Exam Review: Applications of Integrals

We started off class by talking a little bit on how sushi is made. It's a roll, but when sliced (hondamitsubushi!) its made into cross-sections. By knowing the area of these cross sections, we can find the total volume of the initial roll if we have the length. This class was pretty much dedicated to applications of integrals for our exam review.

We started with:





This was easily found with the use of a calculator. But if not given use to a calculator, it's good to know how to solve this question.








We found the antiderivative of the given function over the interval of 0 to pi and replaced all the values of x with the values of the interval to find the difference in area. Now if no calculator is given, it's simple, just simply simplify it to the simplest form. Then repeat for the other question.

Our next question had to deal with the same functions as the last, but we have to find the integral equation of the volume of the shape over the same interval of 0 to pi, but when it's generated over the x-axis, and then the we found the volume when generated over the y-axis.





Now we cut this shape into tiny cylinders and find the volume of each cylinder, given by V = (pi)r²h but because we are dealing with 2 different functions, and finding the difference in the volumes, each graph has its own radius. The volume is then simply stated as V = pi(R² - r²)h, we use an infinite amount of cross sections, and because of this the height is quiet small, its so small that we call it dx. Now we throw the functions' radius' into the equation and voila.

When revolving around the y-axis, it's the same deal, but with the use of a different equation. Note: I'm not sure what it is but I'll edit it in when I find out.


Our next question involved trapezoid sums.





The question shows a table with 9 sets of information when x is various values. We need to find the difference of area between the 2 given functions. Using a trapezoidal approximation, we find the upper limits and lower limits of f(x) and find the differences between these limits of g(x) and find the average. Quite simple.


Next question deals with finding the area of the region of a graph.





Now because the graph has a root between the given interval, we split the region into 2 seperate intervals. One finding the area of the region between the root at -1.100457 and -1.5 (I'm not sure why we're finding the area seemingly backwards on the interval) and we add this to the area of the function in the interval -1.100457 to 1.5

This is all we did in today's class. Review questions are in today's slides. Our exam is in 2 days and good luck to everyone. Next scribe will be .:. J + ME .:. or better known as Jamie.

Today's Slides: May 4

Here they are ...

AP Calculus AB May 4, 2009

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3rd Last Class Before the Exam

WOW! Just 2 more classes until the big exam. That was very fast.

Today we reviewed Techniques of integration.
The first one we looked at was the U-Substitution technique. This can be used when you are antidifferentiating a function that is composed of a factor that is a composite function and the other factor is the derivative of the inner function of the composite function.



I did this question on the smartboard. First I chose a part of the function to be substituted for U. I saw that the denominator and the exponent of e was root x, so I chose that one.
So I let u equal root x and differentiated u as shown in the image above. I also solved for solved for x using the substitution, just in case I needed it. So I substituted all values necessary and I got an antiderivative. Great we found an antiderivative, how do we know if it correct? Just differentiated the antiderivative that you found and if you get the function that was to be antidifferentiated, then you got the right antiderivative.



This question is similar to the previous one but this one is an integral. What is the difference you ask? An antiderivative will give you a function, an integral will give you a number. Even though they are different, they are also related.
To answer this question, just antidifferentiate the function and substitute the limits into the antiderivative to find the answer. This is the definition of the second part of the Fundamental Theorem of Calculus.



Oh no! There is an antiderivative that we've never seen before. What is the antiderivative of cosecant, let alone the antiderivative of cosecant squared. I can't remember how we thought of this but Mr K got us to differentiate Cotangent. Using the Quotient Rule because cotangent can be written as cosine divided by sine as shown in the image above using abbreviations. Don't try to memorize the antiderivative/derivative relationship. Just understand that you can easily build it when needed.

Mr. K began talking about the antiderivative of x to the power of some number times e to the x. He said that it was a telescopic function which means as you evaluate it, the function will get longer, but things can be reduced or factored in such a way that it will shrink like a telescope.

I was playing around with some examples and I discovered the pattern. It is such a cool antiderivative. I'll give an example:



I got to this part by using integration by parts. So f = x³, my f' = 3x², g' = e^x, and g = e^x. Notice that the second term is still an antiderivative that we cannot find easily, but it is a factor of 2 functions, so let's use integration by parts again. So we'll get:



Once again we cant antidifferentiate one of the functions so let's integrate by parts once more:



Now we can finally evaluate that antiderivative. Once you do that you'll notice that there is a common factor of e^x. So let's factor that out and you might see something completely amazing. I'll show you:



Okay look at the polynomial in the brackets. When have you seen those specific terms before? If you think and look carefully the consecutive terms are derivatives of the previous term. I tried this again with another power of x and I got the same pattern. So when you are asked to find the antiderivative of x to the power of some number times e^x, the antiderivative will be e^x times the power of x in the function being antidifferentiated minus the derivative of the previous function until you reach a constant. Don't forget the plus c.



tn-1 means the previous term.

Sorry if my algebra was bad on the image above. But if you try it yourself, it really is a beautiful thing :P

The next few slides on the slide show posted by Mr K is for homework. Study for the exam on Wednesday. Next Scribe will be Francis.

Here is the video of the Scribe:



These guys have very funny podcasts.

That's all folks.

Todays Slides: May 1

Here they are ...

AP Calculus AB May 1, 2009

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The Scuba Steve Dillema

I knew I forgot to do something yesterday. And that was check to blog -__-"

Sorry that this is very late, some urgent things came up and I wasn't able to get home until 1, so I'll just give a quick run down of what happened since everyone was in class.

Anyways, we worked on another question, this one being about Scuba Steve. There was a lot of information, so the question had to be read carefully to fully understand it.



So as you can see, to solve it, we found that the hypotenuse of the triangle at the beginning is 30 root two. we then found the remaining area near the end of the distance to be x since we do't know how long that distance was. Because triangle is a triangle with two sides equal, the top of it would be 30, thus the 170 - x (it's 170 because 200 - 30)

So that makes the second part of the equation for part a. The next slope we found one of the sides to be 45 feet. So then we found the hypotenuse to be root x squared + 2025. And now we can build our equation.

*Red is what Paul stated. Blue = edited work.
The RDT triangle is an example of how to change from rates, to distances, to times.

The next part is done by benchmen (Yes, he's a plural. That's not weird), and he found the derivative of the function, and set it to zero to find x.

Unfortunately, this is as far as we got as we ran out of time.
Next scribe will be Ben

The Truth about Bruce Lee and Chuck Norris

Some real fighting, none of that ninja fighting bullet dodging special effects haha.

Today's Slides: April 30

Here they are ...

AP Calculus AB April 30, 2009

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Today's Slides: April 29

Here they are ...

AP Calculus AB April 29, 2009

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