Higher Engineering Mathematics

436 INTEGRAL CALCULUS

Problem 4. Use the mid-ordinate 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. Hence the mid-ordinates

y 1 ,y 2 ,y 3 andy 4 occur at 1.25, 1.75, 2.25 and

2.75. Corresponding values of

2

√

x

are shown in

the following table.

x

2

√

x

1.25 1.7889

1.75 1.5119

2.25 1.3333

2.75 1.2060

From equation (2):

∫ 3

1

2

√

x

dx≈(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 of

2

√

x

are shown in the

following table.

x

2

√

x

1.125 1.8856

1.375 1.7056

1.625 1.5689

1.875 1.4606

2.125 1.3720

2.375 1.2978

2.625 1.2344

2.875 1.1795

From equation (2):

∫ 3

1

2

√

x

dx≈(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. 4

0

e

−x^2

(^3) dx, correct to
4 significant figures, using the mid-ordinate rule
with 6 intervals.
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-ordinates at
0.20, 0.60, 1.00, 1.40, 1.80 and 2.20. Corresponding
values of e
−x^2
(^3) are shown in the following table.
x e
−x^2
3
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^2
(^3) dx≈(0.40)[0. 98676 + 0. 88692

• 
  
  
  
  0. 71653 + 0. 52031
  
  
  
  


• 
  
  
  
  0. 33960 + 0 .19922]
     =1.460, correct to
     4 significant figures