Planar motion (with integrals)

WEBVTT
Kind: captions
Language: en

00:00:00.777 --> 00:00:01.831
- [Instructor] A particle moving

00:00:01.831 --> 00:00:04.273
in the XY plane has velocity vector given

00:00:04.273 --> 00:00:07.031
by V of T is equal to
all of this business,

00:00:07.031 --> 00:00:09.261
and so using this notation, it just means

00:00:09.261 --> 00:00:12.498
that the X component of
velocity as a function

00:00:12.498 --> 00:00:14.368
of the time is one over T plus seven,

00:00:14.368 --> 00:00:16.023
and the Y component of velocity

00:00:16.023 --> 00:00:18.023
as a function of time is T to the fourth

00:00:18.023 --> 00:00:20.743
for time T greater than or equal to zero.

00:00:20.743 --> 00:00:22.703
At T equals one, the particle is

00:00:22.703 --> 00:00:25.683
at the point three comma four.

00:00:25.683 --> 00:00:28.393
So the first part is what is the magnitude

00:00:28.393 --> 00:00:30.549
of the displacement of the particle

00:00:30.549 --> 00:00:34.189
between time T equals
one and T equals three?

00:00:34.189 --> 00:00:35.860
And then we need to
figure out its position,

00:00:35.860 --> 00:00:37.116
and we need to round to the nearest 10th.

00:00:37.116 --> 00:00:38.574
So like always, pause this video,

00:00:38.574 --> 00:00:40.391
and I think you'll have
to use a calculator,

00:00:40.391 --> 00:00:43.967
but pause this video and try
to work through it on your own.

00:00:43.967 --> 00:00:47.354
So we've done questions
like this in one dimension,

00:00:47.354 --> 00:00:48.870
but now we're doing it in two dimensions,

00:00:48.870 --> 00:00:50.744
but the key is is to just break it up

00:00:50.744 --> 00:00:52.390
into the component dimensions.

00:00:52.390 --> 00:00:54.598
So what we really want to do
is let's find the displacement

00:00:54.598 --> 00:00:57.785
in the X direction, so
really just the change in X,

00:00:57.785 --> 00:00:59.199
and then let's just find the displacement

00:00:59.199 --> 00:01:01.653
in the vertical direction
or our change in Y,

00:01:01.653 --> 00:01:02.851
and then we can use those,

00:01:02.851 --> 00:01:04.993
essentially using the Pythagorean theorem,

00:01:04.993 --> 00:01:08.011
to find that the magnitude
of the total displacement,

00:01:08.011 --> 00:01:10.705
and also, if we know the
change in X and change in Y,

00:01:10.705 --> 00:01:12.704
we just add the change in X to the three,

00:01:12.704 --> 00:01:14.283
and we add the change in Y to the four

00:01:14.283 --> 00:01:16.102
to find the particle's position

00:01:16.102 --> 00:01:18.019
at time T equals three.

00:01:18.992 --> 00:01:20.010
So let's figure it out.

00:01:20.010 --> 00:01:24.494
So change in X from T equals
one to T equals three,

00:01:24.494 --> 00:01:26.638
well, that's just gonna be the integral

00:01:26.638 --> 00:01:29.644
of the rate function in the X direction

00:01:29.644 --> 00:01:32.511
from time equal one to time equals three,

00:01:32.511 --> 00:01:35.708
so in the X direction, we
have one over T plus seven.

00:01:35.708 --> 00:01:38.605
That's our X velocity
as a function of time,

00:01:38.605 --> 00:01:40.688
one over T plus seven DT,

00:01:42.664 --> 00:01:44.774
and what is this going to be equal to?

00:01:44.774 --> 00:01:47.236
Well, you do might want
to do u-substitution

00:01:47.236 --> 00:01:49.122
if you're unfamiliar,
but you might recognize

00:01:49.122 --> 00:01:52.558
that the derivative of T
plus seven is just one,

00:01:52.558 --> 00:01:53.391
so you could think of this

00:01:53.391 --> 00:01:55.768
as one times one over T plus seven,

00:01:55.768 --> 00:01:58.250
and so we really can just
take the antiderivative then

00:01:58.250 --> 00:02:00.536
with respect to T plus seven,

00:02:00.536 --> 00:02:03.508
so you get the natural
log of the absolute value

00:02:03.508 --> 00:02:04.841
of T plus seven,

00:02:06.346 --> 00:02:09.943
and we are going to evaluate that at three

00:02:09.943 --> 00:02:11.528
and then subtract from that.

00:02:11.528 --> 00:02:13.259
It evaluated at one,

00:02:13.259 --> 00:02:14.580
so this is going to be the natural log

00:02:14.580 --> 00:02:16.184
of the absolute value of 10,

00:02:16.184 --> 00:02:19.876
which is just the natural log of 10

00:02:19.876 --> 00:02:23.020
minus the natural log of
the absolute value of eight,

00:02:23.020 --> 00:02:24.819
which is just the natural log of eight,

00:02:24.819 --> 00:02:28.893
which is equal to the
natural log of 10 over eight,

00:02:28.893 --> 00:02:30.659
just using our logarithm properties,

00:02:30.659 --> 00:02:34.903
which is equal to the natural log of 1.25,

00:02:34.903 --> 00:02:36.001
so I could get my calculator out

00:02:36.001 --> 00:02:37.921
in a second to calculate that.

00:02:37.921 --> 00:02:38.842
Actually, let's just...

00:02:38.842 --> 00:02:40.286
Well, I'll do that in a second,

00:02:40.286 --> 00:02:42.572
and then, let's figure
out our change in Y.

00:02:42.572 --> 00:02:44.784
Our change in Y, once again,

00:02:44.784 --> 00:02:47.458
we're gonna take the
integral from one to three.

00:02:47.458 --> 00:02:51.985
That's the time over which
we're thinking about the change,

00:02:51.985 --> 00:02:55.067
and then what is the Y
component of our velocity?

00:02:55.067 --> 00:02:58.082
Well, it's T to the fourth DT.

00:02:58.082 --> 00:02:59.925
Well, this is going to be...

00:02:59.925 --> 00:03:03.778
Take the reverse power rule
T to the fifth over five

00:03:03.778 --> 00:03:05.196
at three and one,

00:03:05.196 --> 00:03:09.628
so this is three to the fifth
over five is 243 over five

00:03:09.628 --> 00:03:13.048
minus one to the fifth
over five minus 1/5,

00:03:13.048 --> 00:03:14.435
so this is equal

00:03:14.435 --> 00:03:15.852
to 242 over five,

00:03:17.737 --> 00:03:20.198
which is what, 48.4?

00:03:20.198 --> 00:03:21.031
48.4.

00:03:22.667 --> 00:03:24.414
Now let me get my calculator out

00:03:24.414 --> 00:03:26.831
for this natural log of 1.25.

00:03:27.667 --> 00:03:28.500
1.25,

00:03:30.455 --> 00:03:31.745
natural log,

00:03:31.745 --> 00:03:33.269
and we'll just round
to two decimal places,

00:03:33.269 --> 00:03:34.852
approximately 0.22,

00:03:35.801 --> 00:03:38.301
so this is approximately 0.22.

00:03:39.930 --> 00:03:42.419
So I figured out our change
in X and our change in Y,

00:03:42.419 --> 00:03:43.992
and actually, just from that,

00:03:43.992 --> 00:03:46.189
we can answer the second
part of the question first.

00:03:46.189 --> 00:03:48.653
What is the particle's
position at T equals three?

00:03:48.653 --> 00:03:52.066
Well, it's going to be our
position at T equals one,

00:03:52.066 --> 00:03:53.777
where to each of the components

00:03:53.777 --> 00:03:55.957
we add the respective change,

00:03:55.957 --> 00:03:57.161
so we would add...

00:03:57.161 --> 00:04:00.355
So this would be three plus our change

00:04:00.355 --> 00:04:02.248
in X from T equals one

00:04:02.248 --> 00:04:03.947
to T equals three,

00:04:03.947 --> 00:04:07.144
and it would be four plus our change in Y.

00:04:07.144 --> 00:04:10.448
So this is going to be equal
to three plus our change in X.

00:04:10.448 --> 00:04:13.213
Well, that's going to
be approximately 3.22,

00:04:13.213 --> 00:04:18.093
and four plus our change
in Y, that is, what, 52.4.

00:04:18.093 --> 00:04:20.510
This right over here is 52.4,

00:04:23.051 --> 00:04:25.453
but we still have to
answer the first question.

00:04:25.453 --> 00:04:27.381
What is the magnitude of the displacement?

00:04:27.381 --> 00:04:28.818
Well, it's the Pythagorean theorem.

00:04:28.818 --> 00:04:30.725
I'll draw a very rough
sketch of what's going on.

00:04:30.725 --> 00:04:33.793
Sometimes it's useful to visualize it.

00:04:33.793 --> 00:04:37.246
So our initial position
is at three comma four,

00:04:37.246 --> 00:04:38.913
so three comma four,

00:04:40.613 --> 00:04:42.365
so we're right over there,

00:04:42.365 --> 00:04:45.260
and we figured out our
change in X isn't much.

00:04:45.260 --> 00:04:47.741
So our change in X is a positives 0.22,

00:04:47.741 --> 00:04:51.009
so our change in X, we're
barely moving in that direction,

00:04:51.009 --> 00:04:53.996
and our change in Y is 48.4,

00:04:53.996 --> 00:04:56.757
so we have a dramatic change.

00:04:56.757 --> 00:05:00.675
It really goes off the charts
over here in that direction,

00:05:00.675 --> 00:05:03.429
but if we wanted to add them together,

00:05:03.429 --> 00:05:04.986
if we want to add those vectors together,

00:05:04.986 --> 00:05:08.957
you could shift over your
change in Y right over here

00:05:08.957 --> 00:05:11.139
and then find the hypotenuse.

00:05:11.139 --> 00:05:13.748
The length of the hypotenuse
would be the magnitude

00:05:13.748 --> 00:05:15.939
of the entire displacement,

00:05:15.939 --> 00:05:18.443
and so let's do that.

00:05:18.443 --> 00:05:19.961
So the magnitude

00:05:19.961 --> 00:05:23.409
of the displacement is
going to be the square root

00:05:23.409 --> 00:05:25.576
of our change in X squared

00:05:27.711 --> 00:05:31.742
plus our change in Y squared once again.

00:05:31.742 --> 00:05:33.509
This is just the Pythagorean theorem,

00:05:33.509 --> 00:05:35.105
and what is this going to be?

00:05:35.105 --> 00:05:38.044
I'll get my calculator out again for this.

00:05:38.044 --> 00:05:40.036
Okay, that was our change in X.

00:05:40.036 --> 00:05:41.453
Let me square it,

00:05:42.517 --> 00:05:45.850
and then plus we are going to have 48.4,

00:05:47.164 --> 00:05:48.914
8.4, squared is equal

00:05:51.572 --> 00:05:53.199
to this right over here,

00:05:53.199 --> 00:05:55.403
and then we take the square root of that,

00:05:55.403 --> 00:05:57.656
so this is going to be...

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See if we...

00:05:59.263 --> 00:06:02.278
Right over there, and there you go.

00:06:02.278 --> 00:06:05.130
The magnitude of our
total displacement is 48,

00:06:05.130 --> 00:06:08.766
if we round to the nearest 10th, 48.4.

00:06:08.766 --> 00:06:11.266
So this is approximately 48.4,

00:06:14.081 --> 00:06:15.306
and we're done.

00:06:15.306 --> 00:06:17.866
Now, one thing that you might be noting is

00:06:17.866 --> 00:06:20.809
it looks like our total
displacement, 48.4, is the same

00:06:20.809 --> 00:06:23.224
as our change in Y.

00:06:23.224 --> 00:06:25.891
Now, the reason why it
came out this way is

00:06:25.891 --> 00:06:28.937
because our change in Y was exactly 48.4

00:06:28.937 --> 00:06:29.770
while the magnitude

00:06:29.770 --> 00:06:32.459
of our displacement was
slightly more than 48.4,

00:06:32.459 --> 00:06:35.842
but when we round to the
nearest 10th, we got to 48.4.

00:06:35.842 --> 00:06:38.355
The reason why they are so
close is because our change

00:06:38.355 --> 00:06:39.855
in X was so small.

00:06:40.904 --> 00:06:43.774
We're talking about
0.22 as the change in X,

00:06:43.774 --> 00:06:46.237
and our change in Y was so much

00:06:46.237 --> 00:06:49.558
that our hypotenuse was
only slightly longer

00:06:49.558 --> 00:06:53.684
than our change in Y, so
that's why we got this result

00:06:53.684 --> 00:06:56.060
for this particular instance.

00:06:56.060 --> 00:06:58.367
In general, you're going
to see the magnitude

00:06:58.367 --> 00:07:01.380
of the displacement is going to be larger

00:07:01.380 --> 00:07:05.223
than the magnitude of either our change

00:07:05.223 --> 00:07:07.336
in X or our change in Y alone.


Approximating limits using tables

WEBVTT
Kind: captions
Language: en

00:00:00.145 --> 00:00:01.748
- This video we're going to try to get

00:00:01.748 --> 00:00:04.058
a sense of what the limit
as x approaches three

00:00:04.058 --> 00:00:06.463
of x to the third minus three x squared

00:00:06.463 --> 00:00:09.166
over five x minus 15

00:00:09.166 --> 00:00:10.253
is.

00:00:10.253 --> 00:00:11.799
Now when I say get a sense,

00:00:11.799 --> 00:00:14.843
we're gonna do that by seeing what

00:00:14.843 --> 00:00:17.882
values for this expression
we get as x gets

00:00:17.882 --> 00:00:20.289
closer and closer to three.

00:00:20.289 --> 00:00:22.113
Now one thing that you
might wanna try out is,

00:00:22.113 --> 00:00:24.412
well what happens to this expression

00:00:24.412 --> 00:00:26.566
when x is equal to three?

00:00:26.566 --> 00:00:29.184
Well then it's going to be
three to the third power

00:00:29.184 --> 00:00:32.316
minus three times three squared,

00:00:32.316 --> 00:00:34.855
over five times three

00:00:34.855 --> 00:00:35.688
minus 15.

00:00:36.542 --> 00:00:39.264
So at x equals three, this
expression's gonna be,

00:00:39.264 --> 00:00:42.689
let's see in the numerator
you have 27 minus 27, zero.

00:00:42.689 --> 00:00:45.601
Over 15 minus 15, over zero.

00:00:45.601 --> 00:00:48.753
So this expression is actually not defined

00:00:48.753 --> 00:00:50.164
at x equals three.

00:00:50.164 --> 00:00:51.516
We get this indeterminate form,

00:00:51.516 --> 00:00:54.026
we get zero over zero.

00:00:54.026 --> 00:00:55.652
But let's see, even though the function,

00:00:55.652 --> 00:00:57.785
or even though the
expression is not defined,

00:00:57.785 --> 00:01:01.930
let's see if we can get a sense
of what the limit might be.

00:01:01.930 --> 00:01:05.210
And to do that, I'm gonna set up a table.

00:01:05.210 --> 00:01:07.710
So let me set up a table here.

00:01:08.818 --> 00:01:11.190
And actually I wanna set up two tables.

00:01:11.190 --> 00:01:14.940
So this is x and this is x to the third

00:01:16.323 --> 00:01:18.735
minus three x squared

00:01:18.735 --> 00:01:19.943
over

00:01:19.943 --> 00:01:21.467
five x

00:01:21.467 --> 00:01:22.300
minus

00:01:22.300 --> 00:01:23.430
15.

00:01:23.430 --> 00:01:24.820
And actually, I'm gonna do that again.

00:01:24.820 --> 00:01:27.158
And I'll tell you why in a second.

00:01:27.158 --> 00:01:31.424
So this is gonna be x and
this is x to the third

00:01:31.424 --> 00:01:34.001
minus three x squared

00:01:34.001 --> 00:01:34.859
over

00:01:34.859 --> 00:01:37.272
five x minus 15.

00:01:37.272 --> 00:01:38.440
The reason why I set up two tables,

00:01:38.440 --> 00:01:39.547
I didn't have to do two tables,

00:01:39.547 --> 00:01:40.848
I could have done it all in one table,

00:01:40.848 --> 00:01:42.268
but hopefully this will make it

00:01:42.268 --> 00:01:44.578
a little bit more intuitive
what I'm trying to do.

00:01:44.578 --> 00:01:48.325
Is on this left table, I'm
gonna, let's try out x values

00:01:48.325 --> 00:01:52.072
that get closer and closer
to three from the left.

00:01:52.072 --> 00:01:54.163
From values that are less than three.

00:01:54.163 --> 00:01:56.635
So for example, you can
go to two point nine

00:01:56.635 --> 00:01:58.484
and figure out what the expression equals

00:01:58.484 --> 00:01:59.824
when x is two point nine.

00:01:59.824 --> 00:02:01.296
But then we can try to get
even a little bit closer

00:02:01.296 --> 00:02:05.726
than that, we could go
to two point nine nine.

00:02:05.726 --> 00:02:06.997
And then we could go
even closer than that.

00:02:06.997 --> 00:02:10.330
We could go to two point nine nine nine.

00:02:11.475 --> 00:02:13.255
And so one way to think about it here

00:02:13.255 --> 00:02:16.241
is as we try to figure out
what this expression equals

00:02:16.241 --> 00:02:18.455
as we get closer and closer to three,

00:02:18.455 --> 00:02:22.215
we're trying to approximate
the limit from the left.

00:02:22.215 --> 00:02:23.048
So limit

00:02:24.621 --> 00:02:25.454
from

00:02:26.490 --> 00:02:27.323
the left.

00:02:28.369 --> 00:02:29.738
Now why do I say the left?

00:02:29.738 --> 00:02:31.973
Well if you think about
this on a coordinate plane,

00:02:31.973 --> 00:02:34.292
these are the x values that
are to the left of three,

00:02:34.292 --> 00:02:36.141
but we're getting closer
and closer and closer.

00:02:36.141 --> 00:02:37.432
We're moving to the right,

00:02:37.432 --> 00:02:39.038
but these are the x values that

00:02:39.038 --> 00:02:42.429
are on the left side of three,
they're less than three.

00:02:42.429 --> 00:02:44.828
But we also, in order
for the limit to exist,

00:02:44.828 --> 00:02:47.693
we have to be approaching the
same thing from both sides.

00:02:47.693 --> 00:02:49.420
From both the left and the right.

00:02:49.420 --> 00:02:51.045
So we could also try to approximate

00:02:51.045 --> 00:02:52.212
the limit from

00:02:53.775 --> 00:02:55.159
the right.

00:02:55.159 --> 00:02:56.926
And so what values would those be?

00:02:56.926 --> 00:02:59.030
Well those would be,
those would be x values

00:02:59.030 --> 00:03:00.656
larger than three.

00:03:00.656 --> 00:03:02.573
So we could say three point one,

00:03:02.573 --> 00:03:03.927
but then we might wanna
get a little bit closer,

00:03:03.927 --> 00:03:05.776
we could go three point zero one.

00:03:05.776 --> 00:03:07.735
But then we might wanna
get even closer to three.

00:03:07.735 --> 00:03:09.953
Three point zero zero

00:03:09.953 --> 00:03:10.797
one.

00:03:10.797 --> 00:03:12.431
And every time we get
closer and closer to three,

00:03:12.431 --> 00:03:15.226
we're gonna get a better
approximation for,

00:03:15.226 --> 00:03:16.872
or we're gonna get a better sense

00:03:16.872 --> 00:03:19.920
of what we are actually approaching.

00:03:19.920 --> 00:03:21.403
So let's get a calculator out and do this.

00:03:21.403 --> 00:03:23.870
And you could keep going,
two nine nine nine nine nine.

00:03:23.870 --> 00:03:26.633
Three point zero zero zero zero one.

00:03:26.633 --> 00:03:28.929
Now one key idea here to point out,

00:03:28.929 --> 00:03:31.062
before I even calculate
what these are going to be,

00:03:31.062 --> 00:03:33.479
sometimes when people say
the limit from both sides,

00:03:33.479 --> 00:03:35.796
or the limit from the left
or the limit from the right,

00:03:35.796 --> 00:03:37.523
they imagine that the limit from the left

00:03:37.523 --> 00:03:39.586
is negative values and
the limit from the right

00:03:39.586 --> 00:03:40.996
are positive values.

00:03:40.996 --> 00:03:43.203
But as you can see here,
the limit from the left

00:03:43.203 --> 00:03:46.645
are to the left of the x
value that you're trying

00:03:46.645 --> 00:03:48.454
to find the limit at.

00:03:48.454 --> 00:03:50.172
So these aren't negative values,

00:03:50.172 --> 00:03:52.477
these are just approaching the three

00:03:52.477 --> 00:03:54.995
right over here from
values less than three.

00:03:54.995 --> 00:03:56.418
This is approaching the three

00:03:56.418 --> 00:03:58.918
from values larger than three.

00:04:00.015 --> 00:04:01.855
So now let's fill out this table,

00:04:01.855 --> 00:04:04.261
and I'm speeding up my
work so that you don't

00:04:04.261 --> 00:04:06.162
have to sit through me typing everything

00:04:06.162 --> 00:04:07.662
into a calculator.

00:04:08.833 --> 00:04:10.795
So based on what we're seeing here,

00:04:10.795 --> 00:04:13.688
I would make the estimate that this looks

00:04:13.688 --> 00:04:16.484
like it's approaching one point eight.

00:04:16.484 --> 00:04:18.434
So is this equal to one point eight?

00:04:18.434 --> 00:04:20.168
As I said, in the future, we're gonna

00:04:20.168 --> 00:04:21.998
be able to find this out exactly.

00:04:21.998 --> 00:04:23.310
But if you're not sure about this

00:04:23.310 --> 00:04:26.724
you could try an even closer
and closer and closer value.


Analyzing motion problems: position

WEBVTT
Kind: captions
Language: en

00:00:01.007 --> 00:00:03.492
- [Instructor] Divya received
the following problem.

00:00:03.492 --> 00:00:05.419
A particle moves in a
straight line with velocity

00:00:05.419 --> 00:00:07.370
v of t is equal to the square root

00:00:07.370 --> 00:00:09.622
of three t minus one meters per second,

00:00:09.622 --> 00:00:11.781
where t is time in seconds.

00:00:11.781 --> 00:00:13.082
At t equals two,

00:00:13.082 --> 00:00:15.172
the particle's distance
from the starting point

00:00:15.172 --> 00:00:18.887
was eight meters in
the positive direction.

00:00:18.887 --> 00:00:22.741
What is the particle's position
at t equals seven seconds?

00:00:22.741 --> 00:00:25.690
Which expression should Divya
use to solve the problem?

00:00:25.690 --> 00:00:27.896
So pause this video and have a go at it.

00:00:27.896 --> 00:00:30.659
Alright now let's do this together.

00:00:30.659 --> 00:00:34.560
So we wanna know the particle's position

00:00:34.560 --> 00:00:36.813
at t is equal to seven.

00:00:36.813 --> 00:00:38.020
So what we could do,

00:00:38.020 --> 00:00:41.898
they tell us what our
position is at t equals two,

00:00:41.898 --> 00:00:44.173
so what the position at

00:00:44.173 --> 00:00:46.101
t equals seven would be your position

00:00:46.101 --> 00:00:47.768
at t equals two plus

00:00:49.584 --> 00:00:50.501
your change

00:00:52.091 --> 00:00:53.008
in position

00:00:55.551 --> 00:00:56.968
from t equals two

00:00:59.591 --> 00:01:01.508
to t is equal to seven.

00:01:02.656 --> 00:01:04.909
And there's another word for this.

00:01:04.909 --> 00:01:07.045
You could also call this your displacement

00:01:07.045 --> 00:01:10.045
from t equals two to t equals seven.

00:01:10.992 --> 00:01:13.709
And we know how to think
about displacement.

00:01:13.709 --> 00:01:17.889
Velocity is your rate of
change of displacement.

00:01:17.889 --> 00:01:20.071
And so if you wanna figure
out your displacement,

00:01:20.071 --> 00:01:22.463
your between two times,

00:01:22.463 --> 00:01:24.785
you would integrate the velocity function.

00:01:24.785 --> 00:01:26.527
So this is going to be the integral

00:01:26.527 --> 00:01:29.290
from t equals two to t equals seven

00:01:29.290 --> 00:01:33.167
of our velocity function, v of t, dt.

00:01:33.167 --> 00:01:36.232
This would be our displacement
from time two to time seven.

00:01:36.232 --> 00:01:37.765
So if they said what is
our change in position

00:01:37.765 --> 00:01:39.228
from time two to time seven,

00:01:39.228 --> 00:01:41.271
it would be just this expression.

00:01:41.271 --> 00:01:42.363
But they want us,

00:01:42.363 --> 00:01:43.361
they want us to figure out,

00:01:43.361 --> 00:01:44.731
or they want Divya to figure out,

00:01:44.731 --> 00:01:47.564
what is the particle's position
at t equals seven seconds.

00:01:47.564 --> 00:01:49.096
So what you'd wanna do is your position

00:01:49.096 --> 00:01:50.722
at t equals two,

00:01:50.722 --> 00:01:52.997
and we know what our
position at t equals two is.

00:01:52.997 --> 00:01:55.203
It was eight meters in
the positive direction,

00:01:55.203 --> 00:01:57.548
so we could just call that
positive eight meters.

00:01:57.548 --> 00:01:59.778
So it's going to be eight plus

00:01:59.778 --> 00:02:01.937
your change in position,

00:02:01.937 --> 00:02:03.655
which is going to be your displacement.

00:02:03.655 --> 00:02:05.954
And let's see we can see this choice

00:02:05.954 --> 00:02:08.137
right over there,

00:02:08.137 --> 00:02:09.646
and that's what we would pick.

00:02:09.646 --> 00:02:10.853
This first option,

00:02:10.853 --> 00:02:13.222
v of seven, that just gives us our

00:02:13.222 --> 00:02:15.660
velocity at time seven,

00:02:15.660 --> 00:02:18.818
or at the exactly at seven seconds.

00:02:18.818 --> 00:02:22.951
Or another way, our rate
of change of displacement

00:02:22.951 --> 00:02:24.925
at seven seconds.

00:02:24.925 --> 00:02:26.341
So that's not what we want.

00:02:26.341 --> 00:02:27.572
This one right over here,

00:02:27.572 --> 00:02:29.731
you have your position

00:02:29.731 --> 00:02:31.612
at t equals two,

00:02:31.612 --> 00:02:33.191
but then you have your change in position

00:02:33.191 --> 00:02:36.651
from t equals zero to t equals seven.

00:02:36.651 --> 00:02:39.042
So this doesn't seem, this isn't right.

00:02:39.042 --> 00:02:42.224
And this is your position at time two

00:02:42.224 --> 00:02:43.724
plus your v prime,

00:02:45.126 --> 00:02:47.727
the derivative of velocity
is the acceleration,

00:02:47.727 --> 00:02:49.422
plus your acceleration at time seven,

00:02:49.422 --> 00:02:51.094
so that's definitely not gonna give you

00:02:51.094 --> 00:02:53.880
your the particle's position.

00:02:53.880 --> 00:02:54.958
So we like that second choice.


Analyzing motion problems: total distance traveled

WEBVTT
Kind: captions
Language: en

00:00:00.430 --> 00:00:03.039
- [Instructor] Alexey received
the following problem:

00:00:03.039 --> 00:00:05.028
A particle moves in a straight line

00:00:05.028 --> 00:00:08.038
with velocity v of t is
equal to negative t squared

00:00:08.038 --> 00:00:10.336
plus eight meters per second,

00:00:10.336 --> 00:00:12.915
where t is time in seconds.

00:00:12.915 --> 00:00:15.354
At t is equal to two,
the particle's distance

00:00:15.354 --> 00:00:18.118
from the starting point was five meters.

00:00:18.118 --> 00:00:21.670
What is the total distance
the particle has traveled

00:00:21.670 --> 00:00:25.467
between t equals two and
t equals six seconds?

00:00:25.467 --> 00:00:28.987
Which expression should Alexey
use to solve the problem?

00:00:28.987 --> 00:00:31.697
So we don't actually have to
figure the actual answer out,

00:00:31.697 --> 00:00:32.530
we just have to figure out

00:00:32.530 --> 00:00:34.615
what is the appropriate expression.

00:00:34.615 --> 00:00:36.119
So like always, pause this video

00:00:36.119 --> 00:00:38.232
and see if you can work
through it on your own.

00:00:38.232 --> 00:00:40.019
So now let's tackle this together.

00:00:40.019 --> 00:00:42.438
So the key question is
what is the total distance

00:00:42.438 --> 00:00:45.153
the particle has traveled
between t equals two

00:00:45.153 --> 00:00:46.971
and t equals six?

00:00:46.971 --> 00:00:49.697
So we just care what happens
between those points,

00:00:49.697 --> 00:00:51.432
we don't care that the particle's distance

00:00:51.432 --> 00:00:54.825
from the starting point was
five meters at t equals two.

00:00:54.825 --> 00:00:58.625
So this right over here is
actually unnecessary information.

00:00:58.625 --> 00:01:00.648
So the first thing that
you might wanna think about

00:01:00.648 --> 00:01:02.741
is well maybe distance
is just the integral

00:01:02.741 --> 00:01:03.907
of the velocity function;

00:01:03.907 --> 00:01:05.167
We've seen that multiple times.

00:01:05.167 --> 00:01:07.454
If you want to find the
change in a quantity,

00:01:07.454 --> 00:01:11.254
you just say the starting
time and the ending time

00:01:11.254 --> 00:01:14.411
and then you integrate the rate function.

00:01:14.411 --> 00:01:16.577
So wouldn't it just be that?

00:01:16.577 --> 00:01:18.759
Now we have to be very very careful.

00:01:18.759 --> 00:01:22.144
If the question was
what is the displacement

00:01:22.144 --> 00:01:25.302
for the particle between time equals two

00:01:25.302 --> 00:01:26.500
and time equals six,

00:01:26.500 --> 00:01:28.395
this would have been the correct answer.

00:01:28.395 --> 00:01:31.606
So this would be displacement.

00:01:31.606 --> 00:01:34.106
Displacement from t equals two

00:01:37.033 --> 00:01:38.783
to t is equal to six.

00:01:39.697 --> 00:01:40.937
But they're not saying displacement.

00:01:40.937 --> 00:01:45.028
They're saying total distance
the particle has traveled.

00:01:45.028 --> 00:01:49.326
So this is the total path
length for the particle.

00:01:49.326 --> 00:01:51.619
So one way to think about it,

00:01:51.619 --> 00:01:54.323
you would integrate not
the velocity function,

00:01:54.323 --> 00:01:56.430
if you integrate velocity,
you get displacement,

00:01:56.430 --> 00:02:00.392
instead, you would integrate
the speed function.

00:02:00.392 --> 00:02:01.376
Now what is speed?

00:02:01.376 --> 00:02:03.236
It is the magnitude of velocity

00:02:03.236 --> 00:02:04.740
and in one dimension, it would just be

00:02:04.740 --> 00:02:08.256
the absolute value of
your velocity function.

00:02:08.256 --> 00:02:10.110
And so the absolute value
of the velocity function,

00:02:10.110 --> 00:02:12.234
this would give you,
integrating the speed,

00:02:12.234 --> 00:02:14.887
this would give you the distance.

00:02:14.887 --> 00:02:17.054
Distance from t equals two

00:02:19.222 --> 00:02:21.477
to t is equal to six,

00:02:21.477 --> 00:02:24.577
and let's see, we have that
choice right over here.

00:02:24.577 --> 00:02:25.602
The displacement one here,

00:02:25.602 --> 00:02:27.390
this is an interesting distracter

00:02:27.390 --> 00:02:30.101
but that is not going to be the choice.

00:02:30.101 --> 00:02:33.546
This one right over here, v prime of six,

00:02:33.546 --> 00:02:36.213
that gives you the acceleration.

00:02:37.284 --> 00:02:39.574
If you're taking the derivative
of the velocity function,

00:02:39.574 --> 00:02:41.640
the acceleration at six seconds,

00:02:41.640 --> 00:02:43.266
that's not what we're interested in.

00:02:43.266 --> 00:02:47.939
And this gives you the absolute
difference in velocity,

00:02:47.939 --> 00:02:50.490
when in between time six and time two,

00:02:50.490 --> 00:02:52.845
that's not what we're
trying to figure out either.


Motion problems with integrals: displacement vs. distance

WEBVTT
Kind: captions
Language: en

00:00:01.495 --> 00:00:02.702
- [Instructor] What we're
going to do in this video

00:00:02.702 --> 00:00:04.723
is start thinking about the position

00:00:04.723 --> 00:00:08.299
of an object traveling in one dimension.

00:00:08.299 --> 00:00:10.226
And to get our bearings there,

00:00:10.226 --> 00:00:13.105
I'm going to introduce a few ideas.

00:00:13.105 --> 00:00:16.605
So the first idea is that of displacement.

00:00:17.912 --> 00:00:20.628
So you might use that
word in everyday language,

00:00:20.628 --> 00:00:24.135
and it literally means
your change in position,

00:00:24.135 --> 00:00:26.135
your change in position.

00:00:28.987 --> 00:00:31.263
Now a related idea that
sometimes get confused

00:00:31.263 --> 00:00:33.957
with displacement is a notion of

00:00:33.957 --> 00:00:35.457
distance traveled.

00:00:37.370 --> 00:00:39.135
You might say well doesn't that,

00:00:39.135 --> 00:00:41.549
isn't that just the same
thing as change in position?

00:00:41.549 --> 00:00:44.104
And you will see shortly, no, it isn't

00:00:44.104 --> 00:00:45.892
always the same thing.

00:00:45.892 --> 00:00:50.025
The distance traveled, this
is the total length of path,

00:00:50.025 --> 00:00:51.025
total length

00:00:52.718 --> 00:00:53.551
of path.

00:00:54.808 --> 00:00:56.596
So what are we talking about?

00:00:56.596 --> 00:00:58.221
Well let's say, and
we're going to introduce

00:00:58.221 --> 00:00:59.940
a little bit of calculus now,

00:00:59.940 --> 00:01:03.399
let's say that we have a
particle's velocity function.

00:01:03.399 --> 00:01:06.743
And so let's say our velocity
as a function of time

00:01:06.743 --> 00:01:08.826
is equal to five minus t.

00:01:09.785 --> 00:01:12.502
Now this is a one-dimensional
velocity function.

00:01:12.502 --> 00:01:14.127
Let's say it's just
telling us our velocity

00:01:14.127 --> 00:01:15.799
in the horizontal direction.

00:01:15.799 --> 00:01:17.610
And oftentimes when
something's one dimension,

00:01:17.610 --> 00:01:20.513
people forget well that too
can be a vector quantity.

00:01:20.513 --> 00:01:22.463
In fact this velocity is a vector quantity

00:01:22.463 --> 00:01:25.482
because you could think
of it if it's positive

00:01:25.482 --> 00:01:26.782
it's moving to the right,

00:01:26.782 --> 00:01:28.570
and if it's negative
it's moving to the left.

00:01:28.570 --> 00:01:30.126
So it has a direction.

00:01:30.126 --> 00:01:32.796
And so sometimes you will see
a vector quantity like this

00:01:32.796 --> 00:01:34.352
have a little arrow on it,

00:01:34.352 --> 00:01:37.092
or you will see it bolded,

00:01:37.092 --> 00:01:39.414
or you will see it bolded like that.

00:01:39.414 --> 00:01:40.969
I like to write an arrow in,

00:01:40.969 --> 00:01:42.920
although that's not
always the convention used

00:01:42.920 --> 00:01:44.986
in different classes.

00:01:44.986 --> 00:01:47.007
Now let's plot what this velocity function

00:01:47.007 --> 00:01:48.260
actually looks like,

00:01:48.260 --> 00:01:50.677
and I did that ahead of time.

00:01:52.533 --> 00:01:53.624
So you can see here,

00:01:53.624 --> 00:01:56.248
at time equals zero, let's
say time is in seconds,

00:01:56.248 --> 00:01:58.988
and our velocity's in meters per second.

00:01:58.988 --> 00:02:01.031
So this is meters per
second right over here,

00:02:01.031 --> 00:02:03.191
and this is seconds on this axis.

00:02:03.191 --> 00:02:05.374
At exactly time zero,

00:02:05.374 --> 00:02:09.089
this object is traveling
at five meters per second.

00:02:09.089 --> 00:02:11.179
We can say to the right, it has a velocity

00:02:11.179 --> 00:02:13.965
of positive five meters per second.

00:02:13.965 --> 00:02:17.959
But then it keeps decelerating
at a constant rate,

00:02:17.959 --> 00:02:19.723
so five seconds into it,

00:02:19.723 --> 00:02:21.349
right at five seconds,

00:02:21.349 --> 00:02:23.439
the particle has no velocity,

00:02:23.439 --> 00:02:25.598
and then it starts
having negative velocity,

00:02:25.598 --> 00:02:27.943
which you could interpret as

00:02:27.943 --> 00:02:30.265
moving to the left.

00:02:30.265 --> 00:02:31.751
So let's think about a few things.

00:02:31.751 --> 00:02:34.189
First let's think about
what is the displacement

00:02:34.189 --> 00:02:36.522
over the first five seconds,

00:02:39.391 --> 00:02:40.224
over first

00:02:41.991 --> 00:02:43.074
five seconds?

00:02:45.080 --> 00:02:47.239
Well we've seen already multiple times,

00:02:47.239 --> 00:02:49.863
if you wanna find the change in quantity,

00:02:49.863 --> 00:02:53.880
you can take the integral
of the rate function of it.

00:02:53.880 --> 00:02:57.479
And so velocity is actually
the rate of displacement

00:02:57.479 --> 00:02:58.826
is one way to think about it.

00:02:58.826 --> 00:03:00.869
So displacement over
the first five seconds,

00:03:00.869 --> 00:03:02.564
we can take the integral

00:03:02.564 --> 00:03:04.306
from zero to five,

00:03:04.306 --> 00:03:05.769
zero to five,

00:03:05.769 --> 00:03:07.719
of our velocity function,

00:03:07.719 --> 00:03:10.598
of our velocity function.

00:03:10.598 --> 00:03:12.433
Just like that.

00:03:12.433 --> 00:03:15.103
And we can even calculate
this really fast.

00:03:15.103 --> 00:03:18.632
That would just be this
area right over here,

00:03:18.632 --> 00:03:20.420
which we can just use a
little bit of geometry.

00:03:20.420 --> 00:03:22.998
This is a five by five triangle,

00:03:22.998 --> 00:03:25.506
so five times five is 25, times 1/2,

00:03:25.506 --> 00:03:28.687
remember area of a triangle's
1/2 base times height.

00:03:28.687 --> 00:03:31.020
So this is going to be 12.5,

00:03:32.402 --> 00:03:34.260
and let's see this is going
to be meters per second

00:03:34.260 --> 00:03:37.580
times seconds, so 12.5 meters.

00:03:37.580 --> 00:03:40.297
So that's the change in position

00:03:40.297 --> 00:03:43.547
for that particle over
the first five seconds.

00:03:43.547 --> 00:03:44.755
Wherever it started,

00:03:44.755 --> 00:03:47.959
it's now going to be 12.5
meters to the right of it,

00:03:47.959 --> 00:03:50.769
assuming that positive is to the right.

00:03:50.769 --> 00:03:53.184
Now what about over,

00:03:53.184 --> 00:03:54.351
over the first

00:03:55.924 --> 00:03:57.294
10 seconds?

00:03:57.294 --> 00:03:58.501
Now this gets interesting,

00:03:58.501 --> 00:04:00.312
and I encourage you to pause
your video and think about it.

00:04:00.312 --> 00:04:04.027
What would be the displacement
over the first 10 seconds?

00:04:04.027 --> 00:04:05.699
Well we would just do the same thing,

00:04:05.699 --> 00:04:08.032
the integral from zero to 10

00:04:09.298 --> 00:04:11.574
of our velocity function,

00:04:11.574 --> 00:04:15.614
our one-dimensional velocity function, dt.

00:04:15.614 --> 00:04:18.285
And so that would be the area from here

00:04:18.285 --> 00:04:21.164
all the way to right over there.

00:04:21.164 --> 00:04:23.161
So this entire area.

00:04:23.161 --> 00:04:24.415
But you might appreciate,

00:04:24.415 --> 00:04:26.435
when you're taking a definite integral,

00:04:26.435 --> 00:04:29.035
if we are below the t-axis

00:04:29.035 --> 00:04:30.661
and above the function like this,

00:04:30.661 --> 00:04:32.333
this is gonna be negative area.

00:04:32.333 --> 00:04:34.422
And in fact this area and this area

00:04:34.422 --> 00:04:36.303
are going to exactly cancel out,

00:04:36.303 --> 00:04:39.740
and you're going to get zero meters.

00:04:39.740 --> 00:04:41.876
Now you might be saying how can that be?

00:04:41.876 --> 00:04:43.246
After 10 seconds how do we,

00:04:43.246 --> 00:04:46.474
what why is our displacement
only zero meters?

00:04:46.474 --> 00:04:48.563
This particle's been
moving the entire time.

00:04:48.563 --> 00:04:50.723
Well remember what's going on.

00:04:50.723 --> 00:04:53.858
The first five seconds,
it's moving to the right,

00:04:53.858 --> 00:04:55.181
it's decelerating the whole time,

00:04:55.181 --> 00:04:56.969
and then right at five seconds,

00:04:56.969 --> 00:05:00.406
it has gone 12.5 meters to the right.

00:05:00.406 --> 00:05:03.540
But then it starts, it's velocity
starts becoming negative,

00:05:03.540 --> 00:05:06.582
and the particle starts
moving to the left.

00:05:06.582 --> 00:05:08.393
And so over the next five seconds,

00:05:08.393 --> 00:05:11.319
it actually moves 12.5 meters to the left,

00:05:11.319 --> 00:05:14.221
and then these two things net out.

00:05:14.221 --> 00:05:17.890
So the particle has gone
over 10 seconds 12.5 meters

00:05:17.890 --> 00:05:20.119
to the right and then
12.5 meters to the left,

00:05:20.119 --> 00:05:24.252
and so its change in
position is zero meters.

00:05:24.252 --> 00:05:26.180
It has not changed.

00:05:26.180 --> 00:05:27.387
Now you might start,

00:05:27.387 --> 00:05:29.105
you might start to be appreciating

00:05:29.105 --> 00:05:30.638
what the difference between displacement

00:05:30.638 --> 00:05:32.681
and distance traveled is.

00:05:32.681 --> 00:05:34.353
So distance,

00:05:34.353 --> 00:05:36.489
if you're talking about
your total length of path,

00:05:36.489 --> 00:05:39.067
you don't care as much about direction.

00:05:39.067 --> 00:05:40.924
And so instead of thinking about velocity,

00:05:40.924 --> 00:05:43.757
what we would do is think about speed.

00:05:43.757 --> 00:05:44.840
And speed is,

00:05:45.870 --> 00:05:47.310
you could view

00:05:47.310 --> 00:05:49.771
in this case, especially in
this one-dimensional case,

00:05:49.771 --> 00:05:54.392
this is equal to the
absolute value of velocity.

00:05:54.392 --> 00:05:56.482
Later on when we do multiple dimensions,

00:05:56.482 --> 00:05:59.733
it would be the magnitude
of the velocity function,

00:05:59.733 --> 00:06:01.729
which is what the absolute function,

00:06:01.729 --> 00:06:03.564
which is what the absolute
value function does,

00:06:03.564 --> 00:06:05.236
in one dimension.

00:06:05.236 --> 00:06:07.697
So what would this look
like if we plotted it?

00:06:07.697 --> 00:06:10.948
Well the absolute value
of the velocity function

00:06:10.948 --> 00:06:13.502
would just look like that.

00:06:13.502 --> 00:06:15.615
And so if you want the distance,

00:06:15.615 --> 00:06:18.773
you would find, the distance
traveled I should say,

00:06:18.773 --> 00:06:21.466
you would find the integral
over the appropriate

00:06:21.466 --> 00:06:25.530
change in time of the speed
function right over here,

00:06:25.530 --> 00:06:27.039
which we have graphed.

00:06:27.039 --> 00:06:30.706
So notice, if we want
the distance traveled,

00:06:31.567 --> 00:06:33.355
so I'll just say I'll write it out,

00:06:33.355 --> 00:06:34.772
distance traveled

00:06:36.768 --> 00:06:38.768
over first five seconds,

00:06:40.576 --> 00:06:42.898
first five seconds,

00:06:42.898 --> 00:06:44.036
what would it be?

00:06:44.036 --> 00:06:46.846
Well it would be the
integral from zero to five

00:06:46.846 --> 00:06:49.888
of the absolute value of
our velocity function,

00:06:49.888 --> 00:06:52.837
which is you can just view
it as our speed function

00:06:52.837 --> 00:06:54.926
right over here, dt.

00:06:54.926 --> 00:06:57.248
And so it would be this area,

00:06:57.248 --> 00:07:00.708
which we already know to be 12.5 meters.

00:07:00.708 --> 00:07:02.566
So for the first five seconds,

00:07:02.566 --> 00:07:05.561
your distance and
displacement are consistent.

00:07:05.561 --> 00:07:08.208
Well that's because you have in this case

00:07:08.208 --> 00:07:10.135
the velocity function is positive,

00:07:10.135 --> 00:07:12.574
so the absolute value of it
is still going to be positive.

00:07:12.574 --> 00:07:15.639
But if you think about over

00:07:15.639 --> 00:07:17.635
the first 10 seconds, your distance,

00:07:17.635 --> 00:07:18.552
10 seconds,

00:07:19.934 --> 00:07:21.002
what is it going to be?

00:07:21.002 --> 00:07:22.883
Pause the video and try to think about it.

00:07:22.883 --> 00:07:25.205
Well that's gonna be the
integral from zero to 10

00:07:25.205 --> 00:07:28.293
of the absolute value of
our velocity function,

00:07:28.293 --> 00:07:31.080
which is going to be equal to what?

00:07:31.080 --> 00:07:33.634
It's going to be this area plus this area

00:07:33.634 --> 00:07:35.793
right over here.

00:07:35.793 --> 00:07:39.091
So plus this area right over here.

00:07:39.091 --> 00:07:42.504
And so this is going to
five times five times 1/2

00:07:42.504 --> 00:07:44.594
plus five times five times 1/2,

00:07:44.594 --> 00:07:47.177
which is going to be 25 meters.

00:07:48.657 --> 00:07:51.467
The particle has gone
12.5 meters to the right,

00:07:51.467 --> 00:07:54.486
and then it goes back
12.5 meters to the left.

00:07:54.486 --> 00:07:58.410
Your displacement, your net
change in position is a zero,

00:07:58.410 --> 00:08:00.662
but the total length of path traveled

00:08:00.662 --> 00:08:01.745
is 25 meters.


Analyzing problems involving definite integrals

WEBVTT
Kind: captions
Language: en

00:00:00.915 --> 00:00:02.494
- [Instructor] The population of a town

00:00:02.494 --> 00:00:04.537
grows at a rate of

00:00:04.537 --> 00:00:06.766
r of t is equal to 300

00:00:06.766 --> 00:00:08.349
times e to the 0.3t

00:00:10.203 --> 00:00:11.851
people per year,

00:00:11.851 --> 00:00:14.934
where time, where t is time in years.

00:00:15.775 --> 00:00:18.376
At time t is equal to two

00:00:18.376 --> 00:00:21.441
the town's population is 1,200 people.

00:00:21.441 --> 00:00:25.608
What is the town's population
at t is equal to seven?

00:00:26.874 --> 00:00:29.823
Which expression can we
use to solve the problem?

00:00:29.823 --> 00:00:31.936
So they don't want us to
actually answer the question.

00:00:31.936 --> 00:00:33.724
They just want us to set up the expression

00:00:33.724 --> 00:00:36.998
using some symbols from calculus.

00:00:36.998 --> 00:00:38.299
So why don't you pause this video

00:00:38.299 --> 00:00:41.224
and try to think about it.

00:00:41.224 --> 00:00:43.314
So let's just remind us
what they've given us.

00:00:43.314 --> 00:00:46.658
They've given us the rate
function right over here.

00:00:46.658 --> 00:00:48.825
And so if you want to find

00:00:49.746 --> 00:00:51.697
a change in population

00:00:51.697 --> 00:00:52.780
from one time

00:00:54.483 --> 00:00:56.085
to another time,

00:00:56.085 --> 00:00:58.338
what you could do is you
could take the integral

00:00:58.338 --> 00:01:00.636
of the rate function

00:01:00.636 --> 00:01:02.285
from the starting time,

00:01:02.285 --> 00:01:03.957
t is equal to two years,

00:01:03.957 --> 00:01:05.838
to t is equal to seven years.

00:01:05.838 --> 00:01:08.810
So we're gonna take the
integral of the rate function,

00:01:08.810 --> 00:01:10.435
and what this is going to tell us,

00:01:10.435 --> 00:01:13.361
this is going to tell us
the change in population

00:01:13.361 --> 00:01:16.194
from time two to time seven.

00:01:16.194 --> 00:01:17.610
So this is,

00:01:17.610 --> 00:01:18.910
we could say let me just write this,

00:01:18.910 --> 00:01:21.302
this is the change.

00:01:21.302 --> 00:01:22.881
I'll use delta for change
or I'll just let me

00:01:22.881 --> 00:01:24.297
just write it out,

00:01:24.297 --> 00:01:27.880
change in population,
change in population,

00:01:28.941 --> 00:01:29.858
population.

00:01:31.496 --> 00:01:33.307
But they don't want us to, we don't,

00:01:33.307 --> 00:01:34.746
they don't, they're not asking us

00:01:34.746 --> 00:01:35.768
for the change in population.

00:01:35.768 --> 00:01:37.997
They want us to know what
is the town's population

00:01:37.997 --> 00:01:39.483
if t is equal to seven.

00:01:39.483 --> 00:01:41.109
So what you would want is,

00:01:41.109 --> 00:01:44.568
you would want what your
population is at t equals two

00:01:44.568 --> 00:01:47.169
plus the change in
population from two to seven

00:01:47.169 --> 00:01:49.746
to get you your population at seven.

00:01:49.746 --> 00:01:52.951
So they tell us the population
at time t equals two,

00:01:52.951 --> 00:01:55.041
the town's population is 1,200 people.

00:01:55.041 --> 00:01:58.268
So if you want the population
at t is equal to seven,

00:01:58.268 --> 00:02:00.172
it's going to be

00:02:00.172 --> 00:02:04.468
1,200 plus how whatever the
change in population is.

00:02:04.468 --> 00:02:06.163
You take the integral
of the rate function,

00:02:06.163 --> 00:02:08.137
you are, and this is
the rate of population,

00:02:08.137 --> 00:02:11.132
this integral is gonna give
you the change in population

00:02:11.132 --> 00:02:13.663
from time t equals two to t equals seven.

00:02:13.663 --> 00:02:17.425
So we can see clearly that
is choice D right over here.

00:02:17.425 --> 00:02:20.188
These other choices, we could
look at 'em really quick.

00:02:20.188 --> 00:02:22.812
Choice b is just the change in population.

00:02:22.812 --> 00:02:26.295
Assuming that this, and
this is actually increasing,

00:02:26.295 --> 00:02:28.872
so this would tell us how much
did the population increase

00:02:28.872 --> 00:02:32.959
from t two, from t equals
two to t equals seven.

00:02:32.959 --> 00:02:34.515
So that's not what we want.

00:02:34.515 --> 00:02:36.976
We want what the actual population is.

00:02:36.976 --> 00:02:39.275
This is how much the population increases

00:02:39.275 --> 00:02:42.479
from time zero to time seven.

00:02:42.479 --> 00:02:43.942
Now you might say well wouldn't that be

00:02:43.942 --> 00:02:45.196
the town's population?

00:02:45.196 --> 00:02:46.705
Well that would be the town's population

00:02:46.705 --> 00:02:48.493
if they had no people at time zero,

00:02:48.493 --> 00:02:49.770
but you can't assume that.

00:02:49.770 --> 00:02:52.928
Maybe the town got settled by
10 people or by 1,000 people

00:02:52.928 --> 00:02:54.623
or who knows whatever else.

00:02:54.623 --> 00:02:55.970
So right over there.

00:02:55.970 --> 00:02:59.058
And this is taking the
derivative of the rate function,

00:02:59.058 --> 00:03:01.868
which is, it's actually a little hard

00:03:01.868 --> 00:03:03.122
to think about what is this.

00:03:03.122 --> 00:03:05.490
This is the rate of change of the rate

00:03:05.490 --> 00:03:08.462
at time seven minus the
rate of change of the rate

00:03:08.462 --> 00:03:09.832
at time two.

00:03:09.832 --> 00:03:11.713
So I would rule that one out as well.

00:03:11.713 --> 00:03:14.337
Let's do one more of these.

00:03:14.337 --> 00:03:15.670
So here we have,

00:03:18.911 --> 00:03:20.676
the depth of water in a tank

00:03:20.676 --> 00:03:23.184
is changing at a rate of r of t

00:03:23.184 --> 00:03:26.574
is equal to 0.3t centimeters per minute,

00:03:26.574 --> 00:03:29.430
where t is the time in minutes.

00:03:29.430 --> 00:03:31.287
At time t equals zero,

00:03:31.287 --> 00:03:34.538
the depth of the water is 35 centimeters.

00:03:34.538 --> 00:03:37.371
What is the change in the water's depth

00:03:37.371 --> 00:03:39.454
during the fourth minute?

00:03:40.297 --> 00:03:42.851
So let's pause the video
again and see if you can

00:03:42.851 --> 00:03:44.244
figure this out again,

00:03:44.244 --> 00:03:46.171
or figure out what expression can we use

00:03:46.171 --> 00:03:47.147
to solve the problem,

00:03:47.147 --> 00:03:49.654
the problem being what is the change

00:03:49.654 --> 00:03:51.744
in the water's depth

00:03:51.744 --> 00:03:53.827
during the fourth minute.

00:03:55.181 --> 00:03:58.617
Alright so we've just talked
about if you're trying to find

00:03:58.617 --> 00:04:00.450
the change in a value,

00:04:01.288 --> 00:04:04.120
you could take the integral
of the rate function

00:04:04.120 --> 00:04:06.001
over the appropriate time.

00:04:06.001 --> 00:04:08.778
So we're talking about
during the fourth minute,

00:04:08.778 --> 00:04:10.589
so we definitely wanna take the integral

00:04:10.589 --> 00:04:12.540
of the rate function.

00:04:12.540 --> 00:04:14.096
We just have to think about the bounds.

00:04:14.096 --> 00:04:15.791
And all the choices here
are taking the integral

00:04:15.791 --> 00:04:16.975
of the rate function.

00:04:16.975 --> 00:04:18.368
So really the interesting part is,

00:04:18.368 --> 00:04:20.992
during the fourth minute.

00:04:20.992 --> 00:04:22.826
So let me just draw a
little number line here,

00:04:22.826 --> 00:04:25.845
and we can think about what
the fourth minute looks like.

00:04:25.845 --> 00:04:27.679
Or actually let me just
draw the whole thing.

00:04:27.679 --> 00:04:31.627
So let's say this is r
of t right over there,

00:04:31.627 --> 00:04:33.717
I could say y is equal to r of t,

00:04:33.717 --> 00:04:35.634
and this is, this is t.

00:04:37.269 --> 00:04:40.009
And let's see, the first
minute goes from zero to one,

00:04:40.009 --> 00:04:42.169
second minute goes from one to two,

00:04:42.169 --> 00:04:44.351
third minute goes from two to three,

00:04:44.351 --> 00:04:46.813
fourth minute goes from three to four.

00:04:46.813 --> 00:04:49.204
The rate function, this
actually just looks like

00:04:49.204 --> 00:04:53.105
a straight up linear rate
function, something like this.

00:04:53.105 --> 00:04:55.520
And so what is the fourth minute?

00:04:55.520 --> 00:04:57.935
Well the fourth minute is
the first minute is this one,

00:04:57.935 --> 00:04:59.839
second, third, fourth.

00:04:59.839 --> 00:05:04.599
The fourth minute is going from
minute three to minute four.

00:05:04.599 --> 00:05:08.291
So it's what we wanna do is
the expression that gives us

00:05:08.291 --> 00:05:11.495
this area right over here
under the rate curve.

00:05:11.495 --> 00:05:14.259
Well this is our lower
bound's going to be three,

00:05:14.259 --> 00:05:16.673
and our upper bound is going to be four.

00:05:16.673 --> 00:05:18.160
And so there you have it,

00:05:18.160 --> 00:05:20.064
it is this first choice.

00:05:20.064 --> 00:05:21.526
You might have been
tempted here if you got

00:05:21.526 --> 00:05:22.525
a little bit confused,

00:05:22.525 --> 00:05:24.359
hey maybe the fourth minute
is after we've crossed

00:05:24.359 --> 00:05:26.287
the t is equal to four,

00:05:26.287 --> 00:05:28.260
but no that would be the fifth minute.

00:05:28.260 --> 00:05:29.909
This would tell us,

00:05:29.909 --> 00:05:33.740
this would tell us our change
over the first four minutes,

00:05:33.740 --> 00:05:35.575
not just during the fourth minute.

00:05:35.575 --> 00:05:38.593
And then this, well this
is just gonna be zero.

00:05:38.593 --> 00:05:40.172
If you're taking, this is,

00:05:40.172 --> 00:05:42.332
what is the change in the
value from three to three.

00:05:42.332 --> 00:05:44.282
Well it didn't change at all,

00:05:44.282 --> 00:05:47.393
because it's essentially at an instant,

00:05:47.393 --> 00:05:49.483
there's no time that
passes from three to three.

00:05:49.483 --> 00:05:51.801
So you can rule all of these out.


Worked examples: interpreting definite integrals in context

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:03.456
- [Instructor] Julia's
revenue is r of t thousand

00:00:03.456 --> 00:00:07.066
dollars per month, where t
is the month of the year.

00:00:07.066 --> 00:00:11.832
Julia had made $3,000 in
the first month of the year,

00:00:11.832 --> 00:00:14.076
what does three plus
the definite role from

00:00:14.076 --> 00:00:17.298
one to five of r of t dt equals 19 mean?

00:00:17.298 --> 00:00:18.551
And we have some choices.

00:00:18.551 --> 00:00:20.464
So like always, pause the video and see

00:00:20.464 --> 00:00:22.270
if you can work through it.

00:00:22.270 --> 00:00:24.413
Alright, now let's work
through this together.

00:00:24.413 --> 00:00:27.849
So they tell us that she made
$3,000 in the first month

00:00:27.849 --> 00:00:30.813
and we also see this three
here, so that's interesting.

00:00:30.813 --> 00:00:32.989
Maybe they represent the same thing,

00:00:32.989 --> 00:00:34.577
we don't know for sure yet.

00:00:34.577 --> 00:00:36.609
But let's look at this definite integral.

00:00:36.609 --> 00:00:38.299
The definite integral from one to five

00:00:38.299 --> 00:00:42.038
of r of t dt, this is the
area under this rate curve,

00:00:42.038 --> 00:00:44.721
r of t is the rate atwhich
Julia makes revenue

00:00:44.721 --> 00:00:46.088
on a monthly basis.

00:00:46.088 --> 00:00:48.612
So if you take the area
under that rate curve,

00:00:48.612 --> 00:00:51.281
that's going to give you the net change

00:00:51.281 --> 00:00:54.398
in revenue from month one to month five,

00:00:54.398 --> 00:00:56.862
how much that increased.

00:00:56.862 --> 00:00:59.330
And so if you add that to the amount

00:00:59.330 --> 00:01:01.562
she made in month one, well that tells you

00:01:01.562 --> 00:01:04.524
the total she makes from
essentially time zero

00:01:04.524 --> 00:01:06.648
all the way to month five.

00:01:06.648 --> 00:01:08.949
And they're saying that is equal to 19.

00:01:08.949 --> 00:01:10.870
So let's see which of these choices

00:01:10.870 --> 00:01:12.343
are consistent with that.

00:01:12.343 --> 00:01:14.962
Julia made an additional $19,000 between

00:01:14.962 --> 00:01:16.604
months one and five.

00:01:16.604 --> 00:01:18.795
Choice A would be correct if you didn't

00:01:18.795 --> 00:01:20.379
see this three over here.

00:01:20.379 --> 00:01:22.841
Because just the definite
integral is the additional

00:01:22.841 --> 00:01:25.407
between months one and
five, but that's not

00:01:25.407 --> 00:01:28.745
what this expression says,
it says three plus this

00:01:28.745 --> 00:01:30.439
is equal to 19.

00:01:30.439 --> 00:01:33.338
If it said Julia made
an additional $16,000,

00:01:33.338 --> 00:01:35.365
well that would make
sense because you could

00:01:35.365 --> 00:01:37.500
subtract three from both sides
and you'd get that result,

00:01:37.500 --> 00:01:39.246
but that's not what they're saying.

00:01:39.246 --> 00:01:42.914
Julia made an average
of $19,000 per month.

00:01:42.914 --> 00:01:44.612
Well once again, that's also not right

00:01:44.612 --> 00:01:47.560
because we just said from the beginning,

00:01:47.560 --> 00:01:51.824
from time zero, all the
way until the fifth month,

00:01:51.824 --> 00:01:54.157
she made a total of $19,000.

00:01:55.112 --> 00:01:58.493
Not the average per month is $19,000.

00:01:58.493 --> 00:02:01.511
Julia made $19,000 in the fifth month.

00:02:01.511 --> 00:02:03.749
Once again, this is not just saying what

00:02:03.749 --> 00:02:05.219
happened in the fifth month.

00:02:05.219 --> 00:02:08.118
This is saying, we have the
$3,000 from the first month

00:02:08.118 --> 00:02:11.289
and then we have the additional from month

00:02:11.289 --> 00:02:15.123
between month one and month
five, so that's not that.

00:02:15.123 --> 00:02:16.430
So this better be our choice.

00:02:16.430 --> 00:02:18.949
By the end of the fifth
month, Julia had made

00:02:18.949 --> 00:02:21.191
a total of $19,000.

00:02:21.191 --> 00:02:22.858
Yes that is correct.

00:02:23.819 --> 00:02:27.534
She made $3,000 in month
one and then as we go

00:02:27.534 --> 00:02:30.484
between month one to
the end of month five,

00:02:30.484 --> 00:02:33.712
to the end of the fifth
month, she has made

00:02:33.712 --> 00:02:35.295
a total of $19,000.

00:02:36.778 --> 00:02:39.278
Let's do another one of these.

00:02:42.079 --> 00:02:44.871
So here we're told the function k of t

00:02:44.871 --> 00:02:46.892
gives the amount of ketchup in kilograms

00:02:46.892 --> 00:02:50.559
produced in a sauce
factory by time in hours

00:02:51.930 --> 00:02:53.403
on a given day.

00:02:53.403 --> 00:02:56.573
So this is really quantity
is a function of time,

00:02:56.573 --> 00:03:00.453
it isn't rate, what does
the definite integral

00:03:00.453 --> 00:03:03.682
from zero to four of k
prime of t dt represent?

00:03:03.682 --> 00:03:05.607
Once again, pause the video and see if you

00:03:05.607 --> 00:03:07.185
can work through it.

00:03:07.185 --> 00:03:10.910
Well k of t is the amount of ketchup

00:03:10.910 --> 00:03:12.052
as function of time.

00:03:12.052 --> 00:03:15.120
So k prime of t, that's going to be the

00:03:15.120 --> 00:03:17.571
rate at which our amount of ketchup

00:03:17.571 --> 00:03:19.876
is changing a function of time.

00:03:19.876 --> 00:03:23.041
But once again, when
you're taking the area

00:03:23.041 --> 00:03:26.213
under the rate curve, that
tells you the net change

00:03:26.213 --> 00:03:29.498
in the original quantity
in the amount of ketchup.

00:03:29.498 --> 00:03:33.495
And as the net change between
time zero and time four,

00:03:33.495 --> 00:03:36.552
so let's see which of these
choices match up to that.

00:03:36.552 --> 00:03:39.337
The average rate of change of the ketchup

00:03:39.337 --> 00:03:41.858
production over the first four hours.

00:03:41.858 --> 00:03:45.899
No, that does not tell us
the average rate of change,

00:03:45.899 --> 00:03:47.435
there's other ways to calculate that

00:03:47.435 --> 00:03:50.170
the time it takes to
produce four kilograms

00:03:50.170 --> 00:03:54.337
of ketchup, so does this
represent the time it takes?

00:03:55.364 --> 00:03:57.385
To represent four kilograms of ketchup.

00:03:57.385 --> 00:03:59.911
No, this four is a time right over here.

00:03:59.911 --> 00:04:02.144
This is gonna tell you how much ketchup

00:04:02.144 --> 00:04:05.428
gets produced from time zero to time four.

00:04:05.428 --> 00:04:09.305
The instantaneous rate of
production at t equals four.

00:04:09.305 --> 00:04:11.821
Now this would be k prime of four,

00:04:11.821 --> 00:04:13.739
that's not what this integral represents.

00:04:13.739 --> 00:04:16.034
The amount of ketchup produced over the

00:04:16.034 --> 00:04:19.701
first four hours, yep
that is exactly right.


Interpreting definite integral as net change

WEBVTT
Kind: captions
Language: en

00:00:00.690 --> 00:00:02.045
- [Instructor] In a previous video

00:00:02.045 --> 00:00:05.386
we started to get an
intuition for rate curves

00:00:05.386 --> 00:00:08.334
and what the area under
a rate curve represents.

00:00:08.334 --> 00:00:10.685
So, for example, this rate curve,

00:00:10.685 --> 00:00:13.266
this might represent a speed of a car

00:00:13.266 --> 00:00:17.831
and how a speed of a car is
changing with respect to time.

00:00:17.831 --> 00:00:20.683
And so this shows us that our
rate is actually changing,

00:00:20.683 --> 00:00:23.202
this isn't distance as function of time,

00:00:23.202 --> 00:00:24.886
this is rate as a function of time.

00:00:24.886 --> 00:00:26.939
So, this looks like car is accelerating.

00:00:26.939 --> 00:00:30.049
At time one it is going
ten meters per second

00:00:30.049 --> 00:00:32.752
and at time five, let's
assume that all of these

00:00:32.752 --> 00:00:35.557
are in seconds, so at five seconds,

00:00:35.557 --> 00:00:38.045
it is going 20 meters per second.

00:00:38.045 --> 00:00:40.394
So, it is is accelerating.

00:00:40.394 --> 00:00:43.599
Now the relationship between
the rate function and the area,

00:00:43.599 --> 00:00:45.863
is that if we're able
to figure out this area,

00:00:45.863 --> 00:00:49.681
then that is the change
in distance of the car.

00:00:49.681 --> 00:00:53.848
So rate or speed in this case
is distance per unit time,

00:00:54.702 --> 00:00:57.781
if we're able to figure out
the area under that curve,

00:00:57.781 --> 00:01:01.215
it will actually give us
our change in distance,

00:01:01.215 --> 00:01:03.665
from time one to time five.

00:01:03.665 --> 00:01:05.688
It won't tell us our total distance,

00:01:05.688 --> 00:01:08.097
cause we won't know what
happens before time one,

00:01:08.097 --> 00:01:10.657
if we're not concerned with that area.

00:01:10.657 --> 00:01:14.470
And the intuition for that,
it's a little bit easier

00:01:14.470 --> 00:01:16.496
if you're dealing with rectangles.

00:01:16.496 --> 00:01:18.216
But just think about this,
let's make a rectangle

00:01:18.216 --> 00:01:21.823
that looks like a pretty good
approximation for the area,

00:01:21.823 --> 00:01:26.799
let's say from time one to
time two right over here.

00:01:26.799 --> 00:01:29.939
Well what is this area from
the rectangle represent?

00:01:29.939 --> 00:01:32.784
To figure out the area we
would multiply one second,

00:01:32.784 --> 00:01:36.196
that would be the width
here, times roughly,

00:01:36.196 --> 00:01:38.672
looks like about ten meters per second.

00:01:38.672 --> 00:01:41.370
And so the units here would
be ten meters per second,

00:01:41.370 --> 00:01:44.290
times on second, or ten meters.

00:01:44.290 --> 00:01:47.228
And we know from early
physics or even before,

00:01:47.228 --> 00:01:49.960
that if you multiply a rate times time,

00:01:49.960 --> 00:01:53.123
or a speed times a time,
you're gonna get a distance.

00:01:53.123 --> 00:01:55.685
And so the unit here is in distance,

00:01:55.685 --> 00:01:59.182
as you can see, this area
is going to represent,

00:01:59.182 --> 00:02:02.603
it's gonna be an approximation
for the distance traveled.

00:02:02.603 --> 00:02:05.924
And so, if you wanted
to get an exact version,

00:02:05.924 --> 00:02:09.008
or an exact number for
the distance traveled.

00:02:09.008 --> 00:02:12.006
You would get the exact
area under the curve.

00:02:12.006 --> 00:02:13.904
And we have a notation for that.

00:02:13.904 --> 00:02:17.523
If you want the exact area
under the curve right over here,

00:02:17.523 --> 00:02:20.040
we use definite integral notation.

00:02:20.040 --> 00:02:21.473
This area right over here,

00:02:21.473 --> 00:02:23.829
we can denote as a definite integral

00:02:23.829 --> 00:02:26.496
from one to five, of R of t, dt.

00:02:29.240 --> 00:02:31.047
And once again, what does this represent?

00:02:31.047 --> 00:02:33.551
In this case when our rate is speed?

00:02:33.551 --> 00:02:36.125
This represents this
whole expression reprsents

00:02:36.125 --> 00:02:39.958
our change in distance
from t is equal to one,

00:02:45.816 --> 00:02:47.649
to t is equal to five.

00:02:49.793 --> 00:02:52.149
Now with that context, let's
actually try to do an example

00:02:52.149 --> 00:02:55.997
problem, the type that you
might see on Khan Academy.

00:02:55.997 --> 00:02:58.767
So, this right over here tells us,

00:02:58.767 --> 00:03:02.510
Eden walked at a rate of r
of t kilometers per hour,

00:03:02.510 --> 00:03:05.100
where t is the time in hours.

00:03:05.100 --> 00:03:06.730
Okay, so t is in hours.

00:03:06.730 --> 00:03:09.448
What does the integral,
the definite integral

00:03:09.448 --> 00:03:12.710
from two to three of r
of t dt equals six mean?

00:03:12.710 --> 00:03:14.668
So before I even look at these choices,

00:03:14.668 --> 00:03:17.620
this is saying, so this is
going from t equals two hours

00:03:17.620 --> 00:03:20.806
to t equals three hours and
it's essentially the area

00:03:20.806 --> 00:03:23.376
under the rate curve and here the rate is,

00:03:23.376 --> 00:03:24.714
we're talking about a speed.

00:03:24.714 --> 00:03:29.510
Eden is walking at a certain
number of kilometers per hour.

00:03:29.510 --> 00:03:32.683
So what this means,
that from time two hours

00:03:32.683 --> 00:03:36.850
to time three hours, Eden
walked an extra six kilometers.

00:03:38.246 --> 00:03:41.175
So, let's see which of
these choices match that.

00:03:41.175 --> 00:03:44.258
Eden walked six kilometers each hour?

00:03:45.186 --> 00:03:48.288
It does tell us from time two to three,

00:03:48.288 --> 00:03:51.656
Eden walked six kilometers,
but it doesn't mean,

00:03:51.656 --> 00:03:55.823
but we don't know what happened
from time zero to time one,

00:03:56.948 --> 00:03:59.155
or from time one to time two.

00:03:59.155 --> 00:04:01.740
So I would rule this out.

00:04:01.740 --> 00:04:04.794
Eden walked six kilometers in three hours.

00:04:04.794 --> 00:04:07.142
So this is a common misconception,

00:04:07.142 --> 00:04:10.153
people will look at the top bound

00:04:10.153 --> 00:04:13.649
and say okay, this area represented
by the definite integral

00:04:13.649 --> 00:04:15.738
this tells us how far in total

00:04:15.738 --> 00:04:18.538
we have walked up until that point.

00:04:18.538 --> 00:04:20.992
That is not what this represents.

00:04:20.992 --> 00:04:23.866
This represents the change in distance

00:04:23.866 --> 00:04:26.433
from time two to time three.

00:04:26.433 --> 00:04:28.090
So, I'll rule that out.

00:04:28.090 --> 00:04:31.413
Eden walked six kilometers
during the third hour.

00:04:31.413 --> 00:04:33.473
Yes, that's what we've been talking about.

00:04:33.473 --> 00:04:38.189
From time equal two hours
to time equal three hours,

00:04:38.189 --> 00:04:40.585
Eden walked six kilometers
and you could view that

00:04:40.585 --> 00:04:44.500
as the third hour, going
from time two to time three.

00:04:44.500 --> 00:04:48.066
Eden's rate increased by
six kilometers per hour

00:04:48.066 --> 00:04:50.850
between hours two and three.

00:04:50.850 --> 00:04:54.485
So let's be very clear,
this right over here,

00:04:54.485 --> 00:04:58.665
this isn't a rate, this is
the area under the rate curve,

00:04:58.665 --> 00:05:01.160
this is what this definite
integral is representing.

00:05:01.160 --> 00:05:04.248
And so this isn't telling
us about our rate changing,

00:05:04.248 --> 00:05:07.819
this is telling us how does
the thing that the rate

00:05:07.819 --> 00:05:11.005
is measuring the change
of, how does that change

00:05:11.005 --> 00:05:12.812
from time two to time three?

00:05:12.812 --> 00:05:15.625
So we would rule that out as well.


Interpreting behavior of _ from graph of _'=ÃÂ

WEBVTT
Kind: captions
Language: en

00:00:00.674 --> 00:00:03.811
- [Instructor] Let g of x be
equal to the definite integral

00:00:03.811 --> 00:00:07.196
from zero to x of f of t dt.

00:00:07.196 --> 00:00:10.951
What is an appropriate
calculus-based justification

00:00:10.951 --> 00:00:15.889
for the fact that g is concave
up on the open interval

00:00:15.889 --> 00:00:17.051
from five to 10?

00:00:17.051 --> 00:00:18.846
So concave up.

00:00:18.846 --> 00:00:21.584
So before I even think about
what it means to be concave up,

00:00:21.584 --> 00:00:23.758
let's just make sure we
understand this relationship

00:00:23.758 --> 00:00:25.266
between g and f.

00:00:25.266 --> 00:00:28.121
One way to understand it is
if we took the derivative

00:00:28.121 --> 00:00:30.496
of both sides of this equation,

00:00:30.496 --> 00:00:34.663
we would get that g prime
of x is equal to f of x.

00:00:37.347 --> 00:00:38.308
The derivative of this

00:00:38.308 --> 00:00:40.796
with respect to x would just be f of x.

00:00:40.796 --> 00:00:42.309
In fact, the whole
reason why we introduced

00:00:42.309 --> 00:00:45.359
this variable t here is
this thing right over here

00:00:45.359 --> 00:00:47.553
is actually a function of x

00:00:47.553 --> 00:00:49.312
'cause x is this upper bound.

00:00:49.312 --> 00:00:51.905
And it would've been weird if
we had x as an upper bound,

00:00:51.905 --> 00:00:53.208
or at least confusing,

00:00:53.208 --> 00:00:55.183
and we were also integrating
with respect to x.

00:00:55.183 --> 00:00:57.879
So we just had to pick kind of
another placeholder variable.

00:00:57.879 --> 00:00:58.764
Didn't have to be t.

00:00:58.764 --> 00:01:00.327
It could be alpha, it could be gamma,

00:01:00.327 --> 00:01:02.714
it could be a, b, or
c, whatever we choose,

00:01:02.714 --> 00:01:07.042
but this is still, right over
here, this is a function of x.

00:01:07.042 --> 00:01:08.693
But when you take the
derivative of both sides,

00:01:08.693 --> 00:01:11.731
you realize that the function
f, which is graphed here.

00:01:11.731 --> 00:01:14.938
And if this were the x axis,
then this would be f of x.

00:01:14.938 --> 00:01:16.784
If this is the t axis,

00:01:16.784 --> 00:01:18.945
then this is y is equal to f of t.

00:01:18.945 --> 00:01:22.303
But generally this is the
graph of our function f,

00:01:22.303 --> 00:01:26.071
which you could also view
as the graph of g prime.

00:01:26.071 --> 00:01:28.863
If this is x, this would be g prime of x.

00:01:28.863 --> 00:01:31.799
And so we're thinking about the interval,

00:01:31.799 --> 00:01:34.124
the open interval from five to 10,

00:01:34.124 --> 00:01:36.979
and we have g's derivative graphed here.

00:01:36.979 --> 00:01:39.031
And we wanna know a
calculus-based justification

00:01:39.031 --> 00:01:43.288
from this graph that lets us
know that g is concave up.

00:01:43.288 --> 00:01:46.335
So what does it mean to be concave up?

00:01:46.335 --> 00:01:50.636
Well, that means that your
slope of tangent line,

00:01:50.636 --> 00:01:54.219
of tangent, slope of
tangent is increasing.

00:01:56.685 --> 00:01:58.619
Or another way of thinking about it,

00:01:58.619 --> 00:02:01.119
your derivative is increasing.

00:02:03.066 --> 00:02:04.190
Or another way to think about it,

00:02:04.190 --> 00:02:06.022
if your derivative is
increasing over an interval,

00:02:06.022 --> 00:02:08.868
then you're concave up on that interval.

00:02:08.868 --> 00:02:11.077
And so here we have a
graph of the derivative,

00:02:11.077 --> 00:02:14.159
and it is indeed increasing
over that interval.

00:02:14.159 --> 00:02:16.306
So our calculus-based justification

00:02:16.306 --> 00:02:17.813
that we'd wanna use is that, look,

00:02:17.813 --> 00:02:21.992
f, which is g prime, is
increasing on that interval.

00:02:21.992 --> 00:02:23.651
The derivative is
increasing on that interval,

00:02:23.651 --> 00:02:26.238
which means that the original
function is concave up.

00:02:26.238 --> 00:02:28.095
f is positive on that interval.

00:02:28.095 --> 00:02:32.084
That's not a sufficient
calculus-based justification.

00:02:32.084 --> 00:02:34.294
Because if your derivative is positive,

00:02:34.294 --> 00:02:36.251
that just means your original
function is increasing.

00:02:36.251 --> 00:02:37.084
It doesn't tell you

00:02:37.084 --> 00:02:39.933
that your original function is concave up.

00:02:39.933 --> 00:02:42.853
f is concave up on the interval.

00:02:42.853 --> 00:02:45.081
Well, just because your
derivative is concave up

00:02:45.081 --> 00:02:48.109
doesn't mean that your original
function is concave up.

00:02:48.109 --> 00:02:51.859
In fact, you could have
a situation like this

00:02:52.730 --> 00:02:55.847
where you're concave
up over that interval,

00:02:55.847 --> 00:02:58.292
but for much of that
interval right over here,

00:02:58.292 --> 00:03:02.132
if this was our graph of f or
g prime, we are decreasing.

00:03:02.132 --> 00:03:04.912
And if we're decreasing
over much of that interval,

00:03:04.912 --> 00:03:06.336
then actually on this part

00:03:06.336 --> 00:03:08.998
our original function
would be concave down.

00:03:08.998 --> 00:03:12.277
The graph of g has a cup
U shape on the interval.

00:03:12.277 --> 00:03:14.308
Well, if we had the graph of g,

00:03:14.308 --> 00:03:16.022
this would be a justification,

00:03:16.022 --> 00:03:18.821
but it wouldn't be a
calculus-based justification.

00:03:18.821 --> 00:03:21.291
Let's do more of these.

00:03:21.291 --> 00:03:25.393
So this next one says, so we
have the exact same setup,

00:03:25.393 --> 00:03:27.137
which actually all of
these examples will have.

00:03:27.137 --> 00:03:29.135
g of x is equal to this thing here.

00:03:29.135 --> 00:03:31.216
What is an appropriate
calculus-based justification

00:03:31.216 --> 00:03:35.254
for the fact that g has a relative minimum

00:03:35.254 --> 00:03:36.903
at x equals eight?

00:03:36.903 --> 00:03:39.632
So once again, they've graphed f here,

00:03:39.632 --> 00:03:43.377
which is the same thing
as the derivative of g.

00:03:43.377 --> 00:03:45.403
And so if we have the
graph of the derivative,

00:03:45.403 --> 00:03:48.178
how do we know that we
have a relative minimum

00:03:48.178 --> 00:03:50.124
at x equals eight?

00:03:50.124 --> 00:03:52.545
Well, the fact that we cross,

00:03:52.545 --> 00:03:57.102
that we're at the x-axis,
that y is equal to zero,

00:03:57.102 --> 00:04:00.675
that the derivative is equal
to zero at x equals eight,

00:04:00.675 --> 00:04:02.534
that tells us that the
slope of the tangent line

00:04:02.534 --> 00:04:04.237
of g at that point is zero.

00:04:04.237 --> 00:04:05.539
But that alone does not tell us

00:04:05.539 --> 00:04:07.377
we have a relative minimum point.

00:04:07.377 --> 00:04:09.379
In order to have a relative minimum point,

00:04:09.379 --> 00:04:14.207
our derivative has to cross
from being negative to positive.

00:04:14.207 --> 00:04:15.730
Why is that valuable?

00:04:15.730 --> 00:04:17.852
Because think about if
your derivative goes

00:04:17.852 --> 00:04:19.587
from being negative to positive,

00:04:19.587 --> 00:04:21.503
that means your original function goes

00:04:21.503 --> 00:04:24.577
from decreasing to increasing.

00:04:24.577 --> 00:04:27.104
It goes from decreasing to increasing.

00:04:27.104 --> 00:04:29.508
And so you would have a
relative minimum point.

00:04:29.508 --> 00:04:31.788
And the choice that describes that,

00:04:31.788 --> 00:04:33.052
this is starting to get there,

00:04:33.052 --> 00:04:36.470
but this alone isn't enough
for a relative minimum point.

00:04:36.470 --> 00:04:38.373
f is negative before x equals eight

00:04:38.373 --> 00:04:40.050
and positive after x equals eight.

00:04:40.050 --> 00:04:42.538
That's exactly what we just described.

00:04:42.538 --> 00:04:43.487
Let's see about these.

00:04:43.487 --> 00:04:47.593
f is concave up on the
interval around x equals six.

00:04:47.593 --> 00:04:50.720
Well, x equals six is a
little bit unrelated to that.

00:04:50.720 --> 00:04:54.854
There's an interval in the
graph of g around x equals eight

00:04:54.854 --> 00:04:57.290
where g of eight is the smallest value.

00:04:57.290 --> 00:04:59.766
Well, this would be a justification
for a relative minimum,

00:04:59.766 --> 00:05:01.683
but it is not calculus-based.

00:05:01.683 --> 00:05:04.602
So once again I'll rule
that one out as well.

00:05:04.602 --> 00:05:06.880
Let's do one more of these.

00:05:06.880 --> 00:05:10.851
So same setup, although we
have a different f and g here,

00:05:10.851 --> 00:05:13.361
and we see it every time with the graph.

00:05:13.361 --> 00:05:16.193
What is a appropriate
calculus-based justification

00:05:16.193 --> 00:05:20.372
for the fact that g is
positive on the interval

00:05:20.372 --> 00:05:23.374
from the closed interval from seven to 12?

00:05:23.374 --> 00:05:28.047
So the positive on the closed
interval from seven to 12.

00:05:28.047 --> 00:05:29.097
So this is interesting.

00:05:29.097 --> 00:05:30.160
Let's just remind ourselves.

00:05:30.160 --> 00:05:31.902
Here we're gonna think a little bit deeper

00:05:31.902 --> 00:05:34.011
about what it means to
be this definite integral

00:05:34.011 --> 00:05:35.604
from zero to x.

00:05:35.604 --> 00:05:39.771
So, if we think about what
happens when x is equal to seven.

00:05:42.289 --> 00:05:44.427
When x is equal to seven, or
another way to think about it,

00:05:44.427 --> 00:05:47.448
g of seven is going to be the integral

00:05:47.448 --> 00:05:50.115
from zero to seven of f of t dt.

00:05:51.993 --> 00:05:53.778
And so the integral from zero to seven,

00:05:53.778 --> 00:05:54.867
if this was a t-axis,

00:05:54.867 --> 00:05:58.291
and, once again, t is just
kind of a placeholder variable

00:05:58.291 --> 00:06:00.788
to help us keep this x up here.

00:06:00.788 --> 00:06:04.614
But we're really talking about
this area right over here.

00:06:04.614 --> 00:06:07.154
And because from zero to seven

00:06:07.154 --> 00:06:09.233
this function is above the x-axis,

00:06:09.233 --> 00:06:11.896
this is going to be a positive area.

00:06:11.896 --> 00:06:14.292
This is a positive area.

00:06:14.292 --> 00:06:16.643
And as we go from seven to 12,

00:06:16.643 --> 00:06:19.150
we're not adding anymore area,

00:06:19.150 --> 00:06:21.125
but we're also not taking any away.

00:06:21.125 --> 00:06:24.667
So actually g of seven
all the way to g of 12

00:06:24.667 --> 00:06:26.431
is going to be the same positive value,

00:06:26.431 --> 00:06:28.496
'cause we're not adding anymore value.

00:06:28.496 --> 00:06:30.702
And when I say g of 12,

00:06:30.702 --> 00:06:35.015
g of 12 is going to be
actually equal to g of seven,

00:06:35.015 --> 00:06:37.559
because, once again, no
added area right here,

00:06:37.559 --> 00:06:39.126
positive or negative.

00:06:39.126 --> 00:06:41.081
So let's see which of these choices match.

00:06:41.081 --> 00:06:44.043
For an x value in the
interval from seven to 12,

00:06:44.043 --> 00:06:46.626
the value of f of x is zero.

00:06:48.459 --> 00:06:51.559
That is true, but that doesn't
mean that we were positive.

00:06:51.559 --> 00:06:53.294
For example, before that interval

00:06:53.294 --> 00:06:55.398
if our function did something like this,

00:06:55.398 --> 00:06:57.712
then we would've had negative
area up to that point,

00:06:57.712 --> 00:06:59.218
and so these would be negative values,

00:06:59.218 --> 00:07:00.779
so I would rule that out.

00:07:00.779 --> 00:07:03.261
For any x value in the
interval from seven to 12,

00:07:03.261 --> 00:07:07.428
the closed interval, the
value of g of x is positive.

00:07:08.492 --> 00:07:12.404
For any x value in the
interval from seven to 12,

00:07:12.404 --> 00:07:15.058
the value of g of x is positive.

00:07:15.058 --> 00:07:18.538
That is true, so I like this one.

00:07:18.538 --> 00:07:19.884
Let me see these other ones.

00:07:19.884 --> 00:07:24.051
f is positive over the closed
interval from zero to seven,

00:07:25.846 --> 00:07:29.179
and it is non-negative over seven to 12.

00:07:31.094 --> 00:07:32.602
I like this one as well.

00:07:32.602 --> 00:07:34.740
And actually the reason why I
would rule out this first one,

00:07:34.740 --> 00:07:37.202
this first one has nothing
to do with the derivative

00:07:37.202 --> 00:07:39.317
and so it's not a
calculus-based justification,

00:07:39.317 --> 00:07:41.169
so I would rule that one out.

00:07:41.169 --> 00:07:42.002
This one is good.

00:07:42.002 --> 00:07:44.073
This is the exact rationale
that I was talking about.

00:07:44.073 --> 00:07:46.192
f is positive from zero to seven,

00:07:46.192 --> 00:07:48.195
so it develops all this positive area,

00:07:48.195 --> 00:07:50.328
and it's non-negative over the interval.

00:07:50.328 --> 00:07:54.082
And so we are going to stay
positive this entire time for g,

00:07:54.082 --> 00:07:58.249
which is the area under f and
above the x-axis from zero

00:07:59.553 --> 00:08:01.887
to our whatever x we wanna pick.

00:08:01.887 --> 00:08:04.192
So I like this choice here.

00:08:04.192 --> 00:08:07.525
f is neither concave up nor concave down

00:08:08.457 --> 00:08:11.957
over the closed interval from seven to 12.

00:08:14.098 --> 00:08:16.169
No, that doesn't really help us

00:08:16.169 --> 00:08:19.644
in saying that g is
positive over that interval.

00:08:19.644 --> 00:08:21.570
So there you go, choice C.


_-substitution: definite integrals

WEBVTT
Kind: captions
Language: en

00:00:01.263 --> 00:00:02.795
- [Instructor] What we're
going to do in this video

00:00:02.795 --> 00:00:04.189
is get some practice

00:00:04.189 --> 00:00:07.973
applying u-substitution
to definite integrals.

00:00:07.973 --> 00:00:10.203
So let's say we have the integral,

00:00:10.203 --> 00:00:13.964
so we're gonna go from x
equals one to x equals two,

00:00:13.964 --> 00:00:16.890
and the integral is two x times

00:00:16.890 --> 00:00:20.307
x squared plus one to the third power dx.

00:00:21.905 --> 00:00:24.320
So I already told you that we're
gonna apply u-substitution,

00:00:24.320 --> 00:00:26.898
but it's interesting to be able
to recognize when to use it.

00:00:26.898 --> 00:00:28.570
And the key giveaway here is,

00:00:28.570 --> 00:00:30.752
well I have this x
squared plus one business

00:00:30.752 --> 00:00:32.006
to the third power,

00:00:32.006 --> 00:00:34.305
but then I also have the
derivative of x squared plus one

00:00:34.305 --> 00:00:36.604
which is two x right over here.

00:00:36.604 --> 00:00:38.508
So I can do the substitution.

00:00:38.508 --> 00:00:40.110
I could say,

00:00:40.110 --> 00:00:42.780
u is equal to x squared plus one,

00:00:42.780 --> 00:00:46.356
in which case the derivative
of u with respect to x

00:00:46.356 --> 00:00:50.211
is just two x plus zero or just two x.

00:00:50.211 --> 00:00:52.695
I could write that in differential form.

00:00:52.695 --> 00:00:54.828
A mathematical hand waving
way of thinking about it

00:00:54.828 --> 00:00:57.220
is multiplying both sides by dx.

00:00:57.220 --> 00:01:00.720
And so you get du is equal to two x dx.

00:01:01.655 --> 00:01:03.582
And so at least this part of the integral

00:01:03.582 --> 00:01:05.324
I can rewrite.

00:01:05.324 --> 00:01:06.995
So let me at least write,

00:01:06.995 --> 00:01:08.133
so this is going to be,

00:01:08.133 --> 00:01:09.201
I'll write the integral.

00:01:09.201 --> 00:01:11.639
We're going think about
the bounds in a second.

00:01:11.639 --> 00:01:14.306
So we have u to the third power,

00:01:15.865 --> 00:01:17.948
u, the same orange color,

00:01:19.418 --> 00:01:21.168
u to the third power.

00:01:23.087 --> 00:01:25.989
That's this stuff right over here.

00:01:25.989 --> 00:01:27.156
And then two x

00:01:28.938 --> 00:01:29.771
times dx.

00:01:30.935 --> 00:01:32.909
Remember you could just view this as two x

00:01:32.909 --> 00:01:35.672
times x squared plus one to
the third power times dx.

00:01:35.672 --> 00:01:37.483
So two x times dx,

00:01:37.483 --> 00:01:39.666
well two x times dx, that is du.

00:01:39.666 --> 00:01:41.338
So that and that together,

00:01:41.338 --> 00:01:42.255
that is du.

00:01:43.428 --> 00:01:44.728
Now an interesting question,

00:01:44.728 --> 00:01:46.493
because this isn't an indefinite integral,

00:01:46.493 --> 00:01:47.979
we're not just trying to
find the antiderivative.

00:01:47.979 --> 00:01:49.442
This is a definite integral.

00:01:49.442 --> 00:01:52.414
So what happens to our
bounds of integration?

00:01:52.414 --> 00:01:54.829
Well there's two ways that
you can approach this.

00:01:54.829 --> 00:01:57.267
You can change your bounds of integration.

00:01:57.267 --> 00:02:00.750
Because this one is x
equals one to x equals two.

00:02:00.750 --> 00:02:03.257
But now we're integrating
with respect to u.

00:02:03.257 --> 00:02:06.346
So one way if you want to keep
your bounds of integration,

00:02:06.346 --> 00:02:08.180
or if you wanna keep
this a definite integral

00:02:08.180 --> 00:02:09.434
I guess you could say,

00:02:09.434 --> 00:02:11.988
you would change your bounds
from u is equal to something

00:02:11.988 --> 00:02:14.891
to u is equal to something else.

00:02:14.891 --> 00:02:16.632
And so your bounds of integration,

00:02:16.632 --> 00:02:20.138
let's see when x is
equal to one, what is u?

00:02:20.138 --> 00:02:21.857
Well when x is equal to one,

00:02:21.857 --> 00:02:24.922
you have one squared plus one,

00:02:24.922 --> 00:02:28.846
so you have two, u is equal
to two in that situation.

00:02:28.846 --> 00:02:32.097
When x is equal to two, what is u?

00:02:32.097 --> 00:02:34.395
Well you have two squared which is four

00:02:34.395 --> 00:02:36.694
plus one, which is five,

00:02:36.694 --> 00:02:38.691
so u is equal to five.

00:02:38.691 --> 00:02:40.595
And you won't typically
see someone writing

00:02:40.595 --> 00:02:42.589
u equal two or u equals five.

00:02:42.589 --> 00:02:44.586
It's often just from two to five

00:02:44.586 --> 00:02:46.490
because we're integrating
with respect to u,

00:02:46.490 --> 00:02:49.532
you assume it's u equals
two to u equals five.

00:02:49.532 --> 00:02:52.434
And so we could just rewrite
this as being equal to

00:02:52.434 --> 00:02:54.851
the integral from two to five

00:02:55.964 --> 00:02:57.714
of u to the third du.

00:02:59.075 --> 00:03:00.468
But it's really important to realize

00:03:00.468 --> 00:03:02.465
why we changed our bounds.

00:03:02.465 --> 00:03:04.787
We are now integrating with respect to u,

00:03:04.787 --> 00:03:06.807
and the way we did it is
we used our substitution

00:03:06.807 --> 00:03:07.945
right over here.

00:03:07.945 --> 00:03:09.826
When x equals one, u is equal to two.

00:03:09.826 --> 00:03:10.940
When x is equal to two,

00:03:10.940 --> 00:03:13.564
two squared plus one, u is equal to five.

00:03:13.564 --> 00:03:16.165
And then we can just evaluate
this right over here.

00:03:16.165 --> 00:03:18.650
Let's see this is going to be equal to,

00:03:18.650 --> 00:03:20.623
the antiderivative of u to the third power

00:03:20.623 --> 00:03:23.040
is u to the fourth over four.

00:03:24.524 --> 00:03:28.007
We're gonna evaluate that at five and two.

00:03:28.007 --> 00:03:32.442
And so this is going to be
five to the fourth over four

00:03:32.442 --> 00:03:35.275
minus two to the fourth over four.

00:03:36.575 --> 00:03:38.503
And then we can simplify this if we like,

00:03:38.503 --> 00:03:41.730
but we've just evaluated
this definite integral.

00:03:41.730 --> 00:03:43.402
Now another way to do it

00:03:43.402 --> 00:03:45.235
is to think about the,

00:03:46.722 --> 00:03:50.414
is to try to solve the
indefinite integral in terms of x

00:03:50.414 --> 00:03:52.922
and use u-substitution as an intermediate.

00:03:52.922 --> 00:03:54.919
So one way to think about this is to say,

00:03:54.919 --> 00:03:57.102
well let's just try to evaluate what

00:03:57.102 --> 00:04:00.074
the indefinite integral of two x

00:04:00.074 --> 00:04:04.241
times x squared plus one
to the third power dx is.

00:04:05.809 --> 00:04:07.411
And then whatever this expression

00:04:07.411 --> 00:04:10.128
ends up being algebraically,

00:04:10.128 --> 00:04:13.495
I'm going to evaluate it at x equals two

00:04:13.495 --> 00:04:15.980
and at x equals one.

00:04:15.980 --> 00:04:19.439
And so then you use the
u-substitution right over here,

00:04:19.439 --> 00:04:21.668
and you would get.

00:04:21.668 --> 00:04:24.339
This would simplify using
the same substitution

00:04:24.339 --> 00:04:27.922
as the integral of u,
u to the third power,

00:04:30.724 --> 00:04:34.045
the du, u to the third power du,

00:04:34.045 --> 00:04:35.972
and once again you're going to evaluate

00:04:35.972 --> 00:04:39.548
this whole thing from x equals two

00:04:39.548 --> 00:04:40.570
and then subtract from that

00:04:40.570 --> 00:04:43.237
it evaluated at x equals one.

00:04:44.563 --> 00:04:46.653
And so this is going to be equal to,

00:04:46.653 --> 00:04:49.765
well let's see this is
u to the fourth power

00:04:49.765 --> 00:04:50.598
over four.

00:04:52.412 --> 00:04:55.407
And once again evaluating
it at x equals two

00:04:55.407 --> 00:04:57.799
and then subtracting from
that at x equals one.

00:04:57.799 --> 00:05:00.214
And then now we can just back-substitute.

00:05:00.214 --> 00:05:01.630
We can say hey look,

00:05:01.630 --> 00:05:04.045
u is equal to x squared plus one,

00:05:04.045 --> 00:05:07.714
so this is the same thing
as x squared plus one

00:05:07.714 --> 00:05:10.214
to the fourth power over four,

00:05:11.638 --> 00:05:13.379
and we're now going to evaluate that

00:05:13.379 --> 00:05:16.259
at x equals two and at x equals one.

00:05:16.259 --> 00:05:17.373
And you will notice that you will get

00:05:17.373 --> 00:05:18.488
the exact same thing.

00:05:18.488 --> 00:05:20.043
When you put x equals two here,

00:05:20.043 --> 00:05:22.087
you get two squared plus one which is five

00:05:22.087 --> 00:05:24.966
to the fourth power over
four, that right over there.

00:05:24.966 --> 00:05:28.124
And then minus one squared plus one is two

00:05:28.124 --> 00:05:29.448
to the fourth power over four,

00:05:29.448 --> 00:05:30.748
that right over there.

00:05:30.748 --> 00:05:32.931
So either way you'll get the same result.

00:05:32.931 --> 00:05:34.974
You can either keep it a definite integral

00:05:34.974 --> 00:05:37.133
and then change your bounds of integration

00:05:37.133 --> 00:05:39.153
and express them in terms of u.

00:05:39.153 --> 00:05:40.431
That's one way to do it.

00:05:40.431 --> 00:05:43.310
The other way is to try to evaluate

00:05:43.310 --> 00:05:44.935
the indefinite integral,

00:05:44.935 --> 00:05:47.350
use u-substitution as
an intermediary step,

00:05:47.350 --> 00:05:49.602
then back-substitute
back and then evaluate

00:05:49.602 --> 00:05:50.852
at your bounds.


_-substitution: defining _

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:01.767
- [Tutor] What we're
going to do in this video

00:00:01.767 --> 00:00:03.321
is give ourselves some practice

00:00:03.321 --> 00:00:05.410
in the first step of u substitution,

00:00:05.410 --> 00:00:07.196
which is often the most difficult

00:00:07.196 --> 00:00:09.134
for those who are first learning it

00:00:09.134 --> 00:00:12.240
and that's recognizing when
u substitution is appropriate

00:00:12.240 --> 00:00:14.696
and then defining an appropriate u.

00:00:14.696 --> 00:00:17.150
So let's just start with an example here,

00:00:17.150 --> 00:00:21.101
so let's say we wanna take
the indefinite integral

00:00:21.101 --> 00:00:23.268
of two x plus one times

00:00:24.740 --> 00:00:28.323
the square root of x squared plus x, dx,

00:00:30.490 --> 00:00:33.074
does u substitution apply
here and if it does,

00:00:33.074 --> 00:00:34.532
how would you define the u?

00:00:34.532 --> 00:00:37.341
Pause the video and try
to think about that.

00:00:37.341 --> 00:00:39.139
Well, we just have to remind ourselves,

00:00:39.139 --> 00:00:43.551
that u substitution is really
trying to undo the chain rule,

00:00:43.551 --> 00:00:46.688
if we remind ourselves what
the chain rule tells us,

00:00:46.688 --> 00:00:48.754
it says look, if we have
a composite function,

00:00:48.754 --> 00:00:50.587
let's say f of g of x,

00:00:52.845 --> 00:00:55.531
f of g of x and we take
the derivative of that

00:00:55.531 --> 00:00:57.517
with respect to x,

00:00:57.517 --> 00:00:58.683
that that is going to be equal

00:00:58.683 --> 00:01:01.162
to the derivative of the outside function

00:01:01.162 --> 00:01:03.059
with respect to the inside function,

00:01:03.059 --> 00:01:03.892
so f prime

00:01:05.557 --> 00:01:07.000
of g of x

00:01:07.000 --> 00:01:09.093
times the derivative
of the inside function,

00:01:09.093 --> 00:01:12.490
times the derivative
of the inside function

00:01:12.490 --> 00:01:14.024
and so u substitution is all about

00:01:14.024 --> 00:01:17.249
well, do we see a pattern
like that inside the integral?

00:01:17.249 --> 00:01:21.387
Do we see a potential
inside function, a g of x,

00:01:21.387 --> 00:01:24.493
where I see its derivative
being multiplied?

00:01:24.493 --> 00:01:27.139
Well, we see that over here,

00:01:27.139 --> 00:01:30.370
if I look at x squared plus x,

00:01:30.370 --> 00:01:33.212
if I make that the u, what's
the derivative of that?

00:01:33.212 --> 00:01:37.208
Well, the derivative of x
squared plus x is two x plus one,

00:01:37.208 --> 00:01:39.652
so we should make that substitution,

00:01:39.652 --> 00:01:43.319
if we say u is equal to x squared plus x,

00:01:44.162 --> 00:01:46.162
then we could say du dx,

00:01:47.325 --> 00:01:49.748
the derivative of u with respect to x

00:01:49.748 --> 00:01:52.695
is equal to two x plus one,

00:01:52.695 --> 00:01:55.279
if we treat our differentials
like variables or numbers,

00:01:55.279 --> 00:01:57.296
we can multiply both sides by dx,

00:01:57.296 --> 00:01:59.225
which is a little bit of
hand wavy mathematics,

00:01:59.225 --> 00:02:00.687
but it's appropriate here,

00:02:00.687 --> 00:02:04.187
so we could say two x plus one times dx

00:02:05.945 --> 00:02:07.736
and now what's really interesting here

00:02:07.736 --> 00:02:09.798
is we have our u right over there

00:02:09.798 --> 00:02:13.965
and notice we have our
two x plus one times dx,

00:02:14.862 --> 00:02:17.218
in fact it's not conventional
to see an integral

00:02:17.218 --> 00:02:19.319
rewritten the way I'm about
to write it, but I will,

00:02:19.319 --> 00:02:22.201
I could rewrite this integral,
you should really view

00:02:22.201 --> 00:02:24.427
this is the product of three things,

00:02:24.427 --> 00:02:26.105
oftentimes people will just view the dx

00:02:26.105 --> 00:02:29.132
as somehow part of the integral operator,

00:02:29.132 --> 00:02:30.699
but you could rearrange it,

00:02:30.699 --> 00:02:32.573
this would actually be legitimate,

00:02:32.573 --> 00:02:36.323
you could say the integral
of the square root

00:02:37.338 --> 00:02:41.505
of x squared plus x
times two x plus one dx

00:02:43.307 --> 00:02:45.024
and if you wanted to be really clear,

00:02:45.024 --> 00:02:47.012
you could even put all of those things

00:02:47.012 --> 00:02:49.523
in parentheses or something like that

00:02:49.523 --> 00:02:51.038
and so here,

00:02:51.038 --> 00:02:52.121
this is our u

00:02:53.528 --> 00:02:56.361
and this right over here is our du

00:02:58.355 --> 00:03:00.630
and so we could rewrite this as being

00:03:00.630 --> 00:03:02.380
equal to the integral

00:03:03.688 --> 00:03:05.688
of the square root of u,

00:03:06.559 --> 00:03:09.050
'cause x squared plus x is u,

00:03:09.050 --> 00:03:12.633
times du, which is much
easier to evaluate.

00:03:13.742 --> 00:03:14.898
If you are still confused there,

00:03:14.898 --> 00:03:17.285
you might recognize it if I rewrite this

00:03:17.285 --> 00:03:19.044
as u to the one half power,

00:03:19.044 --> 00:03:21.870
because now we could just
use the reverse power rule

00:03:21.870 --> 00:03:22.803
to evaluate this

00:03:22.803 --> 00:03:25.252
and then we would have
to undo the substitution,

00:03:25.252 --> 00:03:27.892
once we figure out what
this anti derivative is,

00:03:27.892 --> 00:03:30.535
we then would then reverse substitute

00:03:30.535 --> 00:03:33.702
the x expression back in for the u.