Lesson -10
Introduction to Calculus
Area under the curve (estimation)
There are several practical situations where it is necessary to
estimate the area of irregular figures. Example include estimation of
areas of plots of land by surveyors, area of indicator diagrams of steam
engines by engines and area of water planes and transverse sections of
a ship by naval architects.
There are many methods whereby the area of an irregular plane surface.
A surface may be found from.
1.
Trapezoidal rule
2.
Mid-ordinate rule
3.
Simpson's rule
Trapezoidal rule
To find the area under curve and x-axes as shown in Fig. 10.1, the
base DC is divided into a number of equal intervals of with d. This can be
any number the greater the number, the more accurate the result.
Fig.10.
An ordinates y1, y2, y3 . . . .y10 are accurately measured. The
approximation used in this rule is to assume that each strip is equal to the
area of a trapezium AA'D' D
The area of a trapezium is equal to sum of the parallel sides, times one of
second (1/2) perpendicular distance between the parallel sides.
Hence for the first strip, shown in Fig.10.1 the approximate area is:
A1= ½( y1 + y2) *d
For the second strip, the approximate area is
A2= ½( y2 + y3) *d
,
and so on.
Hence the approximate area of
ABCD = A1 + A2 + A3 +. . . A10
1/2(y1 + y2) * d + 1/2(y2 + y3) * d + 1/ 2(y3 + y4)*d + . .1/2 (y9 + y10) *d
=
d* [1/2y1 +1/2 y2 + 1/2 y2 + 1/2 y3 + 1/2 y3 +1/2 y4 + . . .1/2 y9 +
+ 1/2y10 ]
Inside of the bracket only 1/2y1 and 1/2y10 doesn’t has a pair and then
Area = d*[1/2 y1 + y2 + y3 + y4 + . . . +1/2 y10] =
y1  y10
= d*[ 2
+ y2 + y3 + y4 + . . . y9]
Generally, the trapezoidal rule states that the area of an irregular figure is
given by: Area = (width of interval)*[1/2((first + last ordinate) + sum of
remaining ordinates]
Area = d*[
y1  y n
(y2 + y3 + y4 + . . . .yn-1)]
2
(10.1.)
Example 1
An indicator diagram of a steam engine is 9.00 cm long. Seven
evenly spaced ordinates, including the end ordinates are measured
with the following results: 5.10, 4.60, 3.20, 2.70, 2.32, 2.18, 2.06 cm
Find the area of the diagram and the mean pressure in the cylinder, if
the mean pressure Pa is equal to:
Pa = Area of diagram
Base
A sketch of the indicator diagram is shown below
Area = 9.00 ( 5.10  2.06 + 4.60 + 3.20 + 2.70 + 2.32 + 2.18) =
6
2
= 1.5(3.58 + 4.60 + 3.20 + 2.70 + 2.32 + 2,18) =
= 1.5*18.58 = 27.87 cm2
Pa = Area of diagram = 27.87 = 3.09 Nm-2 = 309.7 kNm-2
base
9.00
2.Mid-ordinates rule
Fig. 10.2
To find the area of ABCD of Fig.10.2, the base is divided into number
of equal strips width d. As with the trapezoidal rule, the greater the number
of
intervals used the more accurate the result.). If each strip assumed to be a
rectangular (see AA'D'D in figure above) and area of it is equal to base
multiplied by mid-ordinate yi. Hence, the approximate area
equal to:
Area =y1d + y2d + y3d + . . . + ynd
where
d=
length of DC
number of mid  ordinates
of
ABCD
is
n – number of strips
Area = d*[y1 + y2 + y3 + . . . + yn]
(10.2.)
Generally, the mid-ordinate rule states that the area an irregular figure is
given by:
Area = (width of interval)*(sum of mid-ordinates)
Example 2.
The values of the y ordinates of a curve and their distance x
from the origin are given below. Plot the graph and find the area
under the curve by mid-ordinate rule.
x = 0, 1, 2, 3, 4, 5, and 6
y = 2, 5, 8, 11, 14, 17, and 20
Solution:
Area = d* [y1 + y2 + y3 + . . .+ yn] and
Mid-ordinate =
y n1  y n
2
Using 6 intervals of width 1 the mid-ordinates of the 6 strips are measured.
Each mid-ordinate is equal to average value between yi and yi+1 ordinates.
The area under the curve is given by:
= 1[3.5 + 6.5 + 9.5 + 12.5 + 15.5 + 18.5] = 66 square units.
3. Simpson's rule
Fig.10.3.
To find the area ABCD of Fig.10.3, the base DC must be divided into
an even number of strips of equal width d, thus producing an odd number
of ordinates. The length of each ordinate, y1, y2,
y3
and yn are accurately
measured. Simpson's rule states that the area of the irregular figure ABCD
is given by:
Area = 1/3d[(y1 + y9) +4(y2 + y4 + y6 +y8) +2(y3 + y5 + y7)]
More generally, the area under curve any irregular figure is given by:
Area = 1/3d [( first and last ) + 4( sum of even ) + 2( sum of odd )]
ordinates
ordinates
ordinates
Area =1/3d[(y1 + yn) +4(y2 + y4 + . . . + yn-1) +2(y3 + y5 + . . . + yn-2)]
(10.3)
When estimating areas of irregular figures Simpson's rule is generally
regarded as the most accurate of the approximate methods available.
Example 3
Sketch a semicircle of radius 10 cm. Erect ordinates at intervals
of 2 cm and determine the lengths of ordinates. Determine the area of
the semicircle by using of Simpson's rule. Calculate the percentage
inaccuracy, correct to two decimal places in respect of calculated
from the formula. The semicircle is shown below with the lengths of
the ordinates.
Note: The first and the last ordinates are equal to 0
Solution:
Area =2/3[(0 + 0)+4(6 + 9.15 + 10 + 9.15 + 6)+2(8 + 9.80 + 9.80
+ 8)] = 2 /3([0 + 4) + (40.30) +2(35.60)] =
= 2/3[161.2 + 71.21] = 154.93cm2
The true area jest:
r (10)2/2 = 157.08cm2
Percentage error = 154.93  157.08 x100= -1.36%
157.08
EXERCISES:
1.
Draw the graph of y = 6x + x2 by compiling a table of values for
y from x = 0 to x = 6. Find the area enclosed by the curve, a ordinates x
= 1 and x = 5 and x axis, using:
2.
a)
The trapezoidal rule
b)
The mid-ordinate rule
c)
Simpson's rule
Plot the graph of y = 3x + 6 between x 1 and x = 4. Estimate the
area enclosed by the curve, the ordinates x = 1 and x = 4 and the xaxis by the mid-ordinate rule.
3.
Find the area enclosed by the curve y = 8x + 20 - x2 and the
positive x and y-axis using Simpson's rule.
4.
The shape of a piece of land is shown below. To estimate
the area of the land, measurements at intervals of 100 m are taken
perpendicular to the straight portion with the results shown above, the
dimensions being in metres. Estimate the area of the land in hectares
(1 hectare = 104m2)
5.
An indicator diagram of a steam engine is 12.0cm in length.
Seven evenly spaced ordinates, including the end ordinate, are
measured with the following results: 6.30, 5.73, 4.31, 3.74, 3.42, 3.31,
3.17cm. Make an estimation of the area of the diagram and from it
find the mean pressure in the cylinder, if the pressure scale is 95.0
kNm-2 to 1.00cm.
6.
A stream of width 18.0 is measured for depth at intervals of
3.0m across its width. The results were:
Distance from bank (m)
0
3.0
6.0
9.0 12.0 15.0 18.0
Depth,(m)
0
1.21 2.23 4.14 3.06 1.72 0
Determine the cross sectional area of the flow of water at this point.
7.
A vehicle starts from rest and its velocity is measured every
second for 6.0 seconds.
Time, t (seconds)
0
1.0
2.0
3.0
14.0 5.0
6.0
Velocity, v (ms-1)
0
1.2
2.4
3.7
5.2 6.0
9.2
Using Simpson's rule calculates the distance travelled in 6.0 seconds
(i.e. the area under the v = f (t) graph) and average speed over this
period (i.e. the area-time).
8.
To estimate the amount of earth to be removed when
constructing a cutting, the cross-sectional area at intervals
of
10m
were firstly estimated and the results are shown below:
Cross-sectional area (m2)
0
3.20 4.10 4.90 4.50 2.80 0
Calculate the volume of earth to be excavated.
9.
A tree trunk is 12.0 m in length and has a varying circular cross
section. The cross sectional area at intervals of 2 m measured from
one end are: 0.95
0.48
0.87
0.76
0.66
0.54
0.45 m2. Estimate the volume of the tree trunk.
10.
The areas of seven horizontal cross-sections of a reservoir at
intervals of 20m are:
330 380 440 460 350 270 220m2.
How many litters of water will fill the reservoir? Express the answer in
scientific notation form.
11.
A cooling tower is 20m high. The inside diameter of the tower at
different heights is given in the following table:
Height (m)
0
Diameter (m)
16.00
5.00
10.00
15.00
13.30
10.70
8.60
20.00
8.00
Estimate the capacity of the tower in m3.
12.
The pressure (p) of a gas at a given temperature varies with the
volume (v) as shown below:
v (dm3) 1.00
1.23
1.43
1.77
2.27
3.58
p (Pa)
75
60
45
30
15
1.00
Plot pressure against volume (p-x axis) and determine the work done
in compressing the gas from 3.58dm3 to 1. 0dm3. Measuring the area
under a curve does this? Hint : apply properly units.
13.
A piece of timber has a cross-sectional area A
perpendicular
to
its length at distance d from its end, as shown below:
d ( m)
0
2
4
6
8
10
12
A (m2)
2. 8 4. 7 6. 2 7. 3 7. 1 6. 7 6. 2
Draw a graph of A against d using Simpson's rule with 12 intervals to
find the volume of the piece of timber.
4.
The heat capacity (C) of nitrogen varies with the natural
logarithm (ln) of the absolute temperature (T) is shown below:
ln T 5.70
C (JK-1) 28.8
5.86 5.99
6.11
6.21
6.31
6.40
29.1 29. 4
29. 7
30.0
30.3
30.6
Plot C against Ln T (x-axes) and determine the entropy change between
values of ln T of 5.70 and 6.40 by measuring the area under the curve.
Answers:
1.
30.67 square units
2.
81 square units
3.
266.7 sq. units
4.
16.63ha
5.
50.70cm2 , 401.4kNm-2
6.
38.9m2
7.
22.7m, 3. 78ms-1
8.
202.7m3
9.
8. 0 3m3
10.
4.38 x07litters
11.
2032m3
12.
135W
13.
73.5m3
14.
20.5JK-1