Higher Engineering Mathematics, Sixth Edition

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

,

2 π

9

,

5 π

18

and

π

3

Correspondingvalues 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

)

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 currentihas the

following values at equal intervals of

2.0milliseconds:

Time (ms) Currenti(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 idt.

Use Simpson’s rule to determine the approximate

charge in the 12millisecond 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 )]

=62mC

Now try the following exercise

Exercise 175 Further problemson

Simpson’s rule

In Problems 1 to 5, evaluate the definite integrals

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

sin2xdx (Use 10 intervals)

[1.260]

In Problems 6 and 7 evaluate the definite inte-

grals using (a) integration, (b) the trapezoidal rule,