A textbook of Computer Based Numerical and Statiscal Techniques part 34 docx

Note: Using the formula, the given interval of integration must be divided into an even number of sub-intervals... Note: Using this formula, the given interval of integration must be div

6.5 TRAPEZOIDAL RULE

Putting n =1 in equation (2) and taking the curve y = f(x) through (x0, y0) and (x0, y0) as a polynomial

of degree one so that differences of order higher than one vanish, we get

0

0

1

x h

x

f x dx h y y y y y y y

+

∫ Similarly, for the next sub interval (x0+h x, 0+2h),we get

2

( 1)

x nh

f x dx y y f x dx y y

+ +

−

Adding the above integrals, we get

0

0

2

x nh

x

h

f x dx y y y y y

+

−

∫ which is known as Trapezoidal rule

Putting n = 2 in equation (2 ) and taking the curve through ( ,x y0 0),( ,x y1 1) and ( ,x y2 2) as a polynomial of degree two so that differences of order higher than two vanish, we get

0

0

2

2

1

6

x

+

∫

2

y y y y y y y y y

0

4

2

3

h

f x dx y y y

+

+

∫

0

0

( 2)

3

x nh

h

f x dx y y y

+

+ −

∫ Adding the above integrals, we get

0

0

3

x nh

x

h

f x dx y y y y y y y y

+

∫ which is known as Simpson’s one-third rule

Note: Using the formula, the given interval of integration must be divided into an even number of sub-intervals.

6.7 SIMPSON’S THREE-EIGHT RULE

Putting n = 3 in equation (2) and taking the curve through (x0, y0), (x1, y1), (x2, y2) and (x3, y3) as a polynomial of degree three so that differences of order higher than three vanish, we get

0

0

3

( ) 3

x

f x dx h y y y y

+

∫

3

8

h

y y y y y y y y y y

( 0 1 2 3)

3

8

h

y y y y

0

6

3

3

8

h

f x dx y y y y

+

+

∫

0

0

6

( 3)

3

8

h

+

+ −

∫

Adding the above integrals, we get

0

0

3

8

x nh

x

h

f x dx y y y y y y y y y y y

+

∫

which is known as Simpson’s three-eighth rule

Note: Using this formula, the given interval of integration must be divided into sub-intervals whose

number n is a multiple of 3.

Putting n = 4 in equation (2) and neglecting all differences of order higher than four, we get

0

0

4

0

( )

x

+

(By Newton’s forward interpolation formula)

4 4

0

y

h y y y y y 

2

45

h

2

45

h

x x

h

+

Adding all these integrals from x0 to x0 + nh, where n is a multiple of 4, we get

0

2

45

x nh

x

h

+

∫

This is known as Boole’s rule

Note: Using Boole’s rule, the number of sub-intervals should be taken as a multiple of 4.

Putting n = 6 in equation (2) and neglecting all differences of order higher than six, we get

0

0

0

( )

x

r r r r r

+



( 1)( 2)( 3)( 4)( 5)

6!

r r r r r r

y dr

h ry y   y  r r  y



6 6 0 0

720 7r 2r r 4r 3r r y





= 6 3 9

41 20

11 20

41 840

h yL + y + y + y + y + y + y

h

3

10

h

+41(y4−4y3+6y2−4y1+y0) 11(+ y5−5y4+10y3−10y2+5y1−y0)

+(y6−6y5+15y4−20y3+15y2−6y1+y0)] 341

42≈1

L

NM O QP

( 0 1 2 3 4 5 6)

3

10

h

y y y y y y y

Similarly, 0 ( ) ( )

0

12

6

3

10

h

+

∫ 0

( 6)

3

10

x nh

h

+

∫ Adding the above integrals, we get

0

3

10

x nh

x

h

+

∫

which is known as Weddle’s rule Here n must be a multiple of 6.

Example 1 Use Trapezoidal rule to evaluate 10 1

1+x dx

∫

Sol Let h = 0.125 and y = f(x) = 1

1 x+ , then the values of y are given for the arguments which

are obtained by dividing the interval [0,1] into eight equal parts are given below:

1

1

y

x

=

+ 1.0 0.8889 0.8000 0.7273 0.6667 0.6154 0.5714 0.5333 0.5

Now by Trapezoidal rule

0

1

h

+

∫

0.125 [1 2(0.8889 0.800 0.7273 0.6667 0.6154 0.5714 0.5333) 0.5] 2

0.125 [1.5 2(4.803)]

2

Example 2 Evaluate 01 1 2

dx x

+

(i) Simpson’s 1

3 rule taking

1 4

h=

(ii) Simpson’s 3

8 rule taking

1 6

h=

(iii) Weddle’s rule taking 1

6

h=

Hence compute an approximate value of π in each case.

Sol (i) The values of 2

1 ( ) 1

f x

x

= + at

1 2 3

0, , , , 1

4 4 4

( )

16

17

x

f x

By Simpon’s 1

3 rule

1

2

3 1

+

∫

1 { }16

Also,

1

4 1

dx

x x

+

∫

4

(ii) The values of ( ) 1 2

1

f x

x

= + at x = 0,

1 2 3 4 5 , , , , ,1

( )

1

x

f x

By Simpson’s 3

8 rule

1

2 0

3

8 1

+

∫ =

6

1

36 37

9 10

9 13

36

4 5

FH IK L FHG IKJ+ + RST + + + UVW+ FHG IKJ

Also,

1 2

dx x

π

= +

∫

4

(iii) By Weddle’s rule

1

2 0

3

10 1

+

∫

1 3

6

 

 

Since

1 2

dx x

π

= +

∫

4

π =

Example 3 Evaluate 06 1 2

dx x

+

(i) Trapezoidal Rule

(ii) Simpson’s one-third rule

(iii) Simpson’s three-eighth rule

(iv) Weddle’s rule.

Sol Divide the interval (0, 6) into six parts each of width h = 1.

The value of ( ) 1 2

1

f x

x

= + are given below:

17

1 26

1 37

(i) By Trapezoidal rule,

2 1

+

∫

1 1 1 2 0.5 0.2 0.1 1 1

=  + +  + + + + 

= 1.410798581 Ans

(ii) By Simpson’s one-third rule,

2

3 1

+

∫

1 1 1 4 0.5 0.1 1 2 0.2 1

=  + +  + + +  + 

(iii) By Simpson’s three-eighth rule,

6

2 0

3

8 1

+

∫ = 3 1 1 3 0.5 0.2 1 1 2(0.1)

= 1.357080836 Ans

(iv) By Weddle’s rule,

6

2 0

3

10 1

+

∫

3 1 5(0.5) 0.2 6(0.1) 1 5 1 1

 

=1.373447475 Ans

Example 4 Using Simpson’s one-third rule, find 6 2

0 (1 )

dx x

+

Sol Divide the interval [0,6] into 6 equal parts with 6 0

6

h= −

= 1 The values of 1 2

(1 )

y x

=

each points of sub-divisions are given by

2

1

y

x

=

By Simpson’s one-third rule, we get

6

2

3 (1 )

dx h

y y y y y y y

x = + + + + + + +

∫

1 4(0.25 0.0625 0.02778) 2(0.11111 0.04) 0.02041

= 1[1.02041 4(0.34028) 2(0.15111)]

= 0.89458 Ans

Example 5 Evaluate 4 2

0 1

dx x

x

+

(i) h = 1 (ii) h = 0.5 Compare the rusults with the actual value and indicate the error in both.

Sol (i) Dividing the given interval into 4 equal subintervals (i.e h = 1), the table is as below:

2

1 5

1 10

1 17

Using Boole’s rule,

1

4

0

2

45

h

∫

= 2 1

45 7 1 32

1

2 12

1

5 32

1

10 7

1 17

( ) ( )+ FH IK+ FH IK+ FH IK+ FH IK

L

∴ 4 2

1

dx

x = +

(ii) Dividing the given interval into 8 equal subintervals (i.e h = 0.5), the table is as below:

53

1 17

Using Boole’s rule,

0

2

45

h

∫

= 1.326373

x

1 2

0

4

+

But the actual value is

2

1

dx

x x

+

∫

Error in I result = 1.325818 1.289412 100 2.746%

1.325818

−

Error in II result = 1.325818 1.326373 100 0.0419%

1.325818

−

Example 6 Evaluate the integral 6 3

0 1

dx x

+

Sol Divide the interval [0,6] into 6 equal parts each of width 6 0 1

6

h= − =

The value of 1 3

1

y x

= +

at each points of sub-divisions are given below:

y 1.0000 0.5000 0.1111 0.0357 0.0153 0.0079 0.0046

By Weddle’s Rule, we get

3 0

3

10 1

+

∫

3 1.0000 5(0.5000 0.0079) 0.1111 0.0153 6(0.357) 0.0046 10

3 1.131 5 0.5079 6 0.0357 10

3 1.131 2.5395 0.2142 10

3 (3.8847) 1.1654

10

Example 7 Evaluate the integral 1.5 3

x dx

e −

Sol Dividing the interval [0, 1, 5] into 6 equal parts of each of width 1.5 0 0.25

6

h= − =

and the values of

3 1

x

x y

e

=

− at each points of sub-interval are given by

Now by Weddle’s rule, we get

0

3

10 1

x

dx y y y y y y y

−

∫ 3(0.25[0 5(0.0549 0.7843) 0.1927 0.5820 0.9694 6(0.3777)]

10

= 0.075[1.7441 + 5(0.8392) +6(0.3777)]

= 0.075[1.7441 + 4.196 + 2.2662] = 0.075(8.2063) = 0.6155 Ans

Example 8 Evaluate the integral 5.2

4 logxdx,

Sol Divide the interval [4, 5.2] into 6 equal sub-interval of each width 5.2 4 0.2

6

−

values of y = log x are given below:

y 1.3862 1.4350 1.4816 1.5261 1.5686 1.6094 1.6486

By Weddle’s Rule, we get

3

10

h xdx= y + y +y +y +y + y +y

∫ =3(0.2)10 [1.3862 5(1.4350 1.6094) 1.4816 1.5686 6(1.5261) 1.6486]+ + + + + +

0.6[1.3862 5(3.0444) 1.4816 1.5686 6(1.526) 1.6486]

10

0.6[6.085 5(3.0444) 6(1.5261)]

10

0.6[6.085 15.222 9.1566]

10

=0.6(30.4636) 1.8278=

Example 9 A river is 80 m wide The depth y of the river at a distance ‘x’ from one bank is given by the

following table:

Find the approximate area of cross section of the river using

(i) Boole’s rule.

(ii) Simpson’s one-third rule.

Sol The required area of the cross-section of the river

Here no of sub intervals is 8

(i) Boole’s rule,

∫ 80

0

2

45

h