Area between a curve and the x-axis

WEBVTT
Kind: captions
Language: en

00:00:00.640 --> 00:00:05.510
So we've got the function, f of x is equal
to x squared.

00:00:05.510 --> 00:00:09.870
And what I'm concerned with, is finding
the area under the curve,

00:00:09.870 --> 00:00:13.900
y is equal to f of x, so that's my y axis.

00:00:13.900 --> 00:00:18.630
This is my, my, x axis.

00:00:18.630 --> 00:00:20.720
Then let me draw my function.

00:00:20.720 --> 00:00:22.460
My function looks like this.

00:00:22.460 --> 00:00:24.740
At least in the, in the first quadrant.

00:00:24.740 --> 00:00:26.150
That's where I'll graph it for now.

00:00:26.150 --> 00:00:29.190
I could also graph it, obviously in the
second quadrant.

00:00:29.190 --> 00:00:34.570
But, what I care about is the area under
this curve and above the positive x axis,

00:00:34.570 --> 00:00:41.840
between, between x equals 1 and x equals
4, x equals 4.

00:00:41.840 --> 00:00:44.272
And I'm tired of approximating areas.

00:00:44.272 --> 00:00:50.350
I wanna find the exact area under this
curve above the x axis.

00:00:50.350 --> 00:00:52.780
And the way we denote the exact area under
the

00:00:52.780 --> 00:00:56.990
curve, this little brown shaded area, is
using the definite integral.

00:00:56.990 --> 00:01:02.170
The definite integral from 1 to 4 of f of
x, dx.

00:01:02.170 --> 00:01:04.400
And the way that, or the way I
conceptualize

00:01:04.400 --> 00:01:07.090
where this notation comes from, is we
imagine a

00:01:07.090 --> 00:01:10.650
bunch of infinite, an infinite number of
infinitely thin

00:01:10.650 --> 00:01:13.240
rectangles that we sum up to find this
area.

00:01:13.240 --> 00:01:18.010
Let me draw one of those infinitely thin
rectangles, maybe not so infinitely thin.

00:01:18.010 --> 00:01:20.670
So, let me draw it like this.

00:01:20.670 --> 00:01:23.590
So, that would be one of the rectangles,
that would be another rectangle.

00:01:23.590 --> 00:01:25.390
This should be reminiscent of a Riemann
sum.

00:01:25.390 --> 00:01:27.685
In fact, that's where the Riemann integral
comes from.

00:01:27.685 --> 00:01:30.288
Think of a Riemann sum where you have an
infinite number

00:01:30.288 --> 00:01:33.411
of these rectangles, where the width of
each of the rectangles.

00:01:33.411 --> 00:01:38.546
This is how I conceptualize it, is dx, and
the height of this rectangle is the

00:01:38.546 --> 00:01:43.453
function evaluated at an x that's within
this interval right over here.

00:01:43.453 --> 00:01:46.210
And so, this part right over here is the
area of

00:01:46.210 --> 00:01:49.690
one of those rectangles, and we were
summing them all up.

00:01:49.690 --> 00:01:54.530
And this kind of an elongated s,
reminiscent of a sigma for summing.

00:01:54.530 --> 00:01:58.760
We're summing up the infinite number, of
those infinitely thin rectangles,

00:01:58.760 --> 00:02:02.860
or the areas of those infinitely thin
rectangles between 1 and 4.

00:02:02.860 --> 00:02:05.660
So that's where the notation of the
definite integral comes from.

00:02:05.660 --> 00:02:06.740
But we still haven't done anything.

00:02:06.740 --> 00:02:10.550
We've just written some notation that says
the exact area of the un-

00:02:10.550 --> 00:02:15.260
Between 1 and 4, under the curve f of x,
and above the x axis.

00:02:15.260 --> 00:02:19.150
In order to actually do anything really
productive with this, we have to turn to

00:02:19.150 --> 00:02:22.140
the second fundamental theorem of
calculus, sometimes called

00:02:22.140 --> 00:02:24.970
part two of the fundamental theorem of
calculus.

00:02:24.970 --> 00:02:29.580
>> Which tells us, that if f has an
antiderivative, so if we

00:02:29.580 --> 00:02:35.060
have the antiderivative of f, so f of x is
derivative,

00:02:35.060 --> 00:02:40.505
derivative of some function capital F of
x, or another way of saying it is,

00:02:40.505 --> 00:02:45.824
f, capital F of x is the antiderivative,

00:02:45.824 --> 00:02:51.020
antiderivative of lower case f of x.

00:02:51.020 --> 00:02:52.940
Then I can evaluate this thing, and we do

00:02:52.940 --> 00:02:56.310
a whole video on conceptually
understanding why this makes sense.

00:02:56.310 --> 00:03:00.000
We could evaluate this, by evaluating the
antiderivative

00:03:00.000 --> 00:03:04.220
of f, or an antiderivative of f, at 4.

00:03:04.220 --> 00:03:09.340
And from that, subtract the antiderivative
evaluated at 1.

00:03:09.340 --> 00:03:14.110
So, let's do it for this particular case
right over here.

00:03:14.110 --> 00:03:17.230
So we are taking, I'll just rewrite this
statement.

00:03:17.230 --> 00:03:19.990
Instead of writing f of x, I'll write x
squared.

00:03:19.990 --> 00:03:24.360
So, the definite integral from 1 to 4 of x
squared dx.

00:03:24.360 --> 00:03:27.370
Well, we're just gonna have to figure out
what the antiderivative is.

00:03:27.370 --> 00:03:33.820
So if f of x is equal to x squared, what
is capital F of x equal to?

00:03:33.820 --> 00:03:36.110
What is the antiderivative?

00:03:36.110 --> 00:03:39.210
Well, you might remember from your power
rule, that if you

00:03:39.210 --> 00:03:43.100
take the derivative with respect to x of x
to the

00:03:43.100 --> 00:03:47.440
third, you are going to get 3x squared,
which is pretty

00:03:47.440 --> 00:03:50.050
darn close to x squared except for this
factor of 3.

00:03:50.050 --> 00:03:51.615
So, let's divide both sides by 3.

00:03:52.750 --> 00:03:56.580
Let's divide both sides by 3, and you get
the derivative

00:03:56.580 --> 00:04:00.520
of x to the third divided by 3 is indeed x
squared.

00:04:00.520 --> 00:04:02.310
Or, you can say this is the same thing as
the

00:04:02.310 --> 00:04:07.830
derivative with respect to x of, x to the
third over 3.

00:04:07.830 --> 00:04:08.830
Take the derivative of this.

00:04:08.830 --> 00:04:11.430
It'll be 3 times one third.

00:04:11.430 --> 00:04:13.630
And then you'll decrement the power, it'll
just be x squared.

00:04:13.630 --> 00:04:15.730
So, this right over here, once again, is x
squared.

00:04:15.730 --> 00:04:20.200
It's just equal to, just equal to x
squared.

00:04:20.200 --> 00:04:24.240
So, in this case, our capital F of x, our
antiderivative,

00:04:24.240 --> 00:04:28.540
is x to the third, x to the third over 3.

00:04:28.540 --> 00:04:34.600
And so we just have to evaluate that at 4
and at 1, and sometimes the way we would,

00:04:34.600 --> 00:04:40.070
the, the notation we would use is, we'll
say that the antiderivative is x

00:04:40.070 --> 00:04:44.660
to the third over 3, and we're going to
evaluate it, the one I, I always

00:04:44.660 --> 00:04:46.090
just like to write the numbers up here,

00:04:46.090 --> 00:04:49.260
at 4 and from that subtracted, evaluated
at 1.

00:04:49.260 --> 00:04:51.430
Sometimes you'll see people write a little
line here too,

00:04:51.430 --> 00:04:54.930
we'll say we're evaluate it at 4 and then
at 1.

00:04:54.930 --> 00:04:56.910
But I'll just do it without the line.

00:04:56.910 --> 00:04:59.010
If we're gonna evaluate this thing at 4
and from

00:04:59.010 --> 00:05:02.330
that subtract it, subtract it evaluated at
1, so this

00:05:02.330 --> 00:05:07.740
going to be equal to 4 to the third power
is 64, so it's going to be 64 over 3.

00:05:07.740 --> 00:05:13.330
Let me color code it, this is, this right
over here, is this, right

00:05:13.330 --> 00:05:15.320
over there and then from that, we're

00:05:15.320 --> 00:05:18.500
going to subtract this business evaluated
at one.

00:05:18.500 --> 00:05:23.330
Well, when you evaluate it at 1, you get 1
to the third is one over 3.

00:05:23.330 --> 00:05:24.900
You get one third.

00:05:24.900 --> 00:05:31.100
So just to be clear, this is this right
over there.

00:05:31.100 --> 00:05:33.660
And then we are ready to just subtract
these fractions.

00:05:33.660 --> 00:05:41.140
64 over 3, minus one third, is equal to 63
over 3.

00:05:41.140 --> 00:05:46.850
And 3 goes into 63 exactly, exactly 21
times.

00:05:46.850 --> 00:05:52.680
So, whatever the units are, the area of
this brown area is 21 square units.


Definite integral as the limit of a Riemann sum

WEBVTT
Kind: captions
Language: en

00:00:00.670 --> 00:00:02.390
We've done several
videos already

00:00:02.390 --> 00:00:05.330
where we're approximating
the area under a curve

00:00:05.330 --> 00:00:08.870
by breaking up that
area into rectangles

00:00:08.870 --> 00:00:12.417
and then finding the sum of
the areas of those rectangles

00:00:12.417 --> 00:00:13.250
as an approximation.

00:00:13.250 --> 00:00:15.750
And this was actually
the first example

00:00:15.750 --> 00:00:18.410
that we looked at where
each of the rectangles

00:00:18.410 --> 00:00:19.480
had an equal width.

00:00:19.480 --> 00:00:21.940
So we equally
partitioned the interval

00:00:21.940 --> 00:00:24.790
between our two boundaries
between a and b.

00:00:24.790 --> 00:00:27.210
And the height of the
rectangle was the function

00:00:27.210 --> 00:00:30.570
evaluated at the left
endpoint of each rectangle.

00:00:30.570 --> 00:00:33.300
And we wanted to generalize it
and write it in sigma notation.

00:00:33.300 --> 00:00:35.110
It looked something like this.

00:00:35.110 --> 00:00:36.090
And this was one case.

00:00:36.090 --> 00:00:37.720
Later on, we looked
at a situation

00:00:37.720 --> 00:00:40.130
where you define the
height by the function

00:00:40.130 --> 00:00:42.140
value at the right endpoint
or at the midpoint.

00:00:42.140 --> 00:00:44.190
And then we even
constructed trapezoids.

00:00:44.190 --> 00:00:50.190
And these are all particular
instances of Riemann sums.

00:00:50.190 --> 00:00:54.784
So this right over
here is a Riemann sum.

00:00:54.784 --> 00:00:56.450
And when people talk
about Riemann sums,

00:00:56.450 --> 00:00:57.790
they're talking about
the more general notion.

00:00:57.790 --> 00:00:59.040
You don't have to
just do it this way.

00:00:59.040 --> 00:01:00.300
You could use trapezoids.

00:01:00.300 --> 00:01:02.690
You don't even have to have
equally-spaced partitions.

00:01:02.690 --> 00:01:05.400
I used equally-spaced partitions
because it made things

00:01:05.400 --> 00:01:07.850
a little bit
conceptually simpler.

00:01:07.850 --> 00:01:10.720
And this right here is
a picture of the person

00:01:10.720 --> 00:01:12.320
that Riemann sums
was named after.

00:01:12.320 --> 00:01:14.510
This is Bernhard Riemann.

00:01:14.510 --> 00:01:18.086
And he made many
contributions to mathematics.

00:01:18.086 --> 00:01:19.710
But what he is most
known for, at least

00:01:19.710 --> 00:01:21.626
if you're taking a
first-year calculus course,

00:01:21.626 --> 00:01:23.090
is the Riemann sum.

00:01:23.090 --> 00:01:26.600
And how this is used to
define the Riemann integral.

00:01:26.600 --> 00:01:28.350
Both Newton and
Leibniz had come up

00:01:28.350 --> 00:01:30.030
with the idea of
the integral when

00:01:30.030 --> 00:01:32.760
they had formulated calculus,
but the Riemann integral

00:01:32.760 --> 00:01:35.260
is kind of the most
mainstream formal,

00:01:35.260 --> 00:01:37.500
or I would say
rigorous, definition

00:01:37.500 --> 00:01:39.480
of what an integral is.

00:01:39.480 --> 00:01:42.790
So as you could imagine, this is
one instance of a Riemann sum.

00:01:42.790 --> 00:01:44.730
We have n right over here.

00:01:44.730 --> 00:01:48.710
The larger n is, the better an
approximation it's going to be.

00:01:48.710 --> 00:01:51.130
So his definition of
an integral, which

00:01:51.130 --> 00:01:53.020
is the actual area
under the curve,

00:01:53.020 --> 00:01:55.230
or his definition of a
definite integral, which

00:01:55.230 --> 00:01:58.400
is the actual area under
a curve between a and b

00:01:58.400 --> 00:02:03.310
is to take this Riemann sum,
it doesn't have to be this one,

00:02:03.310 --> 00:02:09.139
take any Riemann sum, and
take the limit as n approaches

00:02:09.139 --> 00:02:09.819
infinity.

00:02:09.819 --> 00:02:11.360
So just to be clear,
what's happening

00:02:11.360 --> 00:02:12.940
when n approaches infinity?

00:02:12.940 --> 00:02:15.130
Let me draw another
diagram here.

00:02:15.130 --> 00:02:18.440
So let's say that's my y-axis.

00:02:18.440 --> 00:02:20.670
This is my x-axis.

00:02:20.670 --> 00:02:23.060
This is my function.

00:02:23.060 --> 00:02:26.300
As n approaches
infinity-- so this is a,

00:02:26.300 --> 00:02:29.510
this is b-- you're just going
to have a ton of rectangles.

00:02:29.510 --> 00:02:32.382
You're just going to get a
ton of rectangles over there.

00:02:32.382 --> 00:02:34.340
And there are going to
become better and better

00:02:34.340 --> 00:02:37.430
approximations for
the actual area.

00:02:37.430 --> 00:02:40.640
And the actual area
under the curve

00:02:40.640 --> 00:02:50.410
is denoted by the integral
from a to b of f of x times dx.

00:02:50.410 --> 00:02:52.936
And you see where
this is coming from

00:02:52.936 --> 00:02:54.310
or how these
notations are close.

00:02:54.310 --> 00:02:56.810
Or at least in my brain,
how they're connected.

00:02:56.810 --> 00:03:04.080
Delta x was the width for
each of these sections.

00:03:04.080 --> 00:03:05.430
This right here is delta x.

00:03:05.430 --> 00:03:08.350
So that is a delta x.

00:03:08.350 --> 00:03:09.390
This is another delta x.

00:03:09.390 --> 00:03:11.320
This is another delta x.

00:03:11.320 --> 00:03:14.300
A reasonable way to
conceptualize what dx is,

00:03:14.300 --> 00:03:17.150
or what a differential
is, is what delta x

00:03:17.150 --> 00:03:20.060
approaches, if it
becomes infinitely small.

00:03:20.060 --> 00:03:22.592
So you can conceptualize
this, and it's not

00:03:22.592 --> 00:03:24.300
a very rigorous way
of thinking about it,

00:03:24.300 --> 00:03:38.070
is an infinitely small-- but
not 0-- infinitely small delta

00:03:38.070 --> 00:03:41.710
x, is one way that you
can conceptualize this.

00:03:41.710 --> 00:03:43.760
So once again, as you
have your function

00:03:43.760 --> 00:03:46.270
times a little small
change in delta x.

00:03:46.270 --> 00:03:47.890
And you are summing,
although you're

00:03:47.890 --> 00:03:52.520
summing an infinite number
of these things, from a to b.

00:03:52.520 --> 00:03:54.020
So I'm going to
leave you there just

00:03:54.020 --> 00:03:54.890
so that you see the connection.

00:03:54.890 --> 00:03:56.348
You know the name
for these things.

00:03:56.348 --> 00:03:58.730
And once again, this
one over here, this

00:03:58.730 --> 00:03:59.890
isn't the only Riemann sum.

00:03:59.890 --> 00:04:02.050
In fact, this is often
called the left Riemann sum

00:04:02.050 --> 00:04:03.250
if you're using it
with rectangles.

00:04:03.250 --> 00:04:04.541
You can do a right Riemann sum.

00:04:04.541 --> 00:04:05.780
You could use the midpoint.

00:04:05.780 --> 00:04:06.660
You could use a trapezoid.

00:04:06.660 --> 00:04:08.951
But if you take the limit of
any of those Riemann sums,

00:04:08.951 --> 00:04:11.080
as n approaches
infinity, then that you

00:04:11.080 --> 00:04:16.160
get as a Riemann
definition of the integral.

00:04:16.160 --> 00:04:18.640
Now so far, we haven't
talked about how

00:04:18.640 --> 00:04:20.440
to actually evaluate this thing.

00:04:20.440 --> 00:04:22.600
This is just a
definition right now.

00:04:22.600 --> 00:04:26.130
And for that we will
do in future videos.


Trapezoidal sums

WEBVTT
Kind: captions
Language: en

00:00:00.730 --> 00:00:04.070
For fun, let's try to
approximate the area

00:00:04.070 --> 00:00:07.700
under the curve y is
equal to the square root,

00:00:07.700 --> 00:00:12.220
the principal root, of x minus
1, between x is equal to 1

00:00:12.220 --> 00:00:15.480
and x is equal to 6.

00:00:15.480 --> 00:00:18.420
So I want to find
this entire area.

00:00:18.420 --> 00:00:22.240
Or I want to at least
approximate this entire area.

00:00:22.240 --> 00:00:27.250
And the way I'll do
it is by setting up

00:00:27.250 --> 00:00:31.040
five trapezoids of equal width.

00:00:31.040 --> 00:00:33.980
So this will be
the left boundary

00:00:33.980 --> 00:00:35.160
of the first trapezoid.

00:00:35.160 --> 00:00:37.830
This will be its right
boundary, which will also

00:00:37.830 --> 00:00:39.710
be the left boundary of
the second trapezoid.

00:00:39.710 --> 00:00:42.370
This will be the right boundary
of the second trapezoid.

00:00:42.370 --> 00:00:45.810
This is the right boundary
of the third trapezoid.

00:00:45.810 --> 00:00:48.980
This'll be the right boundary
of the fourth trapezoid.

00:00:48.980 --> 00:00:51.420
And then finally, this
will be the right boundary

00:00:51.420 --> 00:00:53.370
of the fifth trapezoid.

00:00:53.370 --> 00:00:58.310
And since we're going from 1 to
6, so we're traveling 6 minus 1

00:00:58.310 --> 00:01:02.470
in the x direction, and I want
to split it into five sections,

00:01:02.470 --> 00:01:06.130
the width of each trapezoid is
just going to be equal to 1.

00:01:06.130 --> 00:01:09.700
And so if we say that the width
of a trapezoid is delta x,

00:01:09.700 --> 00:01:15.200
we just can now say the
delta x is equal to 1.

00:01:15.200 --> 00:01:17.220
So let's set up our trapezoids.

00:01:17.220 --> 00:01:21.696
So the first trapezoid is
going to look like that.

00:01:21.696 --> 00:01:22.820
It's going to go like that.

00:01:22.820 --> 00:01:24.861
Actually, it's going to
be a triangle, not really

00:01:24.861 --> 00:01:25.830
a trapezoid.

00:01:25.830 --> 00:01:29.175
Then the second trapezoid
is going to look like this.

00:01:29.175 --> 00:01:30.800
I guess you could
say a trapezoid where

00:01:30.800 --> 00:01:33.770
one of the sides has length
0 turns into a triangle.

00:01:33.770 --> 00:01:38.550
And then the third trapezoid
is going to look like this.

00:01:38.550 --> 00:01:41.390
And then the fourth trapezoid
is going to look like that.

00:01:41.390 --> 00:01:43.930
And then finally, you
have the fifth trapezoid.

00:01:43.930 --> 00:01:46.040
So let's calculate the
area of each of these,

00:01:46.040 --> 00:01:48.660
and then we will have our
approximation for the area

00:01:48.660 --> 00:01:50.180
under the curve.

00:01:50.180 --> 00:01:53.140
So let's do trapezoid--
or I really should say

00:01:53.140 --> 00:01:56.450
triangle-- this first shape,
whatever you want to call it.

00:01:56.450 --> 00:01:59.240
What is the area of
that going to be?

00:01:59.240 --> 00:02:02.120
Well, the area of a
trapezoid-- and you'll

00:02:02.120 --> 00:02:05.950
see this will just turn into
the area of a triangle-- it's

00:02:05.950 --> 00:02:07.973
the average of the
heights of the two

00:02:07.973 --> 00:02:10.139
sides of the trapezoid, the
way we've looked at it--

00:02:10.139 --> 00:02:11.990
or you could say the average
of the heights of the two

00:02:11.990 --> 00:02:14.460
parallel sides, I guess
is the best way to say it.

00:02:14.460 --> 00:02:21.230
So f of 1-- that's the
height here-- plus f of 2,

00:02:21.230 --> 00:02:23.680
all of that over
2, and then we're

00:02:23.680 --> 00:02:25.670
going to multiply it
times our delta x.

00:02:28.262 --> 00:02:30.220
Actually, let me do that
in that same red color

00:02:30.220 --> 00:02:36.090
to show you that this is the
area of that first trapezoid,

00:02:36.090 --> 00:02:39.170
so times delta x.

00:02:39.170 --> 00:02:42.930
And as you see right over here,
if you look at it, the f of 1

00:02:42.930 --> 00:02:44.520
is just going to be 0.

00:02:44.520 --> 00:02:46.540
So you're going to
have f of 2 times--

00:02:46.540 --> 00:02:49.540
so it's going to be this height
times this base times 1/2,

00:02:49.540 --> 00:02:53.250
which is just the
area of a triangle.

00:02:53.250 --> 00:02:54.910
Let's look at the
second trapezoid,

00:02:54.910 --> 00:02:58.290
trapezoid two right over here.

00:02:58.290 --> 00:03:00.740
What is its area going to be?

00:03:00.740 --> 00:03:07.450
Well, it's going to
be f of 2 plus f of 3.

00:03:07.450 --> 00:03:09.100
f of 2 is this height.

00:03:09.100 --> 00:03:10.430
f of 3 is this height.

00:03:10.430 --> 00:03:12.930
So we're taking the average
of those two heights-- divided

00:03:12.930 --> 00:03:14.650
by 2; that's the
average of those two

00:03:14.650 --> 00:03:18.610
heights-- times the
base, times delta x.

00:03:18.610 --> 00:03:22.040
And then let's do
trapezoid three.

00:03:22.040 --> 00:03:24.070
I think you're
getting the idea here.

00:03:24.070 --> 00:03:27.220
Trapezoid three is
going to be f of 3

00:03:27.220 --> 00:03:33.010
plus f of 4 divided
by 2 times delta x.

00:03:33.010 --> 00:03:34.260
And then-- let's see.

00:03:34.260 --> 00:03:35.490
I'm running out of colors.

00:03:35.490 --> 00:03:37.970
This is trapezoid
four right over here.

00:03:37.970 --> 00:03:46.080
So plus f of 4 plus f of 5, all
of that over 2, times delta x.

00:03:46.080 --> 00:03:47.770
And then we have
our last trapezoid,

00:03:47.770 --> 00:03:49.440
which I will do in yellow.

00:03:49.440 --> 00:03:52.740
So this is trapezoid
number five.

00:03:52.740 --> 00:03:55.465
I'll scroll down a little bit,
get some more real estate.

00:03:55.465 --> 00:03:58.090
So it's going to be plus-- I'll
just write the plus over here--

00:03:58.090 --> 00:04:06.770
plus f of 5 plus f of 6
over 2 times our delta x.

00:04:06.770 --> 00:04:09.370
So let's see how we can
simplify this a little bit.

00:04:09.370 --> 00:04:11.850
All of these terms,
we have a 1/2 delta x,

00:04:11.850 --> 00:04:14.010
so let's actually
factor that out.

00:04:14.010 --> 00:04:16.839
So remember, this is our
approximation of our area.

00:04:16.839 --> 00:04:20.561
So area is
approximately-- remember,

00:04:20.561 --> 00:04:22.019
this is just a
rough approximation.

00:04:22.019 --> 00:04:23.180
It's very clear--
actually, you might say,

00:04:23.180 --> 00:04:24.930
this is pretty good
using the trapezoids,

00:04:24.930 --> 00:04:28.180
but it is pretty
clear that we are

00:04:28.180 --> 00:04:29.530
letting go of some of the area.

00:04:29.530 --> 00:04:30.780
We're letting go of that area.

00:04:30.780 --> 00:04:32.450
We're letting go of some
of this right over here.

00:04:32.450 --> 00:04:33.270
You can barely see it.

00:04:33.270 --> 00:04:35.436
Some of this right over
here, you can barely see it.

00:04:35.436 --> 00:04:37.731
But we are-- this is going
to be, it looks like,

00:04:37.731 --> 00:04:39.855
an underestimate, but it
is a decent approximation.

00:04:39.855 --> 00:04:42.159
But let's see if we can
simplify this expression.

00:04:42.159 --> 00:04:43.950
So it's approximately
going to be equal to,

00:04:43.950 --> 00:04:46.815
I'm going to factor
out a delta x over 2.

00:04:50.776 --> 00:04:53.020
And then what I'm
left with-- and I

00:04:53.020 --> 00:04:55.550
will switch to a neutral
color-- if I factor out

00:04:55.550 --> 00:05:00.220
a delta x over 2, then
I have just an f of 1.

00:05:00.220 --> 00:05:04.319
And then I have two f of
2's, so plus 2 times f of 2.

00:05:04.319 --> 00:05:06.610
And I'm doing this because
you might see a formula that

00:05:06.610 --> 00:05:09.330
looks something like this
in your calculus book,

00:05:09.330 --> 00:05:10.820
and it's not some
mysterious thing.

00:05:10.820 --> 00:05:13.980
They just really summed up
the areas of the trapezoids.

00:05:13.980 --> 00:05:18.120
And then we're going to have
two f of 3's-- plus 2 times f

00:05:18.120 --> 00:05:19.240
of 3's.

00:05:19.240 --> 00:05:24.670
Plus we're going to have two f
of 4's-- plus 2 times f of 4.

00:05:24.670 --> 00:05:29.210
And then we're going to have two
f of 5's-- plus 2 times f of 5.

00:05:29.210 --> 00:05:32.980
And then finally, we're going to
have one f of 6-- plus f of 6.

00:05:32.980 --> 00:05:34.360
If you were to
generalize it, you

00:05:34.360 --> 00:05:36.390
have one of the first
endpoint, the function

00:05:36.390 --> 00:05:38.640
evaluated at the first
endpoint, one of it at the very

00:05:38.640 --> 00:05:42.274
last endpoint, and then two
of all of the rest of them.

00:05:42.274 --> 00:05:43.940
But this is just the
area of trapezoids.

00:05:43.940 --> 00:05:46.560
I'm not actually a big fan
of when textbooks write this.

00:05:46.560 --> 00:05:49.180
Because when you see this,
it's hard to visualize

00:05:49.180 --> 00:05:49.870
the trapezoids.

00:05:49.870 --> 00:05:51.800
When you see this,
it's much clearer

00:05:51.800 --> 00:05:53.910
how you might visualize that.

00:05:53.910 --> 00:05:56.620
But with that out of the way,
let's actually evaluate this.

00:05:56.620 --> 00:05:58.540
Lucky for us, the
math is simple.

00:05:58.540 --> 00:06:00.300
Our delta x is just 1.

00:06:02.950 --> 00:06:05.630
And then we just have to
evaluate all of this business.

00:06:05.630 --> 00:06:08.740
f of 1, let's just
remind ourselves

00:06:08.740 --> 00:06:10.160
what our original function was.

00:06:10.160 --> 00:06:13.570
Our original function was
the square root of x minus 1.

00:06:16.150 --> 00:06:19.820
So f of 1 is the square
root of 1 minus 1,

00:06:19.820 --> 00:06:22.470
so that is just going to be 0.

00:06:22.470 --> 00:06:24.760
This expression
right over here is

00:06:24.760 --> 00:06:28.360
going to be 2 times the
square root of 2 minus 1.

00:06:28.360 --> 00:06:30.940
The square root of
2 minus 1 is just 1,

00:06:30.940 --> 00:06:33.360
so this is just going to be 2.

00:06:33.360 --> 00:06:35.840
Actually, let me do it
in that same-- well,

00:06:35.840 --> 00:06:38.050
I'm now using the purple
for a different purpose

00:06:38.050 --> 00:06:39.360
than just the first trapezoid.

00:06:39.360 --> 00:06:40.568
Hopefully, you realized that.

00:06:40.568 --> 00:06:42.280
I was just sticking
with that pen color.

00:06:42.280 --> 00:06:43.930
Then f of 3.

00:06:43.930 --> 00:06:46.479
3 minus 1 is 2--
square root of 2.

00:06:46.479 --> 00:06:48.770
So the function evaluated at
3 is the square root of 2.

00:06:48.770 --> 00:06:52.150
So this is going to be 2
times the square root of 2.

00:06:52.150 --> 00:06:53.720
Then the function
evaluated at 4.

00:06:53.720 --> 00:06:56.740
When you evaluate it at 4,
you get the square root of 3.

00:06:56.740 --> 00:06:59.880
So this is going to be 2
times the square root of 3.

00:06:59.880 --> 00:07:03.700
And then you get 2 times the
square root of 4-- 5 minus 1

00:07:03.700 --> 00:07:04.510
is 4.

00:07:04.510 --> 00:07:07.040
2 times the square
root of 4 is just four.

00:07:07.040 --> 00:07:11.680
And then finally, you get f of
6 is square root of 6 minus 1,

00:07:11.680 --> 00:07:14.770
is the square root of 5.

00:07:14.770 --> 00:07:16.910
And I think we're now
ready to evaluate.

00:07:16.910 --> 00:07:22.610
So let me get my handy TI-85
out and calculate this.

00:07:22.610 --> 00:07:24.210
So it's going to
be-- well I'm just

00:07:24.210 --> 00:07:26.126
going to calculate--
well, I'll just multiply.

00:07:26.126 --> 00:07:29.594
So 0.5 times open
parentheses-- well, it's a 0.

00:07:29.594 --> 00:07:31.760
I'll just write it, just
so you know what I'm doing.

00:07:31.760 --> 00:07:35.520
0 plus 2 plus-- whoops.

00:07:35.520 --> 00:07:36.880
Lost my calculator.

00:07:36.880 --> 00:07:45.490
Plus 2 times the square root of
2 plus 2 times the square root

00:07:45.490 --> 00:07:56.630
of 3 plus 4-- I'm almost done--
plus the square root of 5--

00:07:56.630 --> 00:08:01.570
so let me write
that-- gives me--

00:08:01.570 --> 00:08:05.120
now we are ready for our
drum roll-- it gives me--

00:08:05.120 --> 00:08:07.740
and I'll just round it-- 7.26.

00:08:07.740 --> 00:08:15.230
So the area is
approximately equal to 7.26

00:08:15.230 --> 00:08:19.670
under the curve y is equal
to the square root of x

00:08:19.670 --> 00:08:23.150
minus 1 between x
equals 1 and x equals 6.

00:08:23.150 --> 00:08:26.323
And we did this
using trapezoids.


Rectangular and trapezoidal Riemann approximations

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:02.980
In the last few videos, we've
been approximating the area

00:00:02.980 --> 00:00:05.530
under the curve using
rectangles, where

00:00:05.530 --> 00:00:08.210
the height of each
rectangle was defined

00:00:08.210 --> 00:00:13.330
by the function evaluated
at the left boundary.

00:00:13.330 --> 00:00:16.920
So this would have been
the first rectangle.

00:00:16.920 --> 00:00:21.130
Then the second rectangle
would look something like this.

00:00:21.130 --> 00:00:24.560
And then we'd go all the
way to the nth rectangle

00:00:24.560 --> 00:00:28.812
would look something like that.

00:00:28.812 --> 00:00:30.645
And we saw-- so this
is the first rectangle,

00:00:30.645 --> 00:00:32.228
this is the second
rectangle, and we'd

00:00:32.228 --> 00:00:34.580
go all the way to
the nth rectangle--

00:00:34.580 --> 00:00:37.250
and so we saw that the way that
you would take the sum of all

00:00:37.250 --> 00:00:40.040
of these rectangles in order
to approximate the area is

00:00:40.040 --> 00:00:45.870
that you would get the
sum from i equals 1 to n.

00:00:45.870 --> 00:00:48.780
And so i is essentially a
count of which rectangle

00:00:48.780 --> 00:00:49.897
we're dealing with.

00:00:49.897 --> 00:00:51.480
And what we're going
to do is multiply

00:00:51.480 --> 00:00:53.280
the height times the base.

00:00:53.280 --> 00:00:56.920
So the height of each rectangle,
the height of rectangle one,

00:00:56.920 --> 00:01:00.750
in this case, was the
function evaluated at x0.

00:01:00.750 --> 00:01:04.080
The height of rectangle two was
the function evaluated at x1.

00:01:04.080 --> 00:01:07.700
The height of rectangle
n was the function

00:01:07.700 --> 00:01:10.940
evaluated at x sub n minus 1.

00:01:10.940 --> 00:01:13.170
So the height of
a rectangle i is

00:01:13.170 --> 00:01:17.900
going to be the function
evaluated x sub i minus 1.

00:01:17.900 --> 00:01:22.710
If i is 2, then we're
evaluating it at x1.

00:01:22.710 --> 00:01:27.610
If i was 2, then this would be
the function evaluated at x1.

00:01:27.610 --> 00:01:29.230
So it's the left boundary.

00:01:29.230 --> 00:01:31.370
And then we have to
multiply it times the width.

00:01:31.370 --> 00:01:33.730
And in the last few
videos, and in this video,

00:01:33.730 --> 00:01:36.800
we will assume that all of the
rectangles have equal width.

00:01:36.800 --> 00:01:39.750
Now, we'll call that
equal width delta x.

00:01:39.750 --> 00:01:42.930
And to find it, we just have
to take the total distance

00:01:42.930 --> 00:01:44.430
that we're going
in the x direction.

00:01:44.430 --> 00:01:47.460
So it's going to be
b minus a divided

00:01:47.460 --> 00:01:49.400
by the number of
rectangles we want.

00:01:49.400 --> 00:01:52.170
So it's going to
be times delta x.

00:01:52.170 --> 00:01:54.770
Now, you might imagine that
this is not the only way

00:01:54.770 --> 00:01:57.190
to take the sum
using rectangles,

00:01:57.190 --> 00:01:59.350
or this is not the only
way to take the sum

00:01:59.350 --> 00:02:03.150
or approximate the area using
some type of geometric shape.

00:02:03.150 --> 00:02:06.600
For example, we could
have created rectangles

00:02:06.600 --> 00:02:09.000
where the height is defined
by the rightmost boundary.

00:02:09.000 --> 00:02:10.449
So let's define that.

00:02:10.449 --> 00:02:11.755
So here's our first rectangle.

00:02:14.690 --> 00:02:17.170
And we're defining the
height by the right boundary

00:02:17.170 --> 00:02:17.923
of the rectangle.

00:02:20.650 --> 00:02:23.830
So this right over
here is rectangle one,

00:02:23.830 --> 00:02:26.265
and its height is f of x1.

00:02:29.120 --> 00:02:31.420
And then for this
one right over here,

00:02:31.420 --> 00:02:33.480
we take the right boundary.

00:02:33.480 --> 00:02:35.750
The right boundary
defines that height.

00:02:35.750 --> 00:02:37.370
If we go all the
way to the-- this

00:02:37.370 --> 00:02:40.880
is rectangle two-- if we
go all way to the nth one,

00:02:40.880 --> 00:02:47.240
we use the right
boundary to define

00:02:47.240 --> 00:02:48.930
the height of the rectangle.

00:02:48.930 --> 00:02:51.690
So in this case-- this
is the nth rectangle--

00:02:51.690 --> 00:02:54.610
how would we write this sum?

00:02:54.610 --> 00:02:57.909
Well, it would be
the sum-- which

00:02:57.909 --> 00:03:00.200
is, remember, we're just
trying to approximate the area

00:03:00.200 --> 00:03:03.220
under the curve-- from
i is equal to 1 to n.

00:03:03.220 --> 00:03:06.240
So i is a count of
each of the rectangles.

00:03:06.240 --> 00:03:10.150
And so the height of the
first rectangle is f of x1.

00:03:10.150 --> 00:03:13.530
The height of the nth
rectangle is f of x sub n.

00:03:13.530 --> 00:03:18.490
So this height right over
here is f of x sub n.

00:03:18.490 --> 00:03:21.230
So the height of
the i-th rectangle

00:03:21.230 --> 00:03:24.470
is going to be f of x sub i.

00:03:24.470 --> 00:03:26.100
Whatever the
rectangle number is,

00:03:26.100 --> 00:03:28.097
we take the x sub
that same number

00:03:28.097 --> 00:03:29.430
and evaluate the function there.

00:03:29.430 --> 00:03:30.550
That gives us the height.

00:03:30.550 --> 00:03:33.220
And we multiply
that times delta x.

00:03:33.220 --> 00:03:36.570
So the difference between
this and this here,

00:03:36.570 --> 00:03:39.090
for the i-th
rectangle, we use x sub

00:03:39.090 --> 00:03:40.780
i minus 1, so the left boundary.

00:03:40.780 --> 00:03:45.760
Here we use the right
boundary, f of x sub i.

00:03:45.760 --> 00:03:47.510
Well, we don't
have to stop there.

00:03:47.510 --> 00:03:50.760
Instead, we could use maybe
the midpoint between the two

00:03:50.760 --> 00:03:52.070
boundaries instead.

00:03:52.070 --> 00:03:56.240
So, for example,
over here we could

00:03:56.240 --> 00:04:01.290
we could use the midpoint
between x0 and x1

00:04:01.290 --> 00:04:03.650
to find the height
of the rectangle.

00:04:03.650 --> 00:04:05.340
So this is right over here.

00:04:05.340 --> 00:04:12.830
This is f of x0
plus x1 over 2, just

00:04:12.830 --> 00:04:15.030
the midpoint between
these two points,

00:04:15.030 --> 00:04:18.610
to define the height
of the rectangle.

00:04:18.610 --> 00:04:21.750
So it would look
something like that.

00:04:21.750 --> 00:04:24.210
And the next one, we
would look at the midpoint

00:04:24.210 --> 00:04:27.550
to define the height.

00:04:27.550 --> 00:04:29.780
And we go all the
way to the nth one,

00:04:29.780 --> 00:04:31.860
and we define the
midpoint between its two

00:04:31.860 --> 00:04:33.390
sides of the rectangle.

00:04:33.390 --> 00:04:36.480
So the function
evaluated there tells us

00:04:36.480 --> 00:04:38.710
how high our
rectangle should be.

00:04:42.230 --> 00:04:45.860
And it would look
something like that.

00:04:45.860 --> 00:04:47.780
And so what would
this sum look like?

00:04:47.780 --> 00:04:50.810
Well, once again, we would
count each of our rectangles,

00:04:50.810 --> 00:04:53.840
so i equals 1 to
i equals n. i is

00:04:53.840 --> 00:04:55.330
which rectangle
we're working on.

00:04:55.330 --> 00:04:57.371
So this is the first one,
this is the second one,

00:04:57.371 --> 00:04:58.720
and this is the nth one.

00:04:58.720 --> 00:05:01.840
And the height isn't just
going to be f evaluated

00:05:01.840 --> 00:05:05.840
x sub i minus 1 or
f evaluated x sub i.

00:05:05.840 --> 00:05:08.970
It'll be the function
evaluated at the midpoint

00:05:08.970 --> 00:05:14.690
between the two-- x sub
i minus 1 plus x sub i,

00:05:14.690 --> 00:05:18.200
all of that over two,
and then times delta x.

00:05:18.200 --> 00:05:23.960
The delta x's are the same in
every one of these scenarios.

00:05:23.960 --> 00:05:26.515
Now finally, let's try to
break out of approximating only

00:05:26.515 --> 00:05:28.640
with rectangles and get a
little bit more creative.

00:05:28.640 --> 00:05:32.460
Why don't we try to
approximate with trapezoids?

00:05:32.460 --> 00:05:33.580
So let's try to do that.

00:05:33.580 --> 00:05:38.020
So what we could have here is
the left part of the trapezoid.

00:05:38.020 --> 00:05:42.120
The height is f of x sub 0.

00:05:42.120 --> 00:05:44.760
So this is f of x sub 0.

00:05:44.760 --> 00:05:50.160
And then the right side of
the trapezoid is f of x sub 1.

00:05:53.556 --> 00:05:55.430
And then what would be
the-- here, and let me

00:05:55.430 --> 00:05:56.429
do that for all of them.

00:05:56.429 --> 00:05:58.290
So that would be
the first trapezoid.

00:05:58.290 --> 00:06:00.280
Then the second trapezoid
would look like this.

00:06:00.280 --> 00:06:01.905
This one looks almost
like a rectangle,

00:06:01.905 --> 00:06:04.910
but we assume that the
top isn't completely flat.

00:06:04.910 --> 00:06:07.075
And then we go all the
way to the nth one.

00:06:07.075 --> 00:06:10.020
And this should be clear that
we're dealing with a trapezoid.

00:06:10.020 --> 00:06:16.590
All the way to the nth one
will look something like that.

00:06:16.590 --> 00:06:20.200
So how would we
calculate this area,

00:06:20.200 --> 00:06:22.760
the area of the trapezoid?

00:06:22.760 --> 00:06:25.580
Well, you just have to remember
that the area of a trapezoid

00:06:25.580 --> 00:06:27.820
is just the average of
the heights of the two

00:06:27.820 --> 00:06:30.140
sides times the base.

00:06:30.140 --> 00:06:33.620
So in this case-- and let me
write it out a little bit.

00:06:33.620 --> 00:06:37.270
So the area right
over there is going

00:06:37.270 --> 00:06:40.120
to be the average
of the heights.

00:06:40.120 --> 00:06:46.950
So it's going to be f of
x sub 0 plus f of x sub 1,

00:06:46.950 --> 00:06:49.180
all of that over 2.

00:06:49.180 --> 00:06:53.350
And then we're going to
multiply that times delta x.

00:06:53.350 --> 00:06:57.380
So that would be the area just
of this one right over here.

00:06:57.380 --> 00:06:59.540
We took the average
of the two heights

00:06:59.540 --> 00:07:02.540
and multiplied that
times the base.

00:07:02.540 --> 00:07:05.890
Now, if we wanted the sum of the
areas of all these trapezoids,

00:07:05.890 --> 00:07:07.750
and we wanted to write
in general terms,

00:07:07.750 --> 00:07:10.140
we could just write it's
the sum-- once again,

00:07:10.140 --> 00:07:12.660
we're going to count
the trapezoids.

00:07:12.660 --> 00:07:15.540
So this is the first
trapezoid, this is the second,

00:07:15.540 --> 00:07:17.710
all the way to
the nth trapezoid.

00:07:17.710 --> 00:07:20.020
So it's i equals
1 to i equals n.

00:07:20.020 --> 00:07:21.850
And the height of
each trapezoid,

00:07:21.850 --> 00:07:24.730
we're going to use the function
evaluated at the left boundary,

00:07:24.730 --> 00:07:29.340
x sub i minus 1, the average
of the function evaluated

00:07:29.340 --> 00:07:30.900
at the left boundary
and the function

00:07:30.900 --> 00:07:33.470
evaluated at the right
boundary, and we're

00:07:33.470 --> 00:07:35.020
going to take the
average of that

00:07:35.020 --> 00:07:37.980
and then multiply
that times the base.

00:07:37.980 --> 00:07:39.875
So the whole reason
why I wanted to do

00:07:39.875 --> 00:07:42.000
this is to show you there's
multiple, multiple ways

00:07:42.000 --> 00:07:42.530
of doing this.

00:07:42.530 --> 00:07:44.090
And in fact, if you wanted
to get really general,

00:07:44.090 --> 00:07:45.790
you could even have
different widths.

00:07:45.790 --> 00:07:48.439
But then that gets a
little bit more confusing.

00:07:48.439 --> 00:07:49.980
But really, just to
show you that you

00:07:49.980 --> 00:07:53.140
might see some of this fancy
notation in your calculus book

00:07:53.140 --> 00:07:54.840
or in your precalculus book.

00:07:54.840 --> 00:07:58.399
But all it's doing is summing
up the areas of trapezoids

00:07:58.399 --> 00:08:00.190
and rectangles depending
on whether they're

00:08:00.190 --> 00:08:03.650
using the right boundaries
of the rectangle

00:08:03.650 --> 00:08:06.850
to define the height, the
left boundaries, the midpoint

00:08:06.850 --> 00:08:08.480
of the left and the
right boundaries,

00:08:08.480 --> 00:08:11.809
or they could even
construct trapezoids.


Riemann approximation introduction

WEBVTT
Kind: captions
Language: en

00:00:00.610 --> 00:00:02.500
What we're going to
try to do in this video

00:00:02.500 --> 00:00:07.470
is approximate the
area under the curve y

00:00:07.470 --> 00:00:13.940
is equal to x squared plus 1
between the intervals x equals

00:00:13.940 --> 00:00:15.830
1 and x equals 3.

00:00:15.830 --> 00:00:17.270
And we're going
to approximate it

00:00:17.270 --> 00:00:22.170
by constructing four rectangles
under the curve of equal width.

00:00:22.170 --> 00:00:24.960
So let's first think about what
those rectangles look like, so

00:00:24.960 --> 00:00:28.470
four rectangles of equal width.

00:00:28.470 --> 00:00:32.070
So it looks like that,
like that, and like that.

00:00:32.070 --> 00:00:35.420
And I haven't really defined the
top of the rectangles just yet.

00:00:35.420 --> 00:00:37.410
But let's think about
what those widths have

00:00:37.410 --> 00:00:39.118
to be if they're going
to be equal width.

00:00:39.118 --> 00:00:41.630
And we can call
that width delta x.

00:00:41.630 --> 00:00:43.840
So this distance
right over here,

00:00:43.840 --> 00:00:46.170
we're going to
call that delta x.

00:00:46.170 --> 00:00:49.720
So delta x is going to have
to be the total distance

00:00:49.720 --> 00:00:51.010
that we're traveling in x.

00:00:51.010 --> 00:00:52.450
So we finish at 3.

00:00:52.450 --> 00:00:53.840
We started at 1.

00:00:53.840 --> 00:00:56.600
And we want four
equal-width rectangles.

00:00:56.600 --> 00:00:59.420
So it's going to
be equal to 1/2.

00:00:59.420 --> 00:01:02.660
So for example,
this first interval

00:01:02.660 --> 00:01:05.800
between the boundary between the
first rectangle and the second

00:01:05.800 --> 00:01:07.970
is going to be 1.5.

00:01:07.970 --> 00:01:09.420
Then we go 1/2 to 2.

00:01:09.420 --> 00:01:11.480
Then we go to 2.5.

00:01:11.480 --> 00:01:13.505
And then we go 1/2 to 3.

00:01:13.505 --> 00:01:15.130
Now, let's think
about how we'll define

00:01:15.130 --> 00:01:17.009
the height of the rectangles.

00:01:17.009 --> 00:01:19.300
For the sake of this video--
we'll see in future videos

00:01:19.300 --> 00:01:21.030
that there's other
ways of doing this--

00:01:21.030 --> 00:01:23.910
I'm going to use the left
boundary of the rectangle

00:01:23.910 --> 00:01:26.450
to define the height-- or
the function, I should say.

00:01:26.450 --> 00:01:29.050
I'm going to use the function
evaluated at the left boundary

00:01:29.050 --> 00:01:30.310
to define the height.

00:01:30.310 --> 00:01:33.470
So for example, for
the first rectangle,

00:01:33.470 --> 00:01:38.720
this point right
over here is f of 1.

00:01:38.720 --> 00:01:42.070
And so I will say that
that is the height

00:01:42.070 --> 00:01:45.690
of our first rectangle.

00:01:45.690 --> 00:01:48.940
Then we go over here
to the left boundary

00:01:48.940 --> 00:01:50.450
of the second rectangle.

00:01:50.450 --> 00:01:56.530
We're now looking at the
function evaluated at 1.5.

00:01:56.530 --> 00:01:59.980
So that is f of 1.5.

00:01:59.980 --> 00:02:01.200
That's the height.

00:02:01.200 --> 00:02:05.360
And so we get our
second rectangle.

00:02:05.360 --> 00:02:08.880
Then, we get-- I can keep
going like this-- we get,

00:02:08.880 --> 00:02:11.290
for this third
rectangle, we have

00:02:11.290 --> 00:02:13.445
the function evaluated at 2.

00:02:17.090 --> 00:02:18.280
So that's right over here.

00:02:18.280 --> 00:02:21.100
That's f of 2.

00:02:21.100 --> 00:02:23.660
And so then we get
our third rectangle.

00:02:23.660 --> 00:02:25.950
And then, finally, we
have our fourth rectangle,

00:02:25.950 --> 00:02:28.760
the function evaluated at 2.5.

00:02:28.760 --> 00:02:31.810
So the function evaluated
at 2.5 is the height.

00:02:31.810 --> 00:02:34.160
So this is f of 2.5.

00:02:34.160 --> 00:02:36.430
Remember, in each
of these, I'm just

00:02:36.430 --> 00:02:39.350
looking at the left
boundary of the rectangle

00:02:39.350 --> 00:02:40.750
and evaluating
the function there

00:02:40.750 --> 00:02:43.700
to get the height
of the rectangle.

00:02:43.700 --> 00:02:45.200
Now that I've set
it up in this way,

00:02:45.200 --> 00:02:47.880
what is the total
approximate area

00:02:47.880 --> 00:02:49.970
using the sum of
these rectangles?

00:02:49.970 --> 00:02:52.830
And clearly, this isn't going
to be a perfect approximation.

00:02:52.830 --> 00:02:55.404
I'm giving up on a
bunch of area here.

00:02:55.404 --> 00:02:57.320
Let me see if I can color
that in with a color

00:02:57.320 --> 00:02:58.670
that I have not used.

00:02:58.670 --> 00:02:59.890
So I'm giving up.

00:02:59.890 --> 00:03:01.080
I'm giving up this area.

00:03:01.080 --> 00:03:02.760
I'm giving up this area.

00:03:02.760 --> 00:03:04.590
I'm giving up that area.

00:03:04.590 --> 00:03:06.774
I'm giving up that area there.

00:03:06.774 --> 00:03:08.190
But this is just
an approximation,

00:03:08.190 --> 00:03:10.080
and maybe if I had
many more rectangles,

00:03:10.080 --> 00:03:13.700
it would be a better
approximation.

00:03:13.700 --> 00:03:16.200
So let's figure out what the
areas of each of the rectangles

00:03:16.200 --> 00:03:16.700
are.

00:03:16.700 --> 00:03:18.200
So the area of this
first rectangle

00:03:18.200 --> 00:03:23.170
is going to be the
height, which is f of 1,

00:03:23.170 --> 00:03:26.660
times the base,
which is delta x.

00:03:26.660 --> 00:03:28.300
The area of the
second rectangle is

00:03:28.300 --> 00:03:33.860
going to be the height, which
we already said was f of 1.5,

00:03:33.860 --> 00:03:37.050
times the base, times delta x.

00:03:37.050 --> 00:03:39.110
The height of the
third rectangle

00:03:39.110 --> 00:03:43.270
is going to be the function
evaluated at its left boundary,

00:03:43.270 --> 00:03:49.820
so f of 2-- so plus f of 2
times the base, times delta x.

00:03:49.820 --> 00:03:53.660
And then, finally, the area of
the third rectangle, the height

00:03:53.660 --> 00:03:55.660
is the function
evaluated at 2.5,

00:03:55.660 --> 00:03:57.710
so plus-- that's
a different color

00:03:57.710 --> 00:04:00.710
than what I wanted to use.

00:04:00.710 --> 00:04:02.810
I wanted to use that
orange color-- so

00:04:02.810 --> 00:04:09.030
plus the function evaluated
at 2.5 times the base.

00:04:09.030 --> 00:04:12.300
This is going to be equal
to our approximate area--

00:04:12.300 --> 00:04:16.760
let me make it clear--
approximate area

00:04:16.760 --> 00:04:18.940
under the curve, just the
sum of these rectangles.

00:04:18.940 --> 00:04:20.779
So let's evaluate this.

00:04:20.779 --> 00:04:23.260
So this is going
to be equal to f

00:04:23.260 --> 00:04:26.520
of-- it's going to be equal to
the function evaluated at 1.

00:04:26.520 --> 00:04:31.680
1 squared plus 1 is just 2, so
it's going to be 2 times 1/2.

00:04:31.680 --> 00:04:34.830
Plus the function
evaluated at 1.25.

00:04:34.830 --> 00:04:38.690
1.25 squared is 2.25.

00:04:38.690 --> 00:04:41.910
And then you add 1 to
it, it becomes 3.25.

00:04:41.910 --> 00:04:48.570
So plus 3.25 times 1/2.

00:04:48.570 --> 00:04:51.580
And then we have the
function evaluated at 2.

00:04:51.580 --> 00:04:56.830
Well, 2 squared plus 1 is
5, so it's 5 times 1/2.

00:04:56.830 --> 00:04:59.910
And then finally, you have
the function evaluated at 2.5.

00:04:59.910 --> 00:05:06.760
2.5 squared is 6.25 plus 1.

00:05:06.760 --> 00:05:12.964
So that's 7.25 times 1/2.

00:05:12.964 --> 00:05:14.380
And just to make
the math simpler,

00:05:14.380 --> 00:05:16.660
we can factor out the 1/2.

00:05:16.660 --> 00:05:19.350
So this is going to be
equal to-- write 1/2

00:05:19.350 --> 00:05:33.480
in a neutral color-- 1/2
times 2 plus 3.25 plus 5

00:05:33.480 --> 00:05:43.360
plus 7.25, which is equal
to 1/2 times-- let's

00:05:43.360 --> 00:05:45.140
see if I can do this in my head.

00:05:45.140 --> 00:05:46.500
2 plus 5 is easy.

00:05:46.500 --> 00:05:47.774
That's 7.

00:05:47.774 --> 00:05:49.950
3 plus 7 is 10.

00:05:49.950 --> 00:05:52.680
And then we have 0.25
plus 0.25, so it's

00:05:52.680 --> 00:05:56.910
going to be 10.5 plus 7 is 17.5.

00:05:56.910 --> 00:06:04.460
So 1/2 times 17.5,
which is equal to 8.75,

00:06:04.460 --> 00:06:07.036
Which, once again, gives
us an approximation.

00:06:07.036 --> 00:06:09.260
And clearly, the way I've
drawn it right over here,

00:06:09.260 --> 00:06:11.004
for the function
we're using, it's

00:06:11.004 --> 00:06:13.170
going to be an underestimate,
because we've given up

00:06:13.170 --> 00:06:15.360
all that pink area that
I had colored in before.

00:06:15.360 --> 00:06:17.890
It's an underestimate,
but it's an approximation

00:06:17.890 --> 00:06:19.902
of the area under the curve.

00:06:19.902 --> 00:06:21.360
In the next few
videos, we're going

00:06:21.360 --> 00:06:24.040
to try to generalize
this to situations where

00:06:24.040 --> 00:06:25.540
we have an arbitrary
function and we

00:06:25.540 --> 00:06:28.720
have an arbitrary
number of rectangles.

00:06:28.720 --> 00:06:31.110
And we'll also start--
in videos after that,

00:06:31.110 --> 00:06:34.480
we'll look at rectangles
where we define the height not

00:06:34.480 --> 00:06:36.480
by the left boundary, but
by the right boundary,

00:06:36.480 --> 00:06:37.600
or by the midpoint.

00:06:37.600 --> 00:06:39.630
Or maybe we don't use
rectangles at all.

00:06:39.630 --> 00:06:42.010
Maybe we might use
things like trapezoids.

00:06:42.010 --> 00:06:44.230
Anyway, have fun.


Riemann sums in summation notation

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:03.610
In the last video, we attempted
to approximate the area

00:00:03.610 --> 00:00:07.520
under a curve by constructing
four rectangles of equal width

00:00:07.520 --> 00:00:11.320
and using the left boundary of
each rectangle, the function

00:00:11.320 --> 00:00:14.380
evaluated at the left boundary,
to determine the height,

00:00:14.380 --> 00:00:16.239
and we came up with
an approximation.

00:00:16.239 --> 00:00:18.720
What I want to do in this video
is generalize things a bit

00:00:18.720 --> 00:00:20.428
using the exact same
method, but doing it

00:00:20.428 --> 00:00:22.600
for an arbitrary function
with arbitrary boundaries

00:00:22.600 --> 00:00:25.550
and an arbitrary
number of rectangles.

00:00:25.550 --> 00:00:26.940
So let's do it.

00:00:26.940 --> 00:00:30.140
So I'm going to draw the
diagram as large as I

00:00:30.140 --> 00:00:34.540
can to make things
as clear as possible.

00:00:34.540 --> 00:00:36.480
So that's my y-axis.

00:00:36.480 --> 00:00:42.410
And this right over
here is my x-axis.

00:00:42.410 --> 00:00:44.500
Let me draw an
arbitrary function.

00:00:44.500 --> 00:00:47.600
So let's say my function
looks something like that.

00:00:47.600 --> 00:00:51.510
So that is y is equal to f of x.

00:00:51.510 --> 00:00:53.460
And let me define my boundaries.

00:00:53.460 --> 00:00:58.230
So let's say this right
over here is x equals a.

00:01:00.790 --> 00:01:06.490
And this right over
here is x equals b.

00:01:06.490 --> 00:01:08.230
So this is b.

00:01:08.230 --> 00:01:13.829
And I'm going to
use n rectangles,

00:01:13.829 --> 00:01:15.620
and I'm going to use
the function evaluated

00:01:15.620 --> 00:01:17.410
at the left boundary
of the rectangle

00:01:17.410 --> 00:01:18.740
to determine its height.

00:01:18.740 --> 00:01:22.030
So, for example, this
will be rectangle one.

00:01:22.030 --> 00:01:24.100
I'm going to evaluate
what f of a is.

00:01:27.300 --> 00:01:31.950
So this right over
here is f of a.

00:01:31.950 --> 00:01:33.880
And then I'm going to
use that as the height

00:01:33.880 --> 00:01:37.400
of my first rectangle.

00:01:37.400 --> 00:01:39.230
So just like that.

00:01:39.230 --> 00:01:44.001
So rectangle number
one looks like this.

00:01:44.001 --> 00:01:45.000
And I'll even number it.

00:01:45.000 --> 00:01:47.862
Rectangle one looks
just like that.

00:01:47.862 --> 00:01:49.320
And just to have
a convention here,

00:01:49.320 --> 00:01:55.380
because I'm going to want to
label each of the x values

00:01:55.380 --> 00:01:58.740
at the left boundary, so
we'll say a is equal to x0.

00:01:58.740 --> 00:02:00.890
a is equal to x0.

00:02:00.890 --> 00:02:03.760
So we could also call this
point right over here x0,

00:02:03.760 --> 00:02:05.699
that x value.

00:02:05.699 --> 00:02:07.240
And then we go to
the next rectangle.

00:02:07.240 --> 00:02:09.880
And we could call this one
right over here, this x value,

00:02:09.880 --> 00:02:10.780
we'll call it x1.

00:02:10.780 --> 00:02:13.060
It's the left boundary
of the next rectangle.

00:02:13.060 --> 00:02:16.870
If we evaluate f of x1, we get
this value right over here.

00:02:16.870 --> 00:02:22.312
This right over here is f of
x1, so it tells us our height.

00:02:22.312 --> 00:02:24.270
And we want an equal
width to the previous one.

00:02:24.270 --> 00:02:26.870
We'll think about what the width
is going to be in a second.

00:02:26.870 --> 00:02:32.930
So this right over here
is our second rectangle

00:02:32.930 --> 00:02:34.910
that we're going to use
to approximate the area

00:02:34.910 --> 00:02:35.576
under the curve.

00:02:35.576 --> 00:02:37.320
That's rectangle number two.

00:02:37.320 --> 00:02:39.070
Let's do rectangle number three.

00:02:39.070 --> 00:02:41.720
Well, rectangle number
three, the left boundary,

00:02:41.720 --> 00:02:43.980
we're just going to
call that x sub 2.

00:02:43.980 --> 00:02:48.435
And its height is going
to be f of x sub 2.

00:02:50.557 --> 00:02:53.140
And its width is going to be the
same width as the other ones.

00:02:53.140 --> 00:02:56.100
I'm just eyeballing
it right over here.

00:02:56.100 --> 00:02:58.050
So this is rectangle
number three.

00:03:01.020 --> 00:03:05.130
And we're going to continue
this process all the way

00:03:05.130 --> 00:03:08.890
until we get to
rectangle number n.

00:03:08.890 --> 00:03:14.080
So this is the n-th
rectangle right

00:03:14.080 --> 00:03:17.740
over here, the n-th rectangle.

00:03:17.740 --> 00:03:20.820
And what am I going to label
this point right over here?

00:03:20.820 --> 00:03:22.180
Well, we already see a pattern.

00:03:22.180 --> 00:03:24.840
The left boundary of the
first rectangle is x sub 0.

00:03:24.840 --> 00:03:28.010
The left boundary of the
second rectangle is x sub 1.

00:03:28.010 --> 00:03:30.270
The left boundary of the
third rectangle is x sub 2.

00:03:30.270 --> 00:03:32.650
So the left boundary
of the n-th rectangle

00:03:32.650 --> 00:03:37.110
is going to be x sub n minus 1.

00:03:37.110 --> 00:03:39.340
Whatever the
rectangle number is,

00:03:39.340 --> 00:03:42.770
the left boundary is x
sub that number minus 1.

00:03:42.770 --> 00:03:45.952
And this is just based on the
convention that we've defined.

00:03:45.952 --> 00:03:47.910
Now, the next thing that
we need to do in order

00:03:47.910 --> 00:03:50.160
to actually calculate
this area is think about

00:03:50.160 --> 00:03:51.590
what is the width?

00:03:51.590 --> 00:03:54.960
So let's call the width of
any of these rectangles--

00:03:54.960 --> 00:03:57.915
and for these purposes, or
the purpose of this example,

00:03:57.915 --> 00:03:59.540
I'm going to assume
that it's constant,

00:03:59.540 --> 00:04:01.140
although you can do these
sums where you actually

00:04:01.140 --> 00:04:02.880
vary the width of the rectangle.

00:04:02.880 --> 00:04:04.600
But then it gets a
little bit fancier.

00:04:04.600 --> 00:04:06.290
So I want it equal width.

00:04:06.290 --> 00:04:08.645
So I want delta x
to be equal width.

00:04:08.645 --> 00:04:10.270
And to think about
what that has to be,

00:04:10.270 --> 00:04:12.390
we just have to think, what's
the total width that we're

00:04:12.390 --> 00:04:12.890
covering?

00:04:12.890 --> 00:04:17.244
Well, the total distance here
is going to be b minus a,

00:04:17.244 --> 00:04:19.660
and we're just going to divide
by the number of rectangles

00:04:19.660 --> 00:04:21.743
that we want, the number
of sections that we want.

00:04:21.743 --> 00:04:23.660
So we want to divide by n.

00:04:23.660 --> 00:04:26.560
So if we assume this
is true, and then we

00:04:26.560 --> 00:04:31.250
assume that a is
equal to x0, and then

00:04:31.250 --> 00:04:40.610
x1 is equal to x0 plus delta x,
x2 is equal to x1 plus delta x,

00:04:40.610 --> 00:04:49.280
and we go all the way to xn is
equal to xn minus 1 plus delta

00:04:49.280 --> 00:04:52.330
x, then we've essentially set
up this diagram right over here.

00:04:52.330 --> 00:04:56.630
b is actually going
to be equal xn.

00:04:56.630 --> 00:04:57.400
So this is xn.

00:04:57.400 --> 00:05:01.180
It's equal to xn
minus 1 plus delta x.

00:05:01.180 --> 00:05:04.540
So now I think we've set up
all of the notation and all

00:05:04.540 --> 00:05:06.890
the conventions in order
to actually calculate

00:05:06.890 --> 00:05:09.700
the area, or our
approximation of the area.

00:05:09.700 --> 00:05:15.030
So our approximation,
approximate area,

00:05:15.030 --> 00:05:17.737
is going to be equal to what?

00:05:17.737 --> 00:05:20.070
Well, it's going to be the
area of the first rectangle--

00:05:20.070 --> 00:05:21.153
so let me write this down.

00:05:21.153 --> 00:05:24.410
So it's going to be rectangle
one-- so the area of rectangle

00:05:24.410 --> 00:05:33.330
one-- so rectangle one plus
the area of rectangle two

00:05:33.330 --> 00:05:35.750
plus the area of rectangle
three-- I think you get

00:05:35.750 --> 00:05:46.140
the point here-- plus all the
way to the area of rectangle n.

00:05:46.140 --> 00:05:47.700
And so what are
these going to be?

00:05:47.700 --> 00:05:52.400
Rectangle one is going to be
its height, which is f of x0

00:05:52.400 --> 00:05:53.080
or f of a.

00:05:53.080 --> 00:05:55.270
Either way. x0 and a
are the same thing.

00:05:55.270 --> 00:05:58.680
So it's f of a
times our delta x,

00:05:58.680 --> 00:06:00.690
times our width, our
height times our width.

00:06:00.690 --> 00:06:05.330
So times delta-- actually
I can write as f of x0,

00:06:05.330 --> 00:06:09.750
I wanted to write-- f
of x0 times delta x.

00:06:09.750 --> 00:06:11.460
What is our height
of rectangle two?

00:06:11.460 --> 00:06:14.555
It's f of x1 times delta x.

00:06:19.010 --> 00:06:20.930
What's our area of
rectangle three?

00:06:20.930 --> 00:06:25.020
It's f of x2 times delta x.

00:06:25.020 --> 00:06:27.120
And then we go all
the way to our area.

00:06:27.120 --> 00:06:29.370
We're taking all the sums,
all the way to rectangle n.

00:06:29.370 --> 00:06:30.470
What's its area?

00:06:30.470 --> 00:06:33.685
It's f of x sub n minus 1.

00:06:33.685 --> 00:06:35.860
Actually, that's a
different shade of orange.

00:06:35.860 --> 00:06:37.360
I'll use that same shade.

00:06:37.360 --> 00:06:43.276
It is f of x sub n
minus 1 times delta x.

00:06:43.276 --> 00:06:43.900
And we're done.

00:06:43.900 --> 00:06:46.050
We've written it in
a very general way.

00:06:46.050 --> 00:06:47.852
But to really make
us comfortable

00:06:47.852 --> 00:06:49.310
with the various
forms of notation,

00:06:49.310 --> 00:06:50.660
especially the types
of notation you

00:06:50.660 --> 00:06:52.920
might see when people are
talking about approximating

00:06:52.920 --> 00:06:55.250
the areas or sums
in general, I'm

00:06:55.250 --> 00:06:58.400
going to use the
traditional sigma notation.

00:06:58.400 --> 00:07:01.680
So another way we could
write this, as the sum,

00:07:01.680 --> 00:07:06.854
this is equal to the
sum from-- and remember,

00:07:06.854 --> 00:07:09.020
this is just based on the
conventions that I set up.

00:07:09.020 --> 00:07:14.222
I'll let i count which rectangle
we're in, from i equals 1 to n.

00:07:14.222 --> 00:07:16.180
And then we're going to
look at each rectangle.

00:07:16.180 --> 00:07:18.740
So the first rectangle,
that's rectangle one.

00:07:18.740 --> 00:07:24.050
So it's going to be f of-- well,
if we're in the i-th rectangle,

00:07:24.050 --> 00:07:26.440
then the left boundary
is going to be

00:07:26.440 --> 00:07:34.030
x sub i minus 1 times delta x.

00:07:34.030 --> 00:07:36.990
And so here, right over
here, is a general way

00:07:36.990 --> 00:07:43.580
of thinking about approximating
the area under a curve using

00:07:43.580 --> 00:07:46.270
rectangles, where the
height of the rectangles

00:07:46.270 --> 00:07:48.710
are defined by
the left boundary.

00:07:48.710 --> 00:07:50.420
And this tells us it's
the left boundary.

00:07:50.420 --> 00:07:57.070
And we see for each, if this
is the i-th rectangle right

00:07:57.070 --> 00:08:00.430
over here, if this is rectangle
i, then this right over here

00:08:00.430 --> 00:08:04.550
is x sub i minus 1, and
this height right over here

00:08:04.550 --> 00:08:09.500
is f of x sub i minus 1.

00:08:09.500 --> 00:08:12.400
So that's all we did right
over there times delta x.

00:08:12.400 --> 00:08:15.270
And then you sum of these
from the first rectangle

00:08:15.270 --> 00:08:16.396
all the way to the end.

00:08:16.396 --> 00:08:18.770
So hopefully that makes you
a little bit more comfortable

00:08:18.770 --> 00:08:19.750
with this notation.

00:08:19.750 --> 00:08:21.340
We're not doing
anything different

00:08:21.340 --> 00:08:24.090
than we did in this first
video, which was hopefully

00:08:24.090 --> 00:08:25.580
fairly straightforward for you.

00:08:25.580 --> 00:08:29.180
We have just generalized it
using a little bit more mathy

00:08:29.180 --> 00:08:30.730
notation.


Limits to define continuity

WEBVTT
Kind: captions
Language: en

00:00:00.520 --> 00:00:04.560
What I want to do in this
video is talk about continuity.

00:00:04.560 --> 00:00:06.950
And continuity of a
function is something

00:00:06.950 --> 00:00:10.490
that is pretty easy to
recognize when you see it.

00:00:10.490 --> 00:00:14.569
But we'll also talk about how we
can more rigorously define it.

00:00:14.569 --> 00:00:16.860
So when I talk about it being
pretty easy to recognize,

00:00:16.860 --> 00:00:18.920
let me draw some functions here.

00:00:18.920 --> 00:00:24.300
So let's say this is the
y-axis, that is the x-axis.

00:00:24.300 --> 00:00:29.310
And if I were to
draw a function,

00:00:29.310 --> 00:00:31.540
let's say f of x looks
like something like this.

00:00:34.465 --> 00:00:35.840
And I would say
over the interval

00:00:35.840 --> 00:00:37.631
that I've drawn it, so
it looks like from x

00:00:37.631 --> 00:00:40.320
is equal to 0, to maybe
that point right over there,

00:00:40.320 --> 00:00:42.624
is this function continuous?

00:00:42.624 --> 00:00:43.790
Well you'd say no, it isn't.

00:00:43.790 --> 00:00:45.748
Look, we're over here,
we see the function just

00:00:45.748 --> 00:00:48.420
jumps all of a sudden, from
this point to this point

00:00:48.420 --> 00:00:49.860
right over here.

00:00:49.860 --> 00:00:56.140
This is not continuous.

00:00:56.140 --> 00:00:58.650
And you might even say
we have a discontinuity

00:00:58.650 --> 00:01:00.460
at this value of
x, right over here.

00:01:00.460 --> 00:01:02.635
We would call this
a discontinuity.

00:01:02.635 --> 00:01:04.260
And actually this
type of discontinuity

00:01:04.260 --> 00:01:05.551
is called a jump discontinuity.

00:01:08.554 --> 00:01:10.220
So you would say this
is not continuous.

00:01:10.220 --> 00:01:13.080
It's obvious that these
two things do not connect.

00:01:13.080 --> 00:01:15.010
They don't touch each other.

00:01:15.010 --> 00:01:17.420
Similarly, if you were to
look at a function that

00:01:17.420 --> 00:01:22.910
looked like-- let me draw
another one-- y and x.

00:01:22.910 --> 00:01:25.590
And let's say the function
looks something like this.

00:01:25.590 --> 00:01:29.815
Maybe right over here,
looks like this, and then

00:01:29.815 --> 00:01:32.930
the function is defined to be
this point right over there.

00:01:32.930 --> 00:01:35.420
Is the function continuous
over the interval

00:01:35.420 --> 00:01:37.219
that I've depicted
right over here?

00:01:37.219 --> 00:01:39.010
And you would immediately
say no, it isn't.

00:01:39.010 --> 00:01:40.740
Because right over
at this point,

00:01:40.740 --> 00:01:44.272
the function goes up to
this point, just like this.

00:01:44.272 --> 00:01:52.060
And this kind of discontinuity--
this is the discontinuity--

00:01:52.060 --> 00:01:54.900
is called a removable
discontinuity.

00:01:54.900 --> 00:01:55.656
Removable.

00:01:55.656 --> 00:01:57.780
One could make a reasonable
argument that this also

00:01:57.780 --> 00:01:59.500
looks like a jump,
but this is typically

00:01:59.500 --> 00:02:02.190
categorized as a
removable discontinuity.

00:02:02.190 --> 00:02:04.550
Because if you just
re-define the function

00:02:04.550 --> 00:02:08.070
so it wasn't up here, but
it was right over here,

00:02:08.070 --> 00:02:09.750
then the function is continuous.

00:02:09.750 --> 00:02:14.190
So you can kind of
remove the discontinuity.

00:02:14.190 --> 00:02:17.210
And then finally, if I were
to draw another function-- so

00:02:17.210 --> 00:02:22.040
let me draw another one
right over here-- x, y.

00:02:22.040 --> 00:02:25.790
And ask you, is this one
continuous over the interval

00:02:25.790 --> 00:02:27.010
that I've depicted?

00:02:27.010 --> 00:02:27.780
And you would say, well, look.

00:02:27.780 --> 00:02:29.488
Yeah, it looks all
connected all the way.

00:02:29.488 --> 00:02:31.210
There aren't any
jumps over here.

00:02:31.210 --> 00:02:33.020
No removable
discontinuities over here.

00:02:33.020 --> 00:02:36.590
This one looks continuous.

00:02:36.590 --> 00:02:37.860
And you would be right.

00:02:37.860 --> 00:02:39.980
So that's the general
sense of continuity.

00:02:39.980 --> 00:02:43.110
And you can kind of
spot it when you see it.

00:02:43.110 --> 00:02:46.620
But let's think about a more
rigorous definition of one.

00:02:46.620 --> 00:02:51.220
And since we already have a
rigorous definition of limits,

00:02:51.220 --> 00:02:53.930
the epsilon delta
definition, gives us

00:02:53.930 --> 00:02:56.500
a rigorous definition
for limits.

00:02:56.500 --> 00:02:59.600
It's a definition for limits.

00:02:59.600 --> 00:03:01.710
So we can prove
when a limit exists,

00:03:01.710 --> 00:03:04.530
and what the value
of that limit is.

00:03:04.530 --> 00:03:07.390
Let's use that to create
a rigorous definition

00:03:07.390 --> 00:03:09.580
of continuity.

00:03:09.580 --> 00:03:14.810
So let's think about a function
over some type of an interval.

00:03:14.810 --> 00:03:19.300
So let's say that we have-- so
let me draw another function.

00:03:19.300 --> 00:03:22.540
Let me draw some
type of a function.

00:03:22.540 --> 00:03:25.570
And then we'll see whether
our more rigorous definition

00:03:25.570 --> 00:03:27.610
of continuity
passes muster, when

00:03:27.610 --> 00:03:30.020
we look at all of
these things up here.

00:03:30.020 --> 00:03:34.470
So let me draw an
interval right over here.

00:03:34.470 --> 00:03:36.513
So it's between that x
value and that x value.

00:03:36.513 --> 00:03:38.830
This is the x-axis,
this is the y-axis.

00:03:38.830 --> 00:03:41.975
And let me draw my function
over that interval.

00:03:41.975 --> 00:03:44.770
Over that interval it
looks something like this.

00:03:44.770 --> 00:03:47.520
So we say that a
function is continuous

00:03:47.520 --> 00:03:49.047
at an interior point.

00:03:49.047 --> 00:03:50.630
So an interior point
is a point that's

00:03:50.630 --> 00:03:55.420
not at the edge of my boundary.

00:03:55.420 --> 00:03:57.712
So this is an interior
point for my interval.

00:03:57.712 --> 00:03:59.670
This would be an end
point, and this would also

00:03:59.670 --> 00:04:00.640
be an end point.

00:04:00.640 --> 00:04:02.660
We'd say it's continuous
at an interior point.

00:04:02.660 --> 00:04:08.830
So continuous at interior
point, interior to my interval,

00:04:08.830 --> 00:04:16.930
means that the limit as,
let's say at interior point c,

00:04:16.930 --> 00:04:19.769
so this is the point
x is equal to c.

00:04:19.769 --> 00:04:22.280
We can say that it's continuous
at the interior point

00:04:22.280 --> 00:04:26.535
c if the limit of
our function-- this

00:04:26.535 --> 00:04:27.910
is our function
right over here--

00:04:27.910 --> 00:04:30.550
if the limit of our
function as x approaches c

00:04:30.550 --> 00:04:34.690
is equal to the value
of our function.

00:04:34.690 --> 00:04:36.480
Now does this make sense?

00:04:36.480 --> 00:04:38.400
Well, what we're saying
is, is at that point,

00:04:38.400 --> 00:04:40.600
well this is f of
c right over there.

00:04:40.600 --> 00:04:42.570
And the limit as
we approach that is

00:04:42.570 --> 00:04:44.960
the same thing as the
value of the function.

00:04:44.960 --> 00:04:46.670
Which makes a lot of sense.

00:04:46.670 --> 00:04:47.780
Now let's think about it.

00:04:47.780 --> 00:04:52.110
If these would have somehow
been able to pass for continuous

00:04:52.110 --> 00:04:53.800
in that context.

00:04:53.800 --> 00:04:58.640
Well, over here, let's say that
this is our point c. f of c

00:04:58.640 --> 00:05:00.560
is right over there.

00:05:00.560 --> 00:05:03.030
That is f of c.

00:05:03.030 --> 00:05:05.570
Now is it the case
that the limit

00:05:05.570 --> 00:05:11.040
of f of x, as x approaches
c, is equal to f of c?

00:05:11.040 --> 00:05:14.040
Well, if we take
the limit of f of x

00:05:14.040 --> 00:05:17.580
as x approaches c from
the positive direction,

00:05:17.580 --> 00:05:20.570
it does look like it is f of c.

00:05:20.570 --> 00:05:23.320
It does look like
it's equal to f of c.

00:05:23.320 --> 00:05:25.280
But if we take the
limit-- but this

00:05:25.280 --> 00:05:30.740
does not equal the
limit of f of x

00:05:30.740 --> 00:05:32.989
as x approaches c from
the negative direction.

00:05:32.989 --> 00:05:34.530
As we go from the
negative direction,

00:05:34.530 --> 00:05:36.670
we're not approaching f of c.

00:05:36.670 --> 00:05:40.460
So therefore, this
does not hold up.

00:05:40.460 --> 00:05:43.210
In order for the limit
to be equal to f of c,

00:05:43.210 --> 00:05:44.900
the limit from
both the directions

00:05:44.900 --> 00:05:46.050
needs to be equal to it.

00:05:46.050 --> 00:05:47.520
And this is not the case.

00:05:47.520 --> 00:05:51.030
So this would not pass muster
by our formal definition,

00:05:51.030 --> 00:05:51.660
which is good.

00:05:51.660 --> 00:05:54.830
Because we see visually
this one is not continuous.

00:05:54.830 --> 00:05:56.330
What about this one
right over here?

00:05:56.330 --> 00:05:57.590
And let me re-set it up.

00:05:57.590 --> 00:05:59.600
So let me make sure
that that looks

00:05:59.600 --> 00:06:01.860
like a hole right over there.

00:06:01.860 --> 00:06:04.210
So we see here,
what is the limit?

00:06:04.210 --> 00:06:07.750
The limit-- and
this is our c, right

00:06:07.750 --> 00:06:11.210
over here-- the limit of
f of x as x approaches

00:06:11.210 --> 00:06:13.970
c, let's say that
that is equal to L.

00:06:13.970 --> 00:06:16.640
And so that, we've seen many
limits like this before,

00:06:16.640 --> 00:06:18.374
that's L right over there.

00:06:18.374 --> 00:06:19.790
And it's pretty
clear just looking

00:06:19.790 --> 00:06:24.070
at this is that L
does not equal f of c.

00:06:24.070 --> 00:06:27.190
This right over here is f of c.

00:06:27.190 --> 00:06:29.480
So once again, this
would not pass our test.

00:06:29.480 --> 00:06:31.850
The limit of f of
x as x approaches

00:06:31.850 --> 00:06:35.180
c, which is this right over
here, is not equal to f of c.

00:06:35.180 --> 00:06:38.640
So once again, this
would not pass our test.

00:06:38.640 --> 00:06:41.020
And here, any of the interior
points would pass our test.

00:06:41.020 --> 00:06:43.330
The limit as x
approaches this value

00:06:43.330 --> 00:06:52.520
is equal to the function
evaluated at that point.

00:06:52.520 --> 00:06:54.970
So it seems to be
good for all of those.

00:06:54.970 --> 00:06:56.470
Now let's give a
definition for when

00:06:56.470 --> 00:06:58.740
we're talking about
boundary points.

00:06:58.740 --> 00:07:01.570
So this is continuity
for an interior point.

00:07:01.570 --> 00:07:13.033
And let's think about
continuity at boundary--

00:07:13.033 --> 00:07:14.783
or let me call it
endpoint, actually, that

00:07:14.783 --> 00:07:19.160
would be better-- at endpoint c.

00:07:22.500 --> 00:07:25.450
So let's first consider
just the left endpoint.

00:07:25.450 --> 00:07:31.930
If left endpoint-- so what I'm
talking about, a left endpoint?

00:07:31.930 --> 00:07:39.870
Let me draw my axes,
x-axis, y-axis.

00:07:39.870 --> 00:07:41.590
And let me draw my interval.

00:07:41.590 --> 00:07:43.940
So let's say this is the
left endpoint of my interval,

00:07:43.940 --> 00:07:46.300
this is the right
endpoint of my interval.

00:07:46.300 --> 00:07:51.070
And let me draw the
function over that interval.

00:07:51.070 --> 00:07:54.770
Looks something like this.

00:07:54.770 --> 00:07:56.360
So when we talk about
a left endpoint,

00:07:56.360 --> 00:08:00.490
we're talking our c
being right over here.

00:08:00.490 --> 00:08:02.781
It is the left endpoint.

00:08:02.781 --> 00:08:04.530
So if we're talking
about a left endpoint,

00:08:04.530 --> 00:08:11.410
we are continuous at
c means-- or to say

00:08:11.410 --> 00:08:13.970
that we're continuous at
this left endpoint c--

00:08:13.970 --> 00:08:20.510
that means that the limit f
of x as x approaches c, well,

00:08:20.510 --> 00:08:23.090
we can't even approach see
from the left hand side.

00:08:23.090 --> 00:08:26.300
We have to approach
from the right.

00:08:26.300 --> 00:08:28.930
Is equal to f of c.

00:08:28.930 --> 00:08:30.910
And so this is
really kind of a, we

00:08:30.910 --> 00:08:32.869
can only approach something
from one direction.

00:08:32.869 --> 00:08:34.618
So we can't just say
the limit in general,

00:08:34.618 --> 00:08:36.260
but we can say the
limit from one side.

00:08:36.260 --> 00:08:38.110
So it's really very
similar to what we just

00:08:38.110 --> 00:08:39.720
said for an interior point.

00:08:39.720 --> 00:08:41.740
And we see over here
it is indeed the case,

00:08:41.740 --> 00:08:44.880
as x approaches c, our
function is approaching

00:08:44.880 --> 00:08:46.110
this point right over here.

00:08:46.110 --> 00:08:49.080
Which is the exact
same thing as f of c.

00:08:49.080 --> 00:08:51.587
So we are continuous
at that point.

00:08:51.587 --> 00:08:53.920
What's an example where an
endpoint-- where we would not

00:08:53.920 --> 00:08:55.590
be continuous and an end point?

00:08:55.590 --> 00:09:01.090
Well, I can imagine a graph
that looks something like this.

00:09:01.090 --> 00:09:06.090
So here's our interval,
and maybe our function.

00:09:06.090 --> 00:09:07.910
So at c it looks like that.

00:09:07.910 --> 00:09:09.530
There's a little
hole right there.

00:09:09.530 --> 00:09:11.321
And then it would look
something like that.

00:09:11.321 --> 00:09:12.950
Or there's no hole,
the function just

00:09:12.950 --> 00:09:15.150
has a removable discontinuity
right over there.

00:09:15.150 --> 00:09:17.000
At least visually
it looks like that.

00:09:17.000 --> 00:09:18.980
And you see that this
would not pass the test.

00:09:18.980 --> 00:09:23.430
Because the limit as we approach
c from the positive direction

00:09:23.430 --> 00:09:24.800
is right over here.

00:09:24.800 --> 00:09:25.950
That's the limit.

00:09:25.950 --> 00:09:28.220
But f of c is up here.

00:09:28.220 --> 00:09:30.300
So f of c does not
equal the limit

00:09:30.300 --> 00:09:33.080
as x approaches c from
the positive direction.

00:09:33.080 --> 00:09:37.915
So this would not be continuous.

00:09:37.915 --> 00:09:39.290
And you could
imagine, what do we

00:09:39.290 --> 00:09:41.980
do if we're dealing
with the right endpoint?

00:09:41.980 --> 00:09:51.200
So we say we're continuous
at right endpoint c

00:09:51.200 --> 00:09:57.560
if-- so let me draw that, do
my best attempt to draw it--

00:09:57.560 --> 00:10:03.260
so this is my x-axis,
this is my y-axis.

00:10:03.260 --> 00:10:06.965
Let me draw my interval
that I care about,

00:10:06.965 --> 00:10:08.340
say it looks
something like this.

00:10:08.340 --> 00:10:12.820
A right endpoint means
c is right over there.

00:10:12.820 --> 00:10:16.640
And we can say that we
are continuous at x,

00:10:16.640 --> 00:10:18.800
the function is
continuous at x equals

00:10:18.800 --> 00:10:25.780
c means that the limit of f of
x as x approaches c-- now we

00:10:25.780 --> 00:10:27.130
can't approach it from both sides.

00:10:27.130 --> 00:10:29.130
We can only approach it
from the left hand side.

00:10:29.130 --> 00:10:31.510
As x approaches c from
the negative direction,

00:10:31.510 --> 00:10:33.630
is equal to f of c.

00:10:33.630 --> 00:10:36.090
If we can say that,
if this is true,

00:10:36.090 --> 00:10:39.300
then this implies that we
are continuous at that right

00:10:39.300 --> 00:10:41.820
endpoint, c, and vice versa.

00:10:41.820 --> 00:10:43.230
And a situation where we're not?

00:10:43.230 --> 00:10:46.140
Well you could imagine instead
of this being defined right

00:10:46.140 --> 00:10:47.660
at that point, you
could create, you

00:10:47.660 --> 00:10:49.720
could say the function jumps up.

00:10:49.720 --> 00:10:51.800
Just like we did
right over there.

00:10:51.800 --> 00:10:57.200
So once again, continuity, not
a really hard to fathom idea.

00:10:57.200 --> 00:11:00.080
Whenever you see the function
just all of a sudden jumping,

00:11:00.080 --> 00:11:02.940
or there's kind of a gap in
it, it's a pretty good sense

00:11:02.940 --> 00:11:05.100
that the function is
not connected there.

00:11:05.100 --> 00:11:06.446
It's not continuous.

00:11:06.446 --> 00:11:07.820
But what we did
in this video is,

00:11:07.820 --> 00:11:11.830
we used limits to define
a more rigorous definition

00:11:11.830 --> 00:11:13.590
of continuity.


The graphical relationship between a function & its derivative (part 1)

WEBVTT
Kind: captions
Language: en

00:00:00.630 --> 00:00:03.990
So I've got this crazy
discontinuous function here,

00:00:03.990 --> 00:00:05.800
which we'll call f of x.

00:00:05.800 --> 00:00:11.250
And my goal is to try to draw
its derivative right over here.

00:00:11.250 --> 00:00:13.170
So what I'm going to
need to think about

00:00:13.170 --> 00:00:15.100
is the slope of
the tangent line,

00:00:15.100 --> 00:00:17.210
or the slope at each
point in this curve,

00:00:17.210 --> 00:00:21.290
and then try my best
to draw that slope.

00:00:21.290 --> 00:00:23.490
So let's try to tackle it.

00:00:23.490 --> 00:00:28.680
So right over here at this
point, the slope is positive.

00:00:28.680 --> 00:00:32.060
And actually, it's
a good bit positive.

00:00:32.060 --> 00:00:34.700
And then as we get larger
and larger x's, the slope

00:00:34.700 --> 00:00:39.730
is still positive, but it's less
positive-- and all the way up

00:00:39.730 --> 00:00:44.640
to this point right over
here, where it becomes 0.

00:00:44.640 --> 00:00:46.670
So let's see how I could
draw that over here.

00:00:46.670 --> 00:00:48.200
So over here we
know that the slope

00:00:48.200 --> 00:00:50.941
must be equal to
0-- right over here.

00:00:50.941 --> 00:00:52.440
Remember over here,
I'm going to try

00:00:52.440 --> 00:00:58.550
to draw y is equal
to f prime of x.

00:00:58.550 --> 00:01:00.380
And I'm going to
assume that this

00:01:00.380 --> 00:01:02.300
is some type of a parabola.

00:01:02.300 --> 00:01:05.610
And you'll learn shortly why
I had to make that assumption.

00:01:05.610 --> 00:01:08.460
But let's say that,
so let's see, here

00:01:08.460 --> 00:01:10.070
the slope is quite positive.

00:01:10.070 --> 00:01:13.220
So let's say the slope
is right over here.

00:01:13.220 --> 00:01:15.520
And then it gets less and
less and less positive.

00:01:15.520 --> 00:01:17.510
And I'll assume it does
it in a linear fashion.

00:01:17.510 --> 00:01:20.051
That's why I had to assume that
it's some type of a parabola.

00:01:20.051 --> 00:01:21.930
So it gets less and
less and less positive.

00:01:21.930 --> 00:01:26.980
Notice here, for example,
the slope is still positive.

00:01:26.980 --> 00:01:29.270
And so when you look
at the derivative,

00:01:29.270 --> 00:01:31.440
the slope is still
a positive value.

00:01:31.440 --> 00:01:34.140
But as we get larger and
larger x's up to this point,

00:01:34.140 --> 00:01:36.090
the slope is getting
less and less positive,

00:01:36.090 --> 00:01:37.420
all the way to 0.

00:01:37.420 --> 00:01:41.950
And then the slope is getting
more and more negative.

00:01:41.950 --> 00:01:46.020
And at this point, it seems like
the slope is just as negative

00:01:46.020 --> 00:01:47.650
as it was positive there.

00:01:47.650 --> 00:01:49.100
So at this point
right over here,

00:01:49.100 --> 00:01:52.620
the slope is just
as negative as it

00:01:52.620 --> 00:01:54.660
was positive right over there.

00:01:54.660 --> 00:01:58.740
So it seems like this
would be a reasonable view

00:01:58.740 --> 00:02:04.060
of the slope of the tangent
line over this interval.

00:02:04.060 --> 00:02:07.130
Now let's think about
as we get to this point.

00:02:07.130 --> 00:02:10.330
Here the slope seems constant.

00:02:10.330 --> 00:02:13.990
Our slope is a constant
positive value.

00:02:13.990 --> 00:02:17.550
So once again, our slope here
is a constant positive line.

00:02:17.550 --> 00:02:20.280
Let me be careful here
because at this point,

00:02:20.280 --> 00:02:23.290
our slope won't really be
defined, because our slope,

00:02:23.290 --> 00:02:26.860
you could draw
multiple tangent lines

00:02:26.860 --> 00:02:28.950
at this little pointy point.

00:02:28.950 --> 00:02:33.070
So let me just draw a
circle right over there.

00:02:33.070 --> 00:02:34.910
But then as we get
right over here,

00:02:34.910 --> 00:02:40.920
the slope seems to be positive.

00:02:40.920 --> 00:02:41.980
So let's draw that.

00:02:41.980 --> 00:02:45.820
The slope seems to be
positive, although it's

00:02:45.820 --> 00:02:47.650
not as positive as it was there.

00:02:47.650 --> 00:02:53.560
So the slope looks like it is--
I'm just trying to eyeball it--

00:02:53.560 --> 00:02:56.660
so the slope is a constant
positive this entire time.

00:02:56.660 --> 00:02:59.260
We have a line with a
constant positive slope.

00:02:59.260 --> 00:03:02.500
So it might look
something like this.

00:03:02.500 --> 00:03:08.850
And let me make it clear what
interval I am talking about.

00:03:08.850 --> 00:03:11.120
I want these things to match up.

00:03:11.120 --> 00:03:13.110
So let me do my best.

00:03:13.110 --> 00:03:17.930
So this matches up to that.

00:03:17.930 --> 00:03:21.270
This matches up over here.

00:03:21.270 --> 00:03:25.000
And we just said we have
a constant positive slope.

00:03:25.000 --> 00:03:30.232
So let's say it looks something
like that over this interval.

00:03:30.232 --> 00:03:32.190
And then we look at this
point right over here.

00:03:32.190 --> 00:03:34.380
So right at this point, our
slope is going to be undefined.

00:03:34.380 --> 00:03:36.504
There's no way that you
could find the slope over--

00:03:36.504 --> 00:03:38.372
or this point of discontinuity.

00:03:38.372 --> 00:03:39.830
But then when we
go over here, even

00:03:39.830 --> 00:03:41.860
though the value of our
function has gone down,

00:03:41.860 --> 00:03:43.540
we still have a
constant positive slope.

00:03:43.540 --> 00:03:45.330
In fact, the slope
of this line looks

00:03:45.330 --> 00:03:47.340
identical to the
slope of this line.

00:03:47.340 --> 00:03:49.200
Let me do that in
a different color.

00:03:49.200 --> 00:03:51.720
The slope of this
line looks identical.

00:03:51.720 --> 00:03:53.932
So we're going to continue
at that same slope.

00:03:53.932 --> 00:03:55.640
It was undefined at
that point, but we're

00:03:55.640 --> 00:03:57.205
going to continue
at that same slope.

00:03:59.770 --> 00:04:01.270
And once again,
it's undefined here

00:04:01.270 --> 00:04:03.450
at this point of discontinuity.

00:04:03.450 --> 00:04:06.060
So the slope will look
something like that.

00:04:06.060 --> 00:04:07.310
And then we go up here.

00:04:07.310 --> 00:04:09.720
The value of the
function goes up,

00:04:09.720 --> 00:04:11.260
but now the function is flat.

00:04:11.260 --> 00:04:13.600
So the slope over
that interval is 0.

00:04:13.600 --> 00:04:17.579
The slope over this interval,
right over here, is 0.

00:04:17.579 --> 00:04:21.620
So we could say--
let me make it clear

00:04:21.620 --> 00:04:24.210
what interval I'm
talking about-- the slope

00:04:24.210 --> 00:04:27.790
over this interval is 0.

00:04:27.790 --> 00:04:32.310
And then finally, in
this last section--

00:04:32.310 --> 00:04:37.450
let me do this in orange--
the slope becomes negative.

00:04:37.450 --> 00:04:40.001
But it's a constant negative.

00:04:40.001 --> 00:04:42.000
And it seems actually a
little bit more negative

00:04:42.000 --> 00:04:43.041
than these were positive.

00:04:43.041 --> 00:04:48.270
So I would draw it
right over there.

00:04:48.270 --> 00:04:50.046
So it's a weird
looking function.

00:04:50.046 --> 00:04:51.420
But the whole
point of this video

00:04:51.420 --> 00:04:53.940
is to give you an intuition
for thinking about what

00:04:53.940 --> 00:04:57.630
the slope of this function
might look like at any point.

00:04:57.630 --> 00:04:59.730
And by doing so,
we have essentially

00:04:59.730 --> 00:05:03.541
drawn the derivative
over that interval.


Limits at infinity of quotients with square roots (odd power)

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:09.456
Let's say that f(x) is equal to x over the square root of x^2+1

00:00:09.456 --> 00:00:18.675
and I want to think about the limit of f(x) as x approaches positive infinity

00:00:18.675 --> 00:00:25.289
and the limit of f(x) as x approaches neagative infinity.

00:00:25.289 --> 00:00:28.271
So let's think about what these are going to be.

00:00:28.271 --> 00:00:32.336
Well, once again, and I'm not doing this in an ultra-rigorous way but more in an intuitive way

00:00:32.336 --> 00:00:36.070
is to think about what this function approximately equals

00:00:36.070 --> 00:00:38.706
as we get larger and larger and larger x'es.

00:00:38.706 --> 00:00:44.205
This is the case if we're getting very positive x'es, in very positive infinity direction or very negative.

00:00:44.205 --> 00:00:48.855
Still the absolute value of those x'es are very very very large as we approach positive infinity

00:00:48.855 --> 00:00:50.737
or negative infinity.

00:00:50.737 --> 00:00:54.003
Well, in the numerator we only have only 1 term -- we have this x term --

00:00:54.003 --> 00:00:57.288
-- but in the denominator we have 2 terms under the radical here.

00:00:57.288 --> 00:01:01.404
And as x gets larger and larger and larger either in the positive or the negative direction

00:01:01.404 --> 00:01:05.700
this x squared term is going to really dominate this one

00:01:05.700 --> 00:01:09.205
You can imagine, when x is 1 million, you're going to have a million squared plus one.

00:01:09.205 --> 00:01:13.470
The value of the denominator is going to be dictated by this x squared term.

00:01:13.470 --> 00:01:22.123
So this is going to be approximately equal to x over the square root of x squared.

00:01:22.123 --> 00:01:29.335
This term right over here, the 1 isn't going to matter so much when we get very very very large x'es.

00:01:29.335 --> 00:01:34.671
And this right over here -- x over the square root of x squared or x over the principle root of x squared --

00:01:34.671 --> 00:01:37.739
-- this is going to be equal to x over --

00:01:37.739 --> 00:01:40.374
-- if I square something and then take the principle root --

00:01:40.374 --> 00:01:43.069
-- remember that the principle root is the positive square root of something --

00:01:43.069 --> 00:01:45.373
-- then I'm essentially taking the absolute value of x.

00:01:45.373 --> 00:01:49.737
This is going to be equal to x over the absolute value of x

00:01:49.737 --> 00:01:57.072
for x approaches infinity or for x approaches negative infinity.

00:01:57.072 --> 00:02:00.337
So, another way to say this, another way to restate these limits,

00:02:00.337 --> 00:02:04.803
is as we approach infinity, this limit, we can restate it as the limit,

00:02:04.803 --> 00:02:11.623
this is going to be equal to the limit as x approaches infinity of x over the absolute value of x.

00:02:11.623 --> 00:02:15.289
Now, for positive x'es the absolute value of x is just going to be x.

00:02:15.289 --> 00:02:19.958
This is going to be x divided by x, so this is just going to be 1.

00:02:19.958 --> 00:02:23.621
Similarly, right over here, we take the limit as we go to negative infinity,

00:02:23.621 --> 00:02:29.737
this is going to be the limit of x over the absolute value of x as x approaches negative infinity.

00:02:29.737 --> 00:02:35.003
Remember, the only reason I was able to make this statement is that f(x) and this thing right over here

00:02:35.003 --> 00:02:39.573
become very very similar, you can kind of say converge to each other,

00:02:39.573 --> 00:02:45.538
as x gets very very very large or x gets very very very very negative.

00:02:45.538 --> 00:02:50.969
Now, for negative values of x the absolute value of x is going to be positive,

00:02:50.969 --> 00:02:55.580
x is obviously going to be negative and we're just going to get negative 1.

00:02:55.593 --> 00:03:01.138
And so using this, we can actually try to graph our function.

00:03:01.138 --> 00:03:02.336
So let's try to do that.

00:03:02.336 --> 00:03:06.706
So let's say, that is my y axis,

00:03:06.706 --> 00:03:10.405
this is my x axis,

00:03:10.405 --> 00:03:13.955
and we see that we have 2 horizontal asymptotes.

00:03:13.955 --> 00:03:17.957
We have 1 horizontal asymptote at y=1,

00:03:17.957 --> 00:03:20.070
so let's say this right over here is y=1,

00:03:20.070 --> 00:03:23.602
let me draw that line as dotted line,

00:03:23.602 --> 00:03:25.405
we're going to approach this thing,

00:03:25.405 --> 00:03:29.204
and then we have another horizontal asymptote at y=-1.

00:03:29.204 --> 00:03:37.004
So that might be right over there, y=-1.

00:03:37.004 --> 00:03:41.537
And if we want to plot at least 1 point we can think about what does f(0) equal.

00:03:41.537 --> 00:03:50.289
So, f(0) is going to be equal to 0 over the square root of 0+1, or 0 squared plus 1.

00:03:50.289 --> 00:03:53.540
Well that's all just going to be equal to zero.

00:03:53.540 --> 00:03:57.540
So we have this point, right over here,

00:03:57.540 --> 00:04:02.538
and we know that as x approaches infinity, we're approaching this blue, horizontal asymptote,

00:04:02.538 --> 00:04:06.235
so it might look something like this.

00:04:06.235 --> 00:04:12.669
Let me do it a little bit differently. There you go.

00:04:12.669 --> 00:04:16.289
I'll clean this up. So it might look something like this.

00:04:16.289 --> 00:04:18.337
That's not the color I wanted to use.

00:04:18.337 --> 00:04:25.271
So it might look something like that.

00:04:25.271 --> 00:04:29.592
We get closer and closer to that asymptote as x gets larger and larger

00:04:29.592 --> 00:04:38.138
and then like this -- we get closer and closer to this asymptote as x approaches negative infinity.

00:04:38.138 --> 00:04:39.802
I'm not drawing it so well.

00:04:39.802 --> 00:04:44.206
So that right over there is y=f(x).

00:04:44.206 --> 00:04:48.470
And you can verify this by taking a calculator, trying to plot more points

00:04:48.470 --> 00:04:50.739
or using some type of graphing calculator or something.

00:04:50.739 --> 00:04:54.871
But anyway, I just wanted to tackle another situation we're approaching infinity and or negative infinity

00:04:54.871 --> 00:04:57.404
and we're trying to determine the horizontal asymptotes.

00:04:57.404 --> 00:05:01.070
And remember, the key is just to say what terms dominate

00:05:01.070 --> 00:05:05.956
as x approaches positive infinity or negative infinity.

00:05:05.956 --> 00:05:10.469
To say, well, what is that function going to approach, and it's going to approach this horizontal asymptote

00:05:10.469 --> 00:05:13.240
in the positive direction and this one in the negative.


Limits at infinity of quotients (Part 1)

WEBVTT
Kind: captions
Language: en

00:00:00.640 --> 00:00:04.920
So we have f of x equaling 4x
to the fifth minus 3x squared

00:00:04.920 --> 00:00:08.870
plus 3, all of that over
6x to the fifth minus 100x

00:00:08.870 --> 00:00:10.690
squared minus 10.

00:00:10.690 --> 00:00:12.830
Now, what I want
to think about is--

00:00:12.830 --> 00:00:21.667
what is the limit of f of
x, as x approaches infinity?

00:00:21.667 --> 00:00:23.750
And there are several ways
that you could do this.

00:00:23.750 --> 00:00:26.410
You could actually try to plug
in larger and larger numbers

00:00:26.410 --> 00:00:29.410
for x and see if it seems to
be approaching some value.

00:00:29.410 --> 00:00:31.396
Or you could reason
through this.

00:00:31.396 --> 00:00:33.270
And when I talk about
reasoning through this,

00:00:33.270 --> 00:00:36.140
it's to think about the
behavior of this numerator

00:00:36.140 --> 00:00:39.684
and denominator as x gets
very, very, very large.

00:00:39.684 --> 00:00:41.850
And when I'm talking about
that, what I'm saying is,

00:00:41.850 --> 00:00:43.580
as x gets very, very
large-- let's just

00:00:43.580 --> 00:00:44.610
focus on the numerator.

00:00:44.610 --> 00:00:47.470
As x gets very, very
large, this term right

00:00:47.470 --> 00:00:49.640
over here in the numerator--
4x to the fifth--

00:00:49.640 --> 00:00:51.930
is going to become a much,
much more significant

00:00:51.930 --> 00:00:53.460
than any of these other things.

00:00:53.460 --> 00:00:54.795
Something squaring gets large.

00:00:54.795 --> 00:00:56.670
But something being
raised to the fifth power

00:00:56.670 --> 00:00:59.670
gets raised that
much, much faster.

00:00:59.670 --> 00:01:02.600
Similarly, in the denominator,
this term right over here-,,

00:01:02.600 --> 00:01:05.510
the highest degree
term-- 6x to the fifth--

00:01:05.510 --> 00:01:08.070
is going to grow much, much,
much faster than any of these

00:01:08.070 --> 00:01:08.600
other terms.

00:01:08.600 --> 00:01:11.350
Even though this has 100 as a
coefficient or a negative 100

00:01:11.350 --> 00:01:14.000
as a coefficient, when you take
something to the fifth power,

00:01:14.000 --> 00:01:16.960
it's going to grow so much
faster than x squared.

00:01:16.960 --> 00:01:19.470
So as x gets very,
very, very large,

00:01:19.470 --> 00:01:22.670
this thing is going
to approximate 4x

00:01:22.670 --> 00:01:36.800
to the fifth over 6x to the
fifth for a very large, large x

00:01:36.800 --> 00:01:40.910
Or we could say as x
approaches infinity.

00:01:40.910 --> 00:01:42.877
Now, what could this
be simplified to?

00:01:42.877 --> 00:01:45.210
Well, you have x to the fifth
divided by x to the fifth.

00:01:45.210 --> 00:01:46.730
These are going
to grow together.

00:01:46.730 --> 00:01:49.320
So these you can think
of them as canceling out.

00:01:49.320 --> 00:01:52.130
And so you are left with 2/3.

00:01:52.130 --> 00:01:54.010
So what you could say
is-- the limit of f

00:01:54.010 --> 00:01:57.140
of x, as x approaches infinity,
as x gets larger and larger

00:01:57.140 --> 00:01:58.890
and larger, all of
these other terms

00:01:58.890 --> 00:02:00.480
aren't going to
matter that much.

00:02:00.480 --> 00:02:02.375
And so it's going
to approach 2/3.

00:02:06.820 --> 00:02:08.320
Now, let's look at
the graph and see

00:02:08.320 --> 00:02:09.509
if that actually makes sense.

00:02:09.509 --> 00:02:11.050
What we're actually
saying is that we

00:02:11.050 --> 00:02:13.630
have a horizontal asymptote
at y is equal to 2/3.

00:02:13.630 --> 00:02:15.490
So lets look at the graph.

00:02:15.490 --> 00:02:17.260
So right here is the graph.

00:02:17.260 --> 00:02:19.390
Got it from Wolfram Alpha.

00:02:19.390 --> 00:02:23.440
And we see, indeed, as x gets
larger and larger and larger, f

00:02:23.440 --> 00:02:26.415
of x seems to be approaching
this value that looks right

00:02:26.415 --> 00:02:28.210
at around 2/3.

00:02:28.210 --> 00:02:31.440
So it looks like we have
a horizontal asymptote

00:02:31.440 --> 00:02:32.615
right over here.

00:02:32.615 --> 00:02:34.830
Let me draw that a
little bit neater.

00:02:34.830 --> 00:02:38.860
We have a horizontal
asymptote right at 2/3.

00:02:38.860 --> 00:02:42.340
So let me draw it
as neatly as I can.

00:02:42.340 --> 00:02:48.350
So this right over here
is y is equal to 2/3.

00:02:48.350 --> 00:02:50.540
The limit as x gets
really, really large,

00:02:50.540 --> 00:02:53.370
as it approaches infinity, y
is getting closer and closer

00:02:53.370 --> 00:02:55.310
and closer to 2/3.

00:02:55.310 --> 00:02:57.530
And when we just look
at the graph here,

00:02:57.530 --> 00:02:59.440
it seems like the same
thing is happening

00:02:59.440 --> 00:03:03.820
from the bottom direction, when
x approaches negative infinity.

00:03:03.820 --> 00:03:07.040
So we could say the
limit of f of x,

00:03:07.040 --> 00:03:10.110
as x approaches
negative infinity, that

00:03:10.110 --> 00:03:11.730
also looks like it's 2/3.

00:03:11.730 --> 00:03:13.790
And we can use the
exact same logic.

00:03:13.790 --> 00:03:17.950
When x becomes a very,
very, very negative number,

00:03:17.950 --> 00:03:20.590
as it becomes
further and further

00:03:20.590 --> 00:03:23.620
to the left on the number
line, the only terms

00:03:23.620 --> 00:03:26.800
that are going to matter are
going to be the 4x to the fifth

00:03:26.800 --> 00:03:28.250
and the 6x to the fifth.

00:03:28.250 --> 00:03:30.240
So this is true
for very large x's.

00:03:30.240 --> 00:03:32.580
It's also true for
very negative x's.

00:03:32.580 --> 00:03:36.230
So we could also say, as x
approaches negative infinity,

00:03:36.230 --> 00:03:37.564
this is also true.

00:03:37.564 --> 00:03:39.730
And then, the x to the fifth
over the x to the fifth

00:03:39.730 --> 00:03:40.710
is going to cancel out.

00:03:40.710 --> 00:03:42.970
These are the dominant terms.

00:03:42.970 --> 00:03:45.400
And we're going to
get it equaling 2/3.

00:03:45.400 --> 00:03:49.160
And once again, you see
that in the graph here.

00:03:49.160 --> 00:03:52.610
We have a horizontal asymptote
at y is equal to 2/3.

00:03:52.610 --> 00:03:54.910
We take the limit of f of
x as x approaches infinity,

00:03:54.910 --> 00:03:55.710
we get 2/3.

00:03:55.710 --> 00:03:58.610
And the limit of f of x as x
approaches negative infinity

00:03:58.610 --> 00:03:59.410
is 2/3.

00:03:59.410 --> 00:04:00.910
So in general,
whenever you do this,

00:04:00.910 --> 00:04:02.284
you just have to
think about what

00:04:02.284 --> 00:04:03.900
terms are going to
dominate the rest?

00:04:03.900 --> 00:04:06.072
And focus on those.