Notes, part 4 Arclength, sequences, and improper integrals.

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Presentation transcript:

Notes, part 4 Arclength, sequences, and improper integrals

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:

Examples of the second kind: 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:

Calculate the limit! What is the value of this limit (and hence, of the improper integral )? A. 0 B. 1 C. D. E.

Another improper integral B. C. D. E.

Area between the x-axis and the graph The integral you just worked represents all of the area between the x-axis and the graph of 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:

Another example: What is the value of this limit, in other words, what is A. 0 B. 1 C. 2 D. 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: so this improper integral diverges.

One for you: A. Converge B. Diverge

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).

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

Example Find the arclength of the parabola y = x2 for x between -1 and 1. Since dy / dx = 2x, the element of arclength is so the total length is:

Conclusion So far, we have that the length is To do this integral, we will need a trig substitution. But, appealing to Maple, we get that

Can we do the integral? The arc length integral from before was: This is a trig substitution integral of the second kind: With the identity tan2 + 1 = sec2 in mind, let What about dt ? Since we have that These substitutions transform the integral into This is a tricky integral we need to do by parts!

To integrate Let Then But tan2 = sec2 - 1 , so rewrite the last integral and get

Still going…. 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 So we’ve got this so far for where

We need a triangle! So far, with From the triangle, So

Definite integral: So far, we have Therefore 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 (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 To calculate the surface area, we first need to determine the area of the bands. The one centered at the point (x,0) has radius 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 . So the area of the band centered at (x,0) is equal to: Thus, the total surface area is equal to the integral

The surface area turns out to be: 13 3 p

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:

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

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

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 where r is the root. Sequences for which exists and is finite are called convergent, other sequences are called divergent

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. n 2

Practice The sequence A. Converges to 0 B. Converges to 1 C. Converges to n D. Converges to ln 2 E. Diverges

Another... The sequence 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 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... 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 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?

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).

Let’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:

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

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.

Fractals Fractals are geometric figures constructed as a limit of a sequence of geometric figures. Koch Snowflake Sierpinski Gasket Newton's method fractals