NM Presentation

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http://numericalmethods.eng.usf.edu

NUMERICAL INTEGRATION:

NEWTON COTE’s FORMULAS

GROUP MEMBERSMuhammad Saad Qureshi

Athar Minallah Faizan ShahidArslan AhmedNajam Tariq

2

http://numericalmethods.eng.usf.edu

LIST OF TOPICS TO BE COVERED

Trapezoidal Rule & its Error Analysis Simpson’s 1/3 Rule & its Error AnalysisSimpson’s 3/8 Rule & its Error Analysis

EXAMPLES !

What is Numerical Integration

b

adx)x(fI

Integration:

3

The process of measuring the area under a function plotted on a graph.

Where: f(x) is the integranda= lower limit of integrationb= upper limit of integration

f(x)

a b

b

a

dx)x(f

y

x

Newton cote’s integration

formula

Trapezoidal Rule

SAAD QURESHI…

Basis of Trapezoidal Rule

b

a

dxxfI )( )x(f)x(f n

Trapezoidal Rule is based on the Newton-Cotes Formula that states if one can approximate the integrand as an nth order polynomial…

http://numericalmethods.eng.

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where

nn

nnn xaxa...xaa)x(f

1110and

Basis of Trapezoidal Rule

b

an

b

a)x(f)x(f

Then the integral of that function is approximated by the integral of that nth

order polynomial.


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Trapezoidal Rule assumes n=1, that is, the area under the linear polynomial,

2)b(f)a(f)ab(

b

adx)x(f

Derivation of the Trapezoidal

Rule

7

Method Derived From

Geometry

8

The area under the curve is a trapezoid. The integral

trapezoidofAreadxxfb

a

)(

))((21 widthAvgheight

)ab()a(f)b(f 21

2)b(f)a(f)ab(

Figure 2: Geometric Representation

f(x)

a b

b

a

dx)x(f1

y

x

f1(x)

Example 1The vertical distance covered by a rocket from t=8 to t=30 seconds is given by:

9

30

8

892100140000

1400002000 dtt.t

lnx

a) Use single segment Trapezoidal rule to find the distance covered.

b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).

tEa

10

Solution

2

)()()( bfafabI

a)8a 30b

t.t

ln)t(f 892100140000

1400002000

)(.)(

ln)(f 88982100140000

14000020008

)(.)(

ln)(f 3089302100140000

140000200030

s/m.27177

s/m.67901

2

6790127177830 ..)(I

11

Solution (cont)

m11868

a)

b)The exact value of the above integral is

30

8

892100140000

1400002000 dtt.t

lnx m11061

12

Solution (cont)b) ValueeApproximatValueTrueEt

1186811061

m807

c) The absolute relative true error, , would be t

10011061

1186811061

t %.29597

13

Multiple Segment Trapezoidal Rule

f(x)

a b

y

x

4aba

42 aba

4

3 aba

Figure 4: Multiple (n=4) Segment Trapezoidal Rule

Divide into equal segments as shown in Figure 4. Then the width of each segment is:

n

abh

The integral I is:

b

adx)x(fI

14

Multiple Segment Trapezoidal Rule

b

h)n(a

h)n(a

h)n(a

ha

ha

ha

a

dx)x(fdx)x(f...dx)x(fdx)x(f1

1

2

2

The integral I can be broken into h integrals as:

b

adx)x(f

Applying Trapezoidal rule on each segment gives:

b

adx)x(f

)b(f)iha(f)a(f

nab n

i

1

12

2

15

Example 2The vertical distance covered by a rocket from to seconds is given by:

30

889

21001400001400002000 dtt.

tlnx

a) Use two-segment Trapezoidal rule to find the distance covered.

b) Find the true error, for part (a).c) Find the absolute relative true error, for part (a).

atE

16

Solutiona) The solution using 2-segment Trapezoidal rule is

)b(f)iha(f)a(f

nabI

n

i

1

12

2

2n 8a 30b

2830

n

abh 11

17

Solution (cont)

)(f)iha(f)(f

)(I

i3028

22830 12

1

)(f)(f)(f 3019284

22

6790175484227177422 .).(.

m11266

Then:

18

Solution (cont)

30

8

892100140000

1400002000 dtt.t

lnx m11061

b) The exact value of the above integral is

so the true error is

ValueeApproximatValueTrueEt

1126611061


formula

Error Analysis of Trapezoidal Rule

ATHAR MINALLAH…

20

The error of the composite trapezoidal rule is

the difference between the value of the integral and the numerical result

The error can be written as

There exixt a number ξ between a and b

http://numericalmethods.eng.usf.e

du

Error analysis of Trapezoidal rule

21

Example on Error Analysis

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formula

Simpson’s 1/3 Rule

FAIZAN SHAHID…

Simpson's rule also corresponds to the three-point Newton

Cote’s Quadrature Rule. The method is credited to the mathematician Thomas

Simpson (1710–1761) of Leicestershire, England. Kepler used similar formulas over 100 years prior. For this

reason the method is sometimes called Kepler's rule. Simpson’s 1/3rd rule is an extension of Trapezoidal rule and a

further improvement. Instead of approximating the curve by a straight line, we

approximate it by a quadratic or cubic function.

Introduction:

Simpson’s 1/3rd rule is an extension of Trapezoidal rule

where the integrand is approximated by a second order polynomial.

Hence we show below the basic mathematical background by showing that,

Where is a second order polynomial.

Mathematical Explanation:

b

a

b

a

dxxfdxxfI )()( 2

22102 xaxaa)x(f

mathematical explanation: (Simple Or Single Segment

Case)Choose

)),a(f,a( ,baf,ba

22))b(f,b(and

as the three points of the function to evaluate a0, a1 and a2. 2

2102 aaaaa)a(f)a(f 2

2102 2222

baabaaabafbaf

22102 babaa)b(f)b(f

Solving the previous equations for a0, a1 and a2 give

22

22

0 22

4

baba

)a(fb)a(abfbaabf)b(abf)b(faa

221 22

4332

4

baba

)b(bfbabf)a(bf)b(afbaaf)a(afa

222 22

22

baba

)b(fbaf)a(fa

mathematical explanation: (Simple Or Single Segment Case)

Then

b

adx)x(fI 2

b

adxxaxaa 2

210

b

a

xaxaxa

32

3

2

2

10

32

33

2

22

10abaaba)ab(a

Substituting values of a0, a1, a 2 give

)b(fbaf)a(fabdx)x(fb

a 24

62

Since for Simpson’s 1/3rd Rule, the interval [a, b] is broken into 2 segments, the segment width

2abh

)b(fbaf)a(fhdx)x(fb

a 24

32

Hence

Consider the definite integral in interval [1,2].

Solve for using Simpson's 1/3 rule for function by using simple or 2 segment case:

SOLUTION: We know that for two segment case, the Simpson’s 1/3 Rule is known as,

Where, it is given:

Example: (Simple Or two Segment

Case)

Now, we calculate as,

So, putting in the values,

Example: (Simple Or two Segment

Case)

Just like in multiple segment Trapezoidal Rule, one can subdivide the interval [a, b] into n segments and apply Simpson’s 1/3rd Rule repeatedly overevery two segments. Note that n needs to be even. Divide interval [a, b] intoequal segments, hence the segment width. Where,

mathematical explanation: (multiple Segment Case)

nabh

nx

x

b

adx)x(fdx)x(f

0

ax 0 bxn

Apply Simpson’s 1/3rd Rule over each interval,

...)x(f)x(f)x(f)xx(dx)x(fb

a

64 210

02

...)x(f)x(f)x(f)xx(

64 432

24

f(x)

. . .

x0 x2 xn

-2

xn

x

.....dx)x(fdx)x(fdx)x(fx

x

x

x

b

a

4

2

2

0

n

n

n

n

x

x

x

xdx)x(fdx)x(f....

2

2

4

mathematical explanation:

(multiple Segment Case)...)x(f)x(f)x(f)xx(... nnn

nn

64 234

42

64 12

2)x(f)x(f)x(f)xx( nnn

nn

Since hxx ii 22 n...,,,i 42

Then

...)x(f)x(f)x(fhdx)x(fb

a

6

42 210

...)x(f)x(f)x(fh

642 432

...)x(f)x(f)x(fh nnn

642 234

642 12 )x(f)x(f)x(fh nnn

mathematical explanation:

(multiple Segment Case)

b

adx)x(f ...)x(f...)x(f)x(f)x(fh

n 1310 43

)}]()(...)()(2... 242 nn xfxfxfxf

)()(2)(4)(3

2

2

1

10 n

n

evenii

i

n

oddii

i xfxfxfxfh

)()(2)(4)(3

2

2

1

10 n

n

evenii

i

n

oddii

i xfxfxfxfnab

Consider the definite integral in interval [1,2].

Solve for using Simpson's 1/3 rule for function by using 6 segments:

SOLUTION: We know that for multiple segment case, the Simpson’s 1/3 Rule is known as,

Where,

example: (multiple Segment Case)

)()(2)(4)(3

2

2

1

10 n

n

evenii

i

n

oddii

i xfxfxfxfh

We know that,

Calculating points and their function values,


POINTS: FUNCTION VALUES:

Putting the values in the function,


In the Trapezoid rule method, we start with

rectangular area-elements and replace their horizontal-line tops with slanted lines. The area-elements used to approximate, say, the area under the graph of a function and above a closed interval then become trapezoids. Simpson’s method replaces the slanted-line tops with parabolas.

Simpson’s rule approximations usually achieve a given level of accuracy faster. Simpson’s rule is indeed much better than the Trapezoid rule. As n → ∞ it generally converges much more rapidly to the value of the definite integral than does the Trapezoid rule.

The main dis-advantage of the Simpson’s rule is its conceptualization and derivation marginally more difficult.

Advantages & Dis-Advantages:


formula

Simpson’s 3/8 Rule

ARSLAN AHMED…

Formula

using matrixes using lagrange interpolation

Not to discuss here

Derivation

composite simpson 3/8 rule for multiple segments


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Example 1 (Single Simpson rule)

Compute

30

8,8.9

2100000,140000,140ln2000

b

adxx

xI

by using a single segment Simpson 3/8 ruleSolution

In this example:3333.7

3830

3

abh

83


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2667.17788.982100140000

140000ln20008 00

xfx

4629.3723333.158.93333.152100140000

140000ln2000

3333.153333.78

1

01

xf

hxx


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8976.6086666.228.96666.222100140000

140000ln2000

6666.22)3333.7(282

2

02

xf

hxx

6740.901308.9302100140000

140000ln2000

30)3333.7(383

3

03

xf

hxx


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Applying Eq. (12), one has:

3104.11063

6740.9018976.60834629.37232667.1773333.783

I

I

The “exact” answer can be computed as

34.11061exactI


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Example 2 (Multiple segments Simpson rule)

Compute

30

8,8.9

2100000,140000,140ln2000

b

adxx

xI

using Simple multiple segments rule, with number (of ) segments = = 6 (which corresponds to 2

“big” segments).""h n

83

83


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SolutionIn this example, one has (see Eq. 14):

6666.36

830

h

2667.177,8, 00 xfx

6666.11

6666.38;4104.270,6666.11, 0111

hxxwherexfx

3333.152;4629.372,3333.15, 0222 hxxwherexfx

193;7455.484,19, 0333 hxxwherexfx


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6666.224;8976.608,6666.22, 0444 hxxwherexfx

3333.265;9870.746,3333.26, 0555 hxxwherexfx

306;6740.901,30, 0666 hxxwherexfx


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Applying Eq. (17), one obtains:

6740.9012332667.1776666.3

83 33

,..6,3

51

,..5,2

42

,..4,1

n

ii

n

ii

n

ii xfxfxfI

6740.9017455.4842

9870.7464629.37238976.6084104.27032667.1773750.1I

4696.601,11I


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ruleSimpson

hxxhxxhxx

hxxax

31

5714.201429.34844286.171429.33832857.141429.3282

1429.111429.3818

04

03

02

01

0

301429.3787

8571.261429.36867143.231429.3585

07

06

05

hxxhxxhxx


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2667.17788.982100000,140

000,140ln200080

xf

Similarly: 6740.901

9978.7678260.6463909.5362749.4353241.342

5863.2561429.11

7

6

5

4

3

2

1

xfxfxfxfxfxfxf


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For multiple segments segmentsfirstn 41 using Simpson rule, one obtains (See Eq. 19):

1197.4364

3909.5363241.34222749.4355863.25642667.1773

1429.3

243

1

1

22

,...2

31

,...3,101 1

11

I

I

xfxfxfxfhI n

n

ii

n

ii

31


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For multiple segmentsusing Simpson 3/8 rule, one obtains (See Eq. 17):

segmentslastn 32

2748.6697

2749.435!3241.34235863.25632667.1771429.383

23383

2

2

03

,...6,3

21

,...2

12

,...3,102 1

222

I

skipI

xfxfxfxfxfhI n

n

ii

n

ii

n

ii

The mixed (combined) Simpson 1/3 and 3/8 rules give:

3946.061,112748.66971197.436421

IIII


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Remarks:(a) Comparing the truncated error of Simpson 1/3 rule

fabEt

2880

5

With Simple 3/8 rule (See Eq. 13), the latter seems to offer slightly more accurate answer than the former. However, the cost associated with Simpson 3/8 rule (using 3rd order polynomial function) is significant

higher than the one associated with Simpson 1/3 rule (using 2nd order polynomial function).

(18)


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(b) The number of multiple segments that can be used in the conjunction with Simpson 1/3 rule is 2,4,6,8,..

(any even numbers).

n

n

ii

n

ii

nnn

xfxfxfxfhI

xfxfxfxfxfxfxfxfxfhI

2

...6,4,2

1

,...3,102

124322101

243

4.....443

(19)


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However, Simpson 3/8 rule can be used with the number of segments equal to 3,6,9,12,.. (can be

either certain odd or even numbers).

(c) If the user wishes to use, say 7 segments, then the mixed Simpson 1/3 rule (for the first 4 segments),

and Simpson 3/8 rule (for the last 3 segments).

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