John Legend: Success through effort

WEBVTT
Kind: captions
Language: en

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The thing is, every successful person you know

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has failed at what they love to do, before.

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I've been unsuccessful in my career,

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you just didn't know about it,

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because I wasn't famous yet.

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And I was trying to be famous!

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I was trying to get a record deal for six years!

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And I kept getting turned down.

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Any successful person you know,

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even though it seems like

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everything just happened for them,

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it usually was preceded by a bunch of failure.

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Or at least a significant amount.

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But the great ones aren't the ones that don't fail,

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they're the ones that learn from their failure,

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and they keep going.

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And they're persistent, and they're tenacious.

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And for me,

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I was trying to get a record deal from 1998 to 2004.

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And I got turned down by every major record label,

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including the one I'm signed to now.

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And for me that means that,

00:00:51.880 --> 00:00:53.140
even when you hear no,

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you have to keep going through all those no's,

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until you get to yes.

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And with learning something new,

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I think, it's the same principle.

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When I was younger, I didn't really know

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that you could just get better at things.

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Because I think we're so used to that idea that

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you're talented at something,

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and it's sort of fixed, you know?

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The idea that, "oh, he's just a good singer," you know.

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And even with singing, I've gotten better at singing.

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as I've gotten older.

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I wasn't singing in a way that was sustainable,

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and I wasn't breathing right.

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I wasn't doing a bunch of things correctly.

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I realized that there was a ceiling that I couldn't crack through,

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unless I actually studied,

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and worked under a coach to get better.

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And this idea that things are just fixed,

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is proven not to be the case.

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You can get better at things, you can grow.

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And even people that are innately talented

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still need to go beyond that basic talent

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that they have and cultivate it.


The Growth Mindset

WEBVTT
Kind: captions
Language: en

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(interviewer) Imagine
ourselves 400 years ago.

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They had about a 15% literacy rate.

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And I suspect that if you were to talk
to someone who could read back then

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and ask them, "What percentage of the
population do you think is capable of

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reading?" They might have said, " Well,
maybe 20%, 25%."

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Now you fast forward 400 years,
we know that's a wildly pessimistic

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prediction or assumption. But
it's actually 99.99% of the population

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is capable of reading. Well, what
similar blinders do we have on today?

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So, I'm excited to have Carol Dweck
here, one of my personal heroes and

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who strongly influenced a lot of the
work at Kahn Academy.

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Carol, great to have you here.

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(Carol) Pleasure to be here Sal.

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(Sal) So, so, what I like

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to start off with this word:
growth mindset.

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Which I believe you
came up with.

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(Carol) Mmhm.
(Sal) Um, when did you come up

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with that? What was the
motivation and what is it?

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(Carol) A growth mindset is when

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students believe that their abilities
can be developed. A fixed mindset is

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when they think, "I just have a certain
amount and that's it."

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(Sal) This isn't just kind of
feel good talk,

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this is actually based
in science, that you actually,

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the brain actually does grow stronger,
neural connections actually do form

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when you struggle.

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(Carol) In study after

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study we have shown that kids

00:01:18.592 --> 00:01:24.078
who have a growth mindset get
higher grades. It's not a choice

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between the outcome and the mindset.
It shows that, if kids engage deeply and

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effectively in a learning process, their
grades and test scores are a natural

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byproduct. Kids who are praised for
their intelligence, our research shows,

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don't want a challenge afterwards, they
don't want to work hard on something, and

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if they had difficulty, that's it. We find
that when we praise or parents praise

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the process the child engages in, their
hard work, but not just hard work, their

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strategies, their ideas, their focus,
their perseverance, then the student

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learns these are the ingredients
of success. If it gets harder,

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I'll just do these things.
We've already done work

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with you, uh, inserting growth
mindset statements before a math

00:02:28.094 --> 00:02:32.900
problems, and we found together
that kids did better.

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(Sal) So one thing
that I hope is,

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the folks listening to these
videos, go out there and tell

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their parents, tell their children, tell
their peers, tell their teachers, tell

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their students about growth mindset.
Tell them that their brain grows when

00:02:46.271 --> 00:02:48.121
they get a question
wrong, when they

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struggle, when they
look at their errors and

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they say, "Hey, that's an
interesting!" That you shouldn't be

00:02:53.274 --> 00:02:56.352
ashamed of your mistake, that
you should view that as something

00:02:56.352 --> 00:02:59.671
that is interesting; something to
explore. When you do that, you will

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actually, physically, form neural
connections. Your brain will

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actually grow.


You Can Learn Anything

WEBVTT
Kind: captions
Language: en

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(bright music)

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- [Narrator] Nobody's born smart.

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We all start at zero.

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Can't talk, can't walk,

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certainly can't do algebra.

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Adding, reading, writing,

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riding a bike,

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nobody's good at anything at first.

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There was a time when
Einstein couldn't count to 10

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and Shakespeare had to learn his ABCs

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just like the rest of us.

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Thankfully, we're born to learn.

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Slowly, surely, you stumble, slip, crawl,

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fall, and fail, and fall,

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frustrating, confusing,
trying, struggling,

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until one day, you walk.

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One foot in front of the other.

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One idea on top of the next.

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Each wrong answer making your
brain a little bit stronger.

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Failing is just another word for growing

00:01:01.350 --> 00:01:03.260
and you keep going.

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This is learning.

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Knowing that you'll get it

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even if you haven't gotten it yet.

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Because the most
beautiful, complex concepts

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in the whole universe
are built on basic ideas

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that anyone anywhere can understand.

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Whoever you are, wherever you are,

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you only have to know one thing,

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you can learn anything.


Worked example: Area between two polar graphs

WEBVTT
Kind: captions
Language: en

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- [Voiceover] We have
two polar graphs here,

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r is equal to 3 sine
theta and r is equal to

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3 cosine theta and what we want to do

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is find this area shaded in blue.

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That's kind of the overlap
of these two circles.

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So I encourage you to pause

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the video and give it a go.

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All right so I assume
you've tried and what's

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interesting here is we're clearly bounded

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by two different polar graphs.

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And it looks like they
intersect right over here.

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If we eyeball it, it looks
like they're intersecting

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at when theta is equal to pi

00:00:35.396 --> 00:00:37.966
over four, and we can verify that.

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Cosine of pi over four
is the same thing as sine

00:00:41.775 --> 00:00:46.396
of pi over four so it is
indeed the case that these two

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things are going to be
equal to each other.

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Their point of intersection happens at

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theta is equal to pi over four.

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And if that wasn't as
obvious, you'd set these

00:00:58.217 --> 00:01:00.290
two equal to each other
and figure out the thetas

00:01:00.290 --> 00:01:01.941
where this actually happened, but here it

00:01:01.941 --> 00:01:03.519
jumps out at you a little more.

00:01:03.835 --> 00:01:05.881
So this is theta is equal to pi over four.

00:01:06.929 --> 00:01:11.337
And so the key is to
realize is that for theta

00:01:11.337 --> 00:01:16.337
being between zero and pi
over four we're bounded

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by the red circle, we're
bounded by r is equal

00:01:20.914 --> 00:01:24.246
to 3 sine theta and then
as we go from pi over four

00:01:24.246 --> 00:01:27.679
to pi over two we're
bounded by the black circle,

00:01:27.679 --> 00:01:31.309
we're bounded by r is
equal to 3 cosine theta.

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So we can just break up our
area into those two regions.

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So this first area right
over here we already know

00:01:38.276 --> 00:01:41.778
is going to be one half
times the definite integral

00:01:41.778 --> 00:01:46.778
from zero to pi over four
of, what are we bounded by,

00:01:48.484 --> 00:01:53.484
3 sine of theta and we're gonna
square that thing d theta.

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That's the orange region
and then this, I guess you

00:01:58.875 --> 00:02:01.781
could say this blue
region right over here is

00:02:01.781 --> 00:02:05.100
going to be one half times
the definite integral

00:02:05.100 --> 00:02:09.796
and now we're going to go from
pi over four to pi over two

00:02:11.595 --> 00:02:16.595
of 3 cosine theta squared d theta.

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That's this region right over here.

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Now one thing that might jump out to you

00:02:22.218 --> 00:02:23.686
is that they're going to be the same area.

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These two circles they are
symmetric around this line.

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Theta is equal to pi over four.

00:02:31.993 --> 00:02:33.837
So these are going to be the same area.

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So one thing that we
could do is just solve for

00:02:35.773 --> 00:02:38.653
one of these and then
double it and we will get

00:02:38.653 --> 00:02:40.987
the total region that we care about.

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So the total area, and you can

00:02:43.429 --> 00:02:45.421
verify that for yourself if you like,

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but I'm just going to say
the total area, I'm just

00:02:47.170 --> 00:02:48.808
going to double this right over here.

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So the total area if I just double this,

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just the orange expression,
I'm going to get the

00:02:53.174 --> 00:02:58.143
definite integral from
zero to pi over four of...

00:03:00.050 --> 00:03:05.050
nine, three squared is nine
sine squared theta d theta.

00:03:07.090 --> 00:03:09.296
And you could evaluate this by hand,

00:03:09.296 --> 00:03:10.520
you could evaluate this by calculator,

00:03:10.520 --> 00:03:12.309
let's evaluate this analytically.

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So sine square theta is
the same thing as one half

00:03:16.340 --> 00:03:21.340
times one minus cosine of two theta.

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That's a trigonometric
identity that we've seen

00:03:26.968 --> 00:03:29.283
a lot in trigonometry class,

00:03:29.283 --> 00:03:30.747
actually let me just write it up here.

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So sine square theta is
equal to one half times

00:03:35.048 --> 00:03:38.381
one minus cosine of two theta.

00:03:38.782 --> 00:03:42.584
So if we replace this
with this it's going to be

00:03:42.584 --> 00:03:46.072
equal to, let's take the one half out,

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so we're going to get
nine halves times the

00:03:48.581 --> 00:03:53.240
definite integral from
zero to pi over four of

00:03:53.240 --> 00:03:57.284
one minus cosine two theta d theta.

00:03:57.899 --> 00:04:01.190
And so this is going to
be equal to nine halves

00:04:03.189 --> 00:04:06.657
antiderivative of one
is theta and let's see

00:04:06.657 --> 00:04:11.657
cosine two theta, it's
going to be negative sine

00:04:12.136 --> 00:04:17.136
of two theta, negative
sine of two theta over two.

00:04:23.925 --> 00:04:25.776
Actually let me just, negative

00:04:26.606 --> 00:04:30.368
one half sine of two theta.

00:04:30.368 --> 00:04:32.297
And you could do u substitution
and do it like this,

00:04:32.297 --> 00:04:34.197
but this you might be
able to do in your head.

00:04:34.197 --> 00:04:36.944
And you can verify the
derivative of sine two theta

00:04:36.944 --> 00:04:41.018
is going to be two cosine
of two theta and then

00:04:41.018 --> 00:04:45.110
you multiply it times
the negative one half,

00:04:45.110 --> 00:04:47.240
you just get a negative
one right over there.

00:04:48.723 --> 00:04:52.384
And so then we are going to evaluate

00:04:53.237 --> 00:04:57.856
this at pi over four and at zero.

00:04:58.351 --> 00:05:00.937
So if you evaluate it
at, well luckily if you

00:05:00.937 --> 00:05:02.803
evaluate this thing at
zero this whole thing

00:05:02.803 --> 00:05:04.014
is going to be zero so we really just

00:05:04.014 --> 00:05:05.850
have to evaluate it at pi over four.

00:05:05.965 --> 00:05:09.949
So this is going to be
equal to nine halves times

00:05:09.949 --> 00:05:14.949
pi over four minus one
half sine of two times

00:05:19.813 --> 00:05:24.618
pi over four is pi over
two, sine of pi over two.

00:05:24.657 --> 00:05:28.220
Sine of pi over two we
already know is one,

00:05:28.220 --> 00:05:30.873
so it is really pi over
four, so this right over

00:05:30.873 --> 00:05:32.457
here is just going to be equal to one.

00:05:33.114 --> 00:05:38.114
So this is going to be nine halves times,

00:05:38.486 --> 00:05:40.237
we could say pi over
four minus one half or

00:05:40.237 --> 00:05:45.237
we could say pi over
four minus two over four.

00:05:46.662 --> 00:05:49.067
So we could write it like
that or we could multiply

00:05:49.067 --> 00:05:51.762
everything out or we could
say this is going to be

00:05:51.762 --> 00:05:56.762
equal to nine pi minus 18
over eight and we are done.


Worked example: Area enclosed by cardioid

WEBVTT
Kind: captions
Language: en

00:00:00.809 --> 00:00:03.472
- So this darker curve in blue

00:00:03.472 --> 00:00:07.552
is the graph of r is equal
to 1 minus cosine of theta,

00:00:07.552 --> 00:00:09.848
of course we're dealing
in polar coordinates here.

00:00:09.848 --> 00:00:13.121
And what I'm interested in is
to see if we can figure out

00:00:13.121 --> 00:00:17.344
the area enclosed by this curve.

00:00:17.344 --> 00:00:19.135
And I encourage you to pause the video

00:00:19.135 --> 00:00:21.247
and try it on your own.

00:00:21.247 --> 00:00:23.592
Alright, let's work through it together.

00:00:23.592 --> 00:00:26.424
So we've already seen, we've
already given ourselves

00:00:26.424 --> 00:00:28.416
the intuition for the formula,

00:00:28.416 --> 00:00:33.400
that the area enclosed by a polar graph

00:00:33.400 --> 00:00:37.631
is going to be equal to one
half the definite integral

00:00:37.631 --> 00:00:42.631
from our starting theta
to our ending theta,

00:00:43.135 --> 00:00:48.135
from alpha to beta, of r
of theta squared d theta.

00:00:51.063 --> 00:00:55.328
And so we, essentially,
just have to apply this

00:00:55.328 --> 00:00:58.103
to this function right over here.

00:00:58.103 --> 00:01:01.143
So in this case, the area
is going to be equal to

00:01:01.143 --> 00:01:03.871
one half the definite integral.

00:01:03.871 --> 00:01:06.551
Now what's our alpha and what's our beta?

00:01:06.551 --> 00:01:10.127
Well, we're going from theta
is equal to zero radians,

00:01:10.127 --> 00:01:11.551
and we're essentially going all the way--

00:01:11.551 --> 00:01:13.824
When theta is equal to
zero radians is 1 minus 1

00:01:13.824 --> 00:01:15.471
we're right over there.

00:01:15.471 --> 00:01:18.335
And then we go all the way around

00:01:18.335 --> 00:01:21.647
to theta is equal to two pi radians.

00:01:21.647 --> 00:01:24.768
Notice when we're back at two
pi, cosine of two pi is one,

00:01:24.768 --> 00:01:26.542
one minus one is zero again.

00:01:26.542 --> 00:01:28.118
So we get back to that point.

00:01:28.118 --> 00:01:30.478
So we're going from theta
is equal to zero radians

00:01:30.478 --> 00:01:34.671
to theta is equal to two pi radians.

00:01:34.671 --> 00:01:37.470
Now what's r of theta squared?

00:01:37.470 --> 00:01:39.069
Maybe I'll color code this a little bit.

00:01:39.069 --> 00:01:41.190
R of theta squared.

00:01:41.190 --> 00:01:44.831
Well it's just going to be
one minus cosine of theta.

00:01:44.831 --> 00:01:49.135
One minus cosine theta squared.

00:01:49.135 --> 00:01:51.782
And of course we have our d theta.

00:01:51.782 --> 00:01:54.222
We have our d theta.

00:01:54.222 --> 00:01:56.535
And now we just have to
evaluate this integral.

00:01:56.535 --> 00:01:58.750
So once again, at any
point you feel inspired,

00:01:58.750 --> 00:02:01.230
try to evaluate this.

00:02:01.230 --> 00:02:03.046
So let's do this, alright.

00:02:03.046 --> 00:02:06.182
So what I would do...

00:02:06.182 --> 00:02:08.166
So this is going to be equal to one half

00:02:08.166 --> 00:02:10.706
times the definite integral
from zero to two pi.

00:02:10.706 --> 00:02:12.144
And let me expand this out.

00:02:12.144 --> 00:02:17.144
This is going to be one
minus two cosine theta

00:02:19.368 --> 00:02:24.368
plus cosine squared theta, d theta.

00:02:24.528 --> 00:02:26.368
D theta.

00:02:26.368 --> 00:02:28.072
Now I know how to take the
anti-derivative of one,

00:02:28.072 --> 00:02:29.230
I know how to take the anti-derivitive of

00:02:29.230 --> 00:02:32.238
negative cosine of theta,
but cosine squared theta,

00:02:32.238 --> 00:02:33.390
this is a little bit...

00:02:33.390 --> 00:02:35.605
It doesn't jump out at you
that you can just do this,

00:02:35.605 --> 00:02:37.979
use u-substitution or something like that,

00:02:37.979 --> 00:02:41.220
but lucky for us, we have
our trigonometric identities,

00:02:41.220 --> 00:02:42.995
and so we know.

00:02:42.995 --> 00:02:46.306
We know that cosine squared of theta

00:02:46.306 --> 00:02:49.963
is just the same thing
as one half times one

00:02:49.963 --> 00:02:53.283
plus cosine of two theta.

00:02:53.283 --> 00:02:55.011
You learned this in trigonometry class.

00:02:55.011 --> 00:02:57.324
If you didn't, we'll,
you've learned it just now.

00:02:57.324 --> 00:02:59.770
And that's why this is
one of the more useful

00:02:59.770 --> 00:03:03.544
trigonometric identities if
you're finding any type of

00:03:03.544 --> 00:03:06.381
anti-derivative or if
you're integrating anything.

00:03:06.381 --> 00:03:07.334
And so let's do that.

00:03:07.334 --> 00:03:10.846
Let's rewrite this right over here as

00:03:10.846 --> 00:03:15.846
one half times one plus
cosine of two theta.

00:03:18.667 --> 00:03:20.461
And let's see, and maybe we could...

00:03:20.461 --> 00:03:23.469
Yeah, let's just do it like that.

00:03:23.469 --> 00:03:25.252
I guess we could, if we wanted...

00:03:25.252 --> 00:03:26.928
Well, we'll just do it like that.

00:03:26.928 --> 00:03:31.928
So this is going to be equal to one half,

00:03:32.161 --> 00:03:34.865
and then we are going to...

00:03:34.865 --> 00:03:37.545
One half, now let's just
start taking anti-derivatives.

00:03:37.545 --> 00:03:38.979
One half.

00:03:39.589 --> 00:03:43.572
Now the anti-derivative of
one, with respect to theta,

00:03:43.572 --> 00:03:45.896
is just going to be theta.

00:03:45.896 --> 00:03:49.104
The anti-derivative of
negative two cosine of theta,

00:03:49.104 --> 00:03:51.793
well that's just going to be
negative two sine of theta.

00:03:51.793 --> 00:03:54.225
Negative two sine theta.

00:03:54.225 --> 00:03:55.057
You can take the derivative,

00:03:55.057 --> 00:03:56.415
the derivative of sine is cosine

00:03:56.415 --> 00:03:58.903
and the negative two,
it'll just multiple it

00:03:58.903 --> 00:04:00.624
times the derivative of sine of theta

00:04:00.624 --> 00:04:02.545
so it's negative two cosine of theta.

00:04:02.545 --> 00:04:05.233
And then we're going to have, let's see...

00:04:05.233 --> 00:04:06.348
So let me distribute this.

00:04:06.348 --> 00:04:11.260
This is the same thing
as one half plus one half

00:04:11.260 --> 00:04:13.830
cosine of two theta.

00:04:13.830 --> 00:04:16.516
So let's just assume it's this way.

00:04:16.516 --> 00:04:19.380
So the anti-derivative of one half,

00:04:19.380 --> 00:04:21.677
so the anti-derivative of one-half.

00:04:21.677 --> 00:04:23.100
So I'm really looking at
that right over there.

00:04:23.100 --> 00:04:25.693
It's going to be one half theta.

00:04:25.693 --> 00:04:27.084
One half theta.

00:04:27.084 --> 00:04:29.659
And then the anti-derivative of one half

00:04:29.659 --> 00:04:32.705
cosine of two theta, let's see...

00:04:33.612 --> 00:04:35.470
The derivative of sine of two theta

00:04:35.470 --> 00:04:38.629
is two cosine of two theta.

00:04:39.860 --> 00:04:42.274
So the anti-derivative of this is...

00:04:42.274 --> 00:04:44.979
The anti-derivative of
cosine of two theta,

00:04:44.979 --> 00:04:46.779
and you can do u-substitution if you like,

00:04:46.779 --> 00:04:49.128
but you might be able
to do this in your head,

00:04:49.128 --> 00:04:51.848
the anti-derivative of cosine
of two theta is going to be

00:04:51.848 --> 00:04:53.848
one half sine of two theta.

00:04:53.848 --> 00:04:56.071
And then you have this
one half right over here.

00:04:56.071 --> 00:04:58.061
So this is going to be...

00:04:58.061 --> 00:05:00.549
Let me show you what I'm
finding the anti-derivative of.

00:05:00.549 --> 00:05:02.868
Of that right over there, of this,

00:05:02.868 --> 00:05:04.437
and I guess this right over here.

00:05:04.437 --> 00:05:09.437
So this is going to be plus
one fourth sine of two theta.

00:05:12.166 --> 00:05:14.218
And I encourage you to
find the derivative here

00:05:14.218 --> 00:05:16.447
if that last part was
a little bit confusing.

00:05:16.447 --> 00:05:17.855
The derivative of sine of two theta

00:05:17.855 --> 00:05:19.607
is two cosine of two theta,

00:05:19.607 --> 00:05:21.934
two over one fourth is one half,

00:05:21.934 --> 00:05:24.406
you get to one half cosine of two theta.

00:05:24.406 --> 00:05:27.174
And we're going to
evaluate that at two pi,

00:05:27.174 --> 00:05:30.343
at two pi, and at zero.

00:05:30.343 --> 00:05:32.726
So when you evaluate
it, one thing that might

00:05:32.726 --> 00:05:35.377
jump out at you is when
you evaluate this at zero,

00:05:35.377 --> 00:05:37.257
this whole thing,
everything, every term here

00:05:37.257 --> 00:05:39.734
is just going to be zero, so
that simplifies things nicely.

00:05:39.734 --> 00:05:42.843
So we really just have
to take one half of...

00:05:42.843 --> 00:05:44.179
It evaluated at two pi.

00:05:44.179 --> 00:05:48.789
So this is going to be
one half times two pi,

00:05:49.296 --> 00:05:53.329
two pi, and then sine of two pi is zero,

00:05:53.329 --> 00:05:55.104
so that's just going to be zero.

00:05:55.104 --> 00:05:58.199
And then plus one half times two pi,

00:05:58.199 --> 00:06:00.588
so that's going to be plus pi.

00:06:00.588 --> 00:06:03.997
And then sign of two times
two pi sine of four pi,

00:06:03.997 --> 00:06:05.494
that's still going to be zero.

00:06:05.494 --> 00:06:08.110
So this is going to be zero as well,

00:06:08.110 --> 00:06:10.397
and we are almost done.

00:06:10.397 --> 00:06:13.318
So this is going to be
one half times three pi

00:06:13.318 --> 00:06:18.318
or three halves, three
halves pi is the area,

00:06:19.514 --> 00:06:23.514
is the area of this region.


Evaluating definite integral with calculator

WEBVTT
Kind: captions
Language: en

00:00:00.523 --> 00:00:02.222
- [Voiceover] In the last
video, we tried to find

00:00:02.222 --> 00:00:05.948
the area of the region, I
guess this combined area

00:00:05.948 --> 00:00:08.999
between the blue and this
orange, the area, I guess

00:00:08.999 --> 00:00:12.876
the overlap between these two
circles and we came up with

00:00:12.876 --> 00:00:16.612
nine pi minus 18 all of that over eight.

00:00:16.612 --> 00:00:18.316
What I want to do in this
video, you could have also

00:00:18.316 --> 00:00:21.754
used a typical graphing
calculator to come up

00:00:21.754 --> 00:00:24.211
with the same result, and
it would have actually

00:00:24.211 --> 00:00:27.039
evaluated the definite integral.

00:00:27.039 --> 00:00:29.206
So let's see how you do that.

00:00:29.206 --> 00:00:30.706
Now this, what I'm doing
here, you could do this

00:00:30.706 --> 00:00:34.154
for a traditional, what if
you're dealing with Cartesian

00:00:34.154 --> 00:00:35.385
coordinates, rectangular coordinates,

00:00:35.385 --> 00:00:37.033
or for polar coordinates,
cause it's really just about

00:00:37.033 --> 00:00:38.834
evaluating the definite integral.

00:00:38.834 --> 00:00:41.921
So we wanted to evaluate nine
times the definite integral

00:00:41.921 --> 00:00:45.617
from zero to pi over four
sin squared theta d theta,

00:00:45.617 --> 00:00:47.309
so how do I do that?

00:00:47.309 --> 00:00:51.960
Well I can go to second,
calculus, then I do the

00:00:51.960 --> 00:00:55.266
F N INT, that's definite integral.

00:00:56.127 --> 00:00:58.868
So let's use that function,
and then the first thing,

00:00:58.868 --> 00:01:00.203
you want to say "well, what are you taking

00:01:00.203 --> 00:01:01.795
"the definite integral of?"

00:01:01.795 --> 00:01:04.424
And we're taking the
definite integral of...

00:01:07.054 --> 00:01:09.775
Sine, actually I want
the parentheses, sine,

00:01:09.775 --> 00:01:12.178
and I could use any variable
here, as long as I'm

00:01:12.178 --> 00:01:15.732
consistent with what I'm
integrating with respect to.

00:01:15.732 --> 00:01:18.092
So I tend to use just the "x" button,

00:01:18.092 --> 00:01:19.261
because there is an "x" button,

00:01:19.261 --> 00:01:21.168
but we'll just assume that
in this case x is theta.

00:01:21.168 --> 00:01:26.168
So sine of x squared, instead
of sine of theta squared,

00:01:27.290 --> 00:01:29.227
we're once again assuming
that x is equal to theta.

00:01:29.227 --> 00:01:31.234
And then the next one, you specify,

00:01:31.234 --> 00:01:32.437
"Well what's the variable you're taking

00:01:32.437 --> 00:01:34.292
"the integral with respect to?"

00:01:34.292 --> 00:01:36.912
In this case it's x, if
we'd put in a theta here

00:01:36.912 --> 00:01:39.600
then we would want to put
a theta there as well.

00:01:39.600 --> 00:01:41.741
And then you want the
bounds of integration,

00:01:41.741 --> 00:01:43.873
and you should assume
that your calculator,

00:01:43.873 --> 00:01:46.475
or if you're doing this,
if you're in radiant mode,

00:01:46.475 --> 00:01:48.375
or if you're dealing with
radiants, you should assume

00:01:48.375 --> 00:01:51.443
you're in radiant mode, I just
did before I evaluated this.

00:01:51.443 --> 00:01:54.311
We're going between zero and pi over four.

00:01:55.326 --> 00:01:59.394
Zero and pi over four.

00:02:02.500 --> 00:02:04.480
And then, we get...

00:02:05.781 --> 00:02:07.941
So we get this number,

00:02:07.941 --> 00:02:10.460
and then we wanted to
multiply it times nine.

00:02:10.460 --> 00:02:12.578
So my previous answer times nine.

00:02:12.578 --> 00:02:14.088
If I just press times, it does this,

00:02:14.088 --> 00:02:16.363
"Previous answer times" nine,

00:02:16.363 --> 00:02:18.603
is equal to this number,

00:02:18.603 --> 00:02:20.701
one point two eight four two nine.

00:02:20.701 --> 00:02:22.658
So let's verify that
that's the same exact value

00:02:22.658 --> 00:02:26.154
we got when we actually
evaluated the integral by hand.

00:02:26.154 --> 00:02:31.154
So if we take nine, nine pi
minus 18 divided by eight,

00:02:35.909 --> 00:02:40.149
divided by eight, what do we get?

00:02:40.372 --> 00:02:44.002
We get the exact same value.

00:02:44.249 --> 00:02:45.940
So anyway, hopefully that's satisfying,

00:02:45.940 --> 00:02:47.698
that we got the same value either way,

00:02:47.698 --> 00:02:50.286
and a little exposure for
how you might be able to

00:02:50.286 --> 00:02:52.636
evaluate some integrals
using a calculator,

00:02:52.636 --> 00:02:54.923
which can be useful
when you can't actually

00:02:54.923 --> 00:02:56.888
evaluate them analytically.


Area bounded by polar curves

WEBVTT
Kind: captions
Language: en

00:00:00.591 --> 00:00:02.203
- [Voiceover] We now
have a lot of experience

00:00:02.203 --> 00:00:04.482
finding the areas under
curves when we're dealing

00:00:04.482 --> 00:00:06.558
with things in rectangular coordinates.

00:00:06.558 --> 00:00:07.974
So we saw we took the Riemann sums,

00:00:07.974 --> 00:00:09.577
a bunch of rectangles,
we took the limit as we

00:00:09.577 --> 00:00:12.040
had an infinite number of
infinitely thin rectangles

00:00:12.040 --> 00:00:13.469
and we were able to find the area.

00:00:13.920 --> 00:00:16.404
But now let's move on
to polar coordinates.

00:00:16.826 --> 00:00:18.406
And in polar coordinates
I won't say we're finding

00:00:18.406 --> 00:00:21.043
the area under a curve,
but really in this example

00:00:21.043 --> 00:00:23.864
right over here we have
a part of the graph of

00:00:23.864 --> 00:00:26.009
r is equal to f of theta and we've graphed

00:00:26.009 --> 00:00:27.417
it between theta is equal to

00:00:27.417 --> 00:00:29.418
alpha and theta is equal to beta.

00:00:29.967 --> 00:00:33.087
And what I wanna do in
this video is come up with

00:00:33.087 --> 00:00:38.087
a general expression
for this area in blue.

00:00:38.838 --> 00:00:40.902
This area that is bounded,
I guess you could say

00:00:40.902 --> 00:00:45.902
by those angles and the graph
of r is equal to f of theta.

00:00:50.929 --> 00:00:52.880
And I want you to come
up, or at least attempt

00:00:52.880 --> 00:00:55.198
to come up with an expression on your own,

00:00:55.198 --> 00:00:56.949
but I'll give you a
little bit of a hint here.

00:00:57.448 --> 00:00:58.911
When we did it in rectangular coordinates

00:00:58.911 --> 00:01:00.525
we divided things into rectangles.

00:01:00.900 --> 00:01:03.410
Over here rectangles don't
seem as obvious because

00:01:03.410 --> 00:01:05.190
they're all kind of coming to this point,

00:01:05.451 --> 00:01:08.365
but what if we could divide things into

00:01:11.747 --> 00:01:15.116
sectors or I guess we could
say little pie pieces?

00:01:15.735 --> 00:01:18.476
Someone is doing some
serious drilling downstairs.

00:01:18.476 --> 00:01:20.262
I don't if it's picking
up on the microphone.

00:01:20.262 --> 00:01:21.541
But anyway, I will continue.

00:01:22.067 --> 00:01:24.783
So what would happen if
we could divide this into

00:01:24.783 --> 00:01:28.912
a whole series of kind of pie pieces

00:01:33.061 --> 00:01:34.520
and then take the limit as if we had

00:01:34.520 --> 00:01:38.276
an infinite number of pie pieces?

00:01:38.278 --> 00:01:40.225
So we want to find the
area of each of these pie

00:01:40.225 --> 00:01:43.594
pieces and then take the
limit as the pie pieces

00:01:43.594 --> 00:01:46.189
I guess you could say
become infinitely thin

00:01:46.189 --> 00:01:50.591
and we have an infinite number of them.

00:01:50.946 --> 00:01:54.377
And I'll give you one more
hint, for thinking about

00:01:54.377 --> 00:01:56.484
the area of these pie, I guess you could

00:01:56.484 --> 00:01:59.500
say the area of these pie wedges.

00:02:00.074 --> 00:02:03.183
I'll give you another
hint, so if I have a circle

00:02:03.183 --> 00:02:04.978
I'll do my best attempt at a circle.

00:02:05.610 --> 00:02:08.848
Luckily the plumbing or
whatever is going on downstairs

00:02:08.848 --> 00:02:11.669
has stopped for now
allowing me to focus more on

00:02:11.669 --> 00:02:14.346
the calculus, which is
obviously more important.

00:02:14.747 --> 00:02:17.288
All right so if I have
a circle, that's my best

00:02:17.288 --> 00:02:20.252
attempt at a circle, and it's of radius r

00:02:23.030 --> 00:02:25.247
and let me draw a sector of this circle.

00:02:26.158 --> 00:02:29.268
It's a sector of a circle, so
that's obviously r as well.

00:02:29.673 --> 00:02:32.679
And if this angle right
here is theta, what is

00:02:32.679 --> 00:02:37.679
going to be the area of
this sector right over here?

00:02:40.225 --> 00:02:42.474
So that's my hint for you,
think about what this area

00:02:42.474 --> 00:02:45.464
is going to be and we're
assuming theta is in radians.

00:02:45.772 --> 00:02:49.345
Think about what this area
is going to be and then

00:02:49.345 --> 00:02:52.003
see if you can extend
that to what we're trying

00:02:52.003 --> 00:02:54.745
to do here to figure out, somehow

00:02:54.745 --> 00:02:56.548
I'm giving you a hint again.

00:02:56.548 --> 00:03:00.501
Using integration, finding
an expression for this area.

00:03:01.255 --> 00:03:03.127
So I'm assuming you've had a go at it.

00:03:03.127 --> 00:03:05.203
So first let's think about
this, what's the area

00:03:05.203 --> 00:03:09.151
of the entire circle,
well we already know that.

00:03:09.151 --> 00:03:10.881
That's going to be pi r squared,

00:03:11.892 --> 00:03:13.741
formula for the area of a circle.

00:03:14.105 --> 00:03:15.831
And then what's going
to be the area of this?

00:03:16.151 --> 00:03:18.333
Well it's going to be a
fraction of the circle.

00:03:18.882 --> 00:03:23.755
If this is pi, sorry if this
is theta, if we went two pi

00:03:23.755 --> 00:03:25.597
radians that would be the
whole circle so this is

00:03:25.597 --> 00:03:30.014
going to be theta over
two pi of the circle.

00:03:30.591 --> 00:03:35.591
So times theta over two pi would be the

00:03:39.417 --> 00:03:41.678
area of this sector right over here.

00:03:41.678 --> 00:03:43.620
Area of the whole circle
times the proprotion

00:03:43.620 --> 00:03:46.381
of the circle that we've kind of defined

00:03:46.381 --> 00:03:48.159
or that the sector is made up of.

00:03:48.467 --> 00:03:52.125
And so this would give
us, the pis cancel out,

00:03:52.125 --> 00:03:56.181
it would give us one half
r squared times theta.

00:03:57.179 --> 00:04:01.029
Now what happens if instead of theta,

00:04:01.029 --> 00:04:03.568
so let's look at each of these over here.

00:04:04.180 --> 00:04:06.450
So each of these things that I've drawn,

00:04:06.450 --> 00:04:10.249
let's focus on just one of these wedges.

00:04:10.661 --> 00:04:13.771
I will highlight it in orange.

00:04:17.706 --> 00:04:21.060
So instead of the angle
being theta let's just

00:04:21.060 --> 00:04:23.582
assume it's a really,
really, really small angle.

00:04:24.315 --> 00:04:26.067
We'll use a differential
although this is a bit of

00:04:26.067 --> 00:04:27.876
loosey-goosey mathematics
but the important here

00:04:27.876 --> 00:04:29.551
is to give you the
conceptual understanding.

00:04:29.704 --> 00:04:32.847
I could call it a delta
theta and then eventually

00:04:32.847 --> 00:04:35.132
take the limit as our delta
theta approaches zero.

00:04:35.148 --> 00:04:37.180
But just for conceptual
purposes when we have

00:04:37.180 --> 00:04:41.064
a infinitely small or super
small change in theta,

00:04:41.064 --> 00:04:46.064
so let's call that d theta,
and the radius here or I guess

00:04:52.319 --> 00:04:54.068
we could say this length right over here.

00:04:54.068 --> 00:04:55.195
You could view it as the radius of

00:04:55.195 --> 00:04:57.885
at least the arc right at that point.

00:04:58.128 --> 00:05:02.359
It's going to be r as a
function of the thetas that

00:05:02.359 --> 00:05:04.250
we're around right over
here, but we're just going

00:05:04.250 --> 00:05:06.947
to call that our r right over there.

00:05:07.250 --> 00:05:11.060
And so what is going to be the
area of this little sector?

00:05:11.975 --> 00:05:15.098
Well the area of this
little sector is instead of

00:05:15.098 --> 00:05:16.624
my angle being theta I'm calling

00:05:16.624 --> 00:05:19.279
my angle d theta, this
little differential.

00:05:19.690 --> 00:05:21.932
So instead of one half
r squared it's going to

00:05:21.932 --> 00:05:24.300
be, let me do that in a color you can see.

00:05:24.393 --> 00:05:29.393
This area is going to be
one half r squared d theta.

00:05:32.585 --> 00:05:34.101
Notice here the angle
was theta, here the angle

00:05:34.101 --> 00:05:36.493
was d theta, super, super small angle.

00:05:37.225 --> 00:05:40.725
Now if I wanted to take
the sum of all of these

00:05:40.725 --> 00:05:44.302
from theta is equal to alpha
to theta is equal to beta

00:05:44.302 --> 00:05:46.133
and literally there is an
infinite number of these.

00:05:46.133 --> 00:05:47.786
This is an infinitely small angle.

00:05:48.365 --> 00:05:53.365
Well then for the entire
area right over here

00:05:53.855 --> 00:05:56.149
I could just integrate all of these.

00:05:56.646 --> 00:06:01.646
So that's going to be the
integral from alpha to

00:06:03.546 --> 00:06:08.546
beta of one half r
squared d theta where r,

00:06:11.855 --> 00:06:14.196
of course, is a function of theta.

00:06:14.196 --> 00:06:15.510
So you could even write it this way,

00:06:15.510 --> 00:06:18.310
you could write it as
the integral from alpha

00:06:18.310 --> 00:06:23.310
to beta of one half r of
theta squared d theta.

00:06:25.948 --> 00:06:27.798
Just to remind ourselves or assuming r

00:06:27.798 --> 00:06:30.193
is a function of theta in this case.


Partial fraction expansion to evaluate integral

WEBVTT
Kind: captions
Language: en

00:00:00.659 --> 00:00:04.050
- [Voiceover] Try to evaluate
the following integral.

00:00:04.050 --> 00:00:05.618
So assuming you've had a go at it,

00:00:05.618 --> 00:00:06.930
so let's work through this together.

00:00:06.930 --> 00:00:08.353
And if at any point you get inspired,

00:00:08.353 --> 00:00:09.907
always feel free to pause the video

00:00:09.907 --> 00:00:12.851
and continue on with it on your own.

00:00:12.851 --> 00:00:14.417
So the first thing that
might have jumped out at you

00:00:14.417 --> 00:00:15.859
we have a rational expression.

00:00:15.859 --> 00:00:17.843
The degree in the numerator is the same as

00:00:17.843 --> 00:00:19.410
the degree in the nominator,
so maybe a little bit

00:00:19.410 --> 00:00:22.007
of algebraic long division is called for.

00:00:22.267 --> 00:00:23.171
So let's do that.

00:00:23.171 --> 00:00:28.171
Let's take X-squared minus one

00:00:28.188 --> 00:00:32.516
and divide it into X-squared,
put that in a different color,

00:00:32.516 --> 00:00:36.805
divide it into X-squared
plus X minus five.

00:00:36.805 --> 00:00:41.805
So X-squared plus X minus five.

00:00:42.149 --> 00:00:44.020
And so let's look at
the highest degree terms

00:00:44.020 --> 00:00:46.404
how many times does
X-squared go into X-squared?

00:00:46.404 --> 00:00:49.593
It goes one time, let me
write this in a new color.

00:00:49.593 --> 00:00:50.405
Goes one time.

00:00:50.405 --> 00:00:52.405
One times X-squared minus
one is just going to be

00:00:52.405 --> 00:00:55.445
X-squared minus one.

00:00:55.445 --> 00:00:57.477
Now you subtract this
green expression from

00:00:57.477 --> 00:01:00.372
this mauve expression or I could just add

00:01:00.372 --> 00:01:02.276
the negative on it, so let me just take

00:01:02.276 --> 00:01:03.156
the negative of it.

00:01:03.156 --> 00:01:06.214
So it's going to be
negative X-squared plus one

00:01:06.214 --> 00:01:09.959
and we're going to get the X-squared's

00:01:09.959 --> 00:01:11.575
X-square minus X-square is zero,

00:01:11.575 --> 00:01:13.751
so those cancel out and
we're going to be left with

00:01:13.751 --> 00:01:18.751
X and the negative five
plus one is negative four.

00:01:19.734 --> 00:01:22.630
So we have X minus four left over.

00:01:22.630 --> 00:01:24.771
So we can rewrite the
expression that we're trying to

00:01:24.771 --> 00:01:27.324
find the antiderivative of.

00:01:27.324 --> 00:01:32.324
We can rewrite it as one plus X minus four

00:01:38.248 --> 00:01:43.248
over X-squared minus one.

00:01:43.864 --> 00:01:45.192
Maybe I'll do that in that purple color

00:01:45.192 --> 00:01:47.927
since I already used it as the purple.

00:01:47.927 --> 00:01:50.663
Over X-squared minus one.

00:01:50.663 --> 00:01:53.915
So we did one thing, now
we have a lower degree

00:01:53.915 --> 00:01:56.406
in the numerator than we
have in the denominator.

00:01:56.406 --> 00:01:57.928
And obviously this is
fairly straight forward,

00:01:57.928 --> 00:01:59.176
take the antiderivative of,

00:01:59.176 --> 00:02:01.500
but what do we do now?

00:02:01.500 --> 00:02:04.905
It's not clear if we look
at X-squared minus one

00:02:04.905 --> 00:02:06.906
it's derivative would be two-X,

00:02:06.906 --> 00:02:08.715
which is the same degree as this,

00:02:08.715 --> 00:02:10.635
but it's not X minus four,
so it doesn't look like

00:02:10.635 --> 00:02:13.845
you u-substitution it's
going to help us with this.

00:02:13.845 --> 00:02:15.957
So what can we do now?

00:02:15.957 --> 00:02:18.258
Now we can take out another tool in our

00:02:18.258 --> 00:02:22.042
algebraic tool kit, we will do

00:02:22.042 --> 00:02:23.626
partial fraction expansion.

00:02:23.626 --> 00:02:26.241
Which is essentially
writing this as the sum

00:02:26.241 --> 00:02:28.855
of two rational expressions that have

00:02:28.855 --> 00:02:30.721
a lower degree in the denominator.

00:02:30.721 --> 00:02:32.352
So what do I mean by that?

00:02:32.352 --> 00:02:34.226
So this term right over here,

00:02:34.226 --> 00:02:36.016
X minus four over X-squared minus one,

00:02:36.016 --> 00:02:38.001
we can rewrite that as

00:02:38.001 --> 00:02:42.545
X minus four over, instead
of X-squared minus one,

00:02:42.545 --> 00:02:43.539
we can factor this.

00:02:43.539 --> 00:02:46.353
This is X plus one times X minus one.

00:02:46.353 --> 00:02:47.457
So let's write that,

00:02:47.457 --> 00:02:51.728
this is X plus one times X minus one.

00:02:51.728 --> 00:02:53.503
When we think about
partial fraction expansion

00:02:53.503 --> 00:02:55.360
we say okay, can we write this as the sum

00:02:55.360 --> 00:03:00.360
of something, let's call
that A, over X plus one,

00:03:07.104 --> 00:03:10.350
plus something else, let's call that B,

00:03:10.350 --> 00:03:14.063
plus B over X minus one.

00:03:14.063 --> 00:03:15.807
Can we do that?

00:03:15.807 --> 00:03:18.144
And to attempt to do
that, if we had just add

00:03:18.144 --> 00:03:21.120
these two things, what would we get?

00:03:21.120 --> 00:03:23.456
Well we would find a common denominator,

00:03:23.456 --> 00:03:26.048
which would be X plus
one times X minus one,

00:03:26.048 --> 00:03:29.310
and so you would have, if
any of this looks unfamiliar

00:03:29.310 --> 00:03:30.975
I encourage you to review the videos

00:03:30.975 --> 00:03:33.022
on partial fraction
expansion, because that's

00:03:33.022 --> 00:03:35.745
exactly what we're doing right over here.

00:03:35.745 --> 00:03:38.416
But this would be equal to
if you were to add the two

00:03:38.416 --> 00:03:41.280
your common denominator
would be the product.

00:03:41.280 --> 00:03:46.280
So it would be X plus
one times X minus one.

00:03:51.151 --> 00:03:53.309
So the first term I would
multiply the numerator

00:03:53.309 --> 00:03:55.426
and the denominator times X minus one.

00:03:55.426 --> 00:04:00.426
So it would be A times X minus one

00:04:01.408 --> 00:04:03.984
plus B, the second term I'll multiply the

00:04:03.984 --> 00:04:08.984
numerator and the
denominator times X plus one.

00:04:09.554 --> 00:04:10.790
And so what do we get?

00:04:10.790 --> 00:04:15.790
This is going to be equal to AX,

00:04:16.298 --> 00:04:18.523
maybe I'll do this all in one color.

00:04:18.523 --> 00:04:22.571
This is going to be equal to

00:04:22.571 --> 00:04:27.571
AX minus A plus BX plus B

00:04:35.945 --> 00:04:38.483
and then all of that
over this stuff we keep

00:04:38.483 --> 00:04:39.682
writing over and over again.

00:04:39.682 --> 00:04:41.937
Actually, let me just copy and paste this.

00:04:41.937 --> 00:04:46.937
So copy and paste, I can use
that over and over again.

00:04:47.234 --> 00:04:51.058
So we have that over that.

00:04:51.058 --> 00:04:54.754
So let's see, now we
can group the X terms.

00:04:54.754 --> 00:04:59.754
So we can rewrite this as,
so if we take AX plus BX

00:05:01.825 --> 00:05:06.825
that's going to be A plus B times X.

00:05:07.445 --> 00:05:12.445
Then we have a negative A and a B.

00:05:13.173 --> 00:05:17.652
So plus B minus A, and I'll
just put parenthesis around that

00:05:17.652 --> 00:05:21.333
just so I kind of group
these constant terms.

00:05:21.333 --> 00:05:24.437
Then all of that's going to be divided by,

00:05:24.437 --> 00:05:25.943
good thing I copied and pasted that,

00:05:25.943 --> 00:05:29.445
X plus one times X minus one.

00:05:29.445 --> 00:05:33.824
So now this is the crux of
partial fraction expansion.

00:05:33.824 --> 00:05:35.668
We say, okay we kind of went through this

00:05:35.668 --> 00:05:38.101
whole exercise on the thesis
that we could do this,

00:05:38.101 --> 00:05:40.868
that there is some A and
B for which this is true.

00:05:40.868 --> 00:05:43.685
So if there is some A and
B for which this is true,

00:05:43.685 --> 00:05:45.847
then A plus B must be the coefficient

00:05:45.847 --> 00:05:48.340
of the X term right over here.

00:05:48.340 --> 00:05:51.594
So A plus B must be equal to one,

00:05:51.594 --> 00:05:53.396
must be equal to this coefficient.

00:05:53.396 --> 00:05:56.852
And B minus A must be
equal to the constant,

00:05:56.852 --> 00:05:59.095
must be equal to negative four.

00:05:59.095 --> 00:06:01.061
Or if they are then we
will found an A and a B,

00:06:01.061 --> 00:06:02.265
so let's do that.

00:06:02.265 --> 00:06:04.581
I'll do it up here since I have
a little bit of real estate.

00:06:04.581 --> 00:06:09.581
A plus B is going to be equal to one,

00:06:09.764 --> 00:06:12.244
and B minus A or I could write that

00:06:12.244 --> 00:06:17.244
as negative A plus B is
equal to negative four.

00:06:19.765 --> 00:06:21.493
We could add the left hand sides

00:06:21.493 --> 00:06:22.743
and add the right hand sides

00:06:22.743 --> 00:06:24.676
and then the A's would disappear.

00:06:24.676 --> 00:06:28.724
We would get two B is
equal to negative three

00:06:28.724 --> 00:06:32.325
or B is equal to negative three halves.

00:06:32.325 --> 00:06:37.325
We know that A is equal to one minus B,

00:06:39.429 --> 00:06:43.012
which would be equal to
one plus three halves,

00:06:43.012 --> 00:06:44.932
since B is negative three halves,

00:06:44.932 --> 00:06:46.756
which is equal to five halves.

00:06:46.756 --> 00:06:48.128
A is equal to five halves.

00:06:48.128 --> 00:06:51.297
B is equal to negative three halves.

00:06:51.297 --> 00:06:53.328
And just like that we
can rewrite this whole

00:06:53.328 --> 00:06:55.617
integral in a way that
is a little bit easier

00:06:55.617 --> 00:06:58.043
to take the anti or this whole expression

00:06:58.043 --> 00:06:59.819
so it's easier to integrate.

00:06:59.819 --> 00:07:03.962
So it's going to be the integral of

00:07:03.962 --> 00:07:08.962
one plus A over X plus one.

00:07:09.691 --> 00:07:12.859
A is five halves and so I
could just write that as,

00:07:12.859 --> 00:07:14.267
let me write it this way,

00:07:14.267 --> 00:07:18.393
five halves times one over X plus one.

00:07:18.393 --> 00:07:20.538
I wrote it that way because
it's very straight forward

00:07:20.538 --> 00:07:22.874
to take the antiderivative of this.

00:07:22.874 --> 00:07:26.491
Then plus B over X minus one.

00:07:26.491 --> 00:07:29.467
Which is going to be
negative three halves.

00:07:29.467 --> 00:07:34.467
So I'll just write it
as minus three halves

00:07:35.483 --> 00:07:38.810
times one over X minus one.

00:07:38.810 --> 00:07:43.400
That was this right over here, DX.

00:07:43.400 --> 00:07:46.466
Notice all I did is I took this expression

00:07:46.466 --> 00:07:47.921
right over here and I did a little bit

00:07:47.921 --> 00:07:51.249
of partial fraction expansion

00:07:51.249 --> 00:07:55.522
into these two, I guess
you could say, expressions

00:07:55.522 --> 00:07:57.698
or terms right over there.

00:07:57.698 --> 00:08:00.482
It's fairly straight
forward to integrate this.

00:08:00.482 --> 00:08:04.337
Antiderivative of one,
it's just going to be X.

00:08:04.337 --> 00:08:07.153
Antiderivative of five
halves, one over X plus one,

00:08:07.153 --> 00:08:10.786
is going to be plus five halves,

00:08:10.786 --> 00:08:13.762
the natural log of the
absolute value of X plus one.

00:08:13.762 --> 00:08:15.538
We're able to do that
because the derivative

00:08:15.538 --> 00:08:17.352
of X plus one is just
one, so the derivative

00:08:17.352 --> 00:08:19.410
is there so that we can
take the antiderivative

00:08:19.410 --> 00:08:21.572
with respect to X plus one.

00:08:21.572 --> 00:08:23.355
You could also do
u-substitution like we've done

00:08:23.355 --> 00:08:26.577
in previous examples, U
is equal to X plus one.

00:08:26.577 --> 00:08:31.074
And over here, this is going
to be minus three halves

00:08:31.074 --> 00:08:33.570
times the natural log
of the absolute value

00:08:33.570 --> 00:08:36.322
of X minus one, by the
same exact logic with how

00:08:36.322 --> 00:08:38.514
we were able to take the
antiderivative there.

00:08:38.514 --> 00:08:43.249
And of course we cannot
forget our constant.

00:08:43.249 --> 00:08:44.433
And there we have it.

00:08:44.433 --> 00:08:47.394
We've been able to integrate, we were able

00:08:47.394 --> 00:08:51.156
to evaluate this expression.


Dividing expressions to evaluate integral

WEBVTT
Kind: captions
Language: en

00:00:00.539 --> 00:00:01.696
- [Voiceover] See if you can evaluate

00:00:01.696 --> 00:00:04.300
this integral right over here.

00:00:04.300 --> 00:00:05.713
Assuming you had a goal at it

00:00:05.713 --> 00:00:07.152
so let's work through this together.

00:00:07.152 --> 00:00:09.934
You probably realized that
some of the traditional

00:00:09.934 --> 00:00:11.831
techniques that we've already had

00:00:11.831 --> 00:00:13.592
in our tool kits don't seem to be directly

00:00:13.592 --> 00:00:17.727
applicable, u-substitution and others.

00:00:17.727 --> 00:00:19.770
The key here to realize
is we have a rational

00:00:19.770 --> 00:00:23.167
expression here where the
numerator has the same

00:00:23.167 --> 00:00:25.390
degree or higher than the denominator.

00:00:25.390 --> 00:00:26.647
In this case, the numerator

00:00:26.647 --> 00:00:29.115
and the denominator have the same degree.

00:00:29.115 --> 00:00:32.482
Whenever you see something
like that, it's probably a good

00:00:32.482 --> 00:00:36.152
idea to divide the denominator
into the numerator.

00:00:36.152 --> 00:00:37.478
That's what this rational expression

00:00:37.478 --> 00:00:38.643
could be interpreted as.

00:00:38.643 --> 00:00:42.002
X minus five, divided by
negative two x plus two.

00:00:42.002 --> 00:00:44.657
So it's a little bit of
algebraic long division.

00:00:44.657 --> 00:00:46.205
To actually divide negative two x plus two

00:00:46.205 --> 00:00:48.209
into x minus five is to
see if we can rewrite this

00:00:48.209 --> 00:00:51.686
in a way where we can
evaluate the integral.

00:00:51.686 --> 00:00:52.764
So let's do that.

00:00:52.764 --> 00:00:57.518
We're going to take x minus five.

00:00:57.518 --> 00:01:02.518
X minus five, and divide negative
two x plus two into that.

00:01:04.752 --> 00:01:08.678
Negative two x plus two.

00:01:08.678 --> 00:01:10.749
Look at the highest degree terms.

00:01:10.749 --> 00:01:15.235
How many times does
negative two x go into x?

00:01:15.235 --> 00:01:18.866
Let's going to go negative 1/2 times.

00:01:18.866 --> 00:01:23.456
Negative 1/2 times two is negative one.

00:01:23.456 --> 00:01:26.477
Negative 1/2 times negative two x is

00:01:26.477 --> 00:01:28.737
this is going to be positive x.

00:01:28.737 --> 00:01:30.099
Just like that.

00:01:30.099 --> 00:01:32.170
Now we want to subtract
this yellow expression

00:01:32.170 --> 00:01:35.722
from this blue expression,
and so let's just,

00:01:35.722 --> 00:01:37.712
I'll just take the negative
of this and then add.

00:01:37.712 --> 00:01:41.470
So I'm just going to take
the negative of it and add.

00:01:41.470 --> 00:01:45.788
We are left with negative five
plus one is negative four.

00:01:45.788 --> 00:01:48.201
You could say negative two
x plus two goes into x minus

00:01:48.201 --> 00:01:51.494
five, negative 1/2 times
with negative four left over.

00:01:51.494 --> 00:01:56.085
So we can rewrite this integral,
our original integral as,

00:01:56.085 --> 00:02:01.085
we can rewrite it as
negative 1/2 minus four

00:02:06.242 --> 00:02:11.242
over negative two x plus two, d x.

00:02:16.868 --> 00:02:18.487
Now let's see, looks like we can simplify

00:02:18.487 --> 00:02:19.998
this expression a little bit more.

00:02:19.998 --> 00:02:21.655
The numerator and the denominator,

00:02:21.655 --> 00:02:23.122
they're both divisible by two.

00:02:23.122 --> 00:02:25.282
All of these terms are divisible by two.

00:02:25.282 --> 00:02:26.621
Actually, we all have these negatives

00:02:26.621 --> 00:02:29.442
that always unnecessarily
complicate things.

00:02:29.442 --> 00:02:30.800
Let's actually divide the numerator

00:02:30.800 --> 00:02:34.090
and the denominator by negative two.

00:02:34.090 --> 00:02:36.042
What are we going to have then?

00:02:36.042 --> 00:02:37.649
We divide the numerator by negative two.

00:02:37.649 --> 00:02:39.789
If this is negative four, this is

00:02:39.789 --> 00:02:42.436
going to become positive two.

00:02:42.436 --> 00:02:44.363
Then this, if we divide negative two x

00:02:44.363 --> 00:02:46.989
by negative two, that's
just going to become x.

00:02:46.989 --> 00:02:50.994
Then two divided by negative
two is going to be minus one.

00:02:50.994 --> 00:02:52.853
So our original integral...

00:02:52.853 --> 00:02:53.901
This is just all algebra.

00:02:53.901 --> 00:02:55.396
Everything we've done so far is algebra.

00:02:55.396 --> 00:02:56.903
We just rewritten it using a little bit

00:02:56.903 --> 00:02:58.575
of algebraic long division.

00:02:58.575 --> 00:03:02.667
Our original integral has
simplified to negative 1/2,

00:03:02.667 --> 00:03:04.458
and some might argue it's not simplified

00:03:04.458 --> 00:03:07.168
but it's actually much more
useful for finding the integral.

00:03:07.168 --> 00:03:12.168
Negative 1/2 plus two
over x minus one, d x.

00:03:16.376 --> 00:03:19.005
Now, how do we evaluate this?

00:03:19.005 --> 00:03:23.021
Well, the antiderivative

00:03:23.021 --> 00:03:26.072
of negative 1/2 is
pretty straight forward.

00:03:26.072 --> 00:03:31.072
That's just going to be negative 1/2 x,

00:03:31.585 --> 00:03:35.425
plus the entire derivative
of two over x minus one.

00:03:35.425 --> 00:03:36.690
You might reduce here ahead.

00:03:36.690 --> 00:03:39.408
The derivative of x minus one is just one

00:03:39.408 --> 00:03:42.251
so you could say that the
derivative is sitting there.

00:03:42.251 --> 00:03:44.338
We can essentially use
substitution in our heads

00:03:44.338 --> 00:03:47.269
and say, "Okay, let's just
take the entire derivative,

00:03:47.269 --> 00:03:49.684
"we could say with respect
to x minus one which will be

00:03:49.684 --> 00:03:52.456
"the natural log of the
absolute value of x minus one."

00:03:52.456 --> 00:03:53.958
If all of that sounds really confusing,

00:03:53.958 --> 00:03:56.424
I'll let you do the u-substitution.

00:03:56.424 --> 00:04:01.424
If I were just trying to evaluate

00:04:02.849 --> 00:04:06.995
the integral two over x minus one, d x,

00:04:06.995 --> 00:04:10.849
I could see, okay, the derivative
of x minus one is just one

00:04:10.849 --> 00:04:15.217
so I could say u is equal to x minus one,

00:04:15.217 --> 00:04:19.529
and then d u is going to be equal to d x.

00:04:19.529 --> 00:04:23.547
This is going to be, we can
rewrite in terms of u as two,

00:04:23.547 --> 00:04:27.218
I'll just take the constant
out two times the integral

00:04:27.218 --> 00:04:32.218
of one over u, d u, which
we know as two times

00:04:32.487 --> 00:04:37.411
the natural log of the
absolute value of u plus c.

00:04:37.411 --> 00:04:40.689
In this case, we know
that u is x minus one.

00:04:40.689 --> 00:04:41.939
This is equal to two times

00:04:41.939 --> 00:04:46.939
the natural log of x minus one plus c.

00:04:47.463 --> 00:04:49.681
That's what we're going
to have right over here.

00:04:49.681 --> 00:04:53.229
So plus two times the natural log

00:04:53.229 --> 00:04:57.456
of the absolute value
of x minus one plus c.

00:04:57.456 --> 00:04:59.029
The plus c doesn't just
come from this one,

00:04:59.029 --> 00:05:02.626
this general overtaking the
integral of the whole thing.

00:05:02.626 --> 00:05:05.066
It could be some constant
because if we go the other way,

00:05:05.066 --> 00:05:08.090
we take the derivative and
the constant will go away.

00:05:08.090 --> 00:05:12.130
Let me just put the
plus c right over there,

00:05:12.130 --> 00:05:17.130
and we are done.