Integral Infomercial Introduction?

So, Mr.K, like the derivative commercials, we had our commercials as a reply to your introductory derivative commercial (the one where you said "I challenge you" etc.). Are you going to make one up for the integrals too? Because we're just doing the finishing touches to ours and we're not sure on how you want us to "submit" it.

Don't forget, you guys, Mr. Beaumont is interested in seeing our projects! =)

BOB!

First off this unit was not bad, not bad at all. I feel relatively confident about this unit, although there were a couple confusing sections. I'm sure they won't be so bad, especially since I'm studying them right now. Hopefully it's enough! Today the pre-test wasn't so bad. I didn't do so well, but after a quick revising, I realized how simple it really was, and hopefully I will remember this by the time the test comes along. I get confused about the accumulation functions, and how they're related to derivatives, and the derivative of certain functions, but I think overall, I'm going to be alright. Good luck everyone.

Accumulation Functions Cont'd

This scribe post will be short as I hope to fully complete it tomorrow (hopefully) or Saturday (likely). Its like 2am. I think my name is George.

Today we started by reviewing the idea that an accumulation function is a composite function, with one function (the function defining limits) taking place within another function (the function we are finding the integral/area under of).

After that review, we worked on some problems. A good thing to remember is that if the integral is going from a value to your function, you want to use that inverse thingy and make it negative so you're going from your function to a value. This is on Slide 3, the second example. The integral from 2 to x is equal to the negative of the integral from x to 2, which is simpler to solve out.

After that he gave us a question that involved finding the total area under a cubic function. However, we hit a snag because nobody quite remembered how to find the roots of a cubic function! After a long demonstration and reminder about synthetic division, Mr. K gave us the roots. I highly reccomend going over the Gr 11 notes he added to the slides (Slides 6 - 11) if you dont remember synthetic division and or want to brush up.

However, after finding the roots, the solution is simple since we already know how to find the area under the graph via the integral. Be careful, as the integral = SIGNED area, meaning if you take the integral from the left most root to the right most root you'd get the wrong answer because part of that integral is negative and thus that integral would be smaller than the total area that we are looking for. In this case, because the integral from 1  to 2 is negative, when we add them together we add the negative integral instead to get the total area.

That basically sums up today's (or rather, yesterdays...) class, I'll try to expand this post to further explain the ideas we explored (yesterday) today, but as of the moment I am way too tired.

And finally because Im tired and want to go to bed, the next scribe is J-Cap (J+ME).

Good Night, and or good morning.

October 23rd: Return of the Definite Integral (sort of)

Starring: The Riemann sum

King Kong as Mysterious outline

and Parabolic Building as Opening Picture.

Sure to be a hit this fall, don't miss it!

Anyway back to serious business.

Today (or rather, yesterday if you want to be technical about it), we continued on yesterday's class about the definite integral. This time, instead of getting data from a table to find the mean of the upper and lower limits, we were provided with a function f(x) = x^2 and used that instead.

We'll start on slide 3:

Here we sketched the graph and tried to find the mean of the upper and lower limits.

We created two intervals: [1, 3/2] and [3/2, 2]. We tried to find the upper and lower limits, but give up.

On slide 4, we change the equation to f(x) = x +1 to make things easier.

So we find the lower limit using the Riemann sum (described in detail on the next slide) and by using some basic algebra.  The answer is the same because the function is linear, but obviously the whole solving by subtracting rectangles wont work so well with a wavy and unpredictable function.

I'm going to try to explain the algebraic thing, but it'll likely be unclear and using general terms.

Since f(x) = x +1, we know its linear, meaning its a straight line. No bumps, curves or waves. Because we want to find the area beneath the function, we can simply get the values by using an interval, in this case [0,4]. So we make a rectangle from x = 0 to x=4 and it goes as high as the function does at this interval (5). So we have a length and width, and thus an area (20). Now, because we want the lower limit, we dont want this extra space thats above the function (because the function is not rectangular, rather it is trapezoidal). So we subtract the extra space by finding the length and width of it (4) and dividing that area by two (because we want to keep half of it). So we get 20 (the whole area) - 16 (the area around the function on the interval [0,4]) / 2 (because we want the half of it that is still under the function) = 12

Which is the same result as we get when we used the Riemann sum.

On slide 6 we tried some more with the f(x) =  x^2 problem. I believe here we compared the accuracy of using smaller intervals vs larger intervals. 

Slide 7 is just talking about intervals in formulas. Here delta t represents the difference between two values on an interval while delta v is the output difference related to delta t.

Slide 8 is an ad.

Well, sorry this post is so late and mostly likely incomprehensible. The next scribe is benofschool.

Orange.

And because we all love this video now, I give you:

Tom Lehrer!

This is a great song to write blog posts to. It's so catchy.

October 22. The Definite Integral

I guess I'm scribe again since I've been caught by Mr. K. That's alright though, no big deal. At least he's still around. :D

Today we started off my watching a youtube video entitled "Lobachevsky - Tom Lehrer" The video is about a mathematician named Lobachevsky who supposedly plagiarized from a fellow mathematician named Gauss.

Well back to the subject. We started out by graphing some values.


The red bars represent the velocity or speed at the given time intervals as it changes gradually. The orange and green bars represent pretty much another graph, where the change in velocity was instant. Whether the change in speed was gradual or instant, we can't determine so. The curve in between the orange and green is the distance travelled, and this is what we will be trying to find.

We then try to find the lower limit by finding area of the lower limits (found in red bars).

Lower limit = sum of area of least possible #'s.

We multiply each value by the time interval (1).
Lower limit = 1.4(1) + 2.7(1) + 3.5(1) + 4.5(1) + 5(1) = 17.1 ft.
Don't add the last value, because this is the upper limit and it can't get any higher than this (t=5).
Now we calculate the upper limit, which includes the orange and green bars.
Upper limit = 2.7(1) + 3.5(1) + 4.5(1) + 5(1) + 5.7(1) = 21.4 ft.
Now estimate the distance by taking the average of these two areas. (17.1 + 21.4)/2 = 19.3 ft.
If we're talking about limits, the range would be 17.1 ft. to 21.4 ft.
At this point we find the ambivalent area (area of uncertainty) which is pretty much the orange-green bars. Ambivalent area = 1.3 + 0.8 + 1 + 0.5+ 0.2 = 4.3 ft.
We have an estimated area of 4.3 ft. under the curve which doesn't exactly match the previous results we estimated.
Now we decrease the time intervals to be more precise, and to see what happens.

Let's find the lower limit, or left edge of this graph. Left edge = 0.7 + 1.35 + 1.5 + 1.75 + 2.05 + 2.25 + 2.4 + 2.5 + 2.7 = 18.2 ft.

Now we find the upper edge which is pretty much the same, but we exclude the number of least value and include the number of most value. It should equal 20.35 ft.

The distance will be between 18.2 ft. and 20.35 ft.
The mean distance is (18.2 + 20.35) / 2 = 19.3 ft.

The average distance remains the same as our last graph.

Therefore by decreasing the size of the time intervals we have brought both the upper and lower limits' ranges lower.
At 1 sec. interval range = 17.1 ft. and 21.4 ft.
At 0.5 sec. interval range = 18.2 ft. and 20.35 ft.
The mean stayed the same at 19.3 ft.

That was all we did in class, our homework is Exercise 3.1 questions 1, 3, 6, 7 and 9. Hope you enjoyed, if any issues, be sure to tell me and I'll be sure to correct them. The next scribe will be well, not Richard, but Not Paul.