Inter - Bayamon
Lecture
MECN 3500
1
0
Numerical Methods for Engineering
MECN 3500
Professor: Dr. Omar E. Meza Castillo
omeza@bayamon.inter.edu
http://www.bc.inter.edu/facultad/omeza
Department of Mechanical Engineering
Inter American University of Puerto Rico
Bayamon Campus
Inter - Bayamon
Tentative Lectures Schedule
Topic
Lecture
Mathematical Modeling and Engineering Problem Solving
1
Introduction to Matlab
2
Numerical Error
3
Root Finding
4-5-6
System of Linear Equations
7-8
Least Square Curve Fitting
9
Numerical Integration
10
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Ordinary Differential Equations
2
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Inter - Bayamon
Numerical Integration
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Newton-Cotes Integration Formulas
3
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Course Objectives
 To
solve
numerical
problems
appreciate
their
applications
engineering problem solving.
4
and
for
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Introduction

b
a
f ( x)dx  area under the curve between a and b
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• Indefinite integratio n :
 f ( x)dx
b
• Definite integratio n :
 f ( x)dx
a
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Inter - Bayamon
They are based on the strategy of replacing a complicated
function or tabulated data with an approximating function
that is easier to integrate:
b
b
a
a
I   f ( x)dx   f n ( x)dx
where fn(x) is a polynomial of degree n.
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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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Inter - Bayamon
The Trapezoidal Rule
The integral is approximated by a line:
b
b
a
a
I   f ( x)dx   f1 ( x)dx
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 (b  a )
f (a )  f (b)
2
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Example 21.1
Statement: Use the trapezoidal rule to estimate
 0.2  25x  200 x
0.8
2

 675 x 3  900 x 4  400 x 5 dx
0
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Solution:
f (a)  f (b)
I  (b  a)
2
f (0)  f (0.8)
 (0.8  0)
2
0.2  0.232
 0.8
2
 0.1728
 t  89.5%
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The Multiple-Application Trapezoidal Rule
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.
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The areas of individual segments can then be
added to yield the integral for the entire interval.
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The total integral is
I
x1
x2
xn
x0
x1
xn1
 f ( x)dx   f ( x)dx     f ( x)dx
Substituting the trapezoidal rule for each integral:
I h
f ( x0 )  f ( x1 )
f ( xn 1 )  f ( xn )
f ( x1 )  f ( x2 )
h
 h
2
2
2
Grouping terms:
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n 1
I
h

f
(
x
)

2
f
(
x
)

f
(
x
)

0
i
n   (b  a )

2
i 1

n 1
f ( x0 )  2 f ( xi )  f ( xn )
i 1
2n
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Example 21.1
Statement: Use the multiple-application trapezoidal rule
for n = 2 to estimate
 0.2  25x  200 x
0.8
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Solution:
2

 675 x 3  900 x 4  400 x 5 dx
0
f ( x0 )  2 f ( x1 )  f ( x2 )
I  (b  a)
2(2)
f (0)  2 f (0.4)  f (0.8)
 (0.8  0)
4
0.2  2(2.456)  0.232
 0.8
4
 1.0688
 t  34.9%
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Computer Algorithms for the Trapezoidal Rule
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Simpson’s Rules
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.
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Inter - Bayamon
Simpson’s 1/3 Rule
This rule results when a second-order interpolating
polynomial is used.
b
b
a
a
I   f ( x)dx   f 2 ( x)dx
 Let a  x0 and b  x2 ,
 ( x  x1 )( x  x2 )
( x  x0 )( x  x2 )
I  
f ( x0 ) 
f ( x1 )
( x0  x1 )( x0  x2 )
( x1  x0 )( x1  x2 )
x0 
x2
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

( x  x0 )( x  x1 )
f ( x2 ) dx
( x2  x0 )( x2  x1 )

• After integration,
I
h
 f ( x0 )  4 f ( x1 )  f ( x2 ) where
3
h
ba
2
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Example 21.4
Statement: Single Application of Simpson’s 1/3 Rule
f ( x)  0.2  25x  200 x 2  675x3  900 x 4  400 x 5
From a=0 to b=0.8. recall that the exact integral is
1.640533
Solution:
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0.2  4(2.456)  0.232
I  0.8
 1.367467
6
where Et  1.640533  1.367467  0.2730667,  t  16.6%
Which is approximately 5 times more accurate than for a
single application of the trapezoidal rule (Example 21.1)
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Inter - Bayamon
Simpson’s 8/3 Rule
This rule results
polynomial is used.
when
b
b
a
a
a
third-order
interpolating
I   f ( x)dx   f 3 ( x)dx
 This yields ,
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I
3h
 f ( x0 )  3 f ( x1 )  3 f ( x2 )  f ( x3 ) where
8
h
ba
3
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Example 21.6
Statement: Single Application of Simpson’s 3/8 rule to
integrate
f ( x)  0.2  25x  200 x 2  675x3  900 x 4  400 x 5
From a=0 to b=0.8.
Solution:
Simpson’s rule 3/8 requires four equally spaced points:
f (0)  0.2, f (0.2667)  1.432724, f (0.5333)  3.487177, f (0.8)  0.232
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I  0.8
where
0.2  3(1.432724  3.487177)  0.232
 1.519170
8
Et  1.640533  1.519170  0.1213630,  t  7.4%
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Case Studies
Statement: Estimate the cross section area of the stream.
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• Consider this example
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• Trapezoidal rule (h = 4):
0  2(2  4  4  3.4)  0
I  (20  0)
 53.6 m 2
10
• Trapezoidal rule (h = 2):
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0  2(1.8  2  4  4  6  4  3.6  3.4  2.8)  0
I  (20  0)
 63.2 m 2
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Software
Integration with Matlab
• Use quad for functions.
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Homework8  www.bc.inter.edu/facultad/omeza
Omar E. Meza Castillo Ph.D.
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