082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

13.3 Definite and indefinite integrals 451
Substituting forLandC inthe equation forZ 0
( )
µ 0 b Z 0
=
2π ln a
/ ( ) = 1 √ ( )
µ0 b
ln
√2πε r
ε 0 b 2π ε r
ε 0
a
ln
a
For the cable defined in Engineering application 13.4, and substituting for the permeability
of free space, µ 0
= 4π ×10 −7 ,
Z 0
= 1 √
( )
µ 0 1.5 ×10
−3
2π 1.55 ×8.85 ×10 ln =52
−12 0.51 ×10 −3
13.3.1 Useofadummyvariable
Consider the following integrals,I 1
andI 2
:
I 1
=
∫ 1
Then,
[ ] t
3 1
I 1
= =
3
0
0
[ x
3
I 2
=
3
t 2 dt I 2
=
] 1
0
=
∫ 1
0
x 2 dx
( 1
3)
−(0)= 1 3
( 1
3)
−(0)= 1 3
So clearlyI 1
= I 2
. The value ofI 1
does not depend upont, and the value ofI 2
does not
depend uponx. Ingeneral,
I =
∫ b
a
f(t)dt =
∫ b
a
f(x)dx
Because the value ofI is the same, regardless of what the integrating variable may be,
wesayxandt are dummy variables. Indeed we could write
I =
∫ b
a
f(z)dz=
∫ b
a
f(r)dr=
Thenz,randyaredummy variables.
∫ b
a
f(y)dy
EXERCISES13.3
1 Evaluate the following integrals:
∫ 3 ∫ 4
(a) x 3 1
dx (b)
1 1 x dx
∫ 1 ∫ 1
(c) 2 dx (d) e x dx
0
−1
∫ π/3
(e) sint dt
0
∫ π/2
(g) cos3t dt
0
∫ 1.2
(i) tanx dx
1
(f)
(h)
∫ π
sin(t +3)dt
0
∫ 2
cos πt dt
1