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

484 Chapter 16 Further topics in integration
If either (1) or (2) is true the integral is called an improper integral. Evaluation of
improper integrals involves the use of limits.
∫ ∞
1
Example16.5 Evaluate
2 t dt. 2
Solution
∫ ∞
2
1
[−
t dt = 1 ] ∞ 2 t
2
To evaluate − 1 atthe upper limitwe consider lim
t −1 . Clearly the limitis0.Hence,
t→∞ t
∫ ∞
1
(−
t dt=0− 1 )
= 1 2 2 2
2
∫ 1
Example16.6 Evaluate e 2x dx.
−∞
Solution
∫ 1
−∞
[ e
e 2x 2x
dx =
2
] 1
−∞
We need toevaluate lim
x→−∞
∫ 1
−∞
e 2x
. This limitis 0.So,
2
[ ] e
e 2x 2x 1
dx = = e2
2
−∞
2 −0=3.69
Engineeringapplication16.1
Capacitorsinseries
Engineers are often called upon to simplify an electronic circuit in order to make
it easier to analyse. One of the arrangements frequently met is that of two or more
capacitors connected together in series. It is useful to be able to replace this configuration
by a single capacitor with a capacitance value equivalent to that of the
original capacitors in series. Derive an expression for the equivalent capacitance of
two capacitors connected togetherinseries (see Figure 16.1).
Solution
In Engineering application 13.2 we obtained an expression for the voltage across a
capacitor. Thiswas
v = 1 ∫
idt
C