Lecture Note of Simpson’s 1/3 Rule

University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)








Simpson’s 1/3 Rule
Undergraduate Level, 3th Stage



Mr. Waleed Ali Tameemi
Engineer/ College of Engineering/ Babylon University
M.Sc. Civil Engineering/ the University of Kansas/ USA



2016-2017
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3.0 – Simpson’s Rule Method
3.1 – Simpson’s 1/3 Rule
The integration of a function can be approximately calculated using the 1/3 rule of Simpson and as follows:
The number of segments has to be even (n=even #),
?_a^b??f(x)dx=?x/3[y_0+y_n+4?_(i=1)^(n-1)??y_(i_odd )+2?_(i=2)^(n-1)?y_(i_even ) ?]?
?x=(b-a)/n
?=|(True Value-Approximate Value)/(True Value)|×100%
where:
?x: segment length,
n: the number of segments (for Simpson’s 1/3 Rule method, n must be an even number),
?: the error value.
x_0=a y_0=f(x_0)
x_1=x_0+?x y_1=f(x_1)
x_n=x_(n-1)+?x=b y_n=f(x_n)

Ex1: Calculate the value of the given integration.
?_8^30??{2000 ln??[140000/(140000-2100x)?]-9.8x}dx?
The number of segments (n) is equal to 4.
Compare your solution with the exact solution (I=11061).

Solution:
Four segments, n=4, even number
a=8 , b=30
?x=(b-a)/n=(30-8)/4=5.5
y_i=f(x_i )=2000 ln??[140000/(140000-2100x)?]-9.8x_i
i x y
0 x_0=a=8 2000×ln??[140000/(140000-2100×(8))?]-9.8×(8)=177.3

1 8+5.5=13.5 2000×ln??[140000/(140000-2100×(13.5))?]-9.8×(13.5)=320.2
2 13.5+5.5=19 2000×ln??[140000/(140000-2100×(19))?]-9.8×(19)=484.7
3 19+5.5=24.5 2000×ln??[140000/(140000-2100×(24.5))?]-9.8×(24.5)=676.1
4 24.4+5.5=30=b 2000×ln??[140000/(140000-2100×(30))?]-9.8×(30)=901.7

I=?_a^b??f(x)dx=?x/3[y_0+y_n+4?_(i=1)^(n-1)??y_(i_odd )+2?_(i=2)^(n-1)?y_(i_even ) ?]?=5.5/3[177.3+901.7+4×(320.2+676.1)+2×(484.75)]=11078.3

Compare with the true value:
?=|(True Value-Approximate Value)/(True Value)|×100%=|(11061-11078.3)/11061|×100%=0.16%

Ex2: Calculate the value of the given integration.
?_0^10??300x/(1+e^x ) dx?
The number of segments (n) is equal to 4.
Compare your solution with the exact solution (I=246.59).

Solution:
Four segments, n=4, even number
a=0 , b=10
?x=(b-a)/n=(10-0)/4=2.5
y_i=f(x_i )=?_0^10??300x/(1+e^x ) dx?
i x y
0 x_0=a=0 (300×(0))/(1+e^((0)) )=0
1 0+2.5=2.5 (300×(2.5))/(1+e^((2.5)) )=56.89
2 2.5+2.5=5 (300×(5))/(1+e^((5)) )=10.04
3 5+2.5=7.5 (300×(7.5))/(1+e^((7.5)) )=1.24
4 7.5+2.5=10=b (300×(10))/(1+e^((10)) )=0.14

I=?_a^b??f(x)dx=?x/3[y_0+y_n+4?_(i=1)^(n-1)??y_(i_odd )+2?_(i=2)^(n-1)?y_(i_even ) ?]?=2.5/3[0+0.14+4×(56.89+1.24)+2×(10.04)]=210.62

Compare with the true value:
?=|(True Value-Approximate Value)/(True Value)|×100%=|(246.59-210.62)/246.59|×100%=14.59%

Ex3: Calculate the value of the given integration.
?_1^5???(1+x^2 ) dx?
The number of segments (n) is equal to 8.

Solution:

Eight segments (n=8, even #)
a=1 , b=5
?x=(b-a)/n=(5-1)/8=0.5
y_i=f(x_i )=?(1+x^2 )
i x y
0 x_0=a=1 ?(1+?(1)?^2 )=1.41
1 1+0.5=1.5 ?(1+?(1.5)?^2 )=1.80
2 1.5+0.5=2 ?(1+?(2)?^2 )=2.24
3 2+0.5=2.5 ?(1+?(2.5)?^2 )=2.69
4 2.5+0.5=3 ?(1+?(3)?^2 )=3.26
5 3+0.5=3.5 ?(1+?(3.5)?^2 )=3.64
6 3.5+0.5=4 ?(1+?(4)?^2 )=4.12
7 4+0.5=4.5 ?(1+?(4.5)?^2 )=4.61
8 4.5+0.5=5=b ?(1+?(5)?^2 )=5.10

I=?_a^b??f(x)dx=?x/3[y_0+y_n+4?_(i=1)^(n-1)??y_(i_odd )+2?_(i=2)^(n-1)?y_(i_even ) ?]?=0.5/3[1.41+5.10+4×(1.80+2.69+3.64+4.61)+2×(2.24+3.26+4.12)]=12.79
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Homework 13
Use the 1/3 Simpson’s Rule method to estimate the value of the following integration.
?_0^4???xe?^2x dx?
The number of segments (n) is equal 8.
Compare your solution with the exact solution (I=5216.92).
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