Transcript WEEK 5 - University of Pennsylvania

Notes, part 4
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:

e
dx

0
x

and
1
dx
-1  x 2
Examples of the second kind:
1

0
1
dx
x
2
3
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:

e
0
x
b
dx meansthesame thingas lim e dx
b 
0
x
Calculate the limit!
What is the value of this limit

(and hence, of the improper integral
A. 0
B. 1
C.
e
D.
e
E.

b
x
e
 dx )?
0
Another improper integral
1
Recall that
dx  arctanx  C.
2
1 x

1
Whatis the value of 
dx ?
2
1 x
-
A. 0
B. 4

C. 2
D.

E.

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 xaxis and the graph
The other type...
For improper integrals of the other type, we make
the same kind of limit definition:
1

0
1
1
1
dx is defined to be lim 
dx.
a 0 
x
x
a
Another example:
What is the value of this limit, in
1
other words, what is

0
A. 0
B. 1
C. 2
D.
E.
2

1
dx ?
x
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
ln 1  ln a  
0 x dx  alim
0 
so this improper integral diverges.
One for you:
2
3
Does  tan(x)dx convergeor 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
1
dx

dx


for
all
b

1
.
3
4
4
3
1 x  x  7 1 x
3b
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).
Arc length
The length of a curve in the plane is generally difficult
to compute. To do it, you must add up the little “pieces
of arc”, ds. A good approximation to ds is given by the
2
Pythagorean theorem:
 dy 
ds  dx 2  dy 2  1    dx
 dx 
We can use this to find the length of any graph –
provided we can do the integral that results!
Find the arclength of the parabola y = x2 for x between
-1 and 1. Since dy / dx = 2x, the element of arclength is
2
1 4 x dx
so the total length is:
L
1

1
1  4 x dx
2
So far, we have that the length is
L
1

1  4 x dx
2
1
To do this integral, we will need a trig
substitution. But, appealing to Maple, we
get that
1
L
1
1
1  4t dt  5  ln( 5  2)
2
2
Can we do the integral?
1
The arc length integral from before was:

1  4t 2 dt
1
This is a trig substitution integral of the second kind:
With the identity tan2 + 1 = sec2 in mind, let
What about dt ? Since
t  12 tan
4t  tan 
2
2
we have that
dt  12 sec2  d
These substitutions transform the integral into

1  4t dt   sec  d
2
1
2
3
This is a tricky integral we
need to do by parts!
To integrate
Let
sec

d


3
u  sec , dv  sec  d
Then du  sec tan
2
, v  tan
3
2
sec

d


sec

tan


sec

tan
 d


But tan2 = sec2 - 1 , so rewrite the last integral and get
 sec
3
 d  sec  tan    sec  d   sec  d
3
Still going….
 sec
3
 d  sec  tan    sec  d   sec  d
3
It’s remarkable that we’re almost done. The integral
of secant is a known formula, and then you can add
the integral of sec3 to both sides and get
1
1
sec

d


sec

tan


2
2 ln(sec   tan  )  C

3
So we’ve got this so far for 2 1  4t 2 dt where t  12 tan
We need a triangle!
So far,
with
2 1  4t 2 dt  12 sec tan  12 ln(sec  tan )  C
t  12 tan
1  4t 2
From the triangle,

tan  2t , sec  1  4t
So

1  4t 2 dt 
2t
1
4
1
2


1  4t 2 2t   14 ln 1  4t 2  2t  C
Definite integral:
So far, we have



1  4t dt  t 1  4t  ln 1  4t  2t  C
2
1
2
2
1
4
2
Therefore
1

1
1  4 x dx  5 
2
1 ln 
4 

5 2

 52
To get the answer Maple got before, we’d
have to rationalize the numerator inside the
logarithm.
Surface Area
The area of a surface of revolution is calculated in a manner
similar to the volume. The following illustration shows the
paraboloid based on y  x (for x=0..2) that we used before,
together with one of the circular bands that sweep out its
surface area.
To calculate the surface area, we first need to determine
To calculate the surface area
the area of the bands. The one centered at the point (x,0)
has radius
2
2
ds

dx

dy
x and width equal to
.
Since we will be integrating with respect to x (there is a
band for each x), we'll factor the dx out of ds and write
2
 dy 
ds  1    dx . So the area of the band
 dx 
1
dx
centered at (x,0) is equal to: d  2 x 1 
4x
Thus, the total surface area is equal to the integral
  
1
x 4  dx
x
The surface area
turns out to be:
13 
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   
1
 dx  blim
1     cubic units.

1 2
x
1
b
What about the surface area?
This is equal to...


2
1
SA   2 f ( x) 1  f ' ( x) dx  
1  4 dx
x
x
1
1
2
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".
Sequences
The lists of numbers you generate using a numerical
method like Newton's method to get better and better
approximations to the root of an equation are examples of
(mathematical) sequences .
Sequences are infinite lists of numbers, a1 , a2 , a3 ,...
Sometimes it is useful to think of them as functions from
the positive integers into the reals, in other words,
a(1)  a1 , a(2)  a2 , and so forth.
Convergent and Divergent
The feeling we have about numerical methods like the
bisection method, is that if we kept doing it more and
more times, we would get numbers that are closer and
closer to the actual root of the equation. In other words
lim an  r where r is the root.
n
Sequences for which
lim an exists
n 
and is finite are called convergent, other sequences are
called divergent
For example...
The sequence
n
1, 1/2, 1/4, 1/8, 1/16, .... , 1/2 , ... is
convergent (and converges to zero, since
lim 21  0 ), whereas:
n
the sequence 1, 4, 9, 16, .…n2, ... is
divergent.
n
Practice
n 1 
2 3 4
, ...
The sequence  , , , ......,
n2 
3 4 5
A. Converges to 0
B. Converges to 1
C. Converges to n
D. Converges to ln 2
E. Diverges
Another...
 2 3 4

n n 1
The sequence  3 , 4 ,  5 , ......, (1) n  2 , ...


A. Converges to 0
B. Converges to 1
C. Converges to -1
D. Converges to ln 2
E. Diverges
A powerful existence theorem
It is sometimes possible to assert that a sequence is
convergent even if we can't find the limit right away. We
do this by using the least upper bound property of the
real numbers:
If a sequence has the property that a1 <a 2<a 3< .... is called
a "monotonically increasing" sequence. For such a
sequence, either the sequence is bounded (all the terms
are less than some fixed number) or else it increases
without bound to infinity. The latter case is divergent,
and the former must converge to the least upper bound
of the set of numbers {a1, a2, ... } . So if we find some
upper bound, we are guaranteed convergence, even if
we can't find the least upper bound.
Consider the sequence...
2 , 2  2 , 2  2  2 etc.
To get each term from the previous one, you add 2 and
then take the square root.
It is clear that this is a monotonically increasing
sequence. It is convergent because all the terms are less
than 2. To see this, note that if x>2, then
x2  2x  2  x, and so x  2  x. So our terms can't be
greater than 2, since adding 2 and taking the square root
makes our terms bigger, not smaller.
Therefore, the sequence has a limit, by the theorem.
QUESTION:
What is the limit?
2 , 2  2 , 2  2  2 , ...
Newton’s Method
A better way of generating a sequence of numbers that
are (usually) better and better solutions of an equation is
called Newton's method.
In it, you improve a guess at the root by calculating the
place where the tangent line drawn to the graph of f(x) at
the guess intersects the x-axis. Since the tangent line to
the graph of y = f(x) at x = a is y = f(a) + f '(a) (x-a), and
this line hits the x-axis when y=0, we solve for x in the
equation f(a) + f '(a)(x-a) = 0 and get x = a - f(a)/f '(a).
xnew  xold
f ( xold )

f ' ( xold )
Let’s try it
on the same function we used before,
f ( x)  x  2x  2
3
with the guess that the root x1 = 2. Then the next guess is
f ( x1 )
2 9
x2  x1 
 2 
f ' ( x1 )
10 5
This is 1.8. Let's try it again. A calculator helps:
f ( x2 )
f (1.8)
x3  x2 
 1.8 
 1.76995
f ' ( x2 )
f ' (1.8)
We’re already quite close...
with much less work than in the bisection method! One
more time:
f ( x3 )
f (1.76995)
x4  x3 
 1.76995
 1.769292663
f ' ( x3 )
f ' (1.76005)
And according to Maple, the root is
fsolve(f(x)=0);
1.769292354
So with not much work we have the answer to six
significant figures!
Your turn…
Try Newton's method out on the equation
x  2x  3  0
5
First make a reasonable guess, then iterate. Report your
answer when you get two successive iterations to agree to
five decimal places.
Fractals
Fractals are geometric figures constructed as a limit of
a sequence of geometric figures.
Koch Snowflake
Sierpinski Gasket
Newton's method fractals