On Friday's class, Mr. K told us that this was the last class for this unit and that the pretest would be on Monday with the test coming Tuesday. I'm sure he said something encouraging us about scholarships or maybe that was another class. But we did also talk about how it was nearing the conclusion of the first semester. My my, how time flies by.

Alright where was I? Right. Second semester. The second half. The beginning of the end. It scares me that high school is ending...it's also somewhat of a thrill. But the main thing we concentrated on for the moment was our scheduling. Even if it wasn't necessary to come to calculus class because of the extra spares, it was, well encouraged. It was a choice. No one is forced to do anything. Lovely words.

After sorting that out, with a fair amount of people agreeing to come into calculus class, or at least that's the impression I got, we split up into groups and began a workshop type class.

In our groups, we concluded and REVIEWED the "total area under a curve question" from the end of Thursday's class.

Just to jog some memories, we used the method of SYNTHETIC DIVISION to find the roots of this cubic function. After the roots were found, we continued on to find the integral of this function; the SIGNED area under the curve. But seeing as part of one curve was under the x-axis, things got complicated. Just a bit. If we were to add the area of the curve in the interval of [1, 2] it would be adding a negative, which is just subtracting. But that is WRONG since it would be subtracting that small area from the larger area. We are trying to find the TOTAL area. This is why the idea of "SIGNED" is fundamental.

The solution? Take the absolute value of the function, so that all of the values would be positive and over the x-axis, so in the case of adding these areas, these areas would both be positive. In finding it out, the class did it both ways and found the values using a calculator.

Below is the graph we put into our calculators, using the same window so that the function was zoomed.

CALCULATOR FUNCTIONS:

To find the integral of a function, type the function in the Y1 window and press 2nd + calc + 7 [integral function]. It will then ask for a number to act as a lower limit. In this case put -2 then press enter. Upper limit, is 2. Then for the interval you've want the integral to be calculated, the area under the curve will be shaded in black and it will give you a value. We were given 10.66 repeating. This value is incorrect but this teaches us how to use the integral function on our graphing calculators.To get the absolute value of the function, all we have to do is go to the y-variable window and insert the absolute value function on our calculators represented by "abs (". Close the parentheses at the end of the function and integrate using the process above. The value we end up getting is a little more, since the values are positive. This gives us a value approx. 11.83 repeating.

Then we caught up to the material we were supposed to learn that day. Finding the area between 2 curves.

We were given 2 functions. We find the area in a similar fashion as finding the area under one curve except for a few differences.

I apologize if this is really LOOSELY EXPLAINED. I'm not trying to be lazy I swear. I'm doing my best. With the time I can spend on calculus. I can always edit this post afterwards.

But the main thing to know is the main pattern for integrating "in-between functions". That is: TOP FUNCTION - BOTTOM FUNCTION. This is said because the "top" function is usually the function with the larger area.

These functions also have values that make a curve under the x-axis making those values negative. In order to make those positive, again, find the absolute value of those functions and integrate.

This unit is kaput!

THE THINGS WE LEARNED [I know I'm missing something on this list..somehow]:

- definition of an ACCUMULATION FUNCTION and how to use them and how they are involved with integrals.
- 2nd Fundamental Theorem of Calculus [or first] which states that the derivative of an accumulation function [integral] is equal to the original function, f(x)
- integral is the SIGNED area under a curve
- how to integrate/find area between two curves; and a helpful tip in doing so is top function - bottom function.

Finally, the next scribe is Hi, I'm Justus. You're the last one yet again. And I'm picking you. Yet again. D

*éjà vu*, I'm telling you. I know you're really busy lately and if you can't handle it and you need help, I'm willing to do a collaborative slide [somehow] to make things easier for you. But then again, I think you have pre-test day, so that's pretty good. It's entirely up to you. Let me know Monday.

As for this disappointing scribe post.. I'll be back to fix you. I'm not happy with the way I made you. I'll be back. Arnold Schwarzenegger said so.

## 1 comment:

Just use the the Integrals Ch 6 tag. The other one was used for Ch 3.

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