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Lecture Note of Simpson’s 3/8 Rule

الكلية كلية الهندسة     القسم  الهندسة البيئية     المرحلة 3
أستاذ المادة وليد علي حسن       13/03/2017 07:20:41
University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)








Simpson’s 3/8 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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3.2 – Simpson’s 3/8 Rule
The integration of a function can be approximately calculated using the 3/8 rule of Simpson and as follows:
The number of segments has to be odd (n=odd #),
?_a^b??f(x)dx=3?x/8[y_0+y_n+3?_(i=1)^(n-1)?y_i ]?
?x=(b-a)/n
?=|(True Value-Approximate Value)/(True Value)|×100%
where:
?x: segment length,
n: the number of segments (for Simpson’s 3/8 Rule method, n must be an odd 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 3.
Compare your solution with the exact solution (I=11061).

Solution:
Three segments, n=3, odd number
a=8 , b=30
?x=(b-a)/n=(30-8)/3=7.334
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+7.334=15.334 2000×ln??[140000/(140000-2100×(15.334))?]-9.8×(15.334)=444.36
2 15.334+7.334=22.667 2000×ln??[140000/(140000-2100×(22.667))?]-9.8×(22.667)=752.65
3 22.667+7.334=30=b 2000×ln??[140000/(140000-2100×(30))?]-9.8×(30)=901.7

I=?_a^b??f(x)dx=3?x/8[y_0+y_n+3?_(i=1)^(n-1)?y_i ]?=(3×7.334)/8[177.3+901.7+3×(444.36+752.65)]=12843.8

Compare with the true value:
?=|(True Value-Approximate Value)/(True Value)|×100%=|(11061-12843.8)/11061|×100%=16.1%

Ex2: Calculate the value of the given integration.
?_0^10??300x/(1+e^x ) dx?
The number of segments (n) is equal to 3.
Compare your solution with the exact solution (I=246.59).

Solution:
Three segments, n=3, odd number
a=0 , b=10
?x=(b-a)/n=(10-0)/3=3.334
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+3.334=3.334 (300×(3.334))/(1+e^((3.334)) )=34.45
2 3.334+3.334=6.667 (300×(6.667))/(1+e^((6.667)) )=2.54
3 6.667+3.334=10=b (300×(10))/(1+e^((10)) )=0.14

I=?_a^b??f(x)dx=3?x/8[y_0+y_n+3?_(i=1)^(n-1)?y_i ]?=(3×3.334)/8[0+0.14+3×(34.45+2.54)]=138.92

Compare with the true value:
?=|(True Value-Approximate Value)/(True Value)|×100%=|(246.59-138.92)/246.59|×100%=43.67%

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

Solution:

Five segments (n=5, odd #)
a=1 , b=5
?x=(b-a)/n=(5-1)/5=0.8
y_i=f(x_i )=?(1+x^2 )
i x y
0 x_0=a=1 ?(1+?(1)?^2 )=1.41
1 1+0.8=1.8 ?(1+?(1.8)?^2 )=2.06
2 1.8+0.8=2.6 ?(1+?(2.6)?^2 )=2.79
3 2.6+0.8=3.4 ?(1+?(3.4)?^2 )=3.54
4 3.4+0.8=4.2 ?(1+?(4.2)?^2 )=4.32
5 4.2+0.8=5=b ?(1+?(5)?^2 )=5.10

I=?_a^b??f(x)dx=3?x/8[y_0+y_n+3?_(i=1)^(n-1)?y_i ]?=(3×0.8)/8[1.41+5.10+3×(2.06+2.79+3.54+4.32))]=13.39
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Homework 14
Use the 3/8 Simpson’s Rule method to estimate the value of the following integration.
?_0^4???xe?^2x dx?
The number of segments (n) is equal 7.
Compare your solution with the exact solution (I=5216.92).
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