Derivatives of sin x, cos x, tan x, e^x and ln x

WEBVTT
Kind: captions
Language: en

00:00:00.660 --> 00:00:02.570
Let's get some exposure
to the derivatives

00:00:02.570 --> 00:00:03.920
of some of the most
common functions.

00:00:03.920 --> 00:00:05.753
We're not going to prove
them in this video,

00:00:05.753 --> 00:00:07.920
but at least understand
what the derivatives are.

00:00:07.920 --> 00:00:09.836
So first, let's start
with the trig functions.

00:00:09.836 --> 00:00:15.530
If I want to take the derivative
with respect to x of sine of x,

00:00:15.530 --> 00:00:18.571
this is going to be
equal to cosine of x.

00:00:18.571 --> 00:00:21.070
And if you look at their graphs,
it'll make intuitive sense.

00:00:21.070 --> 00:00:22.760
Once again I have
not proved it here,

00:00:22.760 --> 00:00:24.135
but this is a good
thing to know,

00:00:24.135 --> 00:00:26.810
that the derivative of
sine of x is cosine of x.

00:00:26.810 --> 00:00:29.370
Now what about the
derivative of cosine of x?

00:00:29.370 --> 00:00:34.460
What about the derivative with
respect to x of cosine of x?

00:00:34.460 --> 00:00:38.330
Well, this one's going
to be negative sine of x.

00:00:38.330 --> 00:00:40.000
So the derivative
of sine is cosine,

00:00:40.000 --> 00:00:42.640
and the derivative
cosine is negative sine.

00:00:42.640 --> 00:00:49.810
And then finally, the
derivative of tangent

00:00:49.810 --> 00:00:56.670
of x is equal to 1
over cosine squared

00:00:56.670 --> 00:01:01.800
of x, which is equal to
the secant squared of x.

00:01:01.800 --> 00:01:05.220
Once again, these are all
very good things to know.

00:01:05.220 --> 00:01:07.570
Now let's talk a little
bit about exponentials

00:01:07.570 --> 00:01:08.897
and logarithms.

00:01:08.897 --> 00:01:10.480
So the derivative--
and actually, this

00:01:10.480 --> 00:01:13.770
is one of the coolest
results, and it once again

00:01:13.770 --> 00:01:17.590
speaks to how cool e is as
a number, the derivative

00:01:17.590 --> 00:01:19.886
with respect to
x of e to the x--

00:01:19.886 --> 00:01:21.260
we need a drum
roll for this one.

00:01:21.260 --> 00:01:23.520
This is one of the coolest
things in mathematics.

00:01:23.520 --> 00:01:27.505
The derivative of e to
the x is e to the x.

00:01:27.505 --> 00:01:28.630
Now what does that tell us?

00:01:28.630 --> 00:01:30.255
And I have to take
a little pause here,

00:01:30.255 --> 00:01:31.680
because this is
just so exciting.

00:01:31.680 --> 00:01:33.520
So let me graph e to the x.

00:01:33.520 --> 00:01:35.480
So that's my y-axis.

00:01:35.480 --> 00:01:40.620
Let's say that this right
over here is my x-axis.

00:01:40.620 --> 00:01:43.270
So if I have very
negative values of x,

00:01:43.270 --> 00:01:46.720
e to a very negative value,
we are approaching zero.

00:01:46.720 --> 00:01:49.330
And then e to the
0 is 1, so that's

00:01:49.330 --> 00:01:50.670
going to be 1 right over there.

00:01:50.670 --> 00:01:53.210
So it's going to look
something like this.

00:01:53.210 --> 00:01:54.460
And then it's an exponential.

00:01:54.460 --> 00:01:56.940
It's going to go, it's going
to start increasing really,

00:01:56.940 --> 00:01:58.900
really, really,
really, really fast.

00:01:58.900 --> 00:02:01.940
So let's say that's the graph
of y is equal to e to the x.

00:02:01.940 --> 00:02:05.510
What this tells us is that
at any point-- so let's

00:02:05.510 --> 00:02:07.210
say I go right over here.

00:02:07.210 --> 00:02:10.570
I say when x is equal
to 0, e to the 0

00:02:10.570 --> 00:02:15.230
is 1, what's the slope
of the tangent line here?

00:02:15.230 --> 00:02:17.380
Turns out that is also 1.

00:02:17.380 --> 00:02:18.260
Amazing.

00:02:18.260 --> 00:02:24.020
If I go to x is equal
to 1 right over here,

00:02:24.020 --> 00:02:30.406
the function evaluated here gets
us e to the 1 power or just e.

00:02:30.406 --> 00:02:32.780
And what's the slope of the
tangent line right over here?

00:02:32.780 --> 00:02:34.200
It is also e.

00:02:34.200 --> 00:02:37.920
At any point right over here,
the slope of the tangent line

00:02:37.920 --> 00:02:40.780
is equal to the value of
the function at that point.

00:02:40.780 --> 00:02:42.660
This is amazing.

00:02:42.660 --> 00:02:44.870
This is what is so cool about e.

00:02:44.870 --> 00:02:46.730
Anyway, that's not the
point of this video.

00:02:46.730 --> 00:02:49.060
This video is to give
you a catalog of all

00:02:49.060 --> 00:02:51.860
of the derivatives that
you might really need.

00:02:51.860 --> 00:02:54.050
So then finally,
if we're thinking

00:02:54.050 --> 00:02:55.940
about the derivative
with respect

00:02:55.940 --> 00:03:02.890
to x of the natural log of x,
this is going to be equal to--

00:03:02.890 --> 00:03:04.670
and this is also fascinating.

00:03:04.670 --> 00:03:09.450
This is equal to 1 over
x or x to the negative 1.

00:03:09.450 --> 00:03:12.440
So somehow, we have
our natural log

00:03:12.440 --> 00:03:14.690
has kind of inserted
itself into-- when

00:03:14.690 --> 00:03:16.670
you take the derivative,
as filling in the gap

00:03:16.670 --> 00:03:19.915
that the power rule
left vacant, which is,

00:03:19.915 --> 00:03:22.010
is there some function
whose derivative is

00:03:22.010 --> 00:03:23.650
equal to x to the negative 1?

00:03:23.650 --> 00:03:26.000
The power rule gave us
functions whose derivatives

00:03:26.000 --> 00:03:28.160
might be x to the negative
2, x to the negative 3,

00:03:28.160 --> 00:03:30.330
or x to the squared
or x to the fifth.

00:03:30.330 --> 00:03:33.010
But it left the x to
the negative 1 vacant,

00:03:33.010 --> 00:03:34.940
and it's filled by
the natural log of x.

00:03:34.940 --> 00:03:36.250
Now I haven't proved it here.

00:03:36.250 --> 00:03:38.000
I've just catalogued
these for you.

00:03:38.000 --> 00:03:40.525
And then we can use
these in future videos,

00:03:40.525 --> 00:03:43.390
and we'll prove them
in future videos.


Area between curves with multiple boundaries

WEBVTT
Kind: captions
Language: en

00:00:00.520 --> 00:00:04.050
What I want to do in this video
is find the area of this region

00:00:04.050 --> 00:00:07.200
that I'm shading in yellow.

00:00:07.200 --> 00:00:11.140
And what might seem challenging
is that throughout this region,

00:00:11.140 --> 00:00:12.810
I have the same lower function.

00:00:12.810 --> 00:00:14.540
Or I guess the
lower boundary is y

00:00:14.540 --> 00:00:16.870
is equal to x squared
over 4 minus 1.

00:00:16.870 --> 00:00:19.242
But I have a different
upper boundary.

00:00:19.242 --> 00:00:20.700
And the way that
we can tackle this

00:00:20.700 --> 00:00:23.290
is by dividing this
area into two sections,

00:00:23.290 --> 00:00:26.640
or dividing this region into two
regions, the region on the left

00:00:26.640 --> 00:00:28.250
and the region on
the right, where

00:00:28.250 --> 00:00:30.730
for this first region,
which I'll do--

00:00:30.730 --> 00:00:34.310
I'll color even more in
yellow-- for this first region,

00:00:34.310 --> 00:00:35.950
over that entire interval in x.

00:00:35.950 --> 00:00:40.360
And it looks like x is
going between 0 and 1.

00:00:40.360 --> 00:00:44.280
y equals-- when x is equal to
1, this function is equal to 1.

00:00:44.280 --> 00:00:47.220
When x is equal to 1, this
function is also equal to 1.

00:00:47.220 --> 00:00:48.810
So this is the point 1 comma 1.

00:00:48.810 --> 00:00:50.320
That's where they intersect.

00:00:50.320 --> 00:00:53.250
So for this section, this
subregion right over here,

00:00:53.250 --> 00:00:57.420
y equals square root of x is the
upper function the entire time.

00:00:57.420 --> 00:00:59.230
And then we can have
a-- we can set up

00:00:59.230 --> 00:01:02.860
a different-- we can
separately tackle figuring out

00:01:02.860 --> 00:01:04.920
the area of this region.

00:01:04.920 --> 00:01:07.960
From x is equal to 1
to x is equal to 2,

00:01:07.960 --> 00:01:10.890
where y equals 2 minus
x, is the upper function.

00:01:10.890 --> 00:01:12.450
So let's do it.

00:01:12.450 --> 00:01:14.710
So let's first think
about this first region.

00:01:14.710 --> 00:01:17.050
Well, that's going to be
the definite integral from x

00:01:17.050 --> 00:01:19.640
is equal to 0 to
x is equal to 1.

00:01:19.640 --> 00:01:25.120
And our upper function is square
root of x, so square root of x.

00:01:25.120 --> 00:01:28.390
And then from that, we want to
subtract our lower function--

00:01:28.390 --> 00:01:32.320
square root of x minus x
squared over 4 minus 1.

00:01:39.200 --> 00:01:42.400
And then of course,
we have our dx.

00:01:42.400 --> 00:01:46.350
So this right over here, this is
describing the area in yellow.

00:01:46.350 --> 00:01:49.730
And you could imagine it, that
this part right over here,

00:01:49.730 --> 00:01:51.660
the difference between
these two functions

00:01:51.660 --> 00:01:53.164
is essentially this height.

00:01:53.164 --> 00:01:54.580
Let me do it in a
different color.

00:01:57.820 --> 00:01:59.680
And then you
multiply it times dx.

00:01:59.680 --> 00:02:03.390
You get a little
rectangle with width dx.

00:02:03.390 --> 00:02:06.600
And then you do that for each x.

00:02:06.600 --> 00:02:08.860
Each x you get a
different rectangle.

00:02:08.860 --> 00:02:10.650
And then you sum them all up.

00:02:10.650 --> 00:02:14.570
And you take the limit as
your change in x approaches 0.

00:02:14.570 --> 00:02:16.664
So as you get ultra,
ultra thin rectangles,

00:02:16.664 --> 00:02:18.330
and you have an
infinite number of them.

00:02:18.330 --> 00:02:21.060
And that's our definition,
or the Riemann definition

00:02:21.060 --> 00:02:22.820
of what a definite integral is.

00:02:22.820 --> 00:02:25.370
And so this is the area
of the left region.

00:02:25.370 --> 00:02:27.370
And by the exact same
logic, we could figure out

00:02:27.370 --> 00:02:28.972
the area of the right region.

00:02:28.972 --> 00:02:30.680
The right region--
and then we could just

00:02:30.680 --> 00:02:32.127
sum the two things together.

00:02:32.127 --> 00:02:34.210
The right region, we're
going from x is equal to 0

00:02:34.210 --> 00:02:38.530
to x-- sorry, x is equal to
1 to x is equal to 2, 1 to 2.

00:02:38.530 --> 00:02:42.130
The upper function is 2 minus x.

00:02:42.130 --> 00:02:47.220
And from that, we're going to
subtract the lower function,

00:02:47.220 --> 00:02:49.660
which is x squared
over 4 minus 1.

00:02:53.780 --> 00:02:56.060
And now we just
have to evaluate.

00:02:56.060 --> 00:02:58.800
So let's first simplify
this right over here.

00:02:58.800 --> 00:03:02.100
This is equal to the
definite integral

00:03:02.100 --> 00:03:09.220
from 0 to 1 of square root of x
minus x squared over 4 plus 1,

00:03:09.220 --> 00:03:12.020
dx-- I'm going to write
it all in one color now--

00:03:12.020 --> 00:03:18.710
plus the definite integral
from 1 to 2 of 2 minus x,

00:03:18.710 --> 00:03:21.330
minus x squared over 4.

00:03:21.330 --> 00:03:25.330
Then subtracting a negative is
a positive 3-- or a positive 1.

00:03:25.330 --> 00:03:26.650
We could just add it to this 2.

00:03:26.650 --> 00:03:29.330
And so this 2 just becomes a 3.

00:03:29.330 --> 00:03:34.747
I said 2 minus
negative 1 is 3, dx.

00:03:34.747 --> 00:03:36.705
And now we just have to
take the antiderivative

00:03:36.705 --> 00:03:39.310
and evaluate it at 1 and 0.

00:03:39.310 --> 00:03:42.130
So the antiderivative
of this is-- well,

00:03:42.130 --> 00:03:43.480
this is x to the 1/2.

00:03:43.480 --> 00:03:44.730
Increment it by 1.

00:03:44.730 --> 00:03:47.500
Increment the power by
1, you get x to the 3/2,

00:03:47.500 --> 00:03:49.200
and then multiply
by the reciprocal

00:03:49.200 --> 00:03:53.650
of the new exponent-- so
it's 2/3 x to the 3/2.

00:03:53.650 --> 00:03:56.410
Minus-- the antiderivative
of x squared over 4

00:03:56.410 --> 00:04:02.160
is x to the third, divided by 3,
divided by 4, so divided by 12,

00:04:02.160 --> 00:04:03.660
plus x.

00:04:03.660 --> 00:04:05.510
That's the antiderivative of 1.

00:04:05.510 --> 00:04:09.590
We're going to
evaluate it at 1 and 0.

00:04:09.590 --> 00:04:11.640
And then here the
antiderivative is

00:04:11.640 --> 00:04:19.670
going to be 3x minus x
squared over 2 minus x

00:04:19.670 --> 00:04:22.029
to the third over 12.

00:04:22.029 --> 00:04:24.450
Once again, evaluate it
at-- or not once again.

00:04:24.450 --> 00:04:28.460
Now we're going to
evaluate at 2 and 1.

00:04:28.460 --> 00:04:30.610
So over here, you evaluate
all of this stuff at 1.

00:04:30.610 --> 00:04:35.690
You get 2/3 minus 1/12 plus 1.

00:04:35.690 --> 00:04:38.410
And then from that, you
subtract this evaluated at 0.

00:04:38.410 --> 00:04:41.010
But this is just all
0, so you get nothing.

00:04:41.010 --> 00:04:44.260
So this is what the yellow
stuff simplified to.

00:04:44.260 --> 00:04:46.740
And then this purple stuff,
or this magenta stuff,

00:04:46.740 --> 00:04:51.070
or mauve, or whatever color this
is, first you evaluate it at 2.

00:04:51.070 --> 00:04:58.330
You get 6 minus-- let's see,
2 squared over 2 is 2, minus 8

00:04:58.330 --> 00:04:58.920
over 12.

00:05:01.520 --> 00:05:03.650
And then from that,
you're going to subtract

00:05:03.650 --> 00:05:05.460
this evaluated at 1.

00:05:05.460 --> 00:05:13.270
So it's going to be 3 times 1--
that's 3-- minus 1/2 minus 1

00:05:13.270 --> 00:05:14.604
over 12.

00:05:14.604 --> 00:05:16.270
And now what we're
essentially left with

00:05:16.270 --> 00:05:17.889
is adding a bunch of fractions.

00:05:17.889 --> 00:05:19.180
So let's see if we can do that.

00:05:19.180 --> 00:05:20.930
It looks like 12 would
be the most obvious

00:05:20.930 --> 00:05:22.300
common denominator.

00:05:22.300 --> 00:05:29.430
So here you have 8/12
minus 1/12 plus 12/12.

00:05:29.430 --> 00:05:31.310
So this simplifies
to-- what's this?

00:05:31.310 --> 00:05:36.440
This is 19/12, the part
that we have in yellow.

00:05:36.440 --> 00:05:40.190
And then this business,
let me do it in this color.

00:05:40.190 --> 00:05:43.280
So 6 minus 2, this is
just going to be 4.

00:05:43.280 --> 00:05:51.100
So we can write this as
48/12-- that's 4-- minus 8/12.

00:05:51.100 --> 00:05:54.460
And then we're going to have to
subtract a 3, which is 36/12.

00:05:57.170 --> 00:06:02.410
Then we're going to add to
1/2, which is just plus 6/12,

00:06:02.410 --> 00:06:06.030
and then we're
going to add a 1/12.

00:06:06.030 --> 00:06:10.730
So this is all going to simplify
to-- let's see, 48 minus 8

00:06:10.730 --> 00:06:18.410
is 40, minus 36 is 4, plus
6 is 10, plus 1 is 11.

00:06:18.410 --> 00:06:21.614
So this becomes plus 11/12.

00:06:21.614 --> 00:06:23.030
Let me make sure
I did that right.

00:06:23.030 --> 00:06:28.830
48 minus 8 is 40,
minus 36 is 4, 10, 11.

00:06:28.830 --> 00:06:30.040
So that looks right.

00:06:30.040 --> 00:06:31.955
And then we're ready
to add these two.

00:06:31.955 --> 00:06:36.010
19 plus 11 is equal to 30/12.

00:06:36.010 --> 00:06:38.290
Or if we want to simplify
this a little bit,

00:06:38.290 --> 00:06:40.990
we can divide the numerator
and the denominator by 6.

00:06:40.990 --> 00:06:44.960
This is equal to
5/2, or 2 and 1/2.

00:06:44.960 --> 00:06:45.630
And we're done.

00:06:45.630 --> 00:06:50.560
We figured out the area
of this entire region.

00:06:50.560 --> 00:06:53.255
It is 2 and 1/2.


Area between curves

WEBVTT
Kind: captions
Language: en

00:00:00.620 --> 00:00:02.190
What I want to do
in this video is

00:00:02.190 --> 00:00:05.070
find the area of the region
in the first quadrant

00:00:05.070 --> 00:00:08.555
right over here that's
below the graph of y

00:00:08.555 --> 00:00:11.870
is equals to square root of
x, but above the graph of y

00:00:11.870 --> 00:00:12.990
is equal to x squared.

00:00:12.990 --> 00:00:17.835
So we're talking about this
region right over here.

00:00:17.835 --> 00:00:19.460
And we know what the
two endpoints are.

00:00:19.460 --> 00:00:22.750
The two endpoints are when these
two functions equal each other.

00:00:22.750 --> 00:00:25.250
So x squared is
equal to square root

00:00:25.250 --> 00:00:31.240
of x at x is equal to 0
and at x is equal to 1.

00:00:31.240 --> 00:00:33.350
So how do we think about this?

00:00:33.350 --> 00:00:35.270
Well, one way to do it
is just to think about,

00:00:35.270 --> 00:00:37.120
well, what's the
area between y is

00:00:37.120 --> 00:00:39.960
equal to square root
of x and the x-axis?

00:00:39.960 --> 00:00:42.930
So that's going to be, and
we're going from 0 to 1,

00:00:42.930 --> 00:00:49.550
so this is the area under
square root of x from 0 to 1 dx.

00:00:49.550 --> 00:00:53.300
So this is literally denoting
this entire region, right

00:00:53.300 --> 00:00:58.440
over here, all the way up
to this boundary point.

00:00:58.440 --> 00:01:01.910
And then from that,
subtract out this region.

00:01:01.910 --> 00:01:04.680
The area underneath
y is equal to x

00:01:04.680 --> 00:01:06.500
squared, but above the x-axis.

00:01:06.500 --> 00:01:11.610
So from that, we could
subtract from 0 to 1.

00:01:11.610 --> 00:01:13.290
That's the interval
we care about.

00:01:13.290 --> 00:01:17.140
The area under x squared dx.

00:01:17.140 --> 00:01:19.197
And this would be
completely legitimate.

00:01:19.197 --> 00:01:21.530
And you might say, hey, look,
I have the same boundaries

00:01:21.530 --> 00:01:24.470
of integration, I have
dx right over here.

00:01:24.470 --> 00:01:29.960
Couldn't I have also written
this as the definite integral

00:01:29.960 --> 00:01:40.874
from 0 to 1 dx of square
root of x minus x squared?

00:01:40.874 --> 00:01:42.790
And if you asked that,
I would say absolutely.

00:01:42.790 --> 00:01:45.410
And this is actually a different
way of conceptualizing it.

00:01:45.410 --> 00:01:47.000
Here, you're taking the area.

00:01:47.000 --> 00:01:50.050
And instead of, if
you conceptualize dx

00:01:50.050 --> 00:01:54.000
as the width of each
of your rectangles,

00:01:54.000 --> 00:01:55.750
now the height of the
rectangle isn't just

00:01:55.750 --> 00:01:59.270
a function between the x-axis
and the function itself.

00:01:59.270 --> 00:02:01.940
The height is the difference
between the two functions.

00:02:01.940 --> 00:02:04.442
So in this case, this would
be the width of a rectangle,

00:02:04.442 --> 00:02:06.900
and then the height would be
the difference between the two

00:02:06.900 --> 00:02:07.810
functions.

00:02:07.810 --> 00:02:11.600
And then you would have another
rectangle right over there.

00:02:11.600 --> 00:02:14.090
All of them have width
dx, and then the height

00:02:14.090 --> 00:02:15.900
is the difference
between the functions.

00:02:15.900 --> 00:02:18.660
And then as you take the
limit as you have super thin,

00:02:18.660 --> 00:02:21.030
as these rectangles get thinner
and thinner and thinner,

00:02:21.030 --> 00:02:22.738
and you have more and
more and more them,

00:02:22.738 --> 00:02:24.670
you are essentially
approximating the area.

00:02:24.670 --> 00:02:26.510
So this is one way of
conceptualizing it.

00:02:26.510 --> 00:02:29.010
Or just using, I guess
the tools we already

00:02:29.010 --> 00:02:31.780
had, the area under the
curve of square root of x.

00:02:31.780 --> 00:02:34.580
Subtracting from that the area
under the curve of x squared.

00:02:34.580 --> 00:02:38.380
These are completely legitimate
ways of conceptualizing it.

00:02:38.380 --> 00:02:40.680
But now let's evaluate it.

00:02:40.680 --> 00:02:42.440
So what's the antiderivative?

00:02:42.440 --> 00:02:44.250
We know from the second
fundamental theorem

00:02:44.250 --> 00:02:48.710
of calculus, we can say that
this is right over here is just

00:02:48.710 --> 00:02:51.814
going to be the antiderivative
of the square root of x.

00:02:51.814 --> 00:02:53.480
Well, square root of
x is the same thing

00:02:53.480 --> 00:02:56.490
as x to the 1/2 power,
so we just increment it.

00:02:56.490 --> 00:03:01.330
So we say x to the 3/2 power,
and then divide by 3/2.

00:03:01.330 --> 00:03:04.010
So that's the same thing
as multiplying by 2/3.

00:03:04.010 --> 00:03:07.200
And we're going to
evaluate that from 0 to 1.

00:03:07.200 --> 00:03:12.150
And from that, we subtract the
antiderivative of x squared,

00:03:12.150 --> 00:03:15.700
is x to the third over 3.

00:03:15.700 --> 00:03:20.090
And we're going to evaluate
that from 0 to 1 as well.

00:03:20.090 --> 00:03:23.440
Now this first expression,
if we evaluated it at 1.

00:03:23.440 --> 00:03:24.860
2/3 times 1 and 3/2.

00:03:24.860 --> 00:03:26.850
That's just going to be 2/3.

00:03:26.850 --> 00:03:28.680
And then from that,
we're going to subtract

00:03:28.680 --> 00:03:30.640
this thing evaluated
at 0, which is just 0.

00:03:30.640 --> 00:03:32.782
So we're just left with 2/3.

00:03:32.782 --> 00:03:34.990
And then we're going to
subtract this thing evaluated

00:03:34.990 --> 00:03:38.520
at 1, minus 1/3 third.

00:03:38.520 --> 00:03:41.702
And then, also, we're going
to subtract this thing.

00:03:41.702 --> 00:03:43.160
Well, just to be
clear, we're going

00:03:43.160 --> 00:03:46.490
to subtract this thing evaluated
at 1, and then from that,

00:03:46.490 --> 00:03:49.520
it evaluated at 0,
which is just 0.

00:03:49.520 --> 00:03:55.484
But this whole thing simplifies
to just subtracting 1/3.

00:03:55.484 --> 00:03:56.400
And there you have it.

00:03:56.400 --> 00:03:57.190
We get our answer.

00:03:57.190 --> 00:04:00.770
2/3 minus 1/3 is equal to 1/3.

00:04:00.770 --> 00:04:02.590
So this area right over here.

00:04:02.590 --> 00:04:08.630
Whatever units we're using
is 1/3 square units of area.


Integration by parts: Âº___cos(x)dx

WEBVTT
Kind: captions
Language: en

00:00:00.540 --> 00:00:02.720
Let's see if we can use
integration by parts

00:00:02.720 --> 00:00:09.271
to find the antiderivative of
e to the x cosine of x, dx.

00:00:09.271 --> 00:00:11.270
And whenever we talk about
integration by parts,

00:00:11.270 --> 00:00:13.090
we always say, well,
which of these functions--

00:00:13.090 --> 00:00:15.060
we're taking a product
of two of these-- which

00:00:15.060 --> 00:00:17.685
of these functions, either the
x or cosine of x, that if I were

00:00:17.685 --> 00:00:19.520
to take its derivative,
becomes simpler.

00:00:19.520 --> 00:00:21.530
And in this case neither
of them become simpler.

00:00:21.530 --> 00:00:24.030
And neither of them become
dramatically more complicated

00:00:24.030 --> 00:00:26.430
when I take their
antiderivative.

00:00:26.430 --> 00:00:28.520
So here, it's kind of a
toss up which one I assign

00:00:28.520 --> 00:00:31.290
to f of x and which one
I assign to g prime of x.

00:00:31.290 --> 00:00:34.020
And actually, you can solve
this problem either way.

00:00:34.020 --> 00:00:36.270
So let's just assign this one.

00:00:36.270 --> 00:00:39.140
Let's assign f of x
equaling e to the x.

00:00:39.140 --> 00:00:43.130
And let's assign g prime of
x as equaling cosine of x.

00:00:43.130 --> 00:00:44.480
So let me write it down.

00:00:44.480 --> 00:00:49.100
We are saying f of x
is equal to e to the x,

00:00:49.100 --> 00:00:52.300
or f prime of x is
equal to e to the x.

00:00:52.300 --> 00:00:54.630
Derivative of e to the
x is just e to the x.

00:00:54.630 --> 00:00:57.730
And we can say that g-- we're
making the assignment-- g

00:00:57.730 --> 00:01:00.510
prime of x is equal
to cosine of x.

00:01:00.510 --> 00:01:03.950
And the antiderivative
of that g of x is also.

00:01:03.950 --> 00:01:06.010
Or the antiderivative
of cosine of x

00:01:06.010 --> 00:01:10.810
is just going to be
equal to sine of x.

00:01:10.810 --> 00:01:13.220
So now let's apply
integration by parts.

00:01:13.220 --> 00:01:15.440
So this thing is
going to be equal to f

00:01:15.440 --> 00:01:19.250
of x times g of x,
which is equal to e

00:01:19.250 --> 00:01:28.600
to the x times sine of x,
minus the antiderivative of f

00:01:28.600 --> 00:01:32.160
prime of x-- f prime
of x is e to the x.

00:01:32.160 --> 00:01:36.080
e to the x times g of x, which
is once again, sine of x.

00:01:43.210 --> 00:01:45.520
Now, it doesn't look like
we've made a lot of progress,

00:01:45.520 --> 00:01:47.145
now we have an
indefinite integral that

00:01:47.145 --> 00:01:48.560
involves a sine of x.

00:01:48.560 --> 00:01:50.480
So let's see if we can
solve that, let's see

00:01:50.480 --> 00:01:52.740
if we solve this one separately.

00:01:52.740 --> 00:01:55.530
So let's say if we were trying
to find the antiderivative.

00:01:55.530 --> 00:01:59.740
The antiderivative of e
to the x, sine of x dx.

00:01:59.740 --> 00:02:01.070
How could we do that?

00:02:01.070 --> 00:02:05.580
Well, similarly, we can set f
of x as equal to e to the x.

00:02:05.580 --> 00:02:07.440
So, and now this is
we're reassigning,

00:02:07.440 --> 00:02:10.310
although we're happening to make
the exact same reassignment.

00:02:10.310 --> 00:02:13.330
So we're saying f of x
is equal to e to the x.

00:02:13.330 --> 00:02:16.030
f prime of x is equal to
just the derivative of that,

00:02:16.030 --> 00:02:17.750
which is still e to the x.

00:02:17.750 --> 00:02:20.580
And then we could say
g of x, in this case,

00:02:20.580 --> 00:02:22.690
is equal to sine of x.

00:02:22.690 --> 00:02:25.660
We'll put these assignments in
the back of our brain for now.

00:02:25.660 --> 00:02:30.080
And then-- let me make this
clear-- g prime of x, let me,

00:02:30.080 --> 00:02:34.710
woops, there you go--
so we have g prime

00:02:34.710 --> 00:02:40.570
of x is equal to sine
of x, which means

00:02:40.570 --> 00:02:45.010
that its antiderivative
is negative cosine of x.

00:02:45.010 --> 00:02:46.660
Derivative of cosine
is negative sine,

00:02:46.660 --> 00:02:49.380
derivative of negative
cosine is positive sine.

00:02:49.380 --> 00:02:53.370
So once again, let's apply
integration by parts.

00:02:53.370 --> 00:02:56.520
So we have f of x times g of x.

00:02:56.520 --> 00:02:59.870
f of x times g of
x is negative--

00:02:59.870 --> 00:03:02.750
is I'll put the negative
out front-- it's negative e

00:03:02.750 --> 00:03:13.330
to the x times cosine of x,
minus the antiderivative of f

00:03:13.330 --> 00:03:14.460
prime of xg of x.

00:03:14.460 --> 00:03:18.770
F prime of x is e to x.

00:03:18.770 --> 00:03:22.020
And then g of x is
negative cosine of x.

00:03:22.020 --> 00:03:23.912
So I'll put the cosine
of x right over here,

00:03:23.912 --> 00:03:25.620
and then the negative,
we can take it out

00:03:25.620 --> 00:03:27.220
of the integral sign.

00:03:27.220 --> 00:03:29.450
And so we're
subtracting a negative.

00:03:29.450 --> 00:03:32.050
That becomes a positive.

00:03:32.050 --> 00:03:36.270
And of course, we have
our dx right over there.

00:03:36.270 --> 00:03:38.740
And you might say, Sal, we're
not making any progress.

00:03:38.740 --> 00:03:40.160
This thing right
over here, we now

00:03:40.160 --> 00:03:42.820
expressed in terms
of an integral that

00:03:42.820 --> 00:03:44.220
was our original integral.

00:03:44.220 --> 00:03:46.890
We've come back full circle.

00:03:46.890 --> 00:03:48.900
But let's try to do
something interesting.

00:03:48.900 --> 00:03:54.260
Let's substitute back
this-- all right,

00:03:54.260 --> 00:03:56.040
let me write it this way.

00:03:56.040 --> 00:04:02.690
Let's substitute back
this thing up here.

00:04:02.690 --> 00:04:04.610
Or actually, let me
write it a different way.

00:04:04.610 --> 00:04:13.740
Let's substitute this for
this in our original equation.

00:04:13.740 --> 00:04:16.070
And let's see if we got
anything interesting.

00:04:16.070 --> 00:04:17.910
So what we'll get is
our original integral,

00:04:17.910 --> 00:04:19.829
on the left hand side here.

00:04:19.829 --> 00:04:21.950
The indefinite integral
or the antiderivative of e

00:04:21.950 --> 00:04:27.330
to the x cosine of
x dx is equal to e

00:04:27.330 --> 00:04:39.027
to the x sine of x, minus
all of this business.

00:04:39.027 --> 00:04:40.860
So let's just subtract
all of this business.

00:04:40.860 --> 00:04:43.070
We're subtracting all of this.

00:04:43.070 --> 00:04:45.660
So if you subtract negative
e to the x cosine of x,

00:04:45.660 --> 00:04:47.490
it's going to be positive.

00:04:47.490 --> 00:04:51.380
It's going to be positive
e to the x, cosine of x.

00:04:54.440 --> 00:04:56.570
And then remember, we're
subtracting all of this.

00:04:56.570 --> 00:04:58.470
So then we're going to subtract.

00:04:58.470 --> 00:05:08.710
So then we have minus the
antiderivative of e to the x,

00:05:08.710 --> 00:05:10.152
cosine of x,dx.

00:05:13.891 --> 00:05:14.890
Now this is interesting.

00:05:14.890 --> 00:05:18.170
Just remember all we did is, we
took this part right over here.

00:05:18.170 --> 00:05:19.780
We said, we used
integration by parts

00:05:19.780 --> 00:05:21.738
to figure out that it's
the same thing as this.

00:05:21.738 --> 00:05:23.320
So we substituted this back in.

00:05:23.320 --> 00:05:24.840
When you subtracted it.

00:05:24.840 --> 00:05:26.600
When you subtracted
this from this,

00:05:26.600 --> 00:05:29.380
we got this business
right over here.

00:05:29.380 --> 00:05:31.830
Now what's interesting
here is we have essentially

00:05:31.830 --> 00:05:34.139
an equation where we
have our expression,

00:05:34.139 --> 00:05:35.430
our original expression, twice.

00:05:35.430 --> 00:05:37.054
We could even assign
this to a variable

00:05:37.054 --> 00:05:38.950
and essentially solve
for that variable.

00:05:38.950 --> 00:05:41.420
So why don't we
just add this thing

00:05:41.420 --> 00:05:43.025
to both sides of the equation?

00:05:43.025 --> 00:05:43.900
Let me make it clear.

00:05:43.900 --> 00:05:47.800
Let's just add the integral
of e to the x cosine

00:05:47.800 --> 00:05:51.710
of x dx to both sides.

00:05:51.710 --> 00:05:54.600
e to the x, cosine of x, dx.

00:05:54.600 --> 00:05:55.559
And what do you get?

00:05:55.559 --> 00:05:57.100
Well, on the left
hand side, you have

00:05:57.100 --> 00:05:59.520
two times our original integral.

00:05:59.520 --> 00:06:05.440
e to the x, cosine of x, dx is
equal to all of this business.

00:06:05.440 --> 00:06:07.125
Is equal to this.

00:06:07.125 --> 00:06:09.132
I'll copy and paste it.

00:06:09.132 --> 00:06:11.890
So copy and paste.

00:06:11.890 --> 00:06:15.630
It's equal to all of that.

00:06:15.630 --> 00:06:20.250
And then this part, this part
right over here, cancels out.

00:06:20.250 --> 00:06:23.040
And now we can solve for
our original expression.

00:06:23.040 --> 00:06:26.350
The antiderivative of e
to the x cosine of x dx.

00:06:26.350 --> 00:06:29.240
We just have to divide
both sides of this,

00:06:29.240 --> 00:06:31.370
essentially an equation, by 2.

00:06:31.370 --> 00:06:34.220
So if you divide the
left hand side by two,

00:06:34.220 --> 00:06:36.280
you're left with our
original expression.

00:06:36.280 --> 00:06:39.200
The antiderivative of e
to the x cosine of x dx.

00:06:39.200 --> 00:06:42.690
And on the right hand side you
have what it must be equal to.

00:06:42.690 --> 00:06:47.132
e to the x sine of x, plus e
to the x cosine of x over 2.

00:06:47.132 --> 00:06:49.110
And now we want to be
careful because this

00:06:49.110 --> 00:06:51.600
is an antiderivative of
our original expression,

00:06:51.600 --> 00:06:53.090
but it's not the only one.

00:06:53.090 --> 00:06:55.550
We always have to remember,
even though we've worked hard

00:06:55.550 --> 00:06:58.770
and we've done-- we've used
integration by parts twice.

00:06:58.770 --> 00:07:01.010
And we've had to
back substitute in.

00:07:01.010 --> 00:07:05.440
We have to remember we should
still have a constant here.

00:07:05.440 --> 00:07:07.820
So if you take the
derivative of this business,

00:07:07.820 --> 00:07:09.630
regardless of what
the constant is,

00:07:09.630 --> 00:07:12.850
you will get e to
the x cosine of x.

00:07:12.850 --> 00:07:15.476
And it's actually a pretty
neat looking expression.


Integration by parts: Âºx____dx

WEBVTT
Kind: captions
Language: en

00:00:00.600 --> 00:00:03.060
Let's see if we can
take the antiderivative

00:00:03.060 --> 00:00:09.250
of x squared times
e to the x dx.

00:00:09.250 --> 00:00:12.490
Now, the key is to recognize
when you can at least attempt

00:00:12.490 --> 00:00:13.839
to use integration by parts.

00:00:13.839 --> 00:00:15.380
And it might be a
little bit obvious,

00:00:15.380 --> 00:00:18.300
because this video is
about integration by parts.

00:00:18.300 --> 00:00:21.510
But the clue that
integration by parts

00:00:21.510 --> 00:00:24.010
may be applicable
is to say, look,

00:00:24.010 --> 00:00:25.890
I've got a function
that's the product of two

00:00:25.890 --> 00:00:28.650
other functions-- in this
case, x squared and e to the x.

00:00:28.650 --> 00:00:30.438
And integration by
parts can be useful is

00:00:30.438 --> 00:00:32.229
if I can take the
derivative of one of them

00:00:32.229 --> 00:00:33.130
and it becomes simpler.

00:00:33.130 --> 00:00:35.213
And if I take the
antiderivative of the other one,

00:00:35.213 --> 00:00:36.714
it becomes no more complicated.

00:00:36.714 --> 00:00:39.130
So in this case, if I were to
take the antiderivative of x

00:00:39.130 --> 00:00:41.450
squared, it does become simpler.

00:00:41.450 --> 00:00:42.222
It becomes 2x.

00:00:42.222 --> 00:00:44.180
And if I take the
antiderivative of e to the x,

00:00:44.180 --> 00:00:46.530
it doesn't become
any more complicated.

00:00:46.530 --> 00:00:49.580
So let's assign f of x
to be equal to x squared.

00:00:49.580 --> 00:00:51.317
And we want that one
to be the one where

00:00:51.317 --> 00:00:53.150
if I take the derivative
it becomes simpler.

00:00:53.150 --> 00:00:55.358
Because I'm going to have
to take the derivative of f

00:00:55.358 --> 00:00:58.140
of x right over here in the
integration by parts formula.

00:00:58.140 --> 00:01:04.981
And let's assign g prime of
x to be equal to e to the x.

00:01:04.981 --> 00:01:07.480
Because later, I'm going to
have to take its antiderivative,

00:01:07.480 --> 00:01:11.500
and the antiderivative of e to
the x is still just e to the x.

00:01:11.500 --> 00:01:13.270
So let me write this down.

00:01:13.270 --> 00:01:17.600
So we are saying that f of x--
I'll do it right over here--

00:01:17.600 --> 00:01:23.090
f of x is equal to x squared,
in which case f prime of x

00:01:23.090 --> 00:01:25.780
is going to be equal to 2x.

00:01:25.780 --> 00:01:28.000
And I'm not worrying about
the constants right now.

00:01:28.000 --> 00:01:29.670
We'll just add a
constant at the end

00:01:29.670 --> 00:01:31.295
to make sure that
our antiderivative is

00:01:31.295 --> 00:01:33.020
in the most general form.

00:01:33.020 --> 00:01:37.590
And then g prime of x is
equal to e to the x, which

00:01:37.590 --> 00:01:40.240
means its
antiderivative, g of x,

00:01:40.240 --> 00:01:43.100
is still equal to e to the x.

00:01:43.100 --> 00:01:46.510
And now we're ready to apply
the right-hand side right over

00:01:46.510 --> 00:01:47.140
here.

00:01:47.140 --> 00:01:49.100
So all this thing
right over here

00:01:49.100 --> 00:01:53.330
is going to be equal to f
of x, which is x squared--

00:01:53.330 --> 00:01:55.910
and let me put it right
underneath-- x squared times g

00:01:55.910 --> 00:02:03.260
of x, which is e
to the x, minus.

00:02:03.260 --> 00:02:05.455
Let me do that in
that yellow color.

00:02:05.455 --> 00:02:07.200
I want to make the
colors match up--

00:02:07.200 --> 00:02:10.590
minus the antiderivative
of f prime of x.

00:02:10.590 --> 00:02:16.290
Well, f prime of x is
2x, times g of x. g of x

00:02:16.290 --> 00:02:21.598
is e to the x dx.

00:02:25.506 --> 00:02:26.880
So you might say,
hey, Sal, we're

00:02:26.880 --> 00:02:29.260
left with another
antiderivative,

00:02:29.260 --> 00:02:31.620
another indefinite
integral right over here.

00:02:31.620 --> 00:02:34.550
How do we solve this one?

00:02:34.550 --> 00:02:37.000
And as you might
guess, the key might

00:02:37.000 --> 00:02:39.590
be integration by parts again.

00:02:39.590 --> 00:02:40.689
And we're making progress.

00:02:40.689 --> 00:02:42.980
This right over here is a
simpler expression than this.

00:02:42.980 --> 00:02:45.880
Notice we were able to reduce
the degree of this x squared.

00:02:45.880 --> 00:02:48.639
It now is just a 2x.

00:02:48.639 --> 00:02:50.680
And what we can do to
simplify this a little bit.

00:02:50.680 --> 00:02:53.410
Since 2 is just a
scalar-- it's a constant,

00:02:53.410 --> 00:02:56.170
it's multiplying the
function-- we can take that out

00:02:56.170 --> 00:02:57.940
of the integral sign.

00:02:57.940 --> 00:02:59.100
So let's take it this way.

00:02:59.100 --> 00:03:01.430
So let me rewrite it this way.

00:03:01.430 --> 00:03:03.590
We can only do that
with a constant

00:03:03.590 --> 00:03:05.070
that's multiplying the function.

00:03:05.070 --> 00:03:07.790
So let me put the
2 right out here.

00:03:07.790 --> 00:03:09.450
And so now what
we're concerned about

00:03:09.450 --> 00:03:11.470
is finding the integral--
let me write it

00:03:11.470 --> 00:03:18.510
right over here-- the
integral of xe to the x dx.

00:03:18.510 --> 00:03:21.760
And this is another
integration by parts problem.

00:03:21.760 --> 00:03:25.780
And so let's again apply the
same principles of integration

00:03:25.780 --> 00:03:26.280
by parts.

00:03:26.280 --> 00:03:27.654
What, when I take
its derivative,

00:03:27.654 --> 00:03:28.690
is going to get simpler?

00:03:28.690 --> 00:03:31.148
Well, x is going to get simpler
when I take its derivative.

00:03:31.148 --> 00:03:33.550
So now, for the purposes
of integration by parts,

00:03:33.550 --> 00:03:37.320
let's redefine f of x
to be equal to just x.

00:03:37.320 --> 00:03:45.660
And then we can still have g
prime of x equaling e to the x.

00:03:45.660 --> 00:03:48.460
And so in this case, let
me write it all down. f

00:03:48.460 --> 00:03:50.912
of x is equal to x.

00:03:50.912 --> 00:03:54.780
f prime of x is equal to 1.

00:03:54.780 --> 00:03:59.556
g prime of x is
equal to e to the x.

00:03:59.556 --> 00:04:02.830
g of x, just the
antiderivative of this,

00:04:02.830 --> 00:04:04.920
is equal to e to the x.

00:04:04.920 --> 00:04:07.550
So let's apply integration
by parts again.

00:04:07.550 --> 00:04:11.310
So this is going to be equal
to f of x times g of x.

00:04:11.310 --> 00:04:12.823
Now f of x is x.

00:04:15.636 --> 00:04:24.950
g of x is e to the x, minus
the antiderivative of f prime

00:04:24.950 --> 00:04:28.490
of x-- well, that's just 1--
times g of x-- e to the x.

00:04:28.490 --> 00:04:37.530
It's just 1 times e to the x dx.

00:04:37.530 --> 00:04:39.550
And remember, all
I'm doing right now--

00:04:39.550 --> 00:04:41.100
you might have lost
track of things--

00:04:41.100 --> 00:04:44.060
I'm just focused on
this antiderivative.

00:04:44.060 --> 00:04:46.982
That antiderivative is
that antiderivative there.

00:04:46.982 --> 00:04:48.440
If we can figure
out what it is, we

00:04:48.440 --> 00:04:52.220
can then substitute back
into our original expression.

00:04:52.220 --> 00:04:55.020
Now, you might appreciate
integration by parts.

00:04:55.020 --> 00:04:57.370
What does this right
over here simplify to?

00:04:57.370 --> 00:05:01.460
What is the antiderivative
of 1 times e to the x dx?

00:05:01.460 --> 00:05:03.950
Or what is the antiderivative
of 1 times e to the x?

00:05:03.950 --> 00:05:05.575
Well, it's just the
antiderivative of e

00:05:05.575 --> 00:05:07.300
to the x, which is
just e to the x.

00:05:07.300 --> 00:05:15.150
So this simplifies
to x times e to the x

00:05:15.150 --> 00:05:17.720
minus the antiderivative of
e to the x, which is just

00:05:17.720 --> 00:05:21.470
e to the x, so minus e to the x.

00:05:21.470 --> 00:05:25.020
And then we can take this
and substitute it back.

00:05:25.020 --> 00:05:27.640
This is the
antiderivative of this.

00:05:27.640 --> 00:05:29.580
So we can substitute
it back up here

00:05:29.580 --> 00:05:32.817
to figure out the antiderivative
of our original expression.

00:05:32.817 --> 00:05:34.900
So the antiderivative of
our original expression--

00:05:34.900 --> 00:05:36.450
we're getting really
close-- is going

00:05:36.450 --> 00:05:42.060
to be equal to-- I'm going
to use different colors so we

00:05:42.060 --> 00:05:43.720
can keep track of things.

00:05:43.720 --> 00:05:46.520
It's going to be equal
to x squared times

00:05:46.520 --> 00:05:56.400
e to the x minus 2 times
all of this business.

00:05:56.400 --> 00:06:00.190
So minus 2 times-- well,
this antiderivative

00:06:00.190 --> 00:06:01.360
we just figured out is this.

00:06:01.360 --> 00:06:07.140
Minus 2 times xe to
the x minus e to the x.

00:06:07.140 --> 00:06:10.780
And if we want, now is a
good time to put our plus C.

00:06:10.780 --> 00:06:12.870
And of course, we
can simplify this.

00:06:12.870 --> 00:06:16.860
This is equal to x squared.

00:06:16.860 --> 00:06:18.720
I like to keep the same colors.

00:06:18.720 --> 00:06:22.440
This is equal to x
squared e to the x.

00:06:22.440 --> 00:06:24.380
You distribute the negative 2.

00:06:24.380 --> 00:06:31.730
You get minus 2xe to
the x plus 2e to the x,

00:06:31.730 --> 00:06:36.500
and then finally, plus
C. And we're done.

00:06:36.500 --> 00:06:38.210
We figured out
the antiderivative

00:06:38.210 --> 00:06:41.920
of what looked like a kind of
hairy-looking expression using

00:06:41.920 --> 00:06:45.180
integration by parts twice.


Integration by parts intro

WEBVTT
Kind: captions
Language: en

00:00:00.461 --> 00:00:01.960
What we're going
to do in this video

00:00:01.960 --> 00:00:04.180
is review the product
rule that you probably

00:00:04.180 --> 00:00:06.001
learned a while ago.

00:00:06.001 --> 00:00:07.500
And from that, we're
going to derive

00:00:07.500 --> 00:00:10.720
the formula for integration
by parts, which could really

00:00:10.720 --> 00:00:15.819
be viewed as the inverse product
rule, integration by parts.

00:00:15.819 --> 00:00:17.860
So let's say that I start
with some function that

00:00:17.860 --> 00:00:19.950
can be expressed
as the product f

00:00:19.950 --> 00:00:22.940
of x, can be expressed as a
product of two other functions,

00:00:22.940 --> 00:00:25.090
f of x times g of x.

00:00:25.090 --> 00:00:30.210
Now let's take the
derivative of this function,

00:00:30.210 --> 00:00:32.867
let's apply the derivative
operator right over here.

00:00:32.867 --> 00:00:35.200
And this, once again, just a
review of the product rule.

00:00:35.200 --> 00:00:36.890
It's going to be the derivative
of the first function

00:00:36.890 --> 00:00:38.460
times the second function.

00:00:38.460 --> 00:00:40.050
So it's going to be
f-- no, I'm going

00:00:40.050 --> 00:00:44.750
to do that blue color-- it's
going to be f-- that's not

00:00:44.750 --> 00:00:51.870
blue-- it's going to be f
prime of x times g of x times--

00:00:51.870 --> 00:00:57.480
that's not the same
color-- times g of x

00:00:57.480 --> 00:01:02.740
plus the first function times
the derivative of the second,

00:01:02.740 --> 00:01:06.640
plus the first function, f
of x, times the derivative

00:01:06.640 --> 00:01:09.760
of the second.

00:01:09.760 --> 00:01:11.315
This is all a review
right over here.

00:01:11.315 --> 00:01:12.690
The derivative of
the first times

00:01:12.690 --> 00:01:14.770
the second function
plus the first function

00:01:14.770 --> 00:01:17.130
times the derivative
of the second function.

00:01:17.130 --> 00:01:19.100
Now, let's take
the antiderivative

00:01:19.100 --> 00:01:20.970
of both sides of this equation.

00:01:20.970 --> 00:01:22.345
Well if I take
the antiderivative

00:01:22.345 --> 00:01:28.540
of what I have here on the
left, I get f of x times g of x.

00:01:28.540 --> 00:01:30.540
We won't think about
the constant for now.

00:01:30.540 --> 00:01:32.160
We can ignore that for now.

00:01:32.160 --> 00:01:34.010
And that's going to
be equal to-- well

00:01:34.010 --> 00:01:35.830
what's the
antiderivative of this?

00:01:35.830 --> 00:01:39.160
This is going to be the
antiderivative of f prime

00:01:39.160 --> 00:01:50.250
of x times g of x dx
plus the antiderivative

00:01:50.250 --> 00:01:58.200
of f of x g prime of x dx.

00:01:58.200 --> 00:02:01.580
Now, what I want
to do is I'm going

00:02:01.580 --> 00:02:04.910
to solve for this
part right over here.

00:02:04.910 --> 00:02:08.820
And to solve for that, I just
have to subtract this business.

00:02:08.820 --> 00:02:11.530
I just have to subtract this
business from both sides.

00:02:11.530 --> 00:02:15.300
And then if I subtract
that from both sides,

00:02:15.300 --> 00:02:22.190
I'm left with f of x
times g of x minus this,

00:02:22.190 --> 00:02:28.056
minus the antiderivative
of f prime of x

00:02:28.056 --> 00:02:38.630
g of x-- let me do that in
a pink color-- g of x dx

00:02:38.630 --> 00:02:40.620
is equal to what I
wanted to solve for,

00:02:40.620 --> 00:02:51.590
is equal to the antiderivative
of f of x g prime of x dx.

00:02:51.590 --> 00:02:53.090
And to make it a
little bit clearer,

00:02:53.090 --> 00:02:54.650
let me swap sides here.

00:02:54.650 --> 00:02:56.200
So let me copy and paste this.

00:02:56.200 --> 00:03:00.000
So let me copy
and then paste it.

00:03:00.000 --> 00:03:01.230
There you go.

00:03:01.230 --> 00:03:03.450
And then let me copy and
paste the other side.

00:03:03.450 --> 00:03:07.110
So let me copy and paste it.

00:03:07.110 --> 00:03:10.650
So I'm just switching the
sides, just to give it in a form

00:03:10.650 --> 00:03:14.210
that you might be more used
to seeing in a calculus book.

00:03:14.210 --> 00:03:17.290
So this is essentially the
formula for integration

00:03:17.290 --> 00:03:17.870
by parts.

00:03:17.870 --> 00:03:19.400
I will square it off.

00:03:19.400 --> 00:03:21.940
You'll often see it squared
off in a traditional textbook.

00:03:21.940 --> 00:03:23.800
So I will do the same.

00:03:23.800 --> 00:03:25.730
So this right over here
tells us that if we

00:03:25.730 --> 00:03:29.860
have an integral or an
antiderivative of the form

00:03:29.860 --> 00:03:33.050
f of x times the derivative
of some other function,

00:03:33.050 --> 00:03:34.710
we can apply this
right over here.

00:03:34.710 --> 00:03:36.960
And you might say, well this
doesn't seem that useful.

00:03:36.960 --> 00:03:40.430
First I have to identify a
function that's like this.

00:03:40.430 --> 00:03:42.540
And then still I have
an integral in it.

00:03:42.540 --> 00:03:44.390
But what we'll see
in the next video

00:03:44.390 --> 00:03:46.190
is that this can
actually simplify

00:03:46.190 --> 00:03:49.360
a whole bunch of
things that you're

00:03:49.360 --> 00:03:51.870
trying to take the
antiderivative of.


Integration by parts: Âºx_cos(x)dx

WEBVTT
Kind: captions
Language: en

00:00:00.499 --> 00:00:02.810
In the last video, I
claimed that this formula

00:00:02.810 --> 00:00:05.550
would come handy for
solving or for figuring out

00:00:05.550 --> 00:00:07.850
the antiderivative of
a class of functions.

00:00:07.850 --> 00:00:10.040
Let's see if that
really is the case.

00:00:10.040 --> 00:00:13.700
So let's say I want to take
the antiderivative of x

00:00:13.700 --> 00:00:17.591
times cosine of x dx.

00:00:17.591 --> 00:00:19.590
Now if you look at this
formula right over here,

00:00:19.590 --> 00:00:23.230
you want to assign part of this
to f of x and some part of it

00:00:23.230 --> 00:00:24.190
to g prime of x.

00:00:24.190 --> 00:00:27.140
And the question is, well do I
assign f of x to x and g prime

00:00:27.140 --> 00:00:29.270
of x to cosine of x or
the other way around?

00:00:29.270 --> 00:00:32.850
Do I make f of x cosine
of x and g prime of x, x?

00:00:32.850 --> 00:00:34.697
And that thing to
realize is to look

00:00:34.697 --> 00:00:36.530
at the other part of
the formula and realize

00:00:36.530 --> 00:00:38.540
that you're essentially going
to have to solve this right over

00:00:38.540 --> 00:00:39.279
here.

00:00:39.279 --> 00:00:40.820
And here where we
have the derivative

00:00:40.820 --> 00:00:42.880
of f of x times g of x.

00:00:42.880 --> 00:00:44.830
So what you want to
do is assign f of x

00:00:44.830 --> 00:00:46.920
so that the derivative
of f of x is actually

00:00:46.920 --> 00:00:48.620
simpler than f of x.

00:00:48.620 --> 00:00:50.397
And assign g prime
of x that, if you

00:00:50.397 --> 00:00:52.480
were to take its antiderivative,
it doesn't really

00:00:52.480 --> 00:00:54.630
become any more complicated.

00:00:54.630 --> 00:00:57.400
So in this case,
if we assign f of x

00:00:57.400 --> 00:01:01.610
to be equal to x, f prime
of x is definitely simpler,

00:01:01.610 --> 00:01:03.460
f prime of x is equal to 1.

00:01:03.460 --> 00:01:10.030
If we assign g prime of x to be
cosine of x, once again, if we

00:01:10.030 --> 00:01:12.150
take its antiderivative,
that sine of x,

00:01:12.150 --> 00:01:13.610
it's not any more complicated.

00:01:13.610 --> 00:01:15.026
If we did it the
other way around,

00:01:15.026 --> 00:01:17.545
if we set f of x to
be cosine of x, then

00:01:17.545 --> 00:01:18.920
we're taking its
derivative here.

00:01:18.920 --> 00:01:20.990
That's not that much
more complicated.

00:01:20.990 --> 00:01:23.549
But if we set g prime
of x equaling to x

00:01:23.549 --> 00:01:25.340
and then we had to take
its antiderivative,

00:01:25.340 --> 00:01:28.520
we get x squared over 2,
that is more complicated.

00:01:28.520 --> 00:01:30.690
So let me make it
clear over here.

00:01:30.690 --> 00:01:36.310
We are assigning f of
x to be equal to x.

00:01:36.310 --> 00:01:38.710
And that means that
the derivative of f

00:01:38.710 --> 00:01:40.700
is going to be equal to 1.

00:01:40.700 --> 00:01:44.270
We are assigning-- I'll
write it right here-- g

00:01:44.270 --> 00:01:48.840
prime of x to be equal to
cosine of x, which means

00:01:48.840 --> 00:01:55.790
g of x is equal to sine of x,
the antiderivative of cosine

00:01:55.790 --> 00:01:56.660
of x.

00:01:56.660 --> 00:01:59.630
Now let's see, given
these assumptions,

00:01:59.630 --> 00:02:02.570
let's see if we can
apply this formula.

00:02:02.570 --> 00:02:03.980
So this has all of this.

00:02:03.980 --> 00:02:07.730
Let's see, the right-hand
side says f of x times g of x.

00:02:07.730 --> 00:02:09.056
So f of x is x.

00:02:11.603 --> 00:02:13.363
g of x is sine of x.

00:02:16.780 --> 00:02:21.070
And then from that, we are going
to subtract the antiderivative

00:02:21.070 --> 00:02:27.280
of f prime of x-- well,
that's just 1-- times g of x,

00:02:27.280 --> 00:02:34.670
times sine of x dx.

00:02:34.670 --> 00:02:36.950
Now this was a huge
simplification.

00:02:36.950 --> 00:02:40.400
Now I went from trying to solve
the antiderivative of x cosine

00:02:40.400 --> 00:02:43.120
of x to now I just have
to find the antiderivative

00:02:43.120 --> 00:02:44.480
of sine of x.

00:02:44.480 --> 00:02:51.300
And we know the
antiderivative of sine of x dx

00:02:51.300 --> 00:02:55.000
is just equal to
negative cosine of x.

00:02:55.000 --> 00:02:57.460
And of course, we can
throw the plus c in now,

00:02:57.460 --> 00:02:59.240
now that we're pretty
done with taking

00:02:59.240 --> 00:03:00.890
all of our antiderivatives.

00:03:00.890 --> 00:03:03.910
So all of this is
going to be equal to x

00:03:03.910 --> 00:03:12.990
sine of x, x times sine of
x, minus the antiderivative

00:03:12.990 --> 00:03:15.115
of this, which is just
negative cosine of x.

00:03:17.710 --> 00:03:21.660
And then we could throw in a
plus c right at the end of it.

00:03:21.660 --> 00:03:24.050
And doesn't matter if we
subtract a c or add the c.

00:03:24.050 --> 00:03:26.210
We're saying this is
some arbitrary constant

00:03:26.210 --> 00:03:27.660
which could even be negative.

00:03:27.660 --> 00:03:30.490
And so this is all going
to be equal to-- we

00:03:30.490 --> 00:03:36.810
get our drum roll now-- it's
going to be x times sine of x,

00:03:36.810 --> 00:03:40.360
subtract a negative, that
becomes a positive, plus cosine

00:03:40.360 --> 00:03:42.940
of x plus c.

00:03:42.940 --> 00:03:43.904
And we are done.

00:03:43.904 --> 00:03:46.070
We were able to take the
antiderivative of something

00:03:46.070 --> 00:03:49.490
that we didn't know how to take
the antiderivative of before.

00:03:49.490 --> 00:03:51.500
That was pretty interesting.


Indefinite integrals of sin(x), cos(x), and e_

WEBVTT
Kind: captions
Language: en

00:00:00.730 --> 00:00:02.450
I thought I would do
a few more examples

00:00:02.450 --> 00:00:04.991
of taking antiderivatives, just
so we feel comfortable taking

00:00:04.991 --> 00:00:08.810
antiderivatives of all
of the basic functions

00:00:08.810 --> 00:00:10.990
that we know how to
take the derivatives of.

00:00:10.990 --> 00:00:12.989
And on top of that, I
just want to make it clear

00:00:12.989 --> 00:00:15.580
that it doesn't always
have to be functions of x.

00:00:15.580 --> 00:00:17.230
Here we have a function
of t, and we're

00:00:17.230 --> 00:00:20.170
taking the antiderivative
with respect to t.

00:00:20.170 --> 00:00:22.160
And so you would
not write a dx here.

00:00:22.160 --> 00:00:23.240
That is not the notation.

00:00:23.240 --> 00:00:26.637
You'll see why when we
focus on definite integrals.

00:00:26.637 --> 00:00:29.220
So what's the antiderivative of
this business right over here?

00:00:29.220 --> 00:00:32.310
Well, it's going to be the same
thing as the antiderivative

00:00:32.310 --> 00:00:38.680
of sine of t, or the indefinite
integral of sine of t,

00:00:38.680 --> 00:00:41.500
plus the indefinite integral,
or the antiderivative,

00:00:41.500 --> 00:00:42.580
of cosine of t.

00:00:48.220 --> 00:00:50.500
So let's think about what
these antiderivatives are.

00:00:50.500 --> 00:00:52.375
And we already know a
little bit about taking

00:00:52.375 --> 00:00:54.070
the derivatives
of trig functions.

00:00:54.070 --> 00:01:00.380
We know that the derivative
with respect to t of cosine of t

00:01:00.380 --> 00:01:05.910
is equal to negative sine of t.

00:01:05.910 --> 00:01:07.327
So if we want a
sine of t here, we

00:01:07.327 --> 00:01:09.701
would just have to take the
derivative of negative cosine

00:01:09.701 --> 00:01:10.270
t.

00:01:10.270 --> 00:01:12.260
If we take the derivative
of negative cosine t,

00:01:12.260 --> 00:01:15.190
then we get positive sine of t.

00:01:15.190 --> 00:01:17.405
The derivative with
respect to t of cosine t

00:01:17.405 --> 00:01:18.360
is negative sine of t.

00:01:18.360 --> 00:01:19.651
We have the negative out front.

00:01:19.651 --> 00:01:21.730
It becomes positive sine of t.

00:01:21.730 --> 00:01:26.970
So the antiderivative of sine
of t is negative cosine of t.

00:01:26.970 --> 00:01:31.280
So this is going to be equal
to negative cosine of t.

00:01:31.280 --> 00:01:34.200
And then what's the
antiderivative of cosine of t?

00:01:34.200 --> 00:01:36.880
Well, we already know that
the derivative with respect

00:01:36.880 --> 00:01:41.700
to t of sine of t is
equal to cosine of t.

00:01:41.700 --> 00:01:43.690
So cosine of t's
antiderivative is just

00:01:43.690 --> 00:01:48.080
sine of t-- so plus sine of t.

00:01:48.080 --> 00:01:48.770
And we're done.

00:01:48.770 --> 00:01:52.064
We've found the antiderivative.

00:01:52.064 --> 00:01:52.980
Now let's tackle this.

00:01:52.980 --> 00:01:53.896
Now we don't have a t.

00:01:53.896 --> 00:01:57.900
We're taking the indefinite
integral with respect

00:01:57.900 --> 00:01:59.610
to-- actually,
this is a mistake.

00:01:59.610 --> 00:02:01.380
This should be
with respect to a.

00:02:01.380 --> 00:02:03.070
Let me clean this up.

00:02:03.070 --> 00:02:03.970
This should be a da.

00:02:03.970 --> 00:02:05.678
If we were taking this
with respect to t,

00:02:05.678 --> 00:02:08.270
then we would treat all of
these things as just constants.

00:02:08.270 --> 00:02:10.276
But I don't want to
confuse you right now.

00:02:10.276 --> 00:02:11.150
Let me make it clear.

00:02:11.150 --> 00:02:13.284
This is going to be da.

00:02:13.284 --> 00:02:14.950
That's what we are
integrating or taking

00:02:14.950 --> 00:02:17.560
the antiderivative
with respect to.

00:02:17.560 --> 00:02:19.920
So what is this
going to be equal to?

00:02:19.920 --> 00:02:23.920
Well once again, we can rewrite
it as the sum of integrals.

00:02:23.920 --> 00:02:28.010
This is the indefinite
integral of e to the a da,

00:02:28.010 --> 00:02:31.120
so this one right
over here-- a d

00:02:31.120 --> 00:02:35.430
I'll do it in green-- plus
the indefinite integral,

00:02:35.430 --> 00:02:39.580
or the antiderivative,
of 1/a da.

00:02:39.580 --> 00:02:42.862
Now, what is the
antiderivative of e to the a?

00:02:42.862 --> 00:02:45.720
Well, we already know a
little bit about exponentials.

00:02:45.720 --> 00:02:49.450
The derivative with
respect to x of e to the x

00:02:49.450 --> 00:02:51.130
is equal to e to the x.

00:02:51.130 --> 00:02:53.815
That's one of the reasons why
e in the exponential function

00:02:53.815 --> 00:02:55.370
in general is so amazing.

00:02:55.370 --> 00:02:58.930
And if we just replaced
a with x or x with a,

00:02:58.930 --> 00:03:03.626
you get the derivative with
respect to a of e to the a

00:03:03.626 --> 00:03:05.846
is equal to e to the a.

00:03:05.846 --> 00:03:08.050
So the antiderivative
here, the derivative of e

00:03:08.050 --> 00:03:11.220
to the a, the antiderivative
is going to be e to the a.

00:03:11.220 --> 00:03:14.190
And maybe you can shift it
by some type of a constant.

00:03:14.190 --> 00:03:16.690
Oh, and let me
not forget, I have

00:03:16.690 --> 00:03:18.990
to put my constant
right over here.

00:03:18.990 --> 00:03:20.380
I could have a constant factor.

00:03:20.380 --> 00:03:22.360
So let me-- always important.

00:03:22.360 --> 00:03:24.070
Remember the constant.

00:03:24.070 --> 00:03:28.230
So you have a constant
factor right over here.

00:03:28.230 --> 00:03:29.410
Never forget that.

00:03:29.410 --> 00:03:30.710
I almost did.

00:03:30.710 --> 00:03:33.330
So once again, over here,
what's the antiderivative of e

00:03:33.330 --> 00:03:34.440
to the a?

00:03:34.440 --> 00:03:37.626
It is e to the a.

00:03:37.626 --> 00:03:40.410
What's the
antiderivative of 1/a?

00:03:40.410 --> 00:03:42.175
Well, we've seen that
in the last video.

00:03:42.175 --> 00:03:47.080
It is going to be the natural
log of the absolute value of a.

00:03:47.080 --> 00:03:50.860
And then we want to have the
most general antiderivative,

00:03:50.860 --> 00:03:55.050
so there could be a constant
factor out here as well.

00:03:55.050 --> 00:03:56.120
And we are done.

00:03:56.120 --> 00:04:02.937
We found the antiderivative
of both of these expressions.


Antiderivative of hairier expression

WEBVTT
Kind: captions
Language: en

00:00:00.500 --> 00:00:03.630
So our goal in this video is
to take the antiderivative

00:00:03.630 --> 00:00:05.860
of this fairly crazy
looking expression.

00:00:05.860 --> 00:00:08.840
Or another way of saying it is
to find the indefinite integral

00:00:08.840 --> 00:00:11.990
of this crazy
looking expression.

00:00:11.990 --> 00:00:14.040
And the key realization
right over here

00:00:14.040 --> 00:00:17.650
is that this expression is
made up of a bunch of terms.

00:00:17.650 --> 00:00:20.850
And the indefinite integral
of the entire expression

00:00:20.850 --> 00:00:24.980
is going to be equal to the
indefinite integral of each

00:00:24.980 --> 00:00:26.190
of the term.

00:00:26.190 --> 00:00:28.920
So this is going to be equal
to, we could look at this term

00:00:28.920 --> 00:00:33.610
right over here, and just
take the indefinite that,

00:00:33.610 --> 00:00:36.410
7x to the third dx.

00:00:36.410 --> 00:00:38.920
And then from that,
we can subtract

00:00:38.920 --> 00:00:42.816
the indefinite
integral of this thing.

00:00:42.816 --> 00:00:44.190
So we could say
this is, and then

00:00:44.190 --> 00:00:48.780
minus the indefinite integral
of 5 times the square root

00:00:48.780 --> 00:00:51.180
of x dx.

00:00:51.180 --> 00:00:55.570
And then we can look at
this one right over here.

00:00:55.570 --> 00:00:58.790
So then we could say plus
the indefinite integral

00:00:58.790 --> 00:01:01.670
of 18 square root of x.

00:01:01.670 --> 00:01:05.379
Square roots of x,
over x to the third dx.

00:01:05.379 --> 00:01:08.700
And then finally, I'm
running out of colors here,

00:01:08.700 --> 00:01:10.810
finally I need more
colors in my thing.

00:01:10.810 --> 00:01:12.960
We can take the
antiderivative of this.

00:01:12.960 --> 00:01:18.850
So plus the antiderivative of x
to the negative 40th power dx.

00:01:18.850 --> 00:01:20.997
So I've just rewritten this
and color-coded things.

00:01:20.997 --> 00:01:23.080
So let's take the
antiderivative of each of these.

00:01:23.080 --> 00:01:25.714
And you'll see that we'll
be able to do it using our

00:01:25.714 --> 00:01:26.880
whatever we want to call it.

00:01:26.880 --> 00:01:29.530
The inverse of the power
rule, or the anti-power rule,

00:01:29.530 --> 00:01:31.480
whatever you might
want to call it.

00:01:31.480 --> 00:01:34.080
So let's look at the first one.

00:01:34.080 --> 00:01:36.100
So we have-- what
I'm going to do

00:01:36.100 --> 00:01:38.220
is, I'm just going to
find the antiderivative

00:01:38.220 --> 00:01:40.410
without the constant, and just
add the constant at the end.

00:01:40.410 --> 00:01:41.451
For the sake of this one.

00:01:41.451 --> 00:01:44.570
Just to make sure we get the
most general antiderivative.

00:01:44.570 --> 00:01:46.660
So here the exponent is a 3.

00:01:46.660 --> 00:01:48.280
So we can increase it by 1.

00:01:48.280 --> 00:01:50.450
So it's going to
be x to the 4th.

00:01:50.450 --> 00:01:53.145
Let me do that same purple
color, or pink color.

00:01:53.145 --> 00:01:54.770
It's going to be x
to the 4th, or we're

00:01:54.770 --> 00:01:57.040
going to divide by x to the 4th.

00:01:57.040 --> 00:01:59.440
So it's x to the 4th over 4
is the antiderivative of x

00:01:59.440 --> 00:02:00.240
to the 3rd.

00:02:00.240 --> 00:02:03.180
And you just had this scaling
quantity, the seven out front.

00:02:03.180 --> 00:02:05.140
So we can still just
have the seven out front.

00:02:05.140 --> 00:02:07.610
So we get 7x to the 4th over 4.

00:02:07.610 --> 00:02:08.810
Fair enough.

00:02:08.810 --> 00:02:10.520
From that, we're
going to subtract

00:02:10.520 --> 00:02:13.040
the antiderivative of this.

00:02:13.040 --> 00:02:14.870
Now at first this
might not be obvious,

00:02:14.870 --> 00:02:17.260
that you could use our
inverse power rule,

00:02:17.260 --> 00:02:18.945
or anti-power rule here.

00:02:18.945 --> 00:02:20.320
But then you just
need to realize

00:02:20.320 --> 00:02:22.340
that 5 times the
principal root of x

00:02:22.340 --> 00:02:30.980
is the same thing as 5
times x to the 1/2 power.

00:02:30.980 --> 00:02:34.000
And so once again, the
exponent here is 1/2.

00:02:34.000 --> 00:02:36.760
We can increment it by 1.

00:02:36.760 --> 00:02:40.100
So it's going to
be x to the 3/2.

00:02:40.100 --> 00:02:43.330
And then divide by the
incremented exponent.

00:02:43.330 --> 00:02:46.120
So divide by 3/2.

00:02:46.120 --> 00:02:48.130
And of course we had
this 5 out front,

00:02:48.130 --> 00:02:51.130
so we still want to
have the 5 out front.

00:02:51.130 --> 00:02:53.410
Now this next expression
looks even wackier.

00:02:53.410 --> 00:02:55.810
But once again, we can
simplify a little bit.

00:02:55.810 --> 00:02:58.900
This is the same thing-- let
me do it right over here--

00:02:58.900 --> 00:03:03.580
this is the same thing as
18 times x to the 1/2 times

00:03:03.580 --> 00:03:06.160
x to the negative 3 power.

00:03:06.160 --> 00:03:08.480
x to the 3rd in the
denominator is the same thing

00:03:08.480 --> 00:03:10.099
as x to the negative 3.

00:03:10.099 --> 00:03:12.390
We have the same base, we
could just add the exponents.

00:03:12.390 --> 00:03:18.890
So this is going to be equal
to 18 times x to the 2 and 1/2

00:03:18.890 --> 00:03:20.980
power.

00:03:20.980 --> 00:03:22.490
Or another way of
thinking about it,

00:03:22.490 --> 00:03:29.660
this is the same thing as
18 times x to the 5/2 power.

00:03:33.197 --> 00:03:34.030
Did I do that right?

00:03:34.030 --> 00:03:34.530
Yeah.

00:03:34.530 --> 00:03:35.760
Negative 3.

00:03:35.760 --> 00:03:37.494
Oh sorry, this is
negative 2 and 1/2.

00:03:37.494 --> 00:03:38.660
Let me make this very clear.

00:03:38.660 --> 00:03:41.150
And this is going to be
the negative 5/2 power.

00:03:41.150 --> 00:03:45.290
x to the negative 3 is the same
thing as x to the negative 6/2.

00:03:45.290 --> 00:03:50.800
Negative 6/2 plus
1/2 is negative 5/2.

00:03:50.800 --> 00:03:55.560
So once again, we just have
to increment this exponent.

00:03:55.560 --> 00:04:01.160
So negative 5/2 plus 1 is
going to be negative 3/2.

00:04:01.160 --> 00:04:05.600
So you're going to have
x to the negative 3/2.

00:04:05.600 --> 00:04:07.799
And then you divide
by what your exponent

00:04:07.799 --> 00:04:08.840
is when you increment it.

00:04:08.840 --> 00:04:13.010
So divide it by negative 3/2.

00:04:13.010 --> 00:04:15.780
And then you had
the 18 out front.

00:04:15.780 --> 00:04:18.720
And we obviously are going
to have to simplify this.

00:04:18.720 --> 00:04:22.180
And then finally, our
exponent in this term.

00:04:22.180 --> 00:04:24.530
Let me not use that
purple anymore.

00:04:24.530 --> 00:04:27.970
The exponent in this term
right over here is negative 40.

00:04:27.970 --> 00:04:33.380
If we increment it, we get
x to the negative 39 power,

00:04:33.380 --> 00:04:36.480
all of that over negative 39.

00:04:36.480 --> 00:04:40.680
And now we can add
our constant, c.

00:04:40.680 --> 00:04:44.410
And all we need to do is
simplify all of this craziness.

00:04:44.410 --> 00:04:46.730
So the first one is
fairly simplified.

00:04:46.730 --> 00:04:50.440
We can write it as
7/4 x to the 4th.

00:04:50.440 --> 00:04:53.550
Now this term right over
here is essentially negative

00:04:53.550 --> 00:04:55.830
5 divided by 3/2.

00:04:55.830 --> 00:05:01.620
So 5 over 3/2 is
equal to 5 times 2/3,

00:05:01.620 --> 00:05:04.130
which is equal to 10 over 3.

00:05:04.130 --> 00:05:10.060
So this thing right over here
simplifies to negative 10/3 x

00:05:10.060 --> 00:05:11.860
to the 3/2.

00:05:11.860 --> 00:05:14.770
And then we have all
of this craziness.

00:05:14.770 --> 00:05:19.960
Now 18 divided by
negative 3/2 is

00:05:19.960 --> 00:05:24.610
equal to 18 times negative 2/3.

00:05:24.610 --> 00:05:28.340
Which is equal to, well we can
simplify this a little bit,

00:05:28.340 --> 00:05:35.000
this is the same thing
as 6 times negative 2.

00:05:35.000 --> 00:05:37.820
Which is equal to negative 12.

00:05:37.820 --> 00:05:39.280
So this expression
right over here

00:05:39.280 --> 00:05:44.430
is negative 12x to
the negative 3/2.

00:05:44.430 --> 00:05:46.760
And then finally, this
one right over here,

00:05:46.760 --> 00:05:49.210
we can just rewrite it
as, if we want, we could,

00:05:49.210 --> 00:05:52.250
well, we could just
write negative 1/39 x

00:05:52.250 --> 00:05:57.350
to the negative 39 plus c.

00:05:57.350 --> 00:05:58.100
And we're done.

00:05:58.100 --> 00:06:01.420
We've found the indefinite
integral of all this craziness.

00:06:01.420 --> 00:06:03.735
And I encourage you to take
the derivative of this.

00:06:03.735 --> 00:06:05.860
And you can do it using
really just the power rule,

00:06:05.860 --> 00:06:07.220
to take the derivative of this.

00:06:07.220 --> 00:06:10.370
And verify that it does
indeed equal this expression

00:06:10.370 --> 00:06:14.410
that we took the
antiderivative of.


Reverse power rule

WEBVTT
Kind: captions
Language: en

00:00:00.770 --> 00:00:03.720
Let's take the
derivative with respect

00:00:03.720 --> 00:00:08.570
to x of x to the n
plus 1-th power over n

00:00:08.570 --> 00:00:12.460
plus 1 plus some constant c.

00:00:12.460 --> 00:00:14.210
And we're going to
assume here, because we

00:00:14.210 --> 00:00:16.420
want this expression
to be defined,

00:00:16.420 --> 00:00:20.490
we're going to assume that
n does not equal negative 1.

00:00:20.490 --> 00:00:22.740
If it equaled negative
1, we'd be dividing by 0,

00:00:22.740 --> 00:00:24.724
and we haven't defined
what that means.

00:00:24.724 --> 00:00:26.140
So let's take the
derivative here.

00:00:26.140 --> 00:00:29.500
So this is going to be equal
to-- well, the derivative of x

00:00:29.500 --> 00:00:31.800
to the n plus 1
over n plus 1, we

00:00:31.800 --> 00:00:34.120
can just use the
power rule over here.

00:00:34.120 --> 00:00:36.580
So our exponent is n plus 1.

00:00:36.580 --> 00:00:38.360
We can bring it out front.

00:00:38.360 --> 00:00:45.870
So it's going to be n
plus 1 times x to the-- I

00:00:45.870 --> 00:00:48.160
want to use that same color.

00:00:48.160 --> 00:00:52.260
Colors are the hard part--
times x to the-- instead of n

00:00:52.260 --> 00:00:53.944
plus 1, we subtract
1 from the exponent.

00:00:53.944 --> 00:00:55.110
This is just the power rule.

00:00:55.110 --> 00:00:58.790
So n plus 1 minus
1 is going to be n.

00:00:58.790 --> 00:01:01.970
And then we can't forget that we
were dividing by this n plus 1.

00:01:01.970 --> 00:01:04.629
So we have divided by n plus 1.

00:01:04.629 --> 00:01:06.600
And then we have plus c.

00:01:06.600 --> 00:01:09.300
The derivative of a constant
with respect to x-- a constant

00:01:09.300 --> 00:01:12.050
does not change as x
changes, so it is just

00:01:12.050 --> 00:01:15.250
going to be 0, so plus 0.

00:01:15.250 --> 00:01:17.740
And since n is not
equal to negative 1,

00:01:17.740 --> 00:01:19.860
we know that this is
going to be defined.

00:01:19.860 --> 00:01:21.630
This is just going to
be something divided

00:01:21.630 --> 00:01:23.810
by itself, which is
just going to be 1.

00:01:23.810 --> 00:01:28.420
And this whole thing
simplifies to x to the n.

00:01:28.420 --> 00:01:30.170
So the derivative of
this thing-- and this

00:01:30.170 --> 00:01:33.140
is a very general terms--
is equal to x to the n.

00:01:33.140 --> 00:01:37.390
So given that, what is
the antiderivative-- let

00:01:37.390 --> 00:01:39.050
me switch colors here.

00:01:39.050 --> 00:01:48.230
What is the antiderivative
of x to the n?

00:01:48.230 --> 00:01:50.090
And remember, this
is just the kind

00:01:50.090 --> 00:01:51.619
of strange-looking
notation we use.

00:01:51.619 --> 00:01:54.160
It'll make more sense when we
start doing definite integrals.

00:01:54.160 --> 00:01:57.120
But what is the
antiderivative of x to the n?

00:01:57.120 --> 00:01:59.120
And we could say the
antiderivative with respect

00:01:59.120 --> 00:02:00.590
to x, if we want to.

00:02:00.590 --> 00:02:03.255
And another way of calling this
is the indefinite integral.

00:02:07.840 --> 00:02:10.090
Well, we know this
is saying x to the n

00:02:10.090 --> 00:02:11.752
is the derivative of what?

00:02:11.752 --> 00:02:12.960
Well, we just figured it out.

00:02:12.960 --> 00:02:14.120
It's the derivative
of this thing,

00:02:14.120 --> 00:02:15.911
and we've written it
in very general terms.

00:02:15.911 --> 00:02:19.580
We're actually encapsulating
multiple constants here.

00:02:19.580 --> 00:02:21.670
We could have x to
the n plus 1 over n

00:02:21.670 --> 00:02:25.300
plus 1 plus 0, plus 1, plus
2, plus pi, plus a billion.

00:02:25.300 --> 00:02:31.140
So this is going to be equal
to x to the n plus 1 over n

00:02:31.140 --> 00:02:34.110
plus 1 plus c.

00:02:34.110 --> 00:02:35.920
So this is pretty powerful.

00:02:35.920 --> 00:02:40.920
You can kind of view this
as the reverse power rule.

00:02:40.920 --> 00:02:44.560
And it applies
for any n, as long

00:02:44.560 --> 00:02:46.500
as n does not equal negative 1.

00:02:46.500 --> 00:02:47.750
Let me make that very clear.

00:02:47.750 --> 00:02:52.590
n does not equal negative 1.

00:02:52.590 --> 00:02:54.310
Once again, this thing
would be undefined

00:02:54.310 --> 00:02:57.120
if n were equal to negative 1.

00:02:57.120 --> 00:02:58.720
So let's do a couple
of examples just

00:02:58.720 --> 00:03:01.050
to apply this-- you could
call it the reverse power

00:03:01.050 --> 00:03:04.330
rule if you want, or
the anti-power rule.

00:03:04.330 --> 00:03:10.730
So let's take the antiderivative
of x to the fifth power.

00:03:10.730 --> 00:03:13.574
What is the antiderivative
of x to the fifth?

00:03:13.574 --> 00:03:15.240
Well, all we have to
say is, well, look,

00:03:15.240 --> 00:03:16.950
the 5 is equal to the n.

00:03:16.950 --> 00:03:19.610
We just have to increment
the exponent by 1.

00:03:19.610 --> 00:03:24.280
So this is going to be equal
to x to the 5 plus 1 power.

00:03:24.280 --> 00:03:26.772
And then we divide
by that same value.

00:03:26.772 --> 00:03:28.980
Whatever the exponent was
when you increment it by 1,

00:03:28.980 --> 00:03:32.229
we divide by that same
value, divided by 5 plus 1.

00:03:32.229 --> 00:03:33.770
And of course, we
want to encapsulate

00:03:33.770 --> 00:03:35.810
all of the possible
antiderivatives,

00:03:35.810 --> 00:03:37.770
so you put the c
right over there.

00:03:37.770 --> 00:03:43.380
So this is going to be equal to
x to the sixth over 6 plus c.

00:03:43.380 --> 00:03:44.210
And you can verify.

00:03:44.210 --> 00:03:46.370
Take the derivative of
this using the power rule,

00:03:46.370 --> 00:03:49.830
you indeed get x to the fifth.

00:03:49.830 --> 00:03:51.560
Let's try another one.

00:03:51.560 --> 00:03:54.010
Let's try-- now
we'll do it in blue.

00:03:54.010 --> 00:03:58.540
Let's try the antiderivative
of-- let's make it interesting.

00:03:58.540 --> 00:04:06.290
Let's make it 5 times x to
the negative 2 power dx.

00:04:06.290 --> 00:04:08.710
So how would we evaluate this?

00:04:08.710 --> 00:04:10.290
Well, one simplification
you can do--

00:04:10.290 --> 00:04:14.420
and I haven't rigorously
proven it to you just yet--

00:04:14.420 --> 00:04:16.180
but we know that scalars
can go in and out

00:04:16.180 --> 00:04:18.804
of the derivative operator when
you're multiplying by a scalar.

00:04:18.804 --> 00:04:23.140
So this is, indeed, equal to
5 times the antiderivative

00:04:23.140 --> 00:04:26.605
of x to the negative
2 power, dx.

00:04:26.605 --> 00:04:28.710
And now we can just use,
I guess we could call it

00:04:28.710 --> 00:04:30.960
this anti-power rule,
so this is going

00:04:30.960 --> 00:04:38.840
to be equal to 5 times x to
the negative 2 power plus 1

00:04:38.840 --> 00:04:45.310
over the negative 2 power plus
1 plus some constant right

00:04:45.310 --> 00:04:47.320
over here.

00:04:47.320 --> 00:04:52.010
And then we can rewrite this as
5 times negative 2 power plus 1

00:04:52.010 --> 00:04:56.340
is x to the negative 1
over negative 2 plus 1

00:04:56.340 --> 00:04:59.330
is negative 1,
plus some constant.

00:04:59.330 --> 00:05:04.570
And this is equal to 5 times
negative x to the negative 1

00:05:04.570 --> 00:05:06.540
plus some constant.

00:05:06.540 --> 00:05:09.110
And then if we want, we
can distribute the 5.

00:05:09.110 --> 00:05:12.820
So this is equal to negative
5x to the negative 1.

00:05:12.820 --> 00:05:16.340
Now, we could write plus
5 times some constant,

00:05:16.340 --> 00:05:18.460
but this is just an
arbitrary constant.

00:05:18.460 --> 00:05:21.102
So this is still just
an arbitrary constant.

00:05:21.102 --> 00:05:22.560
So maybe we could
[INAUDIBLE] this.

00:05:22.560 --> 00:05:24.809
If you want it to show that
it's a different constant,

00:05:24.809 --> 00:05:27.370
you could say this
is c1, c1, c1.

00:05:27.370 --> 00:05:30.210
You multiply 5 times c1,
you get another constant.

00:05:30.210 --> 00:05:34.304
We could just call that c,
which is equal to 5 times c1.

00:05:34.304 --> 00:05:35.220
But there you have it.

00:05:35.220 --> 00:05:39.640
Negative 5x to the
negative 1 plus c.

00:05:39.640 --> 00:05:42.190
And once again, all of these,
try to evaluate the derivative,

00:05:42.190 --> 00:05:46.550
and you will see that you
get this business, right over

00:05:46.550 --> 00:05:48.100
there.


Antiderivatives and indefinite integrals

WEBVTT
Kind: captions
Language: en

00:00:00.790 --> 00:00:05.380
We know how to take
derivatives of functions.

00:00:05.380 --> 00:00:09.730
If I apply the derivative
operator to x squared,

00:00:09.730 --> 00:00:12.760
I get 2x.

00:00:12.760 --> 00:00:15.450
Now, if I also apply
the derivative operator

00:00:15.450 --> 00:00:20.200
to x squared plus
1, I also get 2x.

00:00:20.200 --> 00:00:25.650
If I apply the derivative
operator to x squared plus pi,

00:00:25.650 --> 00:00:26.810
I also get 2x.

00:00:26.810 --> 00:00:28.800
The derivative of
x squared is 2x.

00:00:28.800 --> 00:00:32.740
Derivative, with respect to x
of pi of a constant, is just 0.

00:00:32.740 --> 00:00:34.560
Derivative, with
respect to x of 1,

00:00:34.560 --> 00:00:36.370
is just a constant, is just 0.

00:00:36.370 --> 00:00:39.380
So once again, this is just
going to be equal to 2x.

00:00:39.380 --> 00:00:41.570
In general, the
derivative, with respect

00:00:41.570 --> 00:00:46.740
to x of x squared
plus any constant,

00:00:46.740 --> 00:00:48.840
is going to be equal to 2x.

00:00:48.840 --> 00:00:51.920
The derivative of x squared,
with respect to x, is 2x.

00:00:51.920 --> 00:00:54.216
Derivative of a constant,
with respect to x,

00:00:54.216 --> 00:00:55.840
a constant does not
change with respect

00:00:55.840 --> 00:00:58.760
to x, so it's just equal to 0.

00:00:58.760 --> 00:01:01.510
So you have-- You apply
the derivative operator

00:01:01.510 --> 00:01:05.280
to any of these
expressions and you get 2x.

00:01:05.280 --> 00:01:07.070
Now, let's go the
other way around.

00:01:07.070 --> 00:01:08.611
Let's think about
the antiderivative.

00:01:12.290 --> 00:01:15.900
And one way to think
about it is we're

00:01:15.900 --> 00:01:18.400
doing the opposite of
the derivative operator.

00:01:18.400 --> 00:01:20.560
The derivative operator,
you get an expression

00:01:20.560 --> 00:01:22.110
and you find it's derivative.

00:01:22.110 --> 00:01:25.170
Now, what we want to do,
is given some expression,

00:01:25.170 --> 00:01:27.763
we want to find what it
could be the derivative of.

00:01:30.430 --> 00:01:32.580
So if someone were
to tell-- or give

00:01:32.580 --> 00:01:38.270
you 2x-- if someone were to
say 2x-- let me write this.

00:01:38.270 --> 00:01:51.420
So if someone were to ask you
what is 2x the derivative of?

00:01:51.420 --> 00:01:55.510
They're essentially asking
you for the antiderivative.

00:01:55.510 --> 00:01:58.820
And so you could say, well, 2x
is the derivative of x squared.

00:02:01.780 --> 00:02:05.085
But you could also say 2x is the
derivative of x squared plus 1.

00:02:05.085 --> 00:02:07.460
You could also say that 2x is
the derivative of x squared

00:02:07.460 --> 00:02:09.769
plus pi, I think you
get the general idea.

00:02:09.769 --> 00:02:12.060
So if you wanted to write it
in the most general sense,

00:02:12.060 --> 00:02:14.070
you would write that
2x is the derivative

00:02:14.070 --> 00:02:17.904
of x squared plus some constant.

00:02:17.904 --> 00:02:19.320
So this is what
you would consider

00:02:19.320 --> 00:02:22.710
the antiderivative of 2x.

00:02:22.710 --> 00:02:24.860
Now, that's all
nice, but this is

00:02:24.860 --> 00:02:27.950
kind of clumsy to have to write
a sentence like this, so let's

00:02:27.950 --> 00:02:31.150
come up with some notation
for the antiderivative.

00:02:31.150 --> 00:02:33.300
And the convention
here is to use

00:02:33.300 --> 00:02:35.530
kind of a strange
looking notation, is

00:02:35.530 --> 00:02:41.390
to use a big elongated s
looking thing like that,

00:02:41.390 --> 00:02:44.080
and a dx around the
function that you're

00:02:44.080 --> 00:02:46.690
trying to take the
antiderivative of.

00:02:46.690 --> 00:02:49.290
So in this case, it would
look something like this.

00:02:49.290 --> 00:02:52.560
This is just saying this is
equal to the antiderivative

00:02:52.560 --> 00:02:56.770
of 2x, and the antiderivative
of 2x, we have already seen,

00:02:56.770 --> 00:03:01.660
is x squared plus c.

00:03:01.660 --> 00:03:03.780
Now, you might be
saying, why do we

00:03:03.780 --> 00:03:07.240
use this type of crazy notation.

00:03:07.240 --> 00:03:10.120
It'll become more obvious when
we study the definite integral

00:03:10.120 --> 00:03:14.500
and areas under curves and
taking sums of rectangles

00:03:14.500 --> 00:03:17.220
in order to approximate
the area of the curve.

00:03:17.220 --> 00:03:19.050
Here, it really
should just be viewed

00:03:19.050 --> 00:03:22.140
as a notation for
antiderivative.

00:03:22.140 --> 00:03:25.890
And this notation right over
here, this whole expression,

00:03:25.890 --> 00:03:35.060
is called the
indefinite integral

00:03:35.060 --> 00:03:38.030
of 2x, which is another
way of just saying

00:03:38.030 --> 00:03:42.010
the antiderivative of 2x.


Area between a curve and the x-axis: negative area

WEBVTT
Kind: captions
Language: en

00:00:00.680 --> 00:00:05.800
What I've got here is the graph
of y is equal to cosine of x.

00:00:05.800 --> 00:00:07.980
What I want to do is
figure out the area

00:00:07.980 --> 00:00:13.010
under the curve y is equal to
f of x, and above the x-axis.

00:00:13.010 --> 00:00:14.850
I'm going to do it
over various intervals.

00:00:14.850 --> 00:00:18.970
So first, let's think about
the area under the curve,

00:00:18.970 --> 00:00:24.715
between x is equal to 0 and
x is equal to pi over 2.

00:00:28.600 --> 00:00:33.024
So we're talking about
this area right over here.

00:00:33.024 --> 00:00:35.440
Well, the way we denote it is
the definite integral from 0

00:00:35.440 --> 00:00:39.800
to pi over 2 of cosine of x dx.

00:00:39.800 --> 00:00:42.320
And remember, all
this is, is kind of,

00:00:42.320 --> 00:00:45.270
it's reminiscent of
taking a sum of a bunch

00:00:45.270 --> 00:00:49.060
of super thin rectangles
with width dx and height

00:00:49.060 --> 00:00:51.751
f of x for each of
those rectangles.

00:00:51.751 --> 00:00:53.250
And then you take
an infinite number

00:00:53.250 --> 00:00:55.390
of those infinitely
thin rectangles.

00:00:55.390 --> 00:00:59.000
And that's kind of what this
notation is trying to depict.

00:00:59.000 --> 00:01:00.490
But we know how to
do this already.

00:01:00.490 --> 00:01:03.360
The second fundamental
theorem of calculus helps us.

00:01:03.360 --> 00:01:06.900
We just have to figure out what
the antiderivative of cosine

00:01:06.900 --> 00:01:08.180
of x is.

00:01:08.180 --> 00:01:11.090
Or what an antiderivative
of cosine of x is.

00:01:11.090 --> 00:01:13.150
Evaluate it at pi
over 2, and from that

00:01:13.150 --> 00:01:15.750
subtract it evaluated at 0.

00:01:15.750 --> 00:01:18.110
So what's the antiderivative
of cosine of x?

00:01:18.110 --> 00:01:20.286
Or what's an antiderivative?

00:01:20.286 --> 00:01:22.160
Well, we know if we take
the derivative-- let

00:01:22.160 --> 00:01:23.900
me write this up here.

00:01:23.900 --> 00:01:29.920
We know that if we take the
derivative of sine of x,

00:01:29.920 --> 00:01:32.250
we get cosine of x.

00:01:32.250 --> 00:01:35.520
So the antiderivative of
cosine of x is sine of x

00:01:35.520 --> 00:01:38.550
Now, why do I keep saying sine
of x is an antiderivative?

00:01:38.550 --> 00:01:40.820
It's not just the
antiderivative.

00:01:40.820 --> 00:01:44.110
Well, I could also
take the derivative

00:01:44.110 --> 00:01:48.490
of sine of x plus any
arbitrary constant,

00:01:48.490 --> 00:01:50.780
and still get cosine of x.

00:01:50.780 --> 00:01:52.530
Because the derivative
of a constant is 0.

00:01:52.530 --> 00:01:55.040
This could be pi, this could
be 5, this could be a million,

00:01:55.040 --> 00:01:57.730
this could be a googol, this
could be any crazy number.

00:01:57.730 --> 00:02:01.180
But the derivative of this is
still going to be cosine x.

00:02:01.180 --> 00:02:04.840
So when I say that we just have
to find an antiderivative, I'm

00:02:04.840 --> 00:02:07.810
just saying, look, we just have
to find one of the derivatives.

00:02:07.810 --> 00:02:10.449
Sine of x is probably the
simplest, because in this case

00:02:10.449 --> 00:02:12.800
the constant is 0.

00:02:12.800 --> 00:02:14.120
So let's evaluate.

00:02:14.120 --> 00:02:16.530
So one way that we
can denote this,

00:02:16.530 --> 00:02:19.590
the antiderivative cosine, or
an antiderivative of cosine

00:02:19.590 --> 00:02:21.900
of x is sine of x.

00:02:21.900 --> 00:02:24.620
And we're going to
evaluate at pi over 2.

00:02:24.620 --> 00:02:27.780
And from that, subtract
and evaluate it at 0.

00:02:27.780 --> 00:02:30.620
So this is going to
be equal to-- let

00:02:30.620 --> 00:02:32.720
me do it right over here.

00:02:32.720 --> 00:02:39.530
This is going to be equal
to sine of pi over 2 minus

00:02:39.530 --> 00:02:44.210
sine of 0, which is equal
to sine of pi over 2 is 1.

00:02:44.210 --> 00:02:45.280
Sine of 0 is 0.

00:02:45.280 --> 00:02:47.630
So it's 1 minus 0 is equal to 1.

00:02:47.630 --> 00:02:51.290
So the area of this region
right over here, this area

00:02:51.290 --> 00:02:55.310
is equal to 1.

00:02:55.310 --> 00:02:57.620
Now let's do
something interesting.

00:02:57.620 --> 00:02:59.970
Let's think about the area.

00:02:59.970 --> 00:03:06.770
Let's think about the area
under the curve between,

00:03:06.770 --> 00:03:10.970
let's say pi over
2 and 3 pi over 2.

00:03:19.430 --> 00:03:21.870
So between here and here.

00:03:21.870 --> 00:03:24.460
So we're talking
about this area.

00:03:24.460 --> 00:03:26.860
We're talking about that
area right over here.

00:03:26.860 --> 00:03:29.760
This is 3 pi over 2.

00:03:29.760 --> 00:03:31.980
So once again, the
way we denote the area

00:03:31.980 --> 00:03:35.670
is the definite integral from pi
over 2 to 3 pi over 2 of cosine

00:03:35.670 --> 00:03:37.600
of x dx.

00:03:37.600 --> 00:03:40.300
The antiderivative-- or an
antiderivative of cosine of x

00:03:40.300 --> 00:03:42.060
is sine of x.

00:03:42.060 --> 00:03:46.490
Evaluated at 3 pi
over 2 and pi over 2.

00:03:46.490 --> 00:03:48.770
So this is going
to be equal to sine

00:03:48.770 --> 00:03:55.330
of 3 pi over 2 minus
sine of pi over 2.

00:03:55.330 --> 00:03:57.480
What's sine of 3 pi over 2?

00:03:57.480 --> 00:04:02.950
If we visualize the unit
circle really fast, 3 pi over 2

00:04:02.950 --> 00:04:06.270
is going all the way 3/4
around the unit circle.

00:04:06.270 --> 00:04:07.490
So it's right over there.

00:04:07.490 --> 00:04:11.340
So sine is the y-coordinate
on that unit circle.

00:04:11.340 --> 00:04:12.400
So it's negative 1.

00:04:12.400 --> 00:04:15.580
So this right over
here is negative 1.

00:04:15.580 --> 00:04:18.279
This right over here,
sine of pi over 2

00:04:18.279 --> 00:04:20.620
is just going
straight up like that.

00:04:20.620 --> 00:04:22.852
So sine of pi over 2 is 1.

00:04:22.852 --> 00:04:23.810
So this is interesting.

00:04:23.810 --> 00:04:27.840
We get negative 1 minus 1,
which is equal to negative 2.

00:04:27.840 --> 00:04:29.760
We've got a negative area here.

00:04:29.760 --> 00:04:32.120
We got a negative area.

00:04:32.120 --> 00:04:33.650
Now how does that make sense?

00:04:33.650 --> 00:04:36.450
We know in the real world,
areas are always positive.

00:04:36.450 --> 00:04:39.780
But what is negative 2
really trying to depict?

00:04:39.780 --> 00:04:42.730
Well, it's trying to
sign it based on the idea

00:04:42.730 --> 00:04:46.110
that now our function
is below the x-axis.

00:04:46.110 --> 00:04:48.310
So we could kind of think
we have an area of 2,

00:04:48.310 --> 00:04:50.570
but it's all below the
x-axis in this case.

00:04:50.570 --> 00:04:53.110
And so it is signed
as negative 2.

00:04:53.110 --> 00:04:56.600
The actual area is 2, but
since it's below the x-axis,

00:04:56.600 --> 00:04:59.160
we get a negative
right over here.

00:04:59.160 --> 00:05:01.220
Now let's do one
more interesting one.

00:05:01.220 --> 00:05:04.890
Let's find the definite
integral from 0

00:05:04.890 --> 00:05:11.310
to 3 pi over 2 of
cosine of x dx.

00:05:11.310 --> 00:05:14.940
Now this is just denoting
this entire area.

00:05:14.940 --> 00:05:19.590
Going from 0 all
the way 3 pi over 2.

00:05:19.590 --> 00:05:22.190
And what do you think
is going to happen?

00:05:22.190 --> 00:05:23.520
Well, let's evaluate it.

00:05:23.520 --> 00:05:30.710
It's going to be sine of 3
pi over 2 minus sine of 0.

00:05:33.950 --> 00:05:37.560
Which is equal to
negative 1 minus 0,

00:05:37.560 --> 00:05:39.890
which is equal to negative 1.

00:05:39.890 --> 00:05:41.300
So what just happened here?

00:05:41.300 --> 00:05:44.450
The area under, all this orange
area that I just outlined,

00:05:44.450 --> 00:05:50.150
is clearly not negative, or any
area isn't a negative number.

00:05:50.150 --> 00:05:51.840
And the area isn't even 1.

00:05:51.840 --> 00:05:53.320
But what just happened here?

00:05:53.320 --> 00:06:00.370
Well, we saw in the first case
that this first area was 1.

00:06:00.370 --> 00:06:03.780
The area of this first
region right over here is 1.

00:06:03.780 --> 00:06:06.610
And then the area of the second
region, we got negative 2.

00:06:11.370 --> 00:06:13.250
And so one way to
interpret it is

00:06:13.250 --> 00:06:19.400
that your net area above
the x-axis is negative 1,

00:06:19.400 --> 00:06:22.880
or another way to say it,
your net area is negative 1.

00:06:22.880 --> 00:06:27.530
So it's taking the 1
region above the x-axis

00:06:27.530 --> 00:06:29.950
and subtracting the 2 below it.

00:06:29.950 --> 00:06:31.560
So what the definite
integral is doing

00:06:31.560 --> 00:06:34.560
when we evaluate it using the
second fundamental theorem

00:06:34.560 --> 00:06:36.760
of calculus, it's
essentially finding

00:06:36.760 --> 00:06:40.629
the net area above the x-axis.

00:06:40.629 --> 00:06:42.170
And if we get a
negative number, that

00:06:42.170 --> 00:06:44.920
means that the net area is act--
that actually most of the area

00:06:44.920 --> 00:06:46.680
is below the x-axis.

00:06:46.680 --> 00:06:50.116
If we get 0, then that
means it all nets out.

00:06:50.116 --> 00:06:51.740
And if you want to
see a case that's 0,

00:06:51.740 --> 00:06:58.570
take the integral from 0 all
the way all the way 2 pi.

00:06:58.570 --> 00:07:01.100
And this will evaluate to 0
because you have an area of 1

00:07:01.100 --> 00:07:03.380
and another area of
1, but it nets out

00:07:03.380 --> 00:07:04.800
with this area of negative 2.

00:07:04.800 --> 00:07:06.070
Let's try that out.

00:07:06.070 --> 00:07:10.460
So if we go from 0 to
2 pi of cosine of x dx,

00:07:10.460 --> 00:07:15.970
this is going to be equal to
sine of 2 pi minus sine of 0,

00:07:15.970 --> 00:07:19.960
which is equal to 0 minus 0,
which is indeed equal to 0.

00:07:19.960 --> 00:07:21.680
Now clearly there was
area here, but all

00:07:21.680 --> 00:07:23.560
of the area that
was above the x-axis

00:07:23.560 --> 00:07:27.134
netted out with all the area
that was below the x-axis.