Higher Engineering Mathematics

(Greg DeLong) #1
NUMERICAL INTEGRATION 439

H

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

,

2 π
9

,

5 π
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

2 π
9

5 π
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

)


1
3


18

)
[(1. 0000 + 0 .8660)+4(0. 9950

+ 0. 9574 + 0 .8969)

+2(0. 9803 + 0 .9286)]

=

1
3


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) Currenti(A)
00
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 idt.

Use Simpson’s rule to determine the approxi-
mate charge in the 12 millisecond period.

From equation (5):

Charge,q=

∫ 12. 0

0

idt≈

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 176 Further problems on
Simpson’s rule

In problems 1 to 5, evaluate the definite inte-
grals usingSimpson’s rule, giving the answers
correct to 3 decimal places.

1.

∫π
2

0


(sinx)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

sinθ
θ

dθ (Use 8 intervals) [0.747]

4.

∫π
2
0

xcosxdx (Use 6 intervals) [0.571]

5.

∫π
3

0

ex

2
sin 2xdx (Use 10 intervals)

[1.260]

In problems 6 and 7 evaluate the definite inte-
grals using (a) integration, (b) the trapezoidal
rule, (c) the mid-ordinate rule, (d) Simpson’s
rule. Give answers correct to 3 decimal places.
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