Higher Engineering Mathematics

NUMERICAL INTEGRATION 435

H

Corresponding values of

1

1 +sinx

are shown in the

table below.

x

1

1 +sinx

0 1.0000

π

12

(or 15◦) 0.79440

π

6

(or 30◦) 0.66667

π

4

(or 45◦) 0.58579

π

3

(or 60◦) 0.53590

5 π

12

(or 75◦) 0.50867

π

2

(or 90◦) 0.50000

From equation (1):

∫ π
2

0

1

1 +sinx

dx≈

(π

12

){ 1

2

(1. 00000 + 0 .50000)

+ 0. 79440 + 0. 66667

+ 0. 58579 + 0. 53590

+ 0. 50867

}

=1.006, correct to 4

significant figures

Now try the following exercise.

Exercise 174 Further problems on the

trapezoidal rule

In Problems 1 to 4, evaluate the definite integrals

using thetrapezoidal rule, giving the answers

correct to 3 decimal places.

1.

∫ 1

0

2

1 +x^2

dx (Use 8 intervals) [1.569]

2.

∫ 3

1

2ln3xdx (Use 8 intervals) [6.979]

3.

∫π

3

0

√

(sinθ)dθ (Use 6 intervals) [0.672]

4.

∫ 1. 4

0

e−x

2

dx (Use 7 intervals) [0.843]

45.3 The mid-ordinate rule

Let a required definite integral be denoted again

by

∫b

aydxand represented by the area under the

graph ofy=f(x) between the limitsx=aandx=b,

as shown in Fig. 45.2.

a

ddd

O

y

y 1 y 2 y 3 yn

b x

y = f(x)

Figure 45.2

With the mid-ordinate rule each interval of width

d is assumed to be replaced by a rectangle of height

equal to the ordinate at the middle point of each

interval, shown asy 1 ,y 2 ,y 3 ,...ynin Fig. 45.2.

Thus

∫b

a

ydx≈dy 1 +dy 2 +dy 3 + ··· +dyn

≈d(y 1 +y 2 +y 3 + ··· +yn)

i.e.the mid-ordinate rule states:

∫b

a

ydx≈(width of interval)

(sum of mid-ordinates)

(2)