The Islamic University of GazaFaculty of Engineering

Civil Engineering Department

Numerical Analysis ECIV 3306

Chapter 22

Integration of Equations

Gauss Quadrature

• Gauss quadrature implements a strategy of positioning any two points on a curve to define a straight line that would balance the positive and negative errors.

• Hence, the area evaluated under this straight line provides an improved estimate of the integral.

Two points Gauss-Legendre Formula• Assume that the two Integration points are xo and x1 such that:

• The object of Gauss quadrature is to determine the equations of the form:

• c0 and c1 are constants, the function arguments x0 and x1 are unknowns…….(4 unknowns)

)()( 1100 xfcxfcI

Two points Gauss-Legendre Formula

• Thus, four unknowns to be evaluated require four conditions.

• If this integration is exact for a constant, 1st order, 2nd order, and 3rd order functions:

1

1

31100

1

1

21100

1

11100

1

11100

0)()(

32)()(

0)()(

21)()(

dxxxfcxfc

dxxxfcxfc

dxxxfcxfc

dxxfcxfc

Two points Gauss-Legendre Formula

• Solving these 4 equations, we can determine c1, c2, x1 and x2.

31

31 ffI

5773503.03

1

5773503.03

1:are pointsn Integratio The

1:are factors weightingThe

1

0

10

x

x

cc

Two points Gauss-Legendre Formula• Since we used limits for the previous integration from –1 to 1

and the actual limits are usually from a to b, then we need first to transform both the function and the integration from the x-system to the xd-system

1ax1bx

d2

abdx

2ab

2abx

f(x)

x

a b

f(xo)

f(x1)

xo x1 -1 1

f()

Higher-Points Gauss-Legendre Formula

)(.....)()()(

)(

nn2211

n

1iii

1

1i

fcfcfcfcI

:points Gauss n usingdfI

Multiple Points Gauss-LegendrePoints Weighting factor Function argument Exact

for 2 1.0 -0.577350269 up to

3rd 1.0 0.577350269 degree

3 0.5555556 -0.774596669 up to 5th

0.8888889 0.0 degree0.5555556 0.774596669

4 0.3478548 -0.861136312 up to 7th

0.6521452 -0.339981044 degree0.6521452 0.3399810440.3478548 0.861136312

6 0.1713245 -0.932469514 up to 11th

0.3607616 -0.661209386 degree0.4679139 -0.2386191860.4679139 0.2386191860.3607616 0.6612093860.1713245 0.932469514

Gauss Quadrature - Example

Find the integral of: f(x) = 0.2 + 25 x – 200 x2 + 675 x3 – 900 x4 + 400 x5

Between the limits 0 to 0.8 using:

– 2 points integration points (ans. 1.822578)

– 3 points integration points (ans. 1.640533)

Improper Integral• Improper integrals can be evaluated by making a change

of variable that transforms the infinite range to one that is finite,

b

A

b A

b

a

a

b

dxxfdxxfdxxf

abdtt

ft

dxxf

)()()(

011)(/1

/12

A

A

dtt

ft

dxxf0

/12

11)(Can be evaluated by Newton-Cotes closed formula

Improper Integral - Examples

• .

• .

• .

dtt

dtt

ttxx

dx

5.0

0

5.0

0 2

2 211

2/11)(1

)2(

dyyedyyedyye yyy

2

22

0

2

0

2 sin sin sin

dttet

dyye ty 2/1

0

2/12

2

2 )/1(sin1 sin

dyyedyyedyye yyy

2

2

2

2

dtet

dyye ty 2/1

0

/13

2 1

Multiple Integration

dydxyxfId

cy

b

ax

),(

• Double integral:

Multiple Integration using Gauss Quadrature Technique

• .

• .

• . ddf

2ab

2cddydxyxfI

1

1

1

1

d

cy

b

ax

),(),(

1ax1bx &

d2

abdx2

ab2

abx

&

1cx1dy &

d2

cddy2

cd2

cdy

&

Multiple Integration using Gauss Quadrature Technique

Now we can use the Gauss Quadrature technique:

If we use two points Gauss Formula:

n

1jjiij

n

1i

fcc2

ab2

cdI ),(

)}],(),({

)},(),({[

31

31fc

31

31fcc

31

31fc

31

31fcc

2ab

2cdI

212

211

2

1jjijj

2

1i

fcc2

ab2

cdI ),(

2

1jj2j1j 3

1fc3

1fcc2

ab2

cdI )],(),([

Double integral - Example

72222),( 22 yxxxyyxT

• Compute the average temperature of a rectangular heated plate which is 8m long in the x direction and 6 m wide in the y direction. The temperature is given as:

• (Use 2 segment applications of the trapezoidal rule in each dimension)

Double integral - Example

6667.58)86/(2816,28163/1

56)86/(2688,2688)2(

)72222( 6

0

8

0

22

avg

avg

TIruleSimpson

TInrulelTrapezoidaMultiple

dxdyyxxxyI

HW: Use two points Gauss formula to solve the problem