Integration
of Piecewise Continuous Functions
Any
function f (x) which is discontinuous at finite number of points in an interval
[a, b] can be made continuous in sub-intervals by breaking the intervals into
these subintervals. If f (x) is discontinuous at points x1, x2,
x3, . . . . xn in (a, b), then we can define subintervals
(a, x1), (x1, x2)…… (xn-1, xn),
(xn, b) such that f (x) is continuous in each of these subintervals.
Such functions are called piecewise continuous functions. For integration of piecewise
continuous function, we integrate f(x) in these sub-intervals and finally add
all the values.
Example.1
Evaluate
the following :
(i).
(ii).
,
(iii).
,
(iv).
,
Here
[.] is the greatest integer function. .
Solution
(i). I =
, we
know cot–1 x ∈ (0, p) " x ∈
R
Thus
cot–1 x = 
Hence
I =
=
30 + cot1 + cot2 + cot3
(iii). Let I = 
(
[sin x + cosx] is
periodic function with period 2p)
[sin
x + cos x] = 
Hence I =
I
= n
(iii). I = 
Hence I = 
(iv). I = 
=
Example.2
Evaluate 
dx, where n is a
natural number greater than 1 and [.] denotes the greatest integer function.
Solution
Let [n e-x] = k, k ≥ 0, k ∈ I
&⇒ k £ n e-x < k + 1
&⇒ 
£ e-x < 
&⇒ ln
< x £ ln
Therefore
=
dx +....+
+
=
(n – 1).ln
+ (n – 2) 
+.....+[ln n – ln
]
=
(n – 1) ln+ (n – 2) ln
+ ... + ln2
=
ln
=
ln
= ln 
TWO USEFUL FORMULAE
1. If n be a positive
integer, then.
=
, when n is even
=
, when n is
odd
2. 
=
when both m and n ∈
even integer
=
when either of m or
n ∈
odd integer
Examples:
(i)
(ii)
(iii)
(iv)
Any
integral reducible to above forms can be solved directly, particularly if the
problem is objective one.
