• Trapezoidal rule

Numerical Methods

dr hab. inż. Katarzyna Zakrzewska, prof. AGH

Lecture 5.

Numerical integration

Outline

• Trapezoidal rule

• Multi-segment trapezoidal rule

• Richardson extrapolation

• Romberg's method

• Simpson's rule

• Gaussian method

Numerical integration - idea

∫

=

a

dx ) x ( f I

X Y

f(x)

a b

∫

a

dx x f ( )

The integral can be approximated by:

∑ Δ

=

i

f c

x

S

1

( )

+1

≤

≤

i

c x

x

Newton–Cotes methods

Newton – Cotes integration belongs to a class of

methods with fixed nodes: function f(x) is interpolated by a polynomial (e.g. Lagrange polynomial)

) x ( f )

x (

f ≈ _n

where:

n n

n

n

( x ) a a x ... a x a x

f =

+

+ +

+

Then, the integral of f(x) can be approximated as an integral of the interpolated function f

(x)

dx x

f dx

x f I

b

a n b

a

∫

∫ ^≈

= ( ) ( )

Trapezoidal rule

The trapezoidal rule

assumes: n = 1, thus:

f

( x ) = a

+ a

x

dx x

a a

dx x

f dx

x f I

b

a b

a b

a

) (

) ( )

( ≈

=

+

= ∫ ∫ ∫

But what is a₀ and a₁ ?

( ) 2

2 2

1 0

a a b

a b

a − + −

=

Now if one chooses, (a,f(a)) i (b,f(b)) as the two points to

approximate f(x) by a straight line from a to b. It follows that:

a a a

a f a

f ( ) =

( ) =

+

b a a

b f b

f ( ) =

( ) =

+

a b

a b f b a a f

−

= ( ) − ( )

0

a f b

a = f ( )− ( )

Trapezoidal rule

trapezoid of

area dx

x f

b

a

∫ ^{( )} ≈

X Y

f(x)

a b

f_I(x)

⎥⎦ ⎤

⎢⎣ ⎡ +

−

∫ =

2

) ( )

) ( (

)

( f a f b

a b

dx x

f

b a

f(a) f(b)

Trapezoidal rule

t t t

v 9 . 8

2100 140000

140000 ln

2000 )

( −

⎥⎦ ⎤

⎢⎣ ⎡

= −

Example 1:

Velocity v(t) of a rocket from t₁=8 s to t₂=30 s is given by:

a) Use the single segment trapezoidal rule to find the distance covered by the rocket from t₁=8 s to t₂=30 s

b) Find the true relative error.

Trapezoidal rule

⎥⎦ ⎤

⎢⎣ ⎡ +

−

≈ 2

) ( )

) (

( f a f b

a b

I ^a = ^{8 s} b = 30 s

t t t

f 9 . 8

2100 140000

140000 ln

2000 )

( −

⎥⎦ ⎤

⎢⎣ ⎡

= −

) 8 ( 8 . ) 9

8 ( 2100 140000

140000 ln

2000 )

8

( ⎥ −

⎦

⎢ ⎤

⎣

⎡

= − f

) 30 ( 8 . ) 9

30 ( 2100 140000

140000 ln

2000 )

30

( ⎥ −

⎦

⎢ ⎤

⎣

⎡

= − f

s 27 m /

.

= 177

s 67 m /

.

= 901 a)

⎥⎦ ⎤

⎢⎣ ⎡ +

−

= 2

67 . 901 27

. ) 177 8

30 (

I = 11868 m

Trapezoidal rule

∫ ⎟⎟

⎠

⎜⎜ ⎞

⎝

⎛ −

⎥⎦ ⎤

⎢⎣ ⎡

= − Δ

8

8 . 2100 9

140000

140000 ln

2000 t dt

x t = 11061 m

b)

11061 100

11868

11061 − ×

=

∈

= 7 .2959 %

The relative error:

Multi-segment trapezoidal rule

n a h b −

=

The true error using a

single segment trapezoidal rule was large. We can

divide the interval from a to b into smaller n segments of equal length – h and apply the trapezoidal rule over each segment:

∫ + ∫ + ∫

∫ +

∫ =

=

+

+ +

+ +

+ a h

h a

h a

h a

h a

h a h

a a b

a

dx x f dx

x f dx

x f dx

x f dx

x f I

2 3

2

4 3

) ( )

( )

( )

( )

(

for n=4

X Y

f(x)

a b

4 a a b−

+ ^a+2b−₄^{a a+3b}₄^−a

∫

dx ) x (

f ⎥⎦ ⎤

⎢⎣ ⎡ +

⎭ ⎬

⎫

⎩ ⎨

⎧ +

− +

= ∑

=

) b ( f )

ih a

( f )

a ( n f

a

b

i 1 1

2 2

∫

∫

∫

∫

− +

− + +

+

+

+ + + +

=

h ) n ( a h

) n ( a

h ) n ( a h

a

h a h

a

a

dx ) x ( f dx

) x ( f ...

dx ) x ( f dx

) x ( f

1 1

2

∫

a

dx ) x ( f

2 ...

) 2 (

) (

2

) (

)

( +

⎥⎦ ⎤

⎢⎣ ⎡ + + +

⎥⎦ +

⎢⎣ ⎤

⎡ + +

= f a h f a h

h h a

f a

h f

⎥⎦ ⎤

⎢⎣ ⎡ + − +

− +

−

+ 2

) ( )

) 1 (

] ( ) 1 (

( [

... f a n h f b

h n

a b

Multi-segment trapezoidal rule

t t t

v 9 . 8

2100 140000

140000 ln

2000 )

( −

⎥⎦ ⎤

⎢⎣ ⎡

= −

Example 2:

Velocity v(t) of a rocket from t₁=8 s to t₂=30 s is given by:

a) Use the complex segment trapezoidal rule to find the distance covered from t₁=8 s to t₂=30 s for n = 2

b) Find the true relative error.

Multi-segment trapezoidal rule

a) ⎥⎦ ⎤

⎢⎣ ⎡ +

⎭ ⎬

⎫

⎩ ⎨

⎧ ∑ +

− +

=

=

( ) ( )

2 ) 2 (

1 1

b f ih

a f a

n f a I b

n i

n = 2 a = 8 s b = 30 s

s

n a

h b 11

2 8 30 − =

− =

=

⎥⎦ ⎤

⎢⎣ ⎡ +

⎭ ⎬

⎫

⎩ ⎨

⎧ ∑ +

− +

=

=

( ) ( 30 )

2 )

8 ) (

2 ( 2

8

30

1

f ih

a f f

I

i

[ ^{( 8 ) 2 ( 19 ) ( 30 )} ]

4

22 f + f + f

= [ ^{177 27 2 484 75 901 67} ]

4

22 . + ( . ) + .

=

11266 m

=

Multi-segment trapezoidal rule

b)

∫ ^⎜ _⎝ ^⎛ _⎢⎣ ^⎡ _{− ⎥⎦} ^{⎤ − ⎟} _⎠ ^⎞

=

8

8 2100 9

140000

140000

2000 . t dt

ln t

x = 11061 m

true value:

The relative error

: 11061 100

11266 11061 − ×

=

∈

= 1 . 8534 %

Multi-segment trapezoidal rule

n ∆x E_t

1 11868 -807 7.296 ---

2 11266 -205 1.853 5.343

3 11153 -91.4 0.8265 1.019

4 11113 -51.5 0.4655 0.3594

5 11094 -33.0 0.2981 0.1669

6 11084 -22.9 0.2070 0.09082 7 11078 -16.8 0.1521 0.05482 8 11074 -12.9 0.1165 0.03560

t %

∈ ∈_a %

Multi-segment trapezoidal rule

The relative error for a simple trapezoidal rule is

b a

), (

"

) f a b

E

( − ζ < ζ <

= 12

3

The relative error in the multi-segment trapezoidal rule is a sum of errors for each segment. The relative error within the first segment is given by:

[ ]

h a

a ), (

"

a f )

h a

E ( + − ζ < ζ < +

=

1

12

) (

"

h f

1 3

12 ζ

=

Estimation of error

By analogy:

[ ]

ih a

h i

a h f

i a ih

E_i a+ − + − ζ_i + − < ζ_i < +

= "( ), ( 1)

12

) ) 1 ( (

)

( ³

) (

"

h f

ζ

= 12

3

for n-th segment :

{ }

[ ]

b h

) n

( a ), (

"

h f ) n

( a

E

b − + − ζ

+ − < ζ

<

= 1

12

1

) (

"

h f

ζ

= 12

3

Estimation of error

The total error in the complex trapezoidal rule is a sum of the errors for a single segment:

∑

=

i

i

t

E

E

1

∑

ζ

=

i

i

) (

"

h f

1 3

12 n

) (

"

f n

) a b

(

n i

∑

=

− ζ

=

2 3

12

Formula:

n

) (

"

f

n i

∑

=

ζ

1

yields an approximate average value of the second derivative in the range of

b x

a < <

2

1 E

∝ α n

Estimation of error

Table below presents the results for the integral

∫

30 8

8 2100 9

140000 140000

2000 . t dt

ln t

as a function of the number of segments n. When n twice increases, the absolute error E_t decreases four times!

Et ∈_t % ∈_a %

n Value

2 11266 -205 1.854 5.343

4 11113 -51.5 0.4655 0.3594

8 11074 -12.9 0.1165 0.03560

16 11065 -3.22 0.02913 0.00401

Estimation of error

Richardson’s extrapolation and Romberg’s method of integration

Richardson’s extrapolation and Romberg’s method of integration constitute an extension of the

trapezoidal method and give better approximation

of the integral by reducing the true error.

n

E

≅ C

n t

TV I E = −

Richardson extrapolation

The true error obtained when using the multi-segment

trapezoidal rule with n segments to approximate an integral is given by:

where: C is an approximate constant of proportionality Since:

true value approximate value using

( ) ^{n TV I}

C

2 2

2 ≅ −

If the number of segments is doubled from n to 2n:

3

2 2n n

n

I I I

TV −

+

≅

Richardson extrapolation

It can be shown that:

( ) ^{n C}

^{≅ TV − I}

( ) ^{n TV I}

C

2 2

2 ≅ −

We get:

t t t

v 9 . 8

2100 140000

140000 ln

2000 )

( −

⎥⎦ ⎤

⎢⎣ ⎡

= −

Example 3:

The velocity v(t) of a rocket from t₁=8 s to t₂=30 is given by:

a) Use Richardson extrapolation rule to find the distance covered for n = 2

b) Find the relative true error

Richardson extrapolation

t %

∈ ∈_a %

n ∆x E_t

1 11868 -807 7.296 ---

2 11266 -205 1.853 5.343

3 11153 -91.4 0.8265 1.019

4 11113 -51.5 0.4655 0.3594

5 11094 -33.0 0.2981 0.1669

6 11084 -22.9 0.2070 0.09082 7 11078 -16.8 0.1521 0.05482 8 11074 -12.9 0.1165 0.03560

Table of results for n = 8 segments (trapezoidal rule)

a)

m

I

= 11266 I

= 11113 m

3

2 2n n

n

I I I

TV −

+

≅ ^{for n=2}

3

2 4 4

I I I

TV −

+

≅ 3

11266 11113

11113 + −

=

11062 m

=

Richardson extrapolation

b)

∫ ^⎟

⎠

⎜ ⎞

⎝

⎛ −

⎥⎦ ⎤

⎢⎣ ⎡

=

−

8

8 2100 9

140000

140000

2000 . t dt

ln t

x = 11061 m

The absolute true error:

11062 11061 −

t

=

E = − 1 m

Richardson extrapolation

The true value:

.

c) The relative error:

11061 100

11062 11061

− ×

=

∈

= 0 .00904 %

Richardson extrapolation

n ∆x (m)

Trapezoidal

rule Trapezoidal rule

∆x (m)

Richardson

extrapolation Richardson extrapolation

12 48

11868 11266 11113 11074

7.296 1.854 0.4655 0.1165

11065-- 11062 11061

0.03616-- 0.009041

0.0000

Comparison of different methods:

t %

∈ ∈_t %

Romberg’s method uses the same pattern as Richardson

extrapolation. However, Romberg used a recursive algorithm for the extrapolation as follows:

3

2 2n n

n

I I I

TV −

+

≅

( ) 3

2 2

2 n n

R n n

I I I

I −

+

= 4

1

2 2

− + −

=

I

I

I

Romberg's method

The true value TV is replaced by the result of the Richardson extrapolation

Note also that the sign is replaced by the sign =

( ) ^I

≅

Esimated true value is given by:

^{TV ≅} ( ) ^I

^{+ Ch}

where: Ch⁴ is the value of the error of approximation

Romberg's method

Another value of integral obtained while doubling the number of segments from 2n to 4n:

( ) 3

2 4

4

4 n n

R n n

I I I

I −

+

=

Estimated true value is given by:

( ) ( ) ( )

15

2

4 4n R n R

n R

I I I

TV −

+

≅

( ) ( )

)

(

−

+

= I

I

I

1 2 4

1 1

1 1

1

≥

− + −

=

I

I

, k I

I

The index k represents the order of extrapolation

k=1 represents the values obtained from the regular trapezoidal rule A general expression for Romberg integration can be written as:

Romberg's method

k=2 represents the values obtained using the true error estimate as O(h²) The value of an integral with for j+ 1 is more accurate than

the value of the integral for j index

t t t

v 9 . 8

2100 140000

140000 ln

2000 )

( −

⎥⎦ ⎤

⎢⎣ ⎡

= −

Example 4:

Velocity v(t) of a rocket from t₁=8 s to t₂=30 s is given by:

a) Use Romberg’s method to find the distance covered. Use n = 1, 2, 4, and 8

b) Find the absolute true error and the relative approximate error

Romberg's method

t %

∈ ∈_a %

Table of results for n = 8 segments (trapezoidal rule)

n ∆x E_t

1 11868 -807 7.296 ---

2 11266 -205 1.853 5.343

3 11153 -91.4 0.8265 1.019

4 11113 -51.5 0.4655 0.3594

5 11094 -33.0 0.2981 0.1669

6 11084 -22.9 0.2070 0.09082 7 11078 -16.8 0.1521 0.05482 8 11074 -12.9 0.1165 0.03560

From this table, the initial values from the trapezoidal rule are:

11868

1 1_,

=

I I

= 11266

11113

3 1_,

=

I I

= 11074

Romberg's method

To get the first order extrapolation values:

3

1 1 2

1 2

1 1

2

, ,

, ,

I I I

I −

+

=

3

11868 11266

11266 −

+

=

Similarly,

3

2 , 1 3

, 1 3

, 1 2

, 2

I I I

I −

+

=

3

11266 11113

11113 + −

=

11062

=

3

3 , 1 4

, 1 4

, 1 3

, 2

I I I

I −

+

=

3

11113 11074

11074 + −

=

11061

=

Romberg's method

For the second order extrapolation:

15

1 2 2

2 2

2 1

3

, ,

, ,

I I I

I −

+

=

15

11065 11062

11062 + −

=

11062

= Similarly,

15

2 , 2 3

, 2 3

, 2 2

, 3

I I I

I −

+

=

15

11062 11061

11061 −

+

=

Romberg's method

For the third order extrapolation

63

1 3 2

3 2

3 1

4

, ,

, ,

I I I

I −

+

=

63

11062 11061

11061 −

+

=

11061 m

=

Romberg's method

11868

11262

11113

11074

11065

11062

11061

11062

11061

11061 1-segment

2-segment 4-segment 8-segment

Appr. 1 Appr. 2 Appr. 3

Improved estimates of the value of an integral using Romberg integration

Romberg's method

Simpson's rule

The trapezoidal rule was based on approximating the integrand by a first order polynomial, and then integrating the polynomial over an interval from a to b. Simpson’s rule assumes that the integrand can be approximated by a second order polynomial.

∫

∫ ^≈

=

a b

a

dx ) x ( f dx

) x ( f

I

where:

2 2 1

0

2

( x ) a a x a x

f = + +

X Y

f(x)

a b

f_I(x)

Simpson's rule

A parabola passing through three points :

)), a ( f , a (

b , f a

b ,

a ⎟

⎠

⎜ ⎞

⎝

⎛ ⎟

⎠

⎜ ⎞

⎝ + ⎛ +

2 2

)) b ( f , b (

2 2 1

0

2

( a ) a a a a a

f )

a (

f = = + +

2 2

1 0

2

2 2 2

2 ⎟

⎠

⎜ ⎞

⎝ + ⎛ +

⎟ ⎠

⎜ ⎞

⎝ + ⎛ +

⎟ =

⎠

⎜ ⎞

⎝

= ⎛ +

⎟ ⎠

⎜ ⎞

⎝

⎛ + a b

b a a a

b a f a

b f a

2 2 1

0

2

( b ) a a b a b f

) b (

f = = + +

Simpson's rule

2 2

2 2

0

2

4 2

b ab

a

) a ( f b ) a ( b abf

abf a )

b ( abf )

b ( f a

a − +

+

⎟ +

⎠

⎜ ⎞

⎝

− ⎛ + +

=

2 1 2

2

4 2 3

2 3 4

b ab

a

) b ( b bf

bf a )

a ( bf )

b ( b af

af a )

a ( af

a − +

⎟ +

⎠

⎜ ⎞

⎝

− ⎛ + +

⎟ +

⎠

⎜ ⎞

⎝

− ⎛ +

−

=

2 2 2

2 2 2 2

b ab

a

) b ( b f

f a )

a ( f

a − +

⎟ ⎠

⎜ ⎞

⎝

⎛ ⎟ +

⎠

⎜ ⎞

⎝

− ⎛ +

=

Coefficients a₀,a₁,a₂ are:

Simpson's rule

≈

∫

a

dx ) x ( f

I

( )

∫ ^{+ +}

=

a

dx x

a x

a

a

b

a

a x a x

x

a ⎥

⎦

⎢ ⎤

⎣

⎡ + +

= 2 3

3 2 2

1 0

3 2

3 3

2 2

2 1 0

a a b

a a b

) a b

(

a −

− + +

−

=

Since:

Simpson's rule

⎥⎦ ⎤

⎢⎣ ⎡ ⎟ +

⎠

⎜ ⎞

⎝ + ⎛ +

= −

∫ ^{f ( x ) dx b a f ( a ) f a b f ( b )}

b

a

4 2

2

6

2 a h b −

=

It follows that:

⎥ ⎦

⎢ ⎤

⎣

⎡ ⎟ +

⎠

⎜ ⎞

⎝ + ⎛ +

∫

^f

^{( x ) dx} = ^h ₃ ^{f ( a ) 4 f a} ₂ ^{b f ( b )}

a

it is called Simpson’s 1/3 rule

Multi-segment Simpson's rule

) ...

x ( f ) x ( f )

x ( ) f

x x

( dx ) x ( f

b

a

⎥⎦ +

⎢⎣ ⎤

⎡ + +

−

∫ =

⁴ ₆

) ...

x ( f ) x ( f )

x ( ) f

x x

( +

⎥⎦ ⎤

⎢⎣ ⎡ + +

−

+ 6

4

2 2 4

f(x)

. . .

x₀ x₂ x_n-2 x_n x

...

dx ) x ( f dx

) x ( f dx

) x ( f

x

x x

x b

a

+ +

= ∫ ∫

∫

2 2

0

∫

∫

−

−

−

+

+

n n

n

x

x x

x

dx ) x ( f dx

) x ( f ....

2 2

4

) ...

x ( f ) x

( f )

x ( ) f

x x

(

...

+

⎥⎦ ⎤

⎢⎣ ⎡ + +

−

+

6

4

4 4 2

⎤

⎡ + +

−

+ f ( x

) 4 f ( x

) f ( x

)

h x

x

−

= 2

n

...,

,

,

i = 2 4

Simpson's rule

) ...

x ( f ) x ( f )

x ( h f

dx ) x ( f

b

a

⎥⎦ +

⎢⎣ ⎤

⎡ + +

∫ = ²

⁴ ₆

) ...

x ( f ) x ( f )

x (

h f +

⎥⎦ ⎤

⎢⎣ ⎡ + +

+ 6

2

4

) ...

x ( f ) x

( f )

x (

h f

+

⎥⎦ ⎤

⎢⎣ ⎡ + +

+

6

2

4

⎥⎦ ⎤

⎢⎣ ⎡ + +

+

6

2 f ( x

) 4 f ( x

) f ( x )

h

Simpson's rule

b

∫

a

dx ) x (

f ^{= h} [ ^{f ( x}

^{) + 4} { ^{f ( x}

^{) + f ( x}

^{) + ... + f ( x}

⁾ } ^{+ ...} ]

3

{ ^{( ) ( ) ... ( )} } ^{( )}]}

2

... + f x

+ f x

+ + f x

+ f x

⎥ ⎥

⎦

⎤

⎢ ⎢

⎣

⎡ + + +

= ∑ ∑

==

−

==

( ) 2 ( ) ( )

4 )

3 (

2 2 1

0 1 n

n

even i

i

i n

odd i

i

i

f x f x

x f x

h f

⎥ ⎥

⎦

⎤

⎢ ⎢

⎣

⎡ + + +

= − ∑ ∑

==

−

==

( ) 2 ( ) ( )

4 )

3 (

2 2 1

0 1 n

n

even i

i

i n

odd i

i

i

f x f x

x f x

n f

a

b

Approximate values of the integral,

using Simpson's rule with multiple segments

n Approximate values E_t ^{|Єt |} 24

68 10

11065.72 11061.64 11061.40 11061.35 11061.34

4.380.30 0.060.01 0.00

0.0396%

0.0027%

0.0005%

0.0001%

0.0000%

Estimation of errors in Simpson's rule

Error for one segment

b a f a b

E

− ζ < ζ <

−

= ( ),

2880 )

(

) (

) f x x

E (

5 0 1 2

2880 − ζ

−

= h f

( ),

1 5 4

90 ζ

−

= )

( ) f

x x

E (

5 2 2 4

2880 − ζ

−

= h f

( ),

2 5 4

90 ζ

−

=

2 1

0

x

x < ζ <

4 2

2

x

x < ζ <

Simpson's rule errors

Error for the multi-segment

) (

) f x

x

E

(

−

ζ

−

=

5 1 2 2

2880 h f ( ),

i )

(

ζ

−

=

90 x

< ζ

< x

True error

∑

=

1 n

i

i

t

E

E ∑

=

ζ

−

=

1

) 4 5 (

) 90 (

n

i

f

h ∑

=

ζ

− −

=

1

) 4 ( 5

5

) 90 (

) (

n

i

f

n a b

n f n

a b

n

i i

∑

ζ

− −

=

2 1

) 4 (

4

5

( )

90 ) (

Simpson's rule errors

) 4 ( 4

5

90 )

( f

n a E

b −

−

=

the average value of the derivative

n f f

n

i

∑

=

ζ

=

2 1

) 4 ( )

4

(

( )

Gauss-Quadrature Method

=

∫

a

dx ) x ( f

I ≈ c

f ( x

) + c

f ( x

)

Gaussian integral is given by:

Points x₁ and x₂, which define the value of the integrand are not fixed (as before), but there are a priori distributed randomly within <a,b>.

constant coefficients

Gauss-Quadrature Method

. x a x

a x

a a

) x (

f =

+

+

+

( )

∫

∫ ⁼

^{+ + +}

a b

a

dx x

a x

a x

a a

dx ) x (

f

b

a

a x a x

a x x

a ⎥

⎦

⎢ ⎤

⎣

⎡ + + +

= 2 3 4

4 3 3

2 2

1 0

( ) _⎟⎟

⎠

⎜⎜ ⎞

⎝

⎛ −

⎟⎟ +

⎠

⎜⎜ ⎞

⎝

⎛ −

⎟⎟ +

⎠

⎜⎜ ⎞

⎝

⎛ −

+

−

= 2 3 4

4 4

3 3

3 2

2 2

1 0

a a b

a a b

a a b

a b

a

There are four unknowns x₁, x₂, c₁, c_2. These are found by

assuming that the formula gives exact results for integrating a general third order polynomial:

Gauss-Quadrature Method

( ) (

)

2 2 2 2

1 0

2 3

1 3 2

1 2 1

1 0

1

a a x a x a x c a a x a x a x

c dx

) x ( f

b

a

+ +

+ +

+ +

+

∫ =

( ) ( ) ( ) (

)

3 1 1 3 2

2 2 2

1 1 2 2

2 1

1 1 2

1

0

c c a c x c x a c x c x a c x c x

a + + + + + + +

=

=

∫

a

dx ) x ( f

I ≈ c

f ( x

) + c

f ( x

)

( )

∫

∫ ⁼

^{+ + +}

a b

a

dx x

a x

a x

a a

dx ) x (

f

( ) _⎟⎟

⎠

⎜⎜ ⎞

⎝

⎛ −

⎟⎟ +

⎠

⎜⎜ ⎞

⎝

⎛ −

⎟⎟ +

⎠

⎜⎜ ⎞

⎝

⎛ −

+

−

= 2 3 4

4 4

3 3

3 2

2 2

1 0

a a b

a a b

a a b

a b

a

Hence:

The formula would then give:

Gauss-Quadrature Method

2

1

c

c a

b − = +

2 2 1

1 2

2

2 a c x c x

b − = +

2 2 2 2

1 1 3

3

3 a c x c x

b − = +

2 2 3

1 1 4

4

4 a c x c x

b − = +

( ) ( ) ( ) (

)

3 1 1 3 2

2 2 2

1 1 2 2

2 1

1 1 2

1

0

c c a c x c x a c x c x a c x c x

a + + + + + + +

( ) _⎟⎟

⎠

⎜⎜ ⎞

⎝

⎛ −

⎟⎟ +

⎠

⎜⎜ ⎞

⎝

⎛ −

⎟⎟ +

⎠

⎜⎜ ⎞

⎝

⎛ −

+

−

= 2 3 4

4 4

3 3

3 2

2 2

1 0

a a b

a a b

a a b

a b

a