Sunday, December 31, 2006

Happy New Year!

Happy New Year! I hope you all had a great holiday. Seniors, I hope all those essays were finished. Juniors, I hope you all enjoyed not having to write them (this year!). I spent the last week having a great time up in Oregon with my family, and if you thought it was cold here…

You will notice when you log on to our blogs to creat a post that we have switched over to the “new Blogger.” It’s supposed to make things easier and run smoother (we’ll see!) but it will require each of you to switch over as well (and create a Google account) before you can edit or create new posts. Nothing serious, but I wanted to give you a heads up before your turn came around and you panicked!

See you all on the 8th!

Tuesday, December 19, 2006

Wednesday's Quiz Topics

Here’s a list of topics that will be covered on this Wednesday’s Quiz.

Quiz – Sections 5.3-4
Fundamental Theorem of Calculus, Part I (Sec. 5.3, #9,13,17)
Fundamental Theorem of Calculus, Part II (Sec. 5.3, #31,41)
Determine general indefinite integrals (+C!) (Sec. 5.4, #9)
Evaluate definite integrals (Sec. 5.4, #19,25,29)
Explain the meaning of a definite integral expression. (Sec. 5.4, #47,48)
Displacement vs. Total Distance Traveled (Sec. 5.4, #55)

Oh, by the way, no calculators on this one...

I’ll be in early Wednesday and I’ll check in tonight online. See you in class!

"Success is a peace of mind which is a direct result of... knowing that you did your best to become the best you are capable of becoming."
-John Wooden

Monday, December 18, 2006

5.4 Indefinite Integrals and the Net Change Theorem

An indefinite integral is basically the antiderivati
ve of the function. It doesn't have upper and lower bounds because that would make it a definite integral. Indefinite integrals need the +C!

Table of Indefinite Integrals

Sample Problem
Find the general indefinite integral

Using the formula:


The Integral of a rate of change is the
net change (displacement for position functions)

Basically this theorem states that the integral of f or F' from a to b is the area between a and b or the difference in area from the postion of F(a) to F(b).

This can be applied to things such as:



So for a velocity function:
To calculate displacement we can use the equation

to calculate total distance traveled we can add the absolute values of the areas of each sector from each x intercept to the next x intercept

Sample Problem
A particle moves along a line so that its velocity at time t is


a) find the displacement from t=[1,4]
b) find the distance traveled during that time period

Finding the displacement:


Finding the total distance traveled during that time period


The total distance traveled and the displacement are the same because the position function does not pass below the x axis therefore there are no negative areas. If there were negative areas the displacement would be a smaller number and the distance would stay the same.


A Lesser Lesson in Indefinite Integrals but helpful nontheless

Magnus You're Up NeXT!!!

Wednesday, December 13, 2006

5.3 The Fundamental Theorem of Calculus Part II

Fundamental Theorem of Calculus Part II

If f is a continuous function on [a,b] then

where F is any antiderivative of f, that is, a function such that F'=f.

Once you find the antiderivative of f(x), you evaluate the end points from a to b and then subtract the antiderivative function.

Lets look at an example.

Once you find the antiderivative, you plug the top number 5 into x and subtract that antiderivative function from an antiderivative function with the number 3 in x.

You can check your answer by using your calculator.
Plug into your calculator.
Go back to the home page. Click Math then 9.
Once fnInt pops up, put fnInt( ,X,3,5). You will get 98/3

Try this problem!

You will see that this antiderivative is:

Thus you plug in 4 into the x and then subtract the antiderivative with 2 in the x value.

This a reminder to Brian to do the next blog!

This website can help out with this concept:

Dixie Chicks

Thursday's Quiz Topics

Here’s a list of topics that will be covered on this Thursday’s Quiz.

Quiz – Sections 5.1-2
Estimate distance traveled from a velocity graph. (5.1, #15)
Express a Riemann sum as a definite integral. (5.2, #17,19)
Evaluate an integral in terms of areas – show your work! (5.2 #37)
Sketch a graph and estimate the area under the curve using RRAM, LRAM or MRAM (5.1, #3)
Evaluate definite integrals based on a graph (5.2, #33)

I’ll be in early Thursday, available after school this afternoon until 4, and I’ll check in tonight online. See you in class!

In the spirit of the holidays:

Yes, Virginia, There is a Santa Claus
By Francis P. Church, first published in The New York Sun in 1897. [See The People’s Almanac, pp. 1358–9.]
We take pleasure in answering thus prominently the communication below, expressing at the same time our great gratification that its faithful author is numbered among the friends of The Sun:

Dear Editor—
I am 8 years old. Some of my little friends say there is no Santa Claus. Papa says, “If you see it in The Sun, it’s so.” Please tell me the truth, is there a Santa Claus?
Virginia O’Hanlon

Virginia, your little friends are wrong. They have been affected by the skepticism of a skeptical age. They do not believe except they see. They think that nothing can be which is not comprehensible by their little minds. All minds, Virginia, whether they be men’s or children’s, are little. In this great universe of ours, man is a mere insect, an ant, in his intellect as compared with the boundless world about him, as measured by the intelligence capable of grasping the whole of truth and knowledge.

Yes, Virginia, there is a Santa Claus. He exists as certainly as love and generosity and devotion exist, and you know that they abound and give to your life its highest beauty and joy. Alas! how dreary would be the world if there were no Santa Claus! It would be as dreary as if there were no Virginias. There would be no childlike faith then, no poetry, no romance to make tolerable this existence. We should have no enjoyment, except in sense and sight. The external light with which childhood fills the world would be extinguished.

Not believe in Santa Claus! You might as well not believe in fairies. You might get your papa to hire men to watch in all the chimneys on Christmas eve to catch Santa Claus, but even if you did not see Santa Claus coming down, what would that prove? Nobody sees Santa Claus, but that is no sign that there is no Santa Claus. The most real things in the world are those that neither children nor men can see. Did you ever see fairies dancing on the lawn? Of course not, but that’s no proof that they are not there. Nobody can conceive or imagine all the wonders there are unseen and unseeable in the world.

You tear apart the baby’s rattle and see what makes the noise inside, but there is a veil covering the unseen world which not the strongest man, nor even the united strength of all the strongest men that ever lived could tear apart. Only faith, poetry, love, romance, can push aside that curtain and view and picture the supernal beauty and glory beyond. Is it all real? Ah, Virginia, in all this world there is nothing else real and abiding.

No Santa Claus! Thank God! he lives and lives forever. A thousand years from now, Virginia, nay 10 times 10,000 years from now, he will continue to make glad the heart of childhood.

About the Exchange
Francis P. Church’s editorial, “Yes Virginia, There is a Santa Claus” was an immediate sensation, and went on to became one of the most famous editorials ever written. It first appeared in the The New York Sun in 1897, almost a hundred years ago, and was reprinted annually until 1949 when the paper went out of business.

Thirty-six years after her letter was printed, Virginia O’Hanlon recalled the events that prompted her letter:
“Quite naturally I believed in Santa Claus, for he had never disappointed me. But when less fortunate little boys and girls said there wasn’t any Santa Claus, I was filled with doubts. I asked my father, and he was a little evasive on the subject.
“It was a habit in our family that whenever any doubts came up as to how to pronounce a word or some question of historical fact was in doubt, we wrote to the Question and Answer column in The Sun. Father would always say, ‘If you see it in the The Sun, it’s so,’ and that settled the matter.
“ ‘Well, I’m just going to write The Sun and find out the real truth,’ I said to father.
“He said, ‘Go ahead, Virginia. I’m sure The Sun will give you the right answer, as it always does.’ ”
And so Virginia sat down and wrote her parents’ favorite newspaper.
Her letter found its way into the hands of a veteran editor, Francis P. Church. Son of a Baptist minister, Church had covered the Civil War for The New York Times and had worked on the The New York Sun for 20 years, more recently as an anonymous editorial writer. Church, a sardonic man, had for his personal motto, “Endeavour to clear your mind of cant.” When controversal subjects had to be tackled on the editorial page, especially those dealing with theology, the assignments were usually given to Church.
Now, he had in his hands a little girl’s letter on a most controversial matter, and he was burdened with the responsibility of answering it.
“Is there a Santa Claus?” the childish scrawl in the letter asked. At once, Church knew that there was no avoiding the question. He must answer, and he must answer truthfully. And so he turned to his desk, and he began his reply which was to become one of the most memorable editorials in newspaper history.
Church married shortly after the editorial appeared. He died in April, 1906, leaving no children.
Virginia O’Hanlon went on to graduate from Hunter College with a Bachelor of Arts degree at age 21. The following year she received her Master’s from Columbia, and in 1912 she began teaching in the New York City school system, later becoming a principal. After 47 years, she retired as an educator. Throughout her life she received a steady stream of mail about her Santa Claus letter, and to each reply she attached an attractive printed copy of the Church editorial. Virginia O’Hanlon Douglas died on May 13, 1971, at the age of 81, in a nursing home in Valatie, N.Y.

Tuesday, December 12, 2006

5.3 The Fundamental Theorem of Calculus

Hey guys this is Izzy. Today in class we learned about the fundamental theorem of calculus. This theorem establishes a connection between the two branches of calculus: differential calculus and integral calculus. In class we covered how if you multiply the function that you are taking the integral of by a certain factor, that you can factor out that number. Example:

This makes it easier to deal with the original funtion first, without any coefficients to clutter up the process.

In class, we saw how the integral of a function can be a function itself, represented by g(x). Expressing the integral of a function as the function g(x) allows one to actually graph and express the area between the curve of a graph and the x-axis in terms of x:

In this graph,

By using this equation, we can find the area between the graph and the x-axis. Let's say we wanted to find the area between x=0 and x=3. You would multiply the change in x, 3, by f(3), 3, and .5, because it's a triangle. So, g(3)=(3)(3)(.5)=4.5 .

Today we also made the amazing discovery that the antiderivative of a function is the same as the integral of the function. Observe:

As you can see from the graph, the area of the shaded region (teal) is:

We established that

To find g'(x), we have to replace t with x and multiply whatever you get by the derivative of x.

Wait a minute, the derivative of the Area function is also x! Thus A'=g'(x). We just proved that the antiderivative of a function is the same as its integral! i'm so happy.

What we just did relates to the Fundamental theorem of Calculus, Part 1, which states that:
If the function f is continuous on [a,b], then the function g defined by

is continuous on [a,b] and differentiable on (a,b), and g'(x)=f(x)

MOVING ON, we now know how to find the derivative of these types of functions. Let's try this problem: Find

Whenever you see these types of problems, you must first look at the upper bound, or any of the bounds that has a variable in it. In the process of calculating the equation for g'(x), you must replace the variable t with the variable bound, and wherever you see dt, replace it with the derivative of the variable bound. In this case, the upper bound has the variable: x^2.
To solve this problem, we have to use the chain rule (make u=x^2):

There you have it.

This a reminder to Ami to do the next blog!

These are a few websites that help out with this concept:

Here are some cool math jokes
Q: What do you get if you divide the cirucmference of a jack-o-lantern by its diameter?
A: Pumpkin Pi!

Q: Why do you rarely find mathematicians spending time at the beach? A: Because they have sine and cosine to get a tan and don't need the sun!

A mathematician is flying non-stop from Edmonton to Frankfurt with AirTransat. The scheduled flying time is nine hours. Some time after taking off, the pilot announces that one engine had to be turned off due to mechanical failure: "Don't worry - we're safe. The only noticeable effect this will have for us is that our total flying time will be ten hours instead of nine." A few hours into the flight, the pilot informs the passengers that another engine had to be turned off due to mechanical failure: "But don't worry - we're still safe. Only our flying time will go up to twelve hours." Some time later, a third engine fails and has to be turned off. But the pilot reassures the passengers: "Don't worry - even with one engine, we're still perfectly safe. It just means that it will take sixteen hours total for this plane to arrive in Frankfurt." The mathematician remarks to his fellow passengers: "If the last engine breaks down, too, then we'll be in the air for twenty-four hours altogether!"

A math student is pestered by a classmate who wants to copy his homework assignment. The student hesitates, not only because he thinks it's wrong, but also because he doesn't want to be sanctioned for aiding and abetting. His classmate calms him down: "Nobody will be able to trace my homework to you: I'll be changing the names of all the constants and variables: a to b, x to y, and so on." Not quite convinced, but eager to be left alone, the student hands his completed assignment to the classmate for copying. After the deadline, the student asks: "Did you really change the names of all the variables?" "Sure!" the classmate replies. "When you called a function f, I called it g; when you called a variable x, I renamed it to y; and when you were writing about the log of x+1, I called it the timber of x+1..."

Monday, December 11, 2006

5.2 The Definite Integral

So there's some notation for the Riemann sum that we did yesterday. If we take, for example, the area under the curve of y=x^2 from x=1 to x=3 like this:

and we want to have 2 rectangles with the sample points being left endpoints, then we would assign some notation. 1 would become x1, and 2 would be x2. To put this in notation we would use a capital sigma to symbolize sum. Underneath the sigma would be the starting point and over the sigma will be the ending point, both using i as the variable. The sum would be the sum of the rectangles, so it would be the height times the width (change in x). Put all this together and the equation would look like this:

To find the answer to this problem, we would plug in 1 for i then two, and add them. So it would be: f(x1)(x2-x1) + f(x2)(x3-x2) = f(1)(2-1) + f(2)(3-2) = (1)(1) + (4)(1) = 5.

We know from 5.1 that as you have more and more rectangles, the estimate of the area becomes more and more accurate. So ideally we would like infinitely many rectangles. We can write this as :

and to simplify this we use the integral symbol, and it would be :

Because as we get more and more boxes and n approaches infinity, the change in x gets closer and closer to zero so eventually the change in x becomes the derivative at x.

Take this graph and equation for example:

Let's find the area underneath the curve, but above the x axis.

Both the equation and the graph suggest that it's a semi-circle with a radius of 5. We could :

But common sense tells us that since we want want the area under the graph of a semi-circle, then we want the area of a semi-circle. So the area would be

This simple way of looking at the problem can be applied in different ways. Let's suppose we wanted to find the area between the curve and the x-axis if f(x) was 2x from x=0 to x=4 and f(x)=8 from x=4 to x=8 and f(x)=3x-16 from 8 to 10, then you could split that graph into two triangles and one rectangle and work out the area from there.

The area would be (1/2)(4)(8) + (6)(8) + (1/2)(6)(2) = 70.

Another topic covered was the Mid-point estimate. Using the mid-point estimate balances out the over and under estimation, bringing us closer to the actual area underneath the graph.

Using the left side makes an overestimation, but using the right side is an underestimation. Using the midpoint, however, balances out the over and underestimation, making it about right.

There are some properties of integrals as well.


2. If f(x)=c=7, and a=3 and b=5, then:




That's it!

Here's a couple of cute calculus pick-up lines:

1. "Hey, if I was sin^2x and you were cos^2x, together we would be one!"

2. "I wish I was your derivative so that I could lie tangent to your curves."

Here's a website about the properties of integrals:

Ismael, you're up next!