Numerical integration 437
With 6 intervals, each will have a width of
π
2
− 0
6
i.e.
π
12
rad (or 15◦) and the ordinates occur at
0 ,
π
12
,
π
6
,
π
4
,
π
3
,
5 π
12
and
π
2
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 173 Further problemson 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)betweenthelimitsx=aandx=b,asshown
in Fig. 45.2.
y 1 y 2 y 3 yn
O ab
yf(x)
x
y
ddd
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 )