Post 21

Ok, so this whole week we've been doing practice AP tests (twice a day if you also have stats). These tests are really hard, and I'm horrible at them. Most of the time I have no clue what is ever going on. Other times, I think I know what's going on and eliminate answers I believe are not correct so I can make an educated guess, and my guesses are always wrong. I have learned that guessing is bad on these tests, especially if you are as horrible as I am with the whole guessing game. I have also learned you can make negative points on these tests if you do guess incorrectly. This week's really been crazy, but B-rob says it'll get better when we have more practice on the AP tests. She says the ask the same questions on every test, so with practice we'll eventually do well. It's the eventually part that bothers me because we have our first test tomorrow.

Anyway, this week was my side of the class's week to answer the problems on the wiki page. I picked number three which was a problem asking for the area beneath a curve and the volume of a solid. I picked this question because I knew I liked area under the curve, and volume wasn't much different. My problem actually had area by washers, which is very similar to area beneath the curve. So I picked this problem without really looking at how I'd solve it. I thought it would be easy enough. When I looked at the problem, I immediatly saw I didn't know how to solve for the bounds, or even integrate for that matter. When I asked for help, I found the problem could be done very easily in my calculator. The problem was I wasn't very sure how to use my calculator to do this problem. When I figured it out, it was a very easy problem to do. When I finished the problem; however, I doubted myself because I came out with crazy decimals. I was told to be confident in my answer because AP questions would not be very pretty. So for this blog, I'll explain how to do area under a curve and volume by disks and washers.

Area under a curve is very simple. In these problems, you will have two graphs that will yield two equations. You will then plug these equations in to the area under a curve formula. The formula is:

Integral from a-b (top eq-bottom eq)

Sometimes the problem will give you bounds, but if not they're easy to find. To find your bounds, you set your equations equal to each other and solve them. Once you find your bounds, just plug into the formula and integrate as you normally would for a definite integral.

For volume, you are usually given a graph, or an equation that you are asked to revolve around something to make it a solid, and you are asked to then find the volume of that solid. Volume can be found in two different ways by doing this, it can be found by using "disks" or "washers."

For example, you would use disks when you only have one equation. When this one equation (one graph) is rotated about a line it will make a solid object with no hole. If a problem asks you to rotate a graph around a horizontal line that is not the x-axis, you are to subtract that line from the equation and solve. For disks the formula is:

pi * Integral from a-b (eq)^2

Ex: Rotate the graph y=x^4 + x^2 from [0,2] about the line y=1

Ok, so for this one, it's asking you to rotate the graph about a horizontal line. This cannot be done, so you will have to subtract it from your equation once it's plugged in, like this.

pi* Integral from 0-2 (x^4 +x^2 -1)^2
Then integrate as you would in a normal definite integral

For disks, you would use the same process except you would have two equations. The formula for disks would be the same as it is for area under the curve, except the two equations would be squared as they are in disks and multiplied by pi as they are in disks.

Ok, some things I don't understand. My problem this week is I don't remember a thing I learned. Every time I went to make an educated guess it was wrong, and I always eliminated the possible answers. If anyone is good at these AP things, can they please help me?