Transcript Numerical integration

Department of Engineering Mathematics and
Physics, Faculty of Engineering
Zagazig University
2009-2010
Numerical solution of linear equations.
Numerical solution of non linear equations.
Numerical integrations.
Curve fitting.
Numerical solution of ordinary differential equations.
Numerical solution of partial differential equations.
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Numerical integration
It is the evaluation of a definite integral
y
y = f(x)
b
I   f(x) dx .
Area
a
By using a numerical method of
0
approximate integration.
a
b
Where f(x)
it is complicated to integrate analytically.
Or it is given by a table.
Or the integral cannot be represented in terms of
finitely elementary functions.
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x
y
y = f(x)
0
b
a
I   f(x) dx 
a
x1 x2
xn-1 b
x
and a = x0, b =xn
h
f(a)  2 f(x 1 )  2 f(x 2 )  ....  2 f(x n 1 )  f(b) .
2
1

i . e . I   f(x) dx  h  [f(a)  f(b)]  f(x 1 )  f(x 2 )  ....  f(x n 1 ).
2

a
b
Where h 
b a
n
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Evaluate
1
I  e
x 2
dx by means of trapezoidal rule with n = 10. For 4D.
0
b a 1
  0.1
n
10
2
Calculate the values of f(x)  e  x
h 
x
0
f(x)
1
1
I   e
0
0.1
0.2
0.9900 0.9608
 x2
dx 
0.3
0.4
0.5
0.6
0.7
0.8
0.9139 0.8521 0.7788 0.6977 0.6126 0.5273
0.9
1.0
0.4449
0.3679
1 1
[ (1  0.3679)  0.9900  0.9608  0.9139  0.8521  0.7788 
10 2
0.6977  0.6126  0.5273  0.4449]  0.7462
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y
y = f(x)
0
b
I   f(x) dx 
a
a x1 x2 x3
x2n-1 b
x
h
f(a)  f(b)  4f(x 1 )  f(x 3 )  ...  f(x 2 n 1 ) 2f(x 2 )  f(x 4 )  ...  f(x 2 n 2 ).
3
Where h 
b a
2n
and a = x0, b = x2n
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Evaluate
π/ 2
I

1  ((sinx) 2 / 4) dx by means of Simpson’s rule with 2n = 6. For 4D.
0
b a π/ 2  0 π
h 


2n
6
12
Calculate the values of f(x)  1  ((sinx) 2 / 4)
π
6
π
4
0.9682
0.9354
0
π
12
f(x)
1
0.9916
 I   f(x) dx 
I 
2
x
b
a
1

3x12
3
4
5
π
3
0.9014
5π
12
0.8756
π
2
0.8660
h
f(a)  f(b)  4f(x 1 )  f(x 3 )  ...  f(x 2 n 1 ) 2f(x 2 )  f(x 4 )  ...  f(x 2 n 2 ).
3
1  0.8660  40.9916  0.9354  0.8756 20.9682  0.9014  1.4674.
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3
 rule
8
b
I   f(x) dx 
a
Where n = 3,
3h
f 0  3 f1  3 f 2  f3 .
8
b a
h
3
b
3h
f 0  5 f1  f 2  6 f3  f 4  5 f5  f 6 .
I   f(x) dx 
10
a
Where n = 6,
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h
b a
6
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Evaluate
0.5
I   e x sin 2 xdx . Using 3  rule .
8
0.2
h 
b a 0.5  0.2

 0.1
3
3
Calculate the values of
I 
f(x)  e x sin 2 x
x
0.2
0.3
0.4
0.5
f(x)
0.4756
0.7622
1.0702
1.3873
3(0.1)
0.4756  3(0.7622  1.0702)  1.3873  0.2760.
8
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Evaluate
0.5
I   e x sin 2 xdx .
Using Weddl’s rule
0.2
b a 0.5  0.2
h 

 0.05
6
6
Calculate the values of f(x)  e x sin 2 x
0
1
2
3
x
0.2
0.25
0.3
0.35
0.4756
f(x)
b
 I   f(x) dx 
a
0.6156
0.7622
0.9142
4
0.4
5
0.45
6
0.5
1.0702
1.2285
1.3873
3h
f 0  5 f1  f 2  6 f3  f 4  5 f5  f 6 .
10
3(0.05)
0.4756  5(0.6156)  0.7622  6(0.9142)  1.0702  5(1.2285)  1.3873  0.2760.
I 
10
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The double integral evaluated as an iterated integral
d b
I    f(x, y) dx dy .
c a
d b
  (  f(x, y) dx) dy .
c
a
b
d
a
c
  (  f(x, y) dy) dx .
We can the do the same, using any integration method.
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Evaluate
2.64.6
I

2
1
4 xy dx dy,
take h = 0.2 in x-direction, k = 0.3 in y-direction. Use
in x- direction and Simpson’s rule in y-direction.
3
 rule
8
Calculate the values of f(x, y)  1
xy
y
x
4
4.2
4.4
4.6
2
0.125
0.1191
0.1136
0.1087
2.3
0.1087
0.1035
0.0988
0.0945
2.6
0.0962
0.0916
0.0874
0.0836
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3
Applying  rule in x-direction
8
For y
=2
3(0.2)
0.125  3(0.1191  0.1136)  0.1087  0.0698.
 I0 
8
For y
=2.3
3(0.2)
0.1087  3(0.1035  0.0988)  0.0945  0.0608
 I1 
8
For y
=2.6 3(0.2)
0.0962  3(0.0916  0.0874)  0.0836  0.0538
 I2 
8
We can put in the
table
2
2.3
2.6
y
f(y)
0.0698
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0.0608
0.0538
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Applying Simpson’s in ydirection
2
y
f(y)
0.0698
2.3
2.6
0.0608
0.0538
k
I  f(a)  f(b)  4f(x 1 )  f(x 3 )  ...  f(x 2 n 1 ) 2f(x 2 )  f(x 4 )  ...  f(x 2 n 2 ).
3
I 
0.3
0.0698  0.0538  4(0.0608)  0.0367.
3
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