Higher Engineering Mathematics, Sixth Edition

440 Higher Engineering Mathematics

y 1 y 2 y 3 y 4 y 2 n 1

a

ddd

b

yf(x)

x

y

O

Figure 45.4

≈

1

3

d[(y 1 +y 2 n+ 1 )+ 4 (y 2 +y 4 +···+y 2 n)

+ 2 (y 3 +y 5 +···+y 2 n− 1 )]

i.e.Simpson’s rule states:

∫b

a

ydx≈

1

3

(

width of

interval

){(

first + last

ordinate

)

+ 4

(

sum of even

ordinates

)

+ 2

(

sum of remaining

odd ordinates

)}

(5)

Note that Simpson’s rule can only be applied when an

even number of intervals is chosen, i.e. an odd number

of ordinates.

Problem 6. Use Simpson’s rule with (a) 4

intervals, (b) 8 intervals, to evaluate

∫ 3

1

2

√

x

dx,

correct to 3 decimal places.

(a) With 4 intervals, each will have a width of

3 − 1

4

,

i.e. 0.5 and the ordinates will occur at 1.0, 1.5,

2.0, 2.5 and 3.0. The values of the ordinates are as

shown in the table of Problem 1(b), page 436.

Thus, from equation (5):

∫ 3

1

2

√

x

dx≈

1

3

( 0. 5 )[( 2. 0000 + 1. 1547 )

+ 4 ( 1. 6330 + 1. 2649 )+ 2 ( 1. 4142 )]

=

1

3

( 0. 5 )[3. 1547 + 11. 5916

+ 2 .8284]

=2.929,correct to 3 decimal places.

(b) With 8 intervals, each will have a width of

3 − 1

8

, i.e. 0.25 and the ordinates occur at 1.00,

1.25, 1.50, 1.75,..., 3.0. The values of the ordi-

nates are as shown in the table in Problem 2,

page 436.

Thus, from equation (5):

∫ 3

1

2

√

x

dx≈

1

3

( 0. 25 )[( 2. 0000 + 1. 1547 )

+ 4 ( 1. 7889 + 1. 5119 + 1. 3333

+ 1. 2060 )+ 2 ( 1. 6330 + 1. 4142

+ 1. 2649 )]

=

1

3

( 0. 25 )[3. 1547 + 23. 3604

+ 8 .6242]

=2.928,correct to 3 decimal places.

It is noted that the latter answer is exactly the same as

that obtained by integration. In general, Simpson’s rule

is regarded as the most accurate of the three approximate

methods used in numerical integration.

Problem 7. Evaluate

∫ π

3

0

√(

1 −

1

3

sin^2 θ

)

dθ,

correct to 3 decimal places, using Simpson’s

rule with 6 intervals.