Numerical integration 441
With 6 intervals, each will have a width of
π
3
− 0
6
i.e.
π
18
rad (or 10 ◦ ), and the ordinates will occur at
0,
π
18
,
π
9
,
π
6
,
2π
9
,
5π
18
and
π
3
Corresponding values of
1 −
1
3
sin 2 θ
are shown in
the table below.
θ
0
π
18
π
9
π
6
(or 10 ◦ ) (or 20 ◦ ) (or 30 ◦ )
1 −
1
3
sin
2
θ
1.0000 0.9950 0.9803 0.9574
θ
2π
9
5π
18
π
3
(or 40 ◦ ) (or 50 ◦ ) (or 60 ◦ )
1 −
1
3
sin
2
θ
0.9286
0.8969
0.8660
From Equation (5)
π
3
0
1 −
1
3
sin
2
θ
dθ
≈
1
3
π
18
[(1.0000 + 0.8660) + 4(0.9950
+ 0.9574 + 0.8969)
+ 2(0.9803 + 0.9286)]
=
1
3
π
18
[1.8660 + 11.3972 + 3.8178]
= 0.994, correct to 3 decimal places.
Problem 8. An alternating current i has the
following values at equal intervals of
2.0 milliseconds:
Time (ms) Current i (A)
0
0
2.0
3.5
4.0
8.2
6.0
10.0
8.0
7.3
10.0
2.0
12.0
0
Charge, q, in millicoulombs, is given by
q =
12.0
0
i dt.
Use Simpson’s rule to determine the approximate
charge in the 12 millisecond period.
From equation (5):
Charge, q =
12.0
0
i dt ≈
1
3
(2.0) [(0 + 0) + 4(3.5
+10.0 + 2.0) + 2(8.2 + 7.3)]
= 62 mC
Now try the following exercise
Exercise 175 Further problems on
Simpson’s rule
In Problems 1 to 5, evaluate the definite integrals
using Simpson’s rule, giving the answers correct
to 3 decimal places.
1.
π
2
0
(sin x) dx (Use 6 intervals) [1.187]
2.
1.6
0
1
1 + θ 4 dθ (Use 8 intervals)
[1.034]
3.
1.0
0.2
sin θ
θ
dθ
(Use 8 intervals)
[0.747]
4.
π
2
0
x cos x dx (Use 6 intervals) [0.571]
5.
π
3
0
e
x 2 sin 2x dx (Use 10 intervals)
[1.260]
In Problems 6 and 7 evaluate the definite integrals using (a) integration, (b) the trapezoidal rule,