Higher Engineering Mathematics

438 INTEGRAL CALCULUS

y

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

a

ddd

y = f(x)

Oxb

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 434.

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 434.

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.