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.

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.

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.

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. :)

i apologize that this blogg took so long. sorry.

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


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:


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)
Volume of a solid of revolution
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:


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.

Some helpful sites for more information:
"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
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