1 simpson’s 1/3 rd rule of integration. 2 what is integration? integration the process of...
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1
Simpson’s 1/3rd Rule of Integration
2
What is Integration?Integration
b
a
dx)x(fI
The process of measuring the area under a curve.
Where:
f(x) is the integrand
a= lower limit of integration
b= upper limit of integration
f(x)
a b
y
x
b
a
dx)x(f
3
Simpson’s 1/3rd Rule
4
Basis of 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.
Hence
b
a
b
a
dx)x(fdx)x(fI 2
Where is a second order polynomial.
)x(f2
22102 xaxaa)x(f
5
Basis of Simpson’s 1/3rd Rule
Choose
)),a(f,a( ,ba
f,ba
22))b(f,b(an
das the three points of the function to evaluate a0, a1 and a2.
22102 aaaaa)a(f)a(f
2
2102 2222
ba
aba
aaba
fba
f
22102 babaa)b(f)b(f
6
Basis of Simpson’s 1/3rd Rule
Solving the previous equations for a0, a1 and a2 give
22
22
02
24
baba
)a(fb)a(abfba
abf)b(abf)b(faa
2212
2433
24
baba
)b(bfba
bf)a(bf)b(afba
af)a(afa
2222
222
baba
)b(fba
f)a(f
a
7
Basis of Simpson’s 1/3rd Rule
Then
b
a
dx)x(fI 2
b
a
dxxaxaa 2210
b
a
xa
xaxa
32
3
2
2
10
32
33
2
22
10
aba
aba)ab(a
8
Basis of Simpson’s 1/3rd Rule
Substituting values of a0, a1, a 2 give
)b(fba
f)a(fab
dx)x(fb
a 24
62
Since for Simpson’s 1/3rd Rule, the interval [a, b] is brokeninto 2 segments, the segment width
2
abh
9
Basis of Simpson’s 1/3rd Rule
)b(fba
f)a(fh
dx)x(fb
a 24
32
Hence
Because the above form has 1/3 in its formula, it is called Simpson’s 1/3rd Rule.
10
Example 1
a) Use Simpson’s 1/3rd Rule to find the approximate value of x
The distance covered by a rocket from t=8 to t=30 is given by
30
8
892100140000
1400002000 dtt.
tlnx
b) Find the true error, tE
c) Find the absolute relative true error, t
11
Solution
a)
30
8
)( dttfx
)b(fba
f)a(fab
x2
46
)(f)(f)(f 3019486
830
67409017455484426671776
22.).(.
m.7211065
12
Solution (cont)
b) The exact value of the above integral is
30
8
892100140000
1400002000 dtt.
tlnx
m.3411061
True Error
72110653411061 ..Et
m.384
13
Solution (cont)a)c) Absolute relative true error,
%.
..t 100
3411061
72110653411061
%.03960
14
Multiple Segment Simpson’s 1/3rd Rule
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Multiple Segment Simpson’s 1/3rd Rule
Just like in multiple segment Trapezoidal Rule, one can subdivide the interval
[a, b] into n segments and apply Simpson’s 1/3rd Rule repeatedly overevery two segments. Note that n needs to be even. Divide interval[a, b] into equal segments, hence the segment width
n
abh
nx
x
b
a
dx)x(fdx)x(f0
where
ax 0 bxn
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Multiple Segment Simpson’s 1/3rd Rule
Apply Simpson’s 1/3rd Rule over each interval,
...)x(f)x(f)x(f
)xx(dx)x(fb
a
64 210
02
...)x(f)x(f)x(f
)xx(
64 432
24
f(x)
. . .
x0 x2 xn-
2
xn
x
.....dx)x(fdx)x(fdx)x(fx
x
x
x
b
a
4
2
2
0
n
n
n
n
x
x
x
x
dx)x(fdx)x(f....2
2
4
17
Multiple Segment Simpson’s 1/3rd Rule
...)x(f)x(f)x(f
)xx(... nnnnn
64 234
42
64 12
2
)x(f)x(f)x(f)xx( nnn
nn
Since
hxx ii 22 n...,,,i 42
18
Multiple Segment Simpson’s 1/3rd Rule
Then
...)x(f)x(f)x(f
hdx)x(fb
a
64
2 210
...)x(f)x(f)x(f
h
64
2 432
...)x(f)x(f)x(f
h nnn
64
2 234
64
2 12 )x(f)x(f)x(fh nnn
19
Multiple Segment Simpson’s 1/3rd Rule
b
a
dx)x(f ...)x(f...)x(f)x(f)x(fh
n 1310 43
)}]()(...)()(2... 242 nn xfxfxfxf
)()(2)(4)(3
2
2
1
10 n
n
evenii
i
n
oddii
i xfxfxfxfh
)()(2)(4)(3
2
2
1
10 n
n
evenii
i
n
oddii
i xfxfxfxfn
ab
a) Use four segment Simpson’s 1/3rd Rule to find the approximate value of x.
b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).
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Example 2Use 4-segment Simpson’s 1/3rd Rule to approximate the distance
tE
covered by a rocket from t= 8 to t=30 as given by
30
8
8.92100140000
140000ln2000 dtt
tx
a
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SolutionUsing n segment Simpson’s 1/3rd Rule,
4
830 h 5.5
So )8()( 0 ftf
)5.58()( 1 ftf )5.13(f
a)
)5.55.13()( 2 ftf )19(f
)5.519()( 3 ftf )5.24(f
)( 4tf )30(f
22
Solution (cont.)
)()(2)(4)(3
2
2
1
10 n
n
evenii
i
n
oddii
i tftftftfn
abx
2
2
3
1)30()(2)(4)8(
)4(3
830
evenii
i
oddii
i ftftff
)30()(2)(4)(4)8(12
22231 ftftftff
23
Solution (cont.)
)30()19(2)5.24(4)5.13(4)8(6
11fffff
6740.901)7455.484(2)0501.676(4)2469.320(42667.1776
11
m64.11061
cont.
24
Solution (cont.)
In this case, the true error is
64.1106134.11061 tE
b)
m30.0
The absolute relative true error
%10034.11061
64.1106134.11061
t
c)
%0027.0
25
Solution (cont.)
Table 1: Values of Simpson’s 1/3rd Rule for Example 2 with multiple segments
n Approximate Value
Et |Єt |
2468
10
11065.7211061.6411061.4011061.3511061.34
4.380.300.060.010.00
0.0396%0.0027%0.0005%0.0001%0.0000%
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Error in the Multiple Segment Simpson’s 1/3rd Rule
The true error in a single application of Simpson’s 1/3rd Rule is given as
bafab
Et
),(2880
)( )4(5
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’s1/3rd Rule is given by
)(f)xx(
E )(1
45
021 2880
),(fh )(
14
5
90
)(f)xx(
E )(2
45
242 2880
),(fh )(
24
5
90
210 xx
422 xx
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Error in the Multiple Segment Simpson’s 1/3rd Rule
)(f)xx(
E i)()i(i
i
45
122
2880),(f
hi
)( 45
90
12
45
42
12 2880 n
)(nnn f
)xx(E ,f
hn
)(
12
45
90
nnn xx, 2
2
2
)4(5
90 nfh
.
.
.
ii)i( xx 212
21
2
4
nnn xx
2
45
2
22880 n
nnn f
)xx(E
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Error in the Multiple Segment Simpson’s 1/3rd Rule
Hence, the total error in Multiple Segment Simpson’s 1/3rd Rule is
2
1
n
iit EE
2
1
)4(5
)(90
n
iif
h
2
1
)4(5
5
)(90
)(n
iif
n
ab
n
f
n
ab
n
ii
2
1
)4(
4
5 )(
90
)(
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Error in the Multiple Segment Simpson’s 1/3rd Rule
The term n
f
n
ii
2
1
)4( )(is an approximate average value of
bxaxf ),()4(
Hence )4(
4
5
90
)(f
n
abEt
where
n
ff
n
ii
2
1
)4(
)4()(