## Tuesday, May 01, 2007

### Improper integrals

Improper Integrals!

These are used when one or both boudries are positive or negative infinity

in other words....

Taking the antiderivative may give us an answer or may not, depending upon whether the graph is convergent or divergent.

~Convergent graphs go towards some known limit
~Divergent graphs go on until infinity and have no known limit

If given a problem that has boundaries of - infinity to + infinity, break it up into two ingtegrals like so...

SAMPLE PROBLEM!

Perpetuities

A Perpetuity is essentially an annuity with no definite end. They are used primarily to calculate how much interest is earned off of a large amount of money or a large endowment. Scholarships from colleges are a good example of these types of problems.

The general equation is

where PV is the present value, m is the number of times per year the amount is calculated, P is the value you wish to earn off of interest, and r is the interest rate.

For Example:

If a school wished to give any one student a \$15,000 scholarship per year and the bank interest rate was 7%, how much money would need to initially be placed in the bank account?

Here's a helpful site that has graphs that depict all of the concepts stated in this section

NICOLE!! You're up Next...if there are any more blogs to do.

The pleasure we obtain from music comes from counting, but counting unconsciously. Music is nothing but unconscious arithmetic. ~Gottfried Wilhelm Leibniz

### Wednesday's Test (our last!)

Here’s a list of topics that will be covered on this Wednesday's Chapter 7 Test

Chapter 7 Test Topics
Integration by parts – natural log
Integration by parts – trig functions
Integration by parts – exponential
Integration by parts – definite integral (natural log)
Improper Integral – exponential
Improper Integral – exponential, substitution
Improper Integral – divergent/convergent
Trapezoidal Rule – match values
Trapezoidal Rule – data table
Simpson’s Rule – data table
Trapezoidal Rule – graph
Simpson’s Rule – graph

That’s it! I’ll be in early on Thursday, and available after school Wednesday…

"Knowledge is of no value unless you put it into practice."
- Anton Chekhov

Don't let this be you...

## Friday, April 27, 2007

### 7.3 Numerical Integration

Numerical Integration

There are two new ways to find the area of a function using reimann sums. Instead of using small rectangles to find the area we are going to use trapezoids and parabolas.

TRAPAZOIDAL RULE
The equation we use to find the area under a function to the x-axis by using trapazoids is...

The first equation shows us the equation of a trapazoid. The second one shows us how to encorporate the equation for the area of a trapazoid to fit the function. As you can see, there is a pattern. You must multiply the first and last y value by one and the rest of the y values by 2. In the third equation you learn that the change in x is the same as a-b over the height.

SIMPSON'S RULE
The equation we use to find the area under a function to the x-axis by using parabolas is...

As you have noticed this equation also has a pattern like the trapazoidal equation. The first and last y values are multiplied by 1 and between these we alternate multiplying by 4 then 2. Make sure you end by multiplying by 4 before the last y value, otherwise the equation will not be accurate.

EXAMPLE PROBLEM
Calculate the area under the curve to 3 decimal places using the trapezoidal rule to approximate the given integral with the specified value of n.

(Trapazoidal Rule)

First we figure out that there are 4 intervals going from 0 to 2. This means that our f(x)'s will be in incriments for .5.
Second we find the y values for these incriments to 2.

x - y
0 - 1.000
.5 - 1.031
1 - 1.414
1.5 - 2.462
2 - 4.123

Now, we plug in these values for the equation, and then solve. :)

KATELYN UR UP NEXTTTTT!!!!!!!!! :) <3
i apologize that this blogg took so long. sorry.

http://mathworld.wolfram.com/NumericalIntegration.html
http://people.hofstra.edu/faculty/Stefan_waner/RealWorld/integral/numint.html

HAHAHAHAHA... i'm sorry i died laughing when i saw this. HAHAHAHA FABULOUS FIGURE!! :)

Labels:

## Tuesday, April 17, 2007

### 7.1 Integration by Parts

Integration by parts is a way to find the integral of 2 functions that are being multiplied together. As we know:

By integrating both sides of the equation, we come up with:

Then, by simple subtraction, we come up with:

Finally, by substituting the following variables, we come to the
Integration by Parts Formula:

By using this formula, we can now do what we already know how to do which is assign variables u and v for funtions to find the integral of a more complicated function.

In order to choose what functions to assign to u and dv, use these guidlines:
1. du is simpler than u
2. dv is easy to integrate

Let's do a Sample Problem:

Evaluate:

First, assign u and v to their functions and then use the Inegration by Parts Formula to solve the equation.

If you have any trouble, these sites may be able to help you understand a little better:
Good Help
Tough Sample Problems to test your skills

KJ, you are up next with 7.3 Numerical Integration

"Who has not been amazed to learn that the function y = e^x, like a phoenix rising from its own ashes, is its own derivative?" - Francois le Lionnais

## Thursday, April 12, 2007

### Monday's Test Topics

Here’s a list of topics that will be covered on this Monday’s Chapter 6 Test

Chapter 6 Test Topics
General Antiderivatives – polynomials
Riemann Sums (areas under curves – right and left endpoints)
Definite Integrals
Indefinite Integrals
Properties of Integrals (symmetries)
Substitution
Volume of a solid of revolution
Areas/Integrals
Business applications – annuities
Business applications – consumer and producer surplus
Business/Economics applications – Lorentz Curves

That’s it! I’ll be in early on Monday, and available after school today…

Math Art: The Beauty and Symmetry of Mathematics.

1 x 8 + 1 = 9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
1234567 x 8 + 7 = 9876543
12345678 x 8 + 8 = 98765432
123456789 x 8 + 9 = 987654321

1 x 9 + 2 = 11
12 x 9 + 3 = 111
123 x 9 + 4 = 1111
1234 x 9 + 5 = 11111
12345 x 9 + 6 = 111111
123456 x 9 + 7 = 1111111
1234567 x 9 + 8 = 11111111
12345678 x 9 + 9 = 111111111
123456789 x 9 +10= 1111111111

9 x 9 + 7 = 88
98 x 9 + 6 = 888
987 x 9 + 5 = 8888
9876 x 9 + 4 = 88888
98765 x 9 + 3 = 888888
987654 x 9 + 2 = 8888888
9876543 x 9 + 1 = 88888888
98765432 x 9 + 0 = 888888888

Brilliant, isn't it? And finally, take a look at this symmetry:

1 x 1 = 1
11 x 11 = 121
111 x 111 = 12321
1111 x 1111 = 1234321
11111 x 11111 = 123454321
111111 x 111111 = 12345654321
1111111 x 1111111 = 1234567654321
11111111 x 11111111 = 123456787654321
111111111 x 111111111 = 12345678987654321

A unique approach to homework:

## Wednesday, April 11, 2007

6.8 Volumes of Solids of Revolution
In this section we are asked to find the volume of a solid created when we revolve a section of the graph around a line, which we call the axis of revolution. This section of the graph can be the area beneath a curve or it can be the area between two curves. The axis of revolution can be the x-axis or the y-axis, or it could be another line, which would be designated in the problem.
Lets start with the easiest volume to find, using the area beneath a curve and the x-axis as our section of the graph and our axis of revolution, respectively.
Lets say we are given the curve

and asked for the volume of the solid created when we revolve the area below this curve from x=2 to x=9 around the x-axis.
The curve, when revolved around the x-axis, will give us a shape similar to a cylinder, and we can find the area of this cylinder by dividing it into tiny sections and adding their areas. The equation for the volume of our solid would look like this:

where r is equal to distance along the y-axis, or f(x), and h is equal to the distance along the x-axis, or dx. If we used infintely small sections to find the volume of our solid, our equation for volume would look something like this:

or:

So, if we plug in our values from the problem we get

and then we can solve for the volume:

so, the volume of the solid of revolution is

There are some other ways a question about solids of revolution can be asked.
For example, instead of revolving around the x-axis, you could be asked to revolve a shape around another line, like x=-2. In this case, you would end up with a hole through the center of your solid, so you have to subtract the volume of that hole. The equation for a problem like that looks like this:
or, more simply:

where R is the distance from the axis of revolution to the curve farthest away from it, and r is the distance from the closest curve to the axis of revolution. When you are only given one curve and asked to revolve it around a specified line, the second curve would be the x-axis. When you are given two curves, you have to determine which is closer to the axis of revolution and work from there.

"Our minds are as different as our faces. We are all traveling to one destination: happiness, but few are going by the same road."
-Charles Caleb Colton
Next Up: Drew

## Monday, March 26, 2007

### Integrals in Business

In the world of business, integrals are used to determine many numbers that may mean the success or the failure of a company.

With integrals and some statistics, we as the CEO's of companies can determine the consumer surplus, the producer surplus, and the market price

where P is the price, and x is the amount of products, the following graphs portray the relationships between price and goods.

for consumer surplus:
The Producer Surplus:
The Marketing Price curve:

A sample problem: given the following information find the producer surplus
Pd = D(x) = -.001x^2 + 250
Ps = S(x) = .0006x^2 + .02x + 100
x = 300
P = 160

PS =

on your calculator: you should put the equation in the parenthesis in the Y1 function and input fnInt(Y1,x,0,300) and you will get a number: 36300 which is the producer surplus because we used the producer surplus function as the Y1.

Future and present value of an Income Stream:
T = time frame
R(t) = rate of income generated
r = Interest rate for reinvestment
this tells you how much you would have if you put your money in a bank and had interest on top of reinvestment.

Present Value function: which is the exact value of the money at the moment is described by
and the Amount Annuity is the future output of your investment and is described by functions
m: payments per year P : amount of payment

Lorentz Curve: Income Distribution
f(x) = proportion of the total income recieved by the poorest 100x% of the population
Domain: [0,1] f(0) = 0 <- if nobody is counted, nobody has money.
Range: [0,1] f(1) = 1 <- if everybody is counted all the money is in one way or another, distributed.

the proportions are shown in between the two extremes of nobody and everybody.

It is always increasing because a person's income cannot be negative.

the GINI index which shows the coefficient of the inequality is
some more reference is here

Thou art next DANICA

## Wednesday, March 21, 2007

### Chapter 6

Section 6
Area Between Two Curves

We have been able to find the area between the graph and the x-axis, but now we will be able to find the area between two graphs on a given interval (a,b). The concept is pretty easy, it just takes a lot of practice. Please note that a major part of this chapter that must be noticed is the concept that the area is based off of the bottom function subtracted from the top function. This is the reason why we must graph the functions or determine through mathematics which function is the correct top function.

There is just 1 concept we must learn, and simply practice it over and over until its correctly performed quite easily.

The above equation helps us find the area between the curves, if f(x) is the above function in the area and bounded on the bottom by g(x).
The net area between the functions will always be positive! Due to the fact before, we were dealing with finding the area underneath the graph and above the x-axis, we were able to have a negative area. However, we will always have a positive area between two curves.
If the graph allows for the opportunity for symmetry, use it, it will save you a great deal of work. Like the below example:
F(x)=x^3-3x+3 and g(x)=x+3
First Step: Graph the function:

Second Step: Set the two functions equal to each other, in order to find the points of intersection of the two graphs
x^3-3x+3=x+3
X^3-4x=0
x(x^2-4)=0
x(x+2)(x-2)=0
so the points of intersection occur at x=0,2,-2. By looking at the graph, we see the area we are trying to find occurs in (-2,0) and (0,2)
Note: you are not able to combine the area to make (-2,2), because that is describing a different area, compared to the area we actually want. Also, since we want the area, using the top function, which is the cubic function, so we must take the linear function and subtract it from the cubic function.

The result of 4 describes the area from (-2,0), but we still need the area from (0,2), however due to symmetry, we know the areas are equal, so by multiplying the 4 by 2, we find the net area between the two graphs to be 8. I also showed this is equal to writing it out longhand and ignoring the symmetry.

The answer given describes the area between the two curves from (0,2), which is 4. By adding the two areas, we find that 4+4=8, which is the net area between f(x) and
g(x). Note: this is also equal to the first area multiplied by two, since the function is symmetric.

Another Example: Find the area between the curves of the following functions:

To find the points of intersection, set the two functions equal to each other, and you find the points of intersection are: 1,2,3. The areas of the graph we want to find are from (1,2) and from (2,3). We find the area from (1,2) first, see below for work. Since from (1,2), the cubic function lies above g(x), we subtract g(x) from the cubic function.

Then we find the area from (2,3) Since from (2,3), the cubic function lies below the graph of g(x), we subtract the cubic function from the function g(x). See below for work.

We then add (1/4)+(1/4)=(1/2)
So the net area between the two graphs is 1/2.
Note: we could have used the symmetry law also, but the work was done to show the problem.
Here are some external links which can prove helpful.
http://www.karlscalculus.org/calc12_3.html

Reminder:
KYONG! You're up next for Chapter 6-7 Applications of the Definite Integral to Business and Economics on Thursday March 22, 2007.

Personalization:
Here's a couple quotes to help get you guys through the week:

My life has been full of terrible misfortunes most of which never happened.
- Michel de Montaigne

The way to gain a good reputation, is to endeavor to be what you desire to appear.
– Socrates

When one door of happiness closes, another opens, but often we look so long at the closed door that we do not see the one that has been opened for us.
- Helen Keller

There is always room at the top.
- Daniel Webster

## Wednesday, March 14, 2007

### Thursday's Quiz Topics

Here’s a list of topics that will be covered on Thursday’s 6.1-4 Quiz

Quiz Topics - 6.1-4
Indefinite integral – substitution (trig functions)
Area under a trig function (special angles!)
Area approximation – data table
Determine original function f given slope and point
Indefinite integral – substitution (natural log)
Definite Integral – special circumstances
Indefinite integral – substitution (natural logs and exponentials)
Indefinite integral – substitution (trig functions and natural logs)
Area approximation – Midpoints
Definite Integral – geometric
Antiderivative – polynomial
Antiderivative – fractional powers
Antiderivative – position/velocity/acceleration
Area approximation – Right endpoints
Definite Integral – geometric
Definite Integral – polynomial
Definite Integral – fractional powers
Definite Integral – trig function
Indefinite Integral – substitution (fractional exponents)
Indefinite Integral substitution (polynomial)

That’s it! I’ll be in early on Thursday!…

I like nonsense, it wakes up the brain cells. Fantasy is a necessary ingredient in living, It's a way of looking at life through the wrong end of a telescope. Which is what I do, And that enables you to laugh at life's realities.
- Dr. Seuss

## Tuesday, March 06, 2007

### Some links for section 6.3

Hi Gang!
Once again, sorry I won’t be around for class. Hopefully the reading isn’t too confusing. To help you out, here are a couple of links to posts your classmates in AB Calculus created when they covered the topics of Area and the definite integral earlier this year. If they make sense, be sure to thank them. If they don’t, ask them questions!

I’m hoping to be back on Friday, but this cough of mine has to go away first. However, I can still answer emails!

Section B:
Laurie L.
:
AREAS and DISTANCES

Kristin H.:
DEFINITE INTEGRAL

Section C:
Kane O.:
AREAS and DISTANCES

Claire C.:
DEFINITE INTEGRAL

“That which does not kill us makes us stronger.” – Friedrich Nietzsche (or at least I hope so…)

And here’s a link someone sent me that’s just darn cute:
Instructions for properly hugging a baby.

## Wednesday, February 28, 2007

### 12.4 Integration of Trigonometric Functions

Finding the Antiderivatives of Trigonometric Functions

Finding the antiderivatives of Trigonometric functions is quite simple. For such problems, you must use (and memorize) the following Integration Formulas:

Some problems may require other formulas such as the following, but you will not be required to memorize them.

Also, if you come across a problem that looks like this:

1. Find the antiderivative

2. Plug in the top number to the antiderivative

3. Plug in the bottom number to the antiderivative

4. Use your answers from steps two and three and find the difference

Example Problem

Question

Evaluate the following integral.

Solution

Step One: Find the antiderivative.

Step Two: Plug in the the top number to the antiderivative.

Step Three: Plug in the bottom number to the antiderivative.

Step Four: Find the difference.