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LectureLecture
1100
Numerical Methods for EngineeringNumerical Methods for Engineering MECN 3500 MECN 3500
Professor: Dr. Omar E. Meza CastilloProfessor: Dr. Omar E. Meza Castilloomeza@bayamon.inter.edu
http://www.bc.inter.edu/facultad/omeza
Department of Mechanical EngineeringDepartment of Mechanical Engineering
Inter American University of Puerto RicoInter American University of Puerto Rico
Bayamon CampusBayamon Campus
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Tentative Lectures ScheduleTentative Lectures Schedule
TopicTopic LectureLecture
Mathematical Modeling and Engineering Problem SolvingMathematical Modeling and Engineering Problem Solving 11
Introduction to MatlabIntroduction to Matlab 22
Numerical ErrorNumerical Error 33
Root FindingRoot Finding 4-5-64-5-6
System of Linear EquationsSystem of Linear Equations 7-87-8
Least Square Curve FittingLeast Square Curve Fitting 99
Numerical IntegrationNumerical Integration 1010
Ordinary Differential Equations Ordinary Differential Equations
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Newton-Cotes Integration FormulasNewton-Cotes Integration Formulas
Numerical IntegrationNumerical Integration
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To solve numerical problems and To solve numerical problems and appreciate their applications for appreciate their applications for engineering problem solving.engineering problem solving.
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Course ObjectivesCourse Objectives
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badxxfb
a and between curve under the area )(
b
a
dxxf
dxxf
)( :nintegratio Definite •
)( :nintegratio Indefinite •
IntroductionIntroduction
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They are based on the strategy of replacing a complicated function or tabulated data with an approximating function that is easier to integrate:
b
a n
b
adxxfdxxfI )()(
where fn(x) is a polynomial of degree n.
f1(x)f2(x)
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Piecewise functions can be used also to approximate the integral.
3 piecewise linear functions to approximate f(x) between a and b.
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Two forms of the Newton-Cotes formulas:
Closed Forms: the data points at the beginning and end of the limits of integration are known.
Open Forms: integration limits extend beyond the range of the data.
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2
)()()(
)()( 1
bfafab
dxxfdxxfIb
a
b
a
The integral is approximated by a line:
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Statement: Use the trapezoidal rule to estimate
Example 21.1 Example 21.1
8.0
0
5432 400900675200252.0 dxxxxxx
%5.89
1728.02
232.02.08.0
2
)8.0()0()08.0(
2
)()()(
t
ff
bfafabI
Solution:
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One way to improve the accuracy of the trapezoidal rule is to divide the integration interval from a to b into a number of segments and apply the method to each segment.
The areas of individual segments can then be added to yield the integral for the entire interval.
The Multiple-Application Trapezoidal RuleThe Multiple-Application Trapezoidal Rule
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Lecture 10Lecture 10MEC
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n
n
x
x
x
x
x
x
dxxfdxxfdxxfI1
2
1
1
0
)()()(
2
)()(
2
)()(
2
)()( 12110 nn xfxfh
xfxfh
xfxfhI
n
xfxfxfabxfxfxf
hI
n
n
ii
n
n
ii 2
)()(2)()()()(2)(
2
1
101
10
The total integral is
Substituting the trapezoidal rule for each integral:
Grouping terms:
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Statement: Use the multiple-application trapezoidal rule for n = 2 to estimate
Example 21.1 Example 21.1
8.0
0
5432 400900675200252.0 dxxxxxx
%9.34
0688.14
232.0)456.2(22.08.0
4
)8.0()4.0(2)0()08.0(
)2(2
)()(2)()( 210
t
fff
xfxfxfabI
Solution:
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on Computer Algorithms for the Trapezoidal RuleComputer Algorithms for the Trapezoidal Rule
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More accurate estimate of an integral is obtained if a high-order polynomial is used to connect the points.
The formulas that result from taking the integrals under such polynomials are called Simpson’s rules.
Simpson’s RulesSimpson’s Rules
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This rule results when a second-order interpolating polynomial is used.
2
where)()(4)(3
)())((
))((
)())((
))(()(
))((
))((
,and Let
)()(
210
21202
10
12101
200
2010
21
20
2
2
0
abhxfxfxf
hI
dxxfxxxx
xxxx
xfxxxx
xxxxxf
xxxx
xxxxI
xbxa
dxxfdxxfI
x
x
b
a
b
a
•After integration,
Simpson’s 1/3 RuleSimpson’s 1/3 Rule
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Statement: Single Application of Simpson’s 1/3 Rule
5432 400900675200252.0)( xxxxxxf
From a=0 to b=0.8. recall that the exact integral is 1.640533
%6.16,2730667.0367467.1640533.1where
367467.16
232.0)456.2(42.08.0
ttE
I
Which is approximately 5 times more accurate than for a single application of the trapezoidal rule (Example 21.1)
Example 21.4 Example 21.4
Solution:
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This rule results when a third-order interpolating polynomial is used.
3
where)()(3)(3)(8
3
,yields This
)()(
3210
3
abhxfxfxfxf
hI
dxxfdxxfIb
a
b
a
Simpson’s 8/3 RuleSimpson’s 8/3 Rule
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Statement: Single Application of Simpson’s 3/8 rule to integrate
5432 400900675200252.0)( xxxxxxf
From a=0 to b=0.8.
Simpson’s rule 3/8 requires four equally spaced points:
232.0)8.0(,487177.3)5333.0(,432724.1)2667.0(,2.0)0( ffff
%4.7,1213630.0519170.1640533.1where
519170.18
232.0)487177.3432724.1(32.08.0
ttE
I
Example 21.6 Example 21.6
Solution:
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Statement: Estimate the cross section area of the stream.
Case StudiesCase Studies
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•Consider this example
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•Trapezoidal rule (h = 4):
•Trapezoidal rule (h = 2):
2
2
m 2.6320
0)8.24.36.3464428.1(20)020(
m 6.5310
0)4.3442(20)020(
I
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Integration with Matlab
•Use quad for functions.
SoftwareSoftware
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Homework8 Homework8 www.bc.inter.edu/facultad/omeza
Omar E. Meza Castillo Ph.D.Omar E. Meza Castillo Ph.D.
3030
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