# Improper integrals - PowerPoint by 2U3cbl0

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```									Improper integrals
These are a special kind of limit. An
improper integral is one where either
the interval of integration is infinite, or
else it includes a singularity of the
function being integrated.
Examples of the first
kind are:
              
1
                1 x2
x
e dx   and            dx
0              -
Examples of the second kind:
2
1                                   3
1

0     x
dx           and           

tan x dx
4

The second of these is subtle because the singularity of tan x
occurs in the interior of the interval of integration rather than at
one of the endpoints.
Same method
No matter which kind of improper integral (or
combination of improper integrals) we are
faced with, the method of dealing with them is
the same:
                                                 b

 e dx means the same thing as lim  e dx
x                                                x
b 
0                                                 0
Calculate the limit!
What is the value of this limit

(and hence, of the improper integral    e  x dx )?
A. 0                      0

B. 1

C.   e
b
D.   e
E.   
Another improper integral

1
Recall that         dx  arctan x  C.
1 x 2


1
What is the value of           dx ?
-
1 x 2

A. 0
D.   
B. 

4
E.

C. 2
The integral you just worked represents all of the area
between the x-axis and the graph of 1 2
1 x

Area between the x-
axis and the graph
The other type...
For improper integrals of the other type, we make
the same kind of limit definition:

1                                         1
1                               1

0     x
dx is defined to be lim 
a 0 
a  x
dx.
Another example:
What is the value of this limit, in
1
1
other words, what is   
0    x
dx ?
A. 0

B. 1

C. 2

D.     2

E.   
A divergent improper integral
It is possible that the limit used to define the
improper integral fails to exist or is infinite. In
this case, the improper integral is said to diverge
. If the limit does exist, then the improper
integral converges. For example:
1
1
 x a 0 
0
dx  lim ln 1  ln a  

so this improper integral diverges.
One for you:
2
3

Does  tan( x)dx converge or diverge?

4

A. Converge

B. Diverge
Sometimes it is possible...
to show that an improper integral converges without
actually evaluating it:
1         1
Since 4             4 for all x  0, we have that
x  x7 x
b                  b
1             1             1
 x 4  x  7 dx   x 4 dx  3  3b3 for all b  1.
1                  1
1

So the limit of the first integral must be finite as b goes to
infinity, because it increases as b does but is bounded
above (by 1/3).
A puzzling example...
Consider the surface obtained by rotating the graph
of y = 1/x for x > 1 around the x-axis:

Let’s calculate the volume contained
inside the surface:


V    dx  b 1  b    cubic units.
1 2
x  lim       1

1
This is equal to...
                                       
2     1
SA   2 f ( x) 1  f ' ( x) dx  
2
1  4 dx
1                             1
x    x
This last integral is difficult (impossible) to evaluate
directly, but it is easy to see that its integrand is bigger

than that of the divergent integral         2

1
x
dx
Therefore it, too is divergent, so the surface has infinite
surface area.
This surface is sometimes called "Gabriel's horn" -- it is a
surface that can be "filled with water" but not "painted".

```
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