09 numerical integration

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Numerical Integration

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Objectives

• The student should be able to– Understand the need for numerical integration– Derive the trapezoidal rule using geometric

insight– Apply the trapezoidal rule– Apply Simpson’s rule

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Need for Numerical Integration!

6

1101

2

1

3

1

231

1

0

231

0

2

x

xxdxxxI

11

0

1

0

1 eedxeI xx

1

0

2

dxeI x

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Area under the graph!

• Definite integrations always result in the area under the graph (in x-y plane)

• Are we capable of evaluating an approximate value for the area?

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Example

• To perform the definite integration of the function between (x0 & x1), we may assume that the area is equal to that of the trapezium:

0101

2

1

0

xxyy

dxxfx

x

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Adding adjacent areas

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

The Trapezoidal Rule

2

2

1212

0101

yyxx

yyxxI

Integrating from x0 to x2:

2

212112101001 yxxyxxyxxyxxI

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

The Trapezoidal Rule

hxxxx 1201

If the points are equidistant

22110 hyhyhyhy

I

210 22

yyyh

I

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Dividing the whole interval into “n” subintervals

n

n

ii yyy

hI

1

10 2

2

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

The Algorithm

• To integrate f(x) from a to b, determine the number of intervals “n”

• Calculate the interval length h=(b-a)/n• Evaluate the function at the points yi=f(xi)

where xi=x0+i*h• Evaluate the integral by performing the

summation

n

n

ii yyy

hI

1

10 2

2

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Note that

X0=a

Xn=b

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Example

• Integrate• Using the trapezoidal

rule• Use 2,3,&4 points and

compare the results

1

0

2dxxI

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Solution

• Using 2 points (n=1), h=(1-0)/(1)=1

• Substituting:

212

1yyI 5.010

2

1I

XY

00

11

2 points, 1 interval

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Solution

• Using 3 points (n=2), h=(1-0)/(2)=0.5

• Substituting:

321 22

5.0yyyI

375.0125.0*202

5.0I

XY

00

0.50.25

11

3 points, 2 interval

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Solution

• Using 4 points (n=3), h=(1-0)/(3)=0.333

• Substituting:

4321 222

333.0yyyyI

3519.01444.0*2111.0*202

333.0I

XY

00

0.330.111

0.6670.444

11

4 points, 3 interval

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Let’s use Interpolation!

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Interpolation!

• If we have a function that needs to be integrated between two points

• We may use an approximate form of the function to integrate!

• Polynomials are always integrable• Why don’t we use a polynomial to

approximate the function, then evaluate the integral

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Example

• To perform the definite integration of the function between (x0 & x1), we may interpolate the function between the two points as a line.

001

010 xx

xx

yyyxf

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Example

• Performing the integration on the approximate function:

1

0

1

0

001

010

x

x

x

x

dxxxxx

yyydxxfI

1

0

0

2

01

010 2

x

x

xxx

xx

yyxyI

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Example

• Performing the integration on the approximate function:

00

20

01

010010

21

01

0110 22

xxx

xx

yyxyxx

x

xx

yyxyI

2

0101

yyxxI

• Which is equivalent to the area of the trapezium!

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

The Trapezoidal Rule

2

0101

yyxxI

2

2

1212

0101

yyxx

yyxxI

Integrating from x0 to x2:

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Simpson’s Rule

Using a parabola to join three adjacent points!

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Quadratic Interpolation

• If we get to interpolate a quadratic equation between every neighboring 3 points, we may use Newton’s interpolation formula:

103021 xxxxbxxbbxf

10102

3021 xxxxxxbxxbbxf

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Integrating

10102

3021 xxxxxxbxxbbxf

2

0

2

0

10102

3021

x

x

x

x

dxxxxxxxbxxbbdxxf

2

0

2

0

10

2

10

3

30

2

21 232

x

x

x

x

xxxx

xxx

bxxx

bxbdxxf

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

After substitutions and manipulation!

210 43

2

0

yyyh

dxxfx

x

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Working with three points!

210 43

2

0

yyyh

dxxfx

x

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

For 4-Intervals

432210 443

4

0

yyyyyyh

dxxfx

x

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

In General: Simpson’s Rule

n

n

ii

n

ii

x

x

yyyyh

dxxfn 2

,..4,2

1

,..3,10 24

30

NOTE: the number of intervals HAS TO BE even

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Example

• Integrate• Using the Simpson

rule• Use 3 points

1

0

2dxxI

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Solution

• Using 3 points (n=2), h=(1-0)/(2)=0.5

• Substituting:

• Which is the exact solution!

210 43

5.0yyyI

3

1125.0*40

3

5.0I

ENEM602 Spring 2007Dr. Eng. Mohammad Tawfik

Homework #7

• Chapter 21, p. 610, numbers:21.5, 21.6, 21.10, 21.11.