Higher Engineering Mathematics

IRREGULAR AREAS, VOLUMES AND MEAN VALUES OF WAVEFORMS 217

C

(d) Simpson’s rule

To determine the areaPQRSof Fig. 20.1:

(i) Divide basePSinto anevennumber of

intervals, each of width d (the greater

the number of intervals, the greater the

accuracy).

(ii) Accurately measure ordinatesy 1 ,y 2 ,y 3 ,

etc.

(iii) AreaPQRS=

d

3

[(y 1 +y 7 )+4(y 2 +y 4 +

y 6 )+2(y 3 +y 5 )]

In general, Simpson’s rule states:

Area=

1

3

(

width of

interval

)[(

first+last

ordinate

)

+ 4

(

sum of even

ordinates

)

+ 2

(

sum of remaining

odd ordinates

)]

Problem 1. A car starts from rest and its speed

is measured every second for 6 s:

Time

t(s)0123456

Speedv

(m/s) 0 2.5 5.5 8.75 12.5 17.5 24.0

Determine the distance travelled in 6 seconds

(i.e. the area under thev/tgraph), by (a) the

trapezoidal rule, (b) the mid-ordinate rule, and

(c) Simpson’s rule.

A graph of speed/time is shown in Fig. 20.3.

(a)Trapezoidal rule(see para. (b) above)

The time base is divided into 6 strips each of

width 1 s, and the length of the ordinates mea-

sured. Thus

area=(1)

[(

0 + 24. 0

2

)

+ 2. 5 + 5. 5

+ 8. 75 + 12. 5 + 17. 5

]

=58.75 m

Figure 20.3

(b)Mid-ordinate rule(see para. (c) above)

The time base is divided into 6 strips each of

width 1 second.

Mid-ordinates are erected as shown in Fig. 20.3

by the broken lines. The length of each mid-

ordinate is measured. Thus

area=(1)[1. 25 + 4. 0 + 7. 0 + 10. 75

+ 15. 0 + 20 .25]

=58.25 m

(c)Simpson’s rule(see para. (d) above)

The time base is divided into 6 strips each of

width 1 s, and the length of the ordinates mea-

sured. Thus

area=^13 (1)[(0+ 24 .0)+4(2. 5 + 8. 75

+ 17 .5)+2(5. 5 + 12 .5)]

=58.33 m

Problem 2. A river is 15 m wide. Soundings

of the depth are made at equal intervals of 3 m

across the river and are as shown below.

Depth (m) 0 2.2 3.3 4.5 4.2 2.4 0

Calculate the cross-sectional area of the flow of

water at this point using Simpson’s rule.