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.

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!

Tom Lehrer!

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

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