WEBVTT
Kind: captions
Language: en
00:00:00.360 --> 00:00:06.100
So you're trying to figure out, So we're doing backwards of derivatives, we're doing integrals.
00:00:06.960 --> 00:00:12.820
And integrals can be used for a variety of things, but one of the main things they can be used for is summations.
00:00:13.360 --> 00:00:16.340
By the way Shannon, how much of that Thanksgiving thing did you get on video?
00:00:16.580 --> 00:00:17.320
Shannon: *none of it*
00:00:17.320 --> 00:00:25.640
Aw, my youtube fame just vanished! Oh no, I guess I'll have to keep this job a little but longer.
00:00:25.680 --> 00:00:26.820
Ok, well you can start whenever you're ready.
00:00:26.820 --> 00:00:27.380
Shannon: * I did*.
00:00:27.720 --> 00:00:31.940
Um, ok, probably just as well.
00:00:32.680 --> 00:00:37.540
Alright, so you can use integrals to do sums, summations.
00:00:37.720 --> 00:00:42.640
Ok, well why would you care about sums? Well you want to add things up, that's what a sum is.
00:00:42.740 --> 00:00:48.300
And lots of times, I want to add up lots of things. For example, you're going to price stocks in finance class.
00:00:48.640 --> 00:00:53.100
Stocks are discrete, you get pieces of stock, and get issued dividends.
00:00:53.680 --> 00:00:57.320
And one way you come up with a long term evaluation of a stock,
00:00:57.680 --> 00:01:04.360
is by sort of extrapolating what you think it's dividends will stream over a period of time and using that to calculate the price.
00:01:04.720 --> 00:01:09.880
So what does that mean? Let's say you have stock in something ordinary, like Verizon.
00:01:10.220 --> 00:01:16.260
Verizon issues a quarterly dividend, and you get a check. If you have 100 shares, you might get, lets say $4.
00:01:17.540 --> 00:01:22.700
And $4 isn't [?] it's not a lot of money, but remember if you have 1,000,000 shares, it starts to turn into real money.
00:01:23.080 --> 00:01:32.260
Um and then you get that ever quarter. You get your share of Verizon. And it's dividend in theory will go up. It will go up with inflation,
00:01:32.540 --> 00:01:34.680
it will go up because Verizon will do better.
00:01:41.060 --> 00:01:43.060
And you want to add the stream of dividends up.
00:01:43.420 --> 00:01:49.280
So G.E. announced today that they're going to reduce their dividend. That's going to dramatically affect G.E's stock prices.
00:01:49.280 --> 00:01:53.380
Because either the dividend is going to go down for a period of time and its going to be lower,
00:01:53.700 --> 00:02:02.120
or, it's just going to have a blip. If it's just a blip, it doesn't really have much of an affect because G.E. is an 120 year old company and probably not going anywhere.
00:02:02.480 --> 00:02:06.060
But, if you think it is going to have problems, that's what you do.
00:02:06.060 --> 00:02:12.020
So, you can add up the stock prices, now because these are what they call discrete,
00:02:12.880 --> 00:02:21.440
you only get them quarterly, you don't need to use an integral. You use an integral when you want to add up an infinite set of these.
00:02:22.400 --> 00:02:27.800
However, you could use an integral, the truth is you could probably get pretty close either way if you're doing General Electric.
00:02:28.440 --> 00:02:29.800
This thing does not want to stay on.
00:02:30.920 --> 00:02:34.840
So how do we add things up to do approximations?
00:02:37.140 --> 00:02:42.720
So one of the problems you have with integrals is you can differentiate anything. You cannot integrate anything.
00:02:43.000 --> 00:02:49.920
There are many functions you can't integrate, and so you use approximations to figure out what the integral would be.
00:02:50.640 --> 00:02:53.800
And one way to do an approximation is by doing a sum of rectangles.
00:02:54.640 --> 00:03:06.580
So as I said last time, suppose you just had a curve, and you want to find the area under the curve and the x axis. So you want to find this region.
00:03:07.580 --> 00:03:17.320
You could say, well, I know that if I take a rectangle, here, and, I find the area that rectangle, then it's less than the area of the region.
00:03:17.960 --> 00:03:26.100
So that's not very good, but I could bracket by doing the rectangle above and the rectangle below and sort of going half way between them.
00:03:26.400 --> 00:03:30.600
And that'd be about right, so I could do better if I did it with more rectangles.
00:03:31.180 --> 00:03:36.360
So I could say, cut this into a pair of rectangles.
00:03:37.000 --> 00:03:44.160
Which means my error is this region. I am not adding up that region.
00:03:45.420 --> 00:03:50.320
You say, ok, instead of a pair of rectangles, let me do 4 rectangles.
00:03:59.480 --> 00:04:03.960
Why don't I cut it into 4 rectangles? Now how do you find the are of a rectangle? The area of a rectangle is base*height.
00:04:04.640 --> 00:04:10.280
So I take this region and cut it into 4 identical intervals.
00:04:11.060 --> 00:04:16.980
And then that will give you the base of of each of these rectangles. So to find the height, the height is the y value.
00:04:17.540 --> 00:04:29.120
So what I do is I take one of the 2 points on the interval, either the left or the right, I'm going to take the left point, and go up to where the curve is and use that to establish the height of the rectangle.
00:04:29.880 --> 00:04:41.440
Now I go over say, for all these intervals, I go here and I go up to the curve and across. Like this.
00:04:42.020 --> 00:04:47.220
Ok? That gives me 4 rectangles. Now if I add up the are of those 4 rectangles,
00:04:47.900 --> 00:04:54.420
It's already close to the area under the curve, we're still off, we will have these little regions, under here, they're less.
00:04:54.800 --> 00:05:03.100
But if I did this again, and I did the rectangles above the curve, I'd probably get close. Now how do I do the rectangles above the curve?
00:05:03.760 --> 00:05:13.300
You take the interval and instead of going to the left side of the interval, going up until you hit the curve and go across, you go to the right side and go across.
00:05:19.340 --> 00:05:25.960
Ok? Now I'm over estimating the area. And if I average the 2, I'm going to start to get pretty close.
00:05:27.360 --> 00:05:38.380
So let's see how we can do that mathematically. And instead of 4 rectangles, of course you could do 8 , they'd be thinner, but then I'd have to add up 8 of them, and then I'd say why not a billion?
00:05:38.980 --> 00:05:44.840
Ok, and then they'd be very very thin and there would be a billion of them, but I'd have very tiny error. Yes?
00:05:50.840 --> 00:05:56.820
So the question is when it's under is it less, but when it's above is it more? That's right, if the curve is going up. If it's going down, it's the other way around.
00:05:57.420 --> 00:06:02.900
So you just, for now, just think about using the left side of each interval and the right side of each interval.
00:06:02.900 --> 00:06:12.440
So when I say interval, so let's pick a curve like y = x^2+3, so first of all what does that look like?
00:06:13.100 --> 00:06:23.100
That's a parabola, and since we're only concerned about the first quadrant, it starts and 3 and just kind of goes up, just like those pictures.
00:06:24.680 --> 00:06:33.240
We can make a better curve, as I mess up the board, like that. Ok?
00:06:34.460 --> 00:06:43.160
And I mean, find the area under this from x=0 to x=4.
00:06:44.140 --> 00:06:46.140
So I want to know what's the area under that.
00:06:46.900 --> 00:06:52.700
Well, this rectangle is 3*4 =12, so it's certainly more than 12.
00:06:54.600 --> 00:07:01.440
And 4^2+3 =19, so it's certainly less than 4*19.
00:07:01.800 --> 00:07:02.700
So it's somewhere in the middle.
00:07:03.680 --> 00:07:07.840
So you say, ok, I'm going to use 4 rectangles. 4 is a nice number, it's not too stressful.
00:07:08.440 --> 00:07:14.480
So I'm going to cut this x-axis at 1,2 and 3.
00:07:19.040 --> 00:07:24.720
Now how am I going to find the rectangles. Well let's see. Well first of all, when x is 1, 1^2+3 =4,
00:07:25.860 --> 00:07:30.000
When x=2, 2^2+3=7,
00:07:30.860 --> 00:07:33.800
When x=3, 3^2+3=12.
00:07:34.480 --> 00:07:36.960
So now I have what will be the heights, ok?
00:07:38.140 --> 00:07:50.680
And you say, alright, and you take this first interval from 0-1, so the width of that rectangle is going to be 1, and the height is if I go up to f(0), and go across and draw a rectangle.
00:07:51.260 --> 00:07:56.300
So it's 1*f(0), which we know if 3, but we'll fill that in in a minute.
00:07:58.340 --> 00:08:00.840
And we say, ok, next rectangle.
00:08:01.520 --> 00:08:08.120
And you go to the left side of the interval from 1-2, which is 1, draw up till you hit the curve and go across.
00:08:09.100 --> 00:08:14.480
So the width again is 1, and this time is at f(1). That's the height of the rectangle.
00:08:14.840 --> 00:08:16.320
You see how I'm getting these numbers?
00:08:17.160 --> 00:08:25.500
Ok? The x value is well I choose to take the interval from 0-4 and cut it into that many pieces.
00:08:26.040 --> 00:08:29.400
And this 1, is the width of each one of those rectangles.
00:08:30.160 --> 00:08:37.740
And the height is found by evaluating the function at the left end point of each of those cuts in the x axis.
00:08:38.780 --> 00:08:43.560
So the next one, you go up to 2 and across,
00:08:47.480 --> 00:08:52.160
And for the last one I go up to 3 and go across.
00:08:55.480 --> 00:09:04.060
Ok? So to get the height of each rectangle, you're going to use the y value and the y value is where you hit the curve.
00:09:08.400 --> 00:09:11.640
Now I add these up. So what is f(0), f(0) is 3,
00:09:14.960 --> 00:09:25.460
f(1)=4, f(2) =7, and f(3)=12.
00:09:26.040 --> 00:09:29.640
And I add that up and I get 26.
00:09:31.440 --> 00:09:34.500
Well, the area's about 26.
00:09:35.400 --> 00:09:41.720
It's bigger than 26, because I know I'm not getting the whole area. I'm missing all of that.
00:09:43.600 --> 00:09:49.460
So how can I do better? Well as I said I can cut it into more rectangles, I could cut it into 8 rectangles,
00:09:50.920 --> 00:10:00.780
or 16 rectangles, they don't have to be multiples of 4. You could cut them in to 20, 30, 50, 100, you could use a computer, you cut it into 100 rectangles, it'll be very accurate.
00:10:01.320 --> 00:10:07.720
You want by tiny bits. Imagine this is cut into 100 of these. Every little rectangle coming closer to the curve.
00:10:07.800 --> 00:10:12.820
As you get closer and closer and zoom in on the curve, it starts to look like a line, it stops looking like a curve.
00:10:14.640 --> 00:10:29.220
You say, ok, but I want to do better with just 4 rectangles. Well, now, let's do it again, and instead of picking the left hand side of each interval, we'll pick the right hand point.
00:10:30.060 --> 00:10:33.440
We got those y values by evaluating points of the function.
00:10:35.500 --> 00:10:45.540
Alright, now let's do rectangles again, but this time let's do the right hand rectangles. So these are called left hand rectangles or left end point rectangles,
00:10:51.100 --> 00:10:54.260
And we abbreviate that LHR,
00:10:58.240 --> 00:11:09.920
And when you use right hand rectangles, well the width of each of these rectangles is still going to be 1, but now I go across at 1, I go to the curve and then go down.
00:11:10.620 --> 00:11:14.740
And that gives you the height. So the first one's width is 1, and the height is f(1).
00:11:18.620 --> 00:11:23.260
Now I repeat, I go up to 2, and go across.
00:11:27.220 --> 00:11:29.220
And I go up to 3 and across,
00:11:32.940 --> 00:11:36.240
And 4 and across.
00:11:38.780 --> 00:11:40.060
So certainly over estimated this time.
00:11:47.100 --> 00:12:03.300
But let's see what that comes out to. f(1) =4, f(2)=7, f(3)=12 and f(4)=19.
00:12:04.600 --> 00:12:09.740
Now I get 42. So using right hand rectangles,
00:12:18.980 --> 00:12:23.360
Or RHR, I get 42.
00:12:24.420 --> 00:12:35.060
So since a good guess would be halfway in between 26 and 42, which is 34, we'll do more on Wednesday,
00:12:35.580 --> 00:12:39.440
And then Wednesday, you can come up with the exact number, but we'll focus on doing these for now.
00:12:40.940 --> 00:12:49.740
Ok? So a good guess is halfway between those 2 numbers. Which, by the way doing halfway between those 2 numbers is doing trapezoids, if you were curious
00:12:51.860 --> 00:12:57.600
We'll worry about that later. So we understand the general principle? So let's have you practice one.
00:12:59.280 --> 00:13:02.860
Let's put this up here so everybody can see it.
00:13:17.900 --> 00:13:21.340
Notice, I pick nice easy numbers, everything's an integer,
00:13:21.720 --> 00:13:27.880
Of course this could get messy, you could go from 0-1, with 4 rectangles and the width of each rectangle would be 1/4.
00:13:28.800 --> 00:13:36.820
You'd have to start plugging in fractions. Very rapidly you say this isn't fun anymore. You switch to a calculator or a computer.
00:13:37.080 --> 00:13:40.600
But we'll do another easy one. Let's say you have the curve
00:13:43.500 --> 00:14:09.480
And we want to estimate the area under the curve using 4 LHR, and 4 RHR.
00:14:10.500 --> 00:14:23.880
And the curve, looks just like our other one. It looks like a parabola, because it is a parabola. Ok? Take a minute, see how you do.
00:14:25.880 --> 00:14:32.400
I should put that in, I mean it's not automatically equal to 0 after all. Alright, let's do this.
00:14:33.700 --> 00:14:39.200
So we want to estimate the area from 0-4, our rectangles will be at 1,2 3 and 4.
00:14:44.120 --> 00:14:46.120
Ok? Just like the last problem.
00:14:54.640 --> 00:14:56.340
That's going to be our 4 rectangles.
00:14:59.460 --> 00:15:05.360
The width of each of those is 1, the height of the first one, so if we're doing LHR,
00:15:06.100 --> 00:15:19.420
So it would be 1*f(0)+1*f(1)+1*f(2)+1*f(3).
00:15:19.420 --> 00:15:28.700
By the way notice, first of all we're multiplying by 1 each time, but you could factor that 1 out. Since they all have the same width you could pull that out.
00:15:29.140 --> 00:15:33.300
Cause we're going to do one in a minute where the width is not 1, and you can simplify the arithmetic.
00:15:36.440 --> 00:15:55.220
So f(0)=1, f(1)=3, f(2)=9, anf f(3)=19.
00:16:01.680 --> 00:16:06.620
So after you get the f values out of the equation 2x^2+1,
00:16:06.940 --> 00:16:18.400
Ok? So I plug in the x coordinates to get the y values. And I use the x coordinates starting at the left most end point and count up until 1 before the right hand end point
00:16:21.880 --> 00:16:23.400
So that's 32.
00:16:25.460 --> 00:16:28.440
And if I wanted to do the right hand rectangles,
00:16:30.640 --> 00:16:36.780
It's the same idea, except now we're getting the rectangles at the right side of the interval, and we go up and across.
00:16:40.420 --> 00:16:41.640
As I said, this will over estimate.
00:16:42.280 --> 00:16:46.760
So we know that the area is more than 32, now we can find what it's less than.
00:16:47.820 --> 00:16:52.420
So this will be 1, well what's the right end of 0-1, 1.
00:16:53.720 --> 00:16:57.440
And what's the right end of 1-2? 2.
00:16:58.920 --> 00:17:03.300
And then 3 and then 4.
00:17:06.780 --> 00:17:20.000
So this is 1*(3)+1(9)+1(19)+1(33)
00:17:24.440 --> 00:17:25.900
That's 64.
00:17:30.640 --> 00:17:33.140
So a good guess would be halfway between those 2, which is 48.
00:17:33.840 --> 00:17:39.300
Ok, now let's do this and make it slightly more difficult. Because I know you guys want it to be more difficult.
00:17:41.660 --> 00:17:50.100
The function y=x^2+x+1
00:17:52.740 --> 00:18:14.580
Estimate the area, from x=2 to x=4, using 4, let's just do LHR for the moment. That's a little messier.
00:18:15.260 --> 00:18:18.660
So x^2+x+1 is still just going to be a parabola shape,
00:18:21.360 --> 00:18:27.280
And now we're going to go from 2 to 4.
00:18:29.300 --> 00:18:30.600
Using 4 rectangles.
00:18:37.300 --> 00:18:39.300
So our x coordinates are more annoying.
00:18:40.560 --> 00:18:49.260
We start at 2, we go to 4, we cut that into 4 pieces, so 4 equal pieces is 2, 5/2, 3,7/2 and 4.
00:18:54.580 --> 00:19:02.900
And then you get each rectangle by taking the left end. So what's the left end of 2 and 5/2? 2, and then go across.
00:19:03.840 --> 00:19:09.900
And then from 5/2 to 3, you take the left side, that's 5/2, and go across.
00:19:10.900 --> 00:19:12.900
And then from 3 to 7/2,
00:19:13.580 --> 00:19:16.200
And then from 7/2 to 4.
00:19:20.480 --> 00:19:22.840
So the width of each of these is 1/2.
00:19:25.520 --> 00:19:42.840
You have 1/2*f(2)+1/2*f(5/2)+1/2*f(3)+1/2*f(7/2)
00:19:50.140 --> 00:19:57.340
And what is f(2)? 2^2+2+1=7.
00:20:00.720 --> 00:20:10.140
f(5/2)=(5/2)^2+5/2+1 = 39/4
00:20:14.600 --> 00:20:17.980
f(3)=3^2+3+1=13
00:20:19.640 --> 00:20:28.520
And f(7/2) = (7/2)^2+(7/2)+1 = 67/4
00:20:36.780 --> 00:20:47.040
So this becomes, (7/2)+(39/8)+(13/2)+(67/8)
00:20:48.520 --> 00:20:57.720
Which is let's see, 186/8.
00:20:59.960 --> 00:21:08.120
Or 93/4. Ok? That's a messy number, right? Now of course you could use a calculator, right and I probably messed up, maybe I didn't.
00:21:09.220 --> 00:21:16.320
Ok? If I give it to you on an exam, you won'e be able to use a calculator. So either I would say stop here,
00:21:18.940 --> 00:21:24.480
Ok? Or we say something like set it up, or we have to give you easy numbers, we'll give you integers.
00:21:25.040 --> 00:21:36.500
I don't think it's unreasonable to ask you to square 1/2, but you know, this could be x^3, or log or e or one of those things. Then you really just can't do it.
00:21:36.920 --> 00:21:38.920
Ok? Then you need to use the computer.
00:21:40.060 --> 00:21:48.660
Yeah so you get the idea. It doesn't have to be integers, it doesn't have to be from 0-4, you could go from 2-2.1.
00:21:48.980 --> 00:21:57.200
You could use 12. Ok? So the width of these can start to get very obnoxious and the y value can get very difficult.
00:21:57.860 --> 00:21:59.860
But let's just have you practice setting one up.
00:22:01.120 --> 00:22:39.740
So if I give you, y=x^3+2x, estimate the area from x=3 to x=5, using 4 LHR and 4 RHR. I only want you to set it up. I just want you to show that you can find the correct pieces.
00:22:40.320 --> 00:22:44.040
Ok? Don't try to calculate it, I don't care what it comes out to.
00:22:45.580 --> 00:22:59.560
Again, x^3+2x is just going up, and we're going from 3-5. So you're going to want to cut that into 4 pieces.
00:23:00.660 --> 00:23:12.440
So what's halfway between 3 and 4? 7/2.
00:23:13.160 --> 00:23:17.980
4+5 is 9, so 9/2, so 3, 3.5, 4, 4.5, and 5.
00:23:18.540 --> 00:23:26.820
So we'll see if you know what you're doing. Ok, so how wide is each of these rectangles? Each of these rectangles is 1/2 wide.
00:23:29.500 --> 00:23:31.260
So to do the LHR,
00:23:38.160 --> 00:24:00.060
You have a width of 1/2*f(3)+1/2*f(7/2)+1/2*f(4)+1/2*f(9/2)
00:24:02.260 --> 00:24:14.920
Ok? That's all I would want to see. I don't need you to work out what those numbers are. Because (9/2)^3+2(9/2) is 881/8.
00:24:15.720 --> 00:24:20.820
I think that's right, I could be wrong. Notice though, you could factor out the 1/2.
00:24:20.920 --> 00:24:30.000
And you would have f(3)+f(7/2)+f(4)+f(9/2)
00:24:31.900 --> 00:24:38.720
Ok? So one thing you could do to simplify these is they're always the same width, so you could factor with width out of the problem, then you only have to write it once.
00:24:39.220 --> 00:24:48.360
So if you were setting up the computer program or the more complex calculation, what happens is at some point you turn this into an integral.
00:24:48.960 --> 00:24:52.940
And then taking this out, makes like a lot easier.
00:24:53.920 --> 00:24:55.920
So if we're doing the right hand rectangles,
00:24:59.060 --> 00:25:01.060
The width again is 1/2,
00:25:02.400 --> 00:25:08.500
But now instead of starting at 3 and finishing at 9/2, I'm going to start at 7/2 and finish at 5.
00:25:08.980 --> 00:25:24.320
So I'm going to have (1/2)[f(7/2)f(3)+f(9/2)+f(5)] and again, I don't really care what the number is.
00:25:24.700 --> 00:25:32.260
Ok? Notice, 3 parts of this are the same, only the left end differs and the right end differs.
00:25:32.520 --> 00:25:34.980
Everything in the middle is the same calculation.
00:25:35.600 --> 00:25:41.140
So once you have one of these, you can see how theres some sort of error built in.
00:25:41.320 --> 00:25:45.960
Because this one is using your lower number to give you an under estimate, and this one is using the higher number.
00:25:46.400 --> 00:25:56.920
Now, somebody asked earlier, so left hand is always less and right hand is always bigger? Well, not necessarily. Depends on what the curve looks like.
00:26:07.080 --> 00:26:12.980
Suppose you had a curve where you're going down. Like that.
00:26:14.100 --> 00:26:16.220
Ok, now when you draw the rectangles,
00:26:18.740 --> 00:26:20.740
Well, bad example, sorry.
00:26:24.940 --> 00:26:27.760
Going down this way. Ok?
00:26:29.320 --> 00:26:35.500
And now, the left hand rectangles are over estimating.
00:26:36.160 --> 00:26:41.560
And the right hand rectangles are giving you rectangles that are under the curve.
00:26:42.080 --> 00:26:48.860
So if the curve was going up from right to left, then the left will under estimate it.
00:26:49.320 --> 00:27:02.380
Ok? And of course, we could have a curve that's doing both, then some of the rectangles will under estimate, and some of the rectangles will over estimate.
00:27:07.340 --> 00:27:10.680
That's why it's nice to have 2 systems, an under and an over.
00:27:11.340 --> 00:27:14.440
Ok? Because when you put them together you can usually get a very accurate estimate.
00:27:14.440 --> 00:27:21.460
But you don't really want to have to always calculate twice, so the other way that you get more accurate is that you just do more rectangles.
00:27:22.400 --> 00:27:24.940
So, let me pick another example,
00:27:33.700 --> 00:27:35.700
To go back to our x^2+3 example,
00:27:40.860 --> 00:27:46.320
Let's say I went from 0-4 but I wanted to use a lot of rectangles. So 8.
00:27:52.820 --> 00:27:55.180
So now each one would have a width of 1/2,
00:27:56.780 --> 00:28:01.980
3/2, 5/2, 7/2,
00:28:12.600 --> 00:28:14.100
So you can see I'm starting to have less error.
00:28:16.860 --> 00:28:20.100
And then this would be, each of these has a width of 1/2,
00:28:24.440 --> 00:28:47.860
And I would have (1/2)[f(0)+f(1/2)+f(1)+f(3/2)+f(2)+f(5/2)+f(3)+f(7/2)] and that would be my left hand rectangles.
00:28:48.440 --> 00:28:49.740
But now I have 8 of them.
00:28:52.560 --> 00:28:54.560
And if I was doing a right hand side,
00:28:58.840 --> 00:29:23.440
I'd have 1/2, thats the width of all of them, 1/2[f(1/2)+f(1)+f(3/2)+f(2)+f(5/2)+f(3)+f(7/2)+f(4)] ok? So they start to have many more terms in common.
00:29:24.020 --> 00:29:27.240
And they're only different by the end term each time. Ok?
00:29:27.800 --> 00:29:37.800
Ok? So this will give me a better estimate just using 8. I could do 16, as I said I could do 100, you could even make a little computer program and do it in excel or your calculator or whatever.
00:29:37.880 --> 00:29:43.540
And you could do n=100, and by 100 you're going to be incredibly accurate.You don't really need to do left and right.
00:29:44.460 --> 00:29:49.520
You could just do left or just do right, and what will happen is algebraically we don't do that in this class,
00:29:50.060 --> 00:29:55.940
You say what if I just had an infinite number of them and they're infinitely thin, because I'm cutting them into an infinite number of pieces.
00:29:56.920 --> 00:30:03.880
So of course you can't use infinity, but you could take a limit, so you could say, let's cut it into n rectangles,
00:30:04.180 --> 00:30:06.000
come up with a formula.
00:30:06.280 --> 00:30:16.660
Ok? the width of each of these is the whole interval divided by n, because there's n of them. The height well, I'd start at some value, cut it into n pieces.
00:30:17.140 --> 00:30:21.720
So I have to add all of those up. So if I took the limit of that, it turns into the integral.
00:30:21.720 --> 00:30:31.080
So that's what we're gonna do. You'll see we'll do more of that next class, that's the magic jump, is going from doing it in discrete pieces of rectangles
00:30:31.640 --> 00:30:33.960
into doing it in an infinite number of pieces
00:30:34.380 --> 00:30:43.440
and getting the integral. Just like with the derivative, we would take smaller and smaller pieces of the curve for the tangent line and secant line,
00:30:44.180 --> 00:30:51.040
and then we'd do an infinitely thin difference between 2 points to do the secant line, and that gives you the tangent line.
00:30:52.200 --> 00:31:00.520
So these 2 are linked. We won't get into that, but the derivative and integral are linked and as you saw, when you're integrating you're doing the anti derivative.
00:31:01.440 --> 00:31:07.180
So this infinity concept links the 2 of them together. This cutting things into infinitely narrow pieces.
00:31:08.080 --> 00:31:13.580
Ok, so I can see you guys are all getting a little sleepy, so we'll pick up on Wednesday.