2.

2 Integrals Whose Integrands Become Infinite

Knew that, if a function f is not bounded on an interval [a,b], then f is
not integrable on [a,b].
However, it is possible to extend the concept of integral to include
functions that are not bounded by defining the integral of such function
as a limit of integrals of bounded functions in an appropriate way.
These limits are also called improper integrals.
If f is continuous on [a,b), but is unbounded at b, i.e fails to have
limit as x approaches b from the left ,
b

 f ( x)dx is defined as
a
b c

 f ( x)dx  lim  f ( x)dx

a
c b 
a

Limit exist-converges, otherwise diverges

Note : a  c  b 1
 Example
Evaluate
1 This integral is improper since the integrand approaches
dx
1.

0 1 x
infinity as x approaches 1(the upper limit) from the left.

Solution
f
1

0 c1

 
1
dx c
dx

c
By definition  lim   lim  2 1  x 0
0 1 x c 1
0 1 x c 1


 lim  2 1  c  2  2
c 1

2
 If f is continuous on (a,b], but is unbounded at a, i.e fails to have
limit as x approaches a from the right ,
b b b

 f ( x)dx is defined as  f ( x)dx  clim

a
a  f ( x)dx 
a c

ac b

Limit exist-converges, otherwise diverges

3
 Example
2
dx
The integral 1 1  x is improper
1c 2
-1
since the integrand approaches -ve
f
infinity as x approaches 1(the lower
limit) from the right.

2
dx 2

 lim   ln 1  x  c
dx
1 1  x  clim 
2
1 1  x


 lim   ln  1  ln 1  c 
c 1
c

c 1
 lim  ln 1  c   
c 1
--Diverges

4
 Suppose f is continuous on [a,b], except at point c in (a,b)
where
f   as x  c  and as x  c  (as in picture)

f
f   at point c

then a c b
b c b

 f ( x)dx
a
converges if both  f ( x)dx and  f(x)dx converge.
a c

Define b c b

 f ( x)dx   f ( x)dx   f ( x)dx

a a c
5
 Examples
Evaluate 1
f ( x) 
1. 4 dx ( x  2) 2 / 3
1  x  2 2 / 3
Solution
1 2 4

4
dx 2
dx
4
dx
1  x  2 2 / 3    
1
( x  2) 2/3
2
( x  2) 2/3
c 4
dx dx
 lim  2 / 3  lim 
c2
1
( x  2) c 2
c
( x  2) 2/3
6
 c
dx
lim    1/ 3 c
lim [3( x 2) ]1
 2/3 
c2
1
( x 2) c2

 lim[3(c  2)  3(1) ]  3
1/ 3 1/ 3
c2
4
dx
lim    1/ 3 4
lim [3( x 2 ) ]c
c2
c
( x  2 ) 2/3 c2 

 lim[3(2)  3(c  2) ]  33 2
1/ 3 1/ 3
c2
4
So dx
1  x  2 2 / 3  3  3 2
3

7
 4
2. dx
1  x  2 2 f ( x) 
1
( x  2) 2

Solution
1 2 4
By definition
4 2 4

 f ( x)dx   f ( x)dx   f ( x)dx

1 1 2
c 4
dx dx
 lim   lim 
c2
1
( x  2) 2
c  2
c
( x  2) 2

8
 c c
dx  1 
lim   lim  
c2
1
( x  2) 2
c  2
 ( x  2) 1
 1 
 lim 
c  2 (c  2)
 1 
 
4
dx
So 1  x  2 2 does not converge.

(We do not have to check the other limit, since

at least one of the improper int. diverge, the
main one diverge).
9
 Sometimes we tend to integrate an improper
integral as proper one. See what happen!!!
2
dx
For example
0  x  1 2 We tend to do as below:

2 2
dx  1 
0  x  1 2   x  1  1  1  2
0

This is nonsense, This function lies above

since the integral x-axis, the integration
cannot be should be positive!!
negative! Why??
10
 Examples OF PAST FINAL EXAMS

1. (a) Sketch the region that represent the following improper integral.
 5
1 1
i)  e x
dx ii ) 
0
x2
dx
2

(b) Show that


x

1 x  4 x  3x  2
4 3 2
dx is divergent.

4
1
2. (a) Evaluate  ( x  2)
1
2
3
dx


 dx
x
(b) i) Evaluate the integral e
1

ii) Using the result obtained in part (b)(i), 

e
 x2
what can you say about the integral dx 11
1
3. (a) Find a positive number a satisfying
a 
dx dx
0 1  x 2  a 1  x 2 .
5
1
(b) Consider the integral 
2 x2
dx.

i) Is this an improper integral? Give your reason.

ii) Compute its value.

iii) Determine whether it converges or diverges.


x2  x  2
(c) Determine whether  4 dx
2
x  x 1
2

converges or
Give reasons
diverges. for your
12
answer.