∫ e e e e e e
dx
e
dx
e
dx
e
dx
e
dx
e
f
x
x
x
x
x
01807 .
0
|
|
2
3
=
−
=
f
f
error
Generalized Simpson’s rule using
piecewise
(composite)
method
y
y=f(x)
jh
a
x
n
a
b
h
n
j
j^
=
−
=
−
=
/ )
(
1
) 2 /
,....( (^1) , 0 x
0 x
a
=
1 x
... 2 x
n x
b
=
2 2
− j
x
1 2
− j
x
... 2
j x
Remarks for the method: z
Application at each subinterval. So
n
has to be
even, i.e. total number of points is
n+
.
z
Points at
j=2, 4,…2j
are used twice.
)
(
90
)
(
2 / 1
) 4 (
5
j
n j
f
h
f
E
ζ
∑
−
=
=
Based on
Extreme Value Theorem,
thus
) ( ) ( 2 ) (
) ( 2 ) ( ) ( 2
) ( ) ( ) ( 4
] , [
(^2) /^1
4
4
] , [
4
] , [
(^2) /^1
4
4
] , [
4 ] , [ 4 4 ] , [
max
min
max
min
max
min
x f f n x f
x f n f x f n
x
f
f
x
f
b a x
j
n j
b a x
b a x
j
n j
b a x
b a x j b a x
∈
=
∈
∈
=
∈
∈
∈
≤
≤
≤
≤
≤
≤
ζ ζ
ζ
Or
Use
Intermediate Value Theorem, (if
there exists
a
μ
for which f(
μ
)=K)
), (
)
(
b
f
K
a
f
≤
≤
∑
=
=
2 /
1
4
4
)
(
2
)
(
n j
j
f
n
f
ζ
μ
)
(
2
90
)
(
4
5
μ f n h f E
×
−
=
)
(
180
)
(
4
4
μ
f
h
a
b
−
−
=
n
a
b
h
)
(
−
=
or
Algorithm (for Composite Simpson’s Rule) Input: Output:
)]
( ) ( 4 ) ( 2 ) ( [ 3
2
1
1
2
1 2
1
2
0
n
n j^
j
n j
j^
x f x f x f x f h I
- ∑ + ∑ + =
=
−
− =
n
b
a
,
,
e
SUM
0
SUM
n
a
b
h
)
(
−
=
Step 1: set
0
=
e
SUM
Step 2:
initialize
0
0
=
SUM
Piecewise
for Trapezoida (2points,
)]
( ) ( [ 2 ) (
1
0
(^20)
x f
x f
h
dx x f x x^
- ≈ ∫ 0 x a
=
1
x
1 − j
x
j
x
n x
b
=
[
]
n
a
b
h
jh
a
xj
n
j
b
a
−
=
= ∈ =
,... 1 ,
0
,
μ
can be even or odd (using 2 points instead of 3 points like
Simpson’s rule which requires
n
to be even or total
data points are odd )
n
2
1 1
μ f h a b b f x f a f h
dx
x
f
b a
n j
j^
− =
[ To be derived by students, including Algorithm]•
Piecewise
for
open
Newton-cotes formula:
Composite Midpoint Rule
∫
−
−
∫
∑
=
−
=
2 1
2
0
2
3
)
)...( 1
(
) 2
(
) ( ) ( ) (
n
b a
n i
n
n
i
i^
dt
n
t
t
t
n
f
h
x
f
a
dx
x
f
ζ
,
1
a
x
=
−
,
1
b
x
n
=
2
)
(
−^ +
=
n
a
b
h
) ( 3 ) ( 2 ) (
3
1
0
f
h
x
hf
dx
x
f
x x^
′ ′
∫
=
−
1
1
x
x
<
<
−
ζ
Midpoint rule.
For
open Newton-cote is
n
if
n
is even
The generalized piecewise formula can be proved
(by student)
as
) ( 6 ) ( 2 ) (
2
2 /
0
2
μ f h a b x f h
dx
x
f
b a
n j
j^
′ ′
−
∫
∑
=
=
,
2
)
(
− +
=
n
a
b
h
where
,
) 1
(
h
j
a
x
j^
=
)
,
(
b a ∈ μ , 1
,......
0 ,
1
−
=
n
j Remarks: 1: Required
n
must be even, or total number of data point
(node) must be odd in order to use Midpoint (e.g. at least
n
=0,
x
-
=a,
x
1
=b and x
0
=x
0
.)
2: Since only a single point is used among three points at eachsubinterval in the
open integrals
, only points at
2j
are used.
3: Total subintervals are
n+
over which only n/2 points are used.
•Example: Use both Composite Simpson’s rule
and Composite Trapezoidal rule to
approximate
with an absolute
error less than 0.00002.
∫
π 0
sin
xdx
Conclusion: (1) with piecewise approach, computation
increases, but not round–off error.
(2) error doesn’t depend on
h