Download - Lecture09 - Numerical Integration (I)

Numerical Differentiation and Integration

Lecture 9:Numerical Integration (I)

MTH2212 – Computational Methods and Statistics

Dr. M. HrairiDr. M. Hrairi MTH2212 - Computational Methods and StatisticsMTH2212 - Computational Methods and Statistics 22

Objectives

Introduction Trapezoidal Rule Multiple-Application Trapezoidal Rule Simpson’s 1/3 Rule Multiple-Application Simpson’s 1/3 Rule


Integration – Graphical representation

The integral is the area under the curve


Integration – Engineering applications

How to take complicated integrals and approximate the area under the curve

Integration needed for analyzing irregularly shaped lines, areas, volumes

Examples:


Newton-Cotes Integration Formulas

The Newton-Cotes formulas are the most common numerical integration schemes.

This involves the replacement of a complicated function or tabulated data with an approximating function that is easy to integrate.

Where fn(x) is a polynomial of order n

b

a nb

adxxfdxxfI )()(

nn

nnn xaxaxaxaaxf

11

2210 ...)(


Newton-Cotes Integration Formulas

Closed and open forms Closed: data points at the beginning and end of the limits of integration are

known Open: integration limits that extend beyond the range of data


Trapezoidal Rule

The Trapezoidal rule is the first of the Newton-Cotes closed integration formulas, corresponding to the case where the polynomial is first order i.e. n = 1

f1(x) is expressed using linear interpolation formula

The area under the straight line is an estimate of the integral of f1(x):

b

a

b

adxxfdxxfI )()( 1

)()()(

)()(1 axab

afbfafxf

dxaxab

afbfafI

b

a

)()()(

)(


Trapezoidal Rule

The result of this integration is

2

)()()(

bfafabI

Trapezoidal rule


Trapezoidal Rule

2

)()()(

bfafabI

Area of a trapezoid = Width X Average Height


Error of the Trapezoidal Rule

Truncation error is expressed as:

An approximation of the second derivative is given by

Approximate error estimate is then:

)('')(12

1 3 fabEt

ab

dxxfff

b

a

)(''

'')(''

'')(12

1 3 fabEa


Example 1

Use Trapezoidal Rule to integrate numerically the

following function:

from a = 0 to b = 0.8

5432 400900675200252.0)( xxxxxxf


Example 1 - Solution

Evaluate f(a) and f(b)

Evaluate the integral

Approximate error estimate

2.0)0( f

232.0)8.0( f

1728.02

232.02.0)08.0(

I

56.2)08.0)(60(12

1))((''

12

1 33 abxfEa

6008.0

)8000108004050400()(''

8.0

0

32

dxxxx

xf



True value of the integral is 1.640533

Et = 1.640533 – 0.1728

= 1.467733 εt = 89.5%

Ea = 2.56

Discrepancy between Et and Ea due to the fact that f >’’(x) is not an accurate approximation of f’’(ξ) for an interval of this size.


Multiple-Application or Composite Trapezoidal Rule

Divide integration interval from a to b into n segments of equal width improve the accuracy of the trapezoidal rule

The total integral can be represented as

Substituting the trapezoidal rule for each integral, we get

By grouping terms, we have

n

n

x

x

x

x

x

xdxxfdxxfdxxfI

1

2

1

1

0

)(...)()(

2

)()(...

2

)()(

2

)()( 12110 nn xfxfh

xfxfh

xfxfhI

)()(2)(

2

1

10 n

n

ii xfxfxf

hI


Multiple-Application Trapezoidal Rule


Error of Multiple-Application Trapezoidal Rule

Truncation error is obtained by summing the individual error for each segment

By estimating the average value of f> ’’(x) for the entire interval

Approximate error estimate for multiple-application of trapezoidal rule is then:

As you double the number of segments error is quartered

n

it f

n

abE

13

3

)(''12

)(

n

ff

n

i

1

)(''''

'')(''1

fnfn

i

''12

)(2

3

fn

abEa


Example 2

Apply two segments in the trapezoidal rule to estimate the integral of:

from a = 0 to b = 0.8

5432 400900675200252.0)( xxxxxxf



We have n = 2

Evaluate the integral

True and Approximate error estimate

4.02

08.0

h

2.0)0( f 456.2)4.0( f 232.0)8.0( f

0688.14

232.0)456.2(22.08.0

I

64.0)60()2(12

)08.0(2

3

aE

Et = 1.640533 – 1.0688 = 0.571733


Simpson’s 1/3 Rule

It corresponds to the case where n = 2 i.e. a polynomial of second order:

Let a = x0 and b = x2 and f2(x) a second order Lagrange polynomial, then

b

a

b

adxxfdxxfI )()( 2

dxxfxxxx

xxxx

xfxxxx

xxxxxf

xxxx

xxxxI

x

x

)())((

))((

)())((

))(()(

))((

))((

21202

10

12101

200

2010

212

0


Simpson’s 1/3 Rule

After integration, we have:

Estimated truncation error

)()(4)(3 210 xfxfxfh

I 2

abh

where

)4(5

2880

)(f

abEa

ab

dxxfff

b

a

)(

)()4(

)4()4(


Example 3

Evaluate the integral of:

from a = 0 to b = 0.8 using Simpson’s 1/3 rule. Compute the truncation errors Et and Ea

5432 400900675200252.0)( xxxxxxf



Evaluate the integral with h = (b-a)/2 = 0.4

True truncation error

Et = 1.640533 – 1.367467 = 0.2730667 εt =

16.6%

Approximate truncation error

3674671232045624203

40)()(4)( 210 ..).(.

.xfxfxf

3

hI

240008.0

)()( 4800021600)(

8.0

0

)4()4()4(

dxxf

xfxxf

2730667.0)2400(2880

)8.0()(

2880

)( 5)4(

5

xfab

Ea


Multiple-Application of Simpson’s 1/3 Rule

Just as the trapezoidal rule, Simpson’s 1/3 rule can be improved by dividing the integration interval into n segments of equal width

The total integral can then be represented by

)()(2)(4)(

3

2

,...6,4,2

1

,...5,3,10 n

n

ji

n

ii xfxfxfxf

hI )4(

4

5

180

)(f

n

abEa


Example 4

Use n = 4 to evaluate the integral of:

from a = 0 to b = 0.8 using Simpson’s 1/3 rule. Compute the truncation errors Et and Ea

5432 400900675200252.0)( xxxxxxf


Example 4 – Solution

The integral using composite simpson’s 1/3 rule

h = (0.8-0)/4 = 0.2

f(0)=0.2 f(0.2)=1.288 f(0.4)=2.456

f(0.6)=3.464 f(0.8)=0.232

εt = 1.0%

Approximate truncation error

623466.1

232.0)464.3(4456.23

2.0456.2)288.1(42.0

3

2.0

I

017067.0)2400()4(180

)8.0(

180

)(4

5)4(

4

5

fn

abEa

Et = 1.640533 – 1.623466 = 0.017067