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Basic Concepts Of Definite Integral Part 6

Posted on - 04-01-2017

Math

IIT JEE

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.

 
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