basic_engineering_mathematics0

192 Basic Engineering Mathematics
(d) Simpson’s rule
To determine the area PQRS of Fig. 25.1:
(i) Divide base PS into an even number of intervals, each of
width d (the greater the number of intervals, the greater
the accuracy).
(ii) Accurately measure ordinates y 1 , y 2 , y 3 , etc.
(iii) Area PQRS
= 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 (width of interval) [(first + last ordinate)
3
+ 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) 0 1 2 3 4 5 6
Speed v (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 the v/t graph), by (a) the trapezoidal rule, (b) the
mid-ordinate rule, and (c) Simpson’s rule.
A graph of speed/time is shown in Fig. 25.3.
(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 measured. Thus
[( ) 0 + 24.0
area = (1)
+ 2.5 + 5.5 + 8.75
2
]
+ 12.5 + 17.5 = 58.75 m
(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. 25.3 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 measured. Thus
area = 1 (1)[(0 + 24.0) + 4(2.5 + 8.75 + 17.5)
3
+ 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.
From para. (d) above,
Speed (m/s)
30
25
20
15
10
5
Fig. 25.3
1.25
Graph of speed/time
2.5
4.0
5.5
7.0
8.75
10.75
12.5
15.0
0 1 2 3
Time (seconds)
17.5
20.25
24.0
4 5 6
Area = 1 (3)[(0 + 0) + 4(2.2 + 4.5 + 2.4) + 2(3.3 + 4.2)]
3
= (1)[0 + 36.4 + 15] = 51.4 m 2
Now try the following exercise
Exercise 91
Further problems on areas of irregular
figures (Answers on page 280)
1. Plot a graph of y = 3x − x 2 by completing a table of
values of y from x = 0tox = 3. Determine the area
enclosed by the curve, the x-axis and ordinate x = 0 and
x = 3 by (a) the trapezoidal rule, (b) the mid-ordinate
rule and (c) by Simpson’s rule.
2. Plot the graph of y = 2x 2 + 3 between x = 0 and x = 4.
Estimate the area enclosed by the curve, the ordinates
x = 0 and x = 4, and the x-axis by an approximate
method.