of Piecewise Continuous Functions
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.
the following :
[.] is the greatest integer function. .
(i). I =, we
know cot–1 x ∈ (0, p) " x ∈
cot–1 x =
30 + cot1 + cot2 + cot3
(iii). Let I =
([sin x + cosx] is
periodic function with period 2p)
x + cos x] =
Hence I =
(iii). I =
Hence I =
(iv). I =
Evaluate dx, where n is a
natural number greater than 1 and [.] denotes the greatest integer function.
Let [n e-x] = k, k ≥ 0, k ∈ I
&⇒ k £ n e-x < k + 1
&⇒ £ e-x <
&⇒ ln< x £ ln
(n – 1).ln+ (n – 2) +.....+[ln n – ln]
(n – 1) ln+ (n – 2) ln+ ... + ln2
TWO USEFUL FORMULAE
1. If n be a positive
, when n is even
, when n is
when both m and n ∈
when either of m or
integral reducible to above forms can be solved directly, particularly if the
problem is objective one.