438 Higher Engineering Mathematics
Problem 4. Use the mid-ordinate rule with (a) 4
intervals, (b) 8 intervals, to evaluate∫ 312
√
xdx,
correct to 3 decimal places.(a) With 4 intervals, each will have a width of3 − 1
4,
i.e. 0.5 and the ordinates will occur at 1.0, 1.5, 2.0,
2.5 and 3.0. Hence the mid-ordinatesy 1 ,y 2 ,y 3
andy 4 occur at 1.25, 1.75, 2.25 and 2.75. Corre-
spondingvalues of2
√
xare shownin thefollowing
table.x2
√
x1.25 1.78891.75 1.5119
2.25 1.33332.75 1.2060From equation (2):
∫ 312
√
xdx≈( 0. 5 )[1. 7889 + 1. 5119
+ 1. 3333 + 1 .2060]
=2.920,correct to 3 decimal places.
(b) With 8 intervals, each will have a width of 0.25
and the ordinates will occur at 1.00, 1.25, 1.50,
1.75,...and thus mid-ordinates at 1.125, 1.375,
1.625, 1.875...
Corresponding values of2
√
xare shown in the
following table.x2
√
x1.125 1.88561.375 1.7056
1.625 1.56891.875 1.46062.125 1.37202.375 1.2978
2.625 1.23442.875 1.1795From equation (2):∫ 312
√
xdx≈( 0. 25 )[1. 8856 + 1. 7056+ 1. 5689 + 1. 4606 + 1. 3720+ 1. 2978 + 1. 2344 + 1 .1795]=2.926,correct to 3 decimal places.As previously, the greater the number of intervals
the nearer the result is to the true value (of 2.928, correct
to 3 decimal places).Problem 5. Evaluate∫ 2. 40e−x^2(^3) dx, correct to 4
significant figures, using the mid-ordinate rule with
6intervals.
With 6 intervals each will have a width of
2. 4 − 0
6
,i.e.
0.40 and the ordinates will occur at 0, 0.40, 0.80, 1.20,
1.60, 2.00 and 2.40 and thus mid-ordinatesat 0.20, 0.60,
1.00, 1.40, 1.80 and 2.20. Correspondingvalues of e
−x^2
3
are shown in the following table.
x e
−x 32
0.20 0.98676
0.60 0.88692
1.00 0.71653
1.40 0.52031
1.80 0.33960
2.20 0.19922
From equation (2):
∫ 2. 4
0
e
−x 32
dx≈( 0. 40 )[0. 98676 + 0. 88692
- 71653 + 0. 52031
- 33960 + 0 .19922]
=1.460,correct to 4 significant figures.
- 33960 + 0 .19922]