Notes, part Notes, part 44

Improper Improper integralsintegrals

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 Examples of the first kind are:first kind are:

dxx

dxe x

0 -21

1 and

Examples of the second kind:Examples of the second kind:

32

4

tan and 11

0

dxxdxx

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 methodSame methodNo 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

x

b

x dxedxe00

lim as thingsame themeans

Calculate the limit!Calculate the limit!What is the value of this limit

(and hence, of the improper integral )?

A. 0

B. 1

C.

D.

E.

ebe

dxe x

0

Another improper integralAnother improper integral

?1

1 of value theisWhat

.arctan1

1 that Recall

-2

2

dxx

Cxdxx

A. 0

B.

C.

4

2

D.

E.

Area between the x-Area between the x-axis and the graphaxis and the graph

The integral you just worked represents all of the area

between the x-axis and the graph of 211x

The other type...The other type...For improper integrals of the other type, we make the same kind of limit definition:

.1

lim be todefined is 1 1

0

1

0

dxx

dxx a

a

Another example:Another example:What is the value of this limit, in

other words, what is

A. 0

B. 1

C. 2

D.

E.

2

?11

0

dxx

A divergent improper integralA 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

00

ln1ln lim1

adxx a

so this improper integral diverges.

One for you:One for you:

3

2

4

diverge?or converge )tan( Does

dxx

A. Converge

B. Diverge

Sometimes it is possible...Sometimes it is possible...

to show that an improper integral converges without actually evaluating it:

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).

.1 allfor 3

11

7

1

thathave we,0 allfor 1

7

1 Since

331

1 144

44

bb

dxx

dxxx

xxxx

b b

Arc lengthArc length

The length of a curve in the plane is generally difficult The length of a curve in the plane is generally difficult to compute. To do it, you must add up the little “pieces to compute. To do it, you must add up the little “pieces of arc”, ds. A good approximation to ds is given by the of arc”, ds. A good approximation to ds is given by the Pythagorean theorem: Pythagorean theorem:

We can use this to find the length of any graph – We can use this to find the length of any graph – provided we can do the integral that results!provided we can do the integral that results!

dxdx

dydydxds

222 1

Find the arclength of the parabola Find the arclength of the parabola y = xy = x22 for for xx between between

-1 and 1. Since -1 and 1. Since dy / dx = dy / dx = 22xx, the element of arclength is, the element of arclength is

so the total length is:so the total length is:

1

1

241 dxxL

dxx241

dxxL

1

1

241

So far, we have that the length is So far, we have that the length is

To do this integral, we will need a trig To do this integral, we will need a trig substitution. But, appealing to Maple, we substitution. But, appealing to Maple, we get thatget that

)25ln(2

1541

1

1

2

dttL

Can we do the integral?Can we do the integral?

The arc length integral from before was:The arc length integral from before was:

This is a trig substitution integral of the second kind:This is a trig substitution integral of the second kind:

With the identity tanWith the identity tan22 + 1 = sec + 1 = sec22 in mind, let in mind, let

What about What about dt dt ? Since we have that? Since we have that

These substitutions transform the integral intoThese substitutions transform the integral into

This is a tricky integral we This is a tricky integral we need to do by parts!need to do by parts!

dtt

1

1

241

22 tan4 t

tan21t

ddt 221 sec

ddtt 3212 sec41

To integrateTo integrate d 3sec

LetLet ddvu 2sec,sec

ThenThen tan,tansec vdu

dd 23 tansectansecsec

But tanBut tan2 2 = sec= sec22 - 1 , so rewrite the last integral and get - 1 , so rewrite the last integral and get

ddd secsectansecsec 33

Still going….Still going….

ddd secsectansecsec 33

It’s remarkable that we’re almost done. The integral It’s remarkable that we’re almost done. The integral of secant is a known formula, and then you can add of secant is a known formula, and then you can add the integral of secthe integral of sec33 to both sides and get to both sides and get

Cd )tanln(sectansecsec 21

213

So we’ve got this so far forSo we’ve got this so far for dtt 2412 wherewhere tan21t

We need a triangle!We need a triangle!So far,So far,

Cdtt )tanln(sectansec412 21

212

withwith tan21t

t2

1

241 t

From the triangle,From the triangle,

241sec,2tan tt

SoSo Cttttdtt 241ln24141 2

412

412

Definite integral:Definite integral:

So far, we haveSo far, we have

Cttttdtt 241ln4141 2412

212

ThereforeTherefore

2525

541 41

1

1

2 lndxx

To get the answer Maple got before, To get the answer Maple got before, we’d have to rationalize the we’d have to rationalize the numerator inside the logarithm.numerator inside the logarithm.

The area of a surface of revolution is calculated in a mannerThe area of a surface of revolution is calculated in a manner

similar to the volume. The following illustration shows thesimilar to the volume. The following illustration shows the

paraboloid based onparaboloid based on (for x=0..2) that we used before, (for x=0..2) that we used before,

together with one of the circular bands that sweep out its together with one of the circular bands that sweep out its surface area. surface area.

Surface AreaSurface Area

xy

To calculate the surface areaTo calculate the surface areaTo calculate the surface area, we first need to determine To calculate the surface area, we first need to determine

the area of the bands. The one centered at the point (the area of the bands. The one centered at the point (xx,0),0)

has radius has radius and width equal to and width equal to . .

Since we will be integrating with respect to Since we will be integrating with respect to xx (there is a (there is a

band for each band for each xx), we'll factor the ), we'll factor the dxdx out of out of dsds and write and write

. So the area of the band. So the area of the band

centered at (centered at (xx,0) is equal to: ,0) is equal to:

Thus, the total surface area is equal to the integralThus, the total surface area is equal to the integral

x 22 dydxds

dxdx

dyds

2

1

dxx

xd4

112

dxx

x1

4

The surface area The surface area turns out to be:turns out to be:

131333

A puzzling example...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:

units. cubic 1 lim 1

1

21

bbx dxV

What about the surface area?What about the surface area?This is equal to...

1

41

2 11

2)('1)(2 dx

xxdxxfxfSA

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

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".

dxx

1

2

SequencesSequences

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, Sometimes it is useful to think of them as functions from the positive integers into the reals, in other words,

,...,, 321 aaa

forth. so and ,a(2) ,a(1) 21 aa

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

where r is the root.

Sequences for which exists

and is finite are called convergent, other sequences are called divergent

Convergent and DivergentConvergent and Divergent

nn

a lim

rann

lim

For example...For example...

The sequence

1, 1/2, 1/4, 1/8, 1/16, .... , 1/2 , ... is convergent (and converges to zero, since

), whereas:

the sequence 1, 4, 9, 16, .…n , ... is divergent.

2

n

0 lim21

n

n

PracticePractice

The sequence

...,2

1,......,

5

4,

4

3,

3

2

n

n

A. Converges to 0

B. Converges to 1

C. Converges to n

D. Converges to ln 2

E. Diverges

Another...Another...

The sequence

...,2

1)1(,......,

5

4,

4

3,

3

2

n

nn

A. Converges to 0

B. Converges to 1

C. Converges to -1

D. Converges to ln 2

E. Diverges

A powerful existence theoremA powerful existence theoremIt 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 a <a <a < .... 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 {a , a , ... } . So if we find some upper bound, we are guaranteed convergence, even if we can't find the least upper bound.

1 2 3

1 2

Consider the sequence...Consider the sequence...

222,22 ,2

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

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.

etc.

.2 so and ,222 xxxxx

QUESTION:QUESTION:

What is the limit?

...,222,22 ,2

Newton’s MethodNewton’s MethodA 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).

)('

)(

old

oldoldnew xf

xfxx

Let’s try itLet’s try it

on the same function we used before,

with the guess that the root x1 = 2. Then the next guess is

This is 1.8. Let's try it again. A calculator helps:

22)( 3 xxxf

5

9

10

22

)('

)(

1

112

xf

xfxx

76995.1)8.1('

)8.1(8.1

)('

)(

2

223

f

f

xf

xfxx

We’re already quite close...We’re already quite close...

with much less work than in the bisection method! One more time:

And according to Maple, the root is

fsolve(f(x)=0);

So with not much work we have the answer to six significant figures!

1.769292354

769292663.1)76005.1('

)76995.1(76995.1

)('

)(

3

334

f

f

xf

xfxx

Your turn…Your turn…

Try Newton's method out on the equation

First make a reasonable guess, then iterate. Report your answer when you get two successive iterations to agree to five decimal places.

0325 xx

FractalsFractals

Fractals are geometric figures constructed as a limit of a sequence of geometric figures.

Koch Snowflake

Sierpinski Gasket

Newton's method fractals