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PROFESSOR: Welcome back
to recitation.

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In this video I want us to work
on some problems looking

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at integrals that may
converge or diverge.

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So these are improper
integrals.

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And we want to know if each of
the integrals below converges

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or diverges.

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And if they converge, I want
you to compute them.

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I want you to actually evaluate
it, find a number

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that is the area under
the curve.

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So that's really remember
what the integral is.

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So we should be able to find
actually the finite number

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that represents that.

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So there are three of them.

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The first one is the
integral from 0 to

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infinity of cosine x dx.

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The second one is the integral
from 0 to 1 of natural log x

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divided by x to the 1/2 dx.

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So that's just square root x.

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Remember it down there.

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And the third one is the
integral from minus 1 to 1 of

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x to the minus 2/3 dx.

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So why don't you take a
while to work on this.

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Pause the video.

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When you're feeling good about
your answers bring

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the video back up.

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I'll be back then to show
you how I do them.

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OK, welcome back.

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Well hopefully you were able
to answer these questions.

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The question again was that
we wanted to know if the

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following intervals--
intervals--

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integrals converged
or diverged.

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And then I wanted us
to actually compute

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them if they converged.

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So again, we're going
to start with the

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integral of cosine x dx.

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Then we'll look at the integral
of natural log

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x over root x dx.

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And then we'll look at the
integral from minus 1 to 1 of

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x to the minus 2/3 dx.

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So let's look at
the first one.

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And I'll rewrite it up
here so we have it.

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OK.

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And this is kind of interesting
because maybe this

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is a little bit different than
what you have seen previously

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in the way of saying this
is potentially improper.

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Because cosine x certainly
doesn't blow up anywhere.

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It's bounded between
minus 1 and 1.

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So the function itself
is not blowing up.

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But let's look at what we get
when we try and evaluate this.

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So if we take this integral, we
know the antiderivative of

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cosine is negative--
no it's just sine.

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Sorry, it's just sine x.

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Right?

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The derivative of
sine is cosine.

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So we get sine x.

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And what we're supposed to do,
is that we're supposed to

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take-- let me rewrite that x.

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That's horrible.

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We're supposed to take the limit
as b goes to infinity of

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sine x evaluated from 0 to b.

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Right?

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And so this is pretty
straightforward.

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Again now it's the limit at b
goes to infinity of sine b,

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because sine of 0 is 0.

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Now here's where we run into
trouble, because this limit

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doesn't exist. Right?

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because as b goes to
infinity,sine b, the function

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sine x-- so now it's really the
function sine b, b is now

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the variable if we think
about it that way--

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as b goes to infinity, sine is
going to oscillate as it

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always does between
minus 1 and 1.

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And it's going to continue
to do that.

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It's not going to approach
a certain value and stay

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arbitrarily close
to that value as

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be goes off to infinity.

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So this limit does not exist.
And maybe what's informative

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is to think about how could this
happen as an integral?

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If we know that the integral
we're looking for is really

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the sined area under
the curve.

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So let me explain briefly
what's happening.

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Let me just draw a quick picture
of cosine and explain

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briefly what's happening.

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So cosine starts off like this.

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Right?

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So what happens if I
wanted to integrate

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cosine x from 0 to infinity.

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I first pick up this
much area.

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Sorry this graph is a little
sloped I realize.

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I pick up this much area and
that's all positive.

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And then as I keep moving over
here to here, I pick up the

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same amount of area,
but it's negative.

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So once I get to here my
integral, it'ss 0.

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The area above and the
area below are equal.

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So then I start the process
again with the value 0.

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And so I accumulate some
negative area.

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Then over here it's the same
amount of positive area, and

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it kills it off.

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So I start off, I have some
positive value, and then it

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becomes less positive
and goes to 0.

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Then I get some negative
value accumulated.

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Then it comes up and
goes to 0 again.

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So the point is as I'm moving
off to infinity, the area is

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oscillating.

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And the area, remember, is
actually what the value of

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sine is at that point.

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The area from under the curve
from 0 to any value for cosine

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x is the value of sign
at that point.

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That's what we're seeing here.

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So the point is that this
integral, even though cosine x

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is a bounded function, the area
is accumulating, then

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disappearing, then becoming
negative, then disappearing,

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then accumulating, then
disappearing.

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So there's no value that
it's approaching.

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It's varying between all these
values over and over again.

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So this is a weird one maybe
where the integral doesn't

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exist. But it actually, it
doesn't exists there.

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So hopefully that makes sense,
and even though it's a little

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different, you understand
the idea.

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So now I'm going to go to b.

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So what is b?

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It's the integral from 0 to
1 of lnx over x to 1/2 dx.

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OK.

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And what I want to point is we
probably want to see why is

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this improper.

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Well at 1, we don't of
a problem because

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natural log of 1 is 0.

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And we can put a 1 down here
and nothing bad happens.

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At 0 we have a problem.

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A natural log of 0 actually
doesn't exist. The limit as x

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goes to 0 is natural log
x is minus infinity.

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And then natural log-- or
sorry, natural log--

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0, evaluating square root
x at 0, gives you 0.

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So we have something going
to minus infinity in the

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numerator and 0 in
the denominator.

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As we're trying to integrate
that function as it goes

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towards that value.

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So we have to figure out kind
of what's going on here.

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So let's not worry about the
bounds at the moment.

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Those are obviously going to
be important at the end.

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But let's figure out
what we get as an

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antiderivative for this.

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OK?

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And we don't worry about the
constant, remember, in the

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antiderivative because we're
going to evaluate.

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So what's the best way
to attack this one?

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Well probably you should see
that you got a natural log and

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you've got a power of x.

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So this is really set up to do
an integration by parts.

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Because remember you
like to take

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derivatives of natural log.

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And powers of x are happy to
be integrated or to take

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derivatives of them.

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The only power of x that's maybe
a little bit annoying to

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integrate is x to the minus 1,
because it's integral, because

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it's antiderivative is natural
log instead of like another

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power of x.

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But, this one is not that case
so it's looking good for an

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immigration by parts.

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So again we said for an
integration by parts, natural

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log you love to take
it's derivative.

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So we're going to let you be
natural log of x and u prime

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be 1 over x.

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And thenv prime is x
to the minus 1/2.

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Right?

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This is x to the 1/2 in the
denominator, so v prime is

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actually x to the minus 1/2.

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So v is going to be x to the 1/2
with a correction factor.

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So it's going to need a
2 in front, I believe.

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Let me double check.

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Derivative of this is
1/2 times 2 is 1 x

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to the minus 1/2.

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So I'm doing OK.

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So now what do I get when I
want to take the integral?

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I take uv. So it's going to
be 2x to the 1/2 lnx--

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and then now I'm going
to put in the bounds

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variated from 0 to 1--

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minus the integral
of vu prime.

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So v is x to the 1/2, and u
prime is x to the minus 1,

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really it's 1/x.

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And so I'm going to keep the 2
in front integral from 0 to 1.

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x to the 1/2 over x is
x to the minus 1/2.

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This will be nice because
we've already taken an

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antiderivative once, so
we know what we get.

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All right, now this one we'll
have to look at, because

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notice that as x goes to 0 we
have to figure out if this has

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a limit, OK?

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We have to figure out
of this has a limit.

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And we'll probably need to use
L'Hopital's rule to do that.

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So let's finish this part
up first. So I'll

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just keep this here.

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0 to 1, that's a 0,
and then minus.

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OK, x to the minus 1/2,
its antiderivative

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was 2x to the 1/2.

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So I have another 2, so
I get a 4x to the 1/2

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evaluated from 0 to 1.

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So this is easy because
here, I just get 4.

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Right?

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This part right here, when I
evaluated it I'll just get 4

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and the minus sign in front.

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So this part is just -4, because
at x equals 1 I get 1

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and at x equals 0 I get 0.

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So that's fine.

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This part is fine.

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This part, when I put in 1 for
x, notice what happens.

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I get a 1 here.

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Natural log of 1 is
0, and so I get 0.

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And so the question really is,
what is the limit as x goes to

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0 of this quantity?

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I know right here again,
I have a minus 4 here.

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And then the question is
what do I have here?

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it?

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It could blow up still.

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It could diverge.

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We're going to see
what happens.

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So let me just put a line.

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It's the same problem, but I
want to distinguish for us.

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So we're actually wanting to
find the limit as b goes to 0

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of 2b to the 1/2 natural log b.

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Right?

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That's what we're
interested in.

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That's the only part we
don't yet understand.

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So let's see what happens.

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Well, if I want to make this
into a L'Hopital's rule thing,

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right now I have 0 times
negative infinity.

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So to put it into a form I
recognize, I'm going to

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rewrite this.

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This is actually equal to the
limit as b goes to 0.

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I'm going to keep the natural
log be up here, and I'm going

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to write this as be to the minus
1/2 in the denominator.

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Let's make sure we understand
what just happened.

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I had a b to the 1/2
in the numerator.

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If I put it to them minus 1/2 in
the denominator it's still

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equal to the same thing.

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Right?

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b to the minus 1/2 in the
denominator is equal to b to

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the 1/2 in the numerator.

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OK?

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Really I'm taking 1 over this,
and then I'm dividing by it .

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That's the way you want
to think about it.

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You want to think about saying
I'm taking 1 over this, and

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I'm taking it in the numerator
and the denominator.

248
00:10:47,720 --> 00:10:50,180
so I end up with it just here.

249
00:10:50,180 --> 00:10:51,320
That might have been
confusing.

250
00:10:51,320 --> 00:10:54,380
The real point is this quantity,
one over this

251
00:10:54,380 --> 00:10:56,980
quantity is equal
to that quality.

252
00:10:56,980 --> 00:10:58,290
And now what's the point?

253
00:10:58,290 --> 00:10:59,870
Why did I bother to do that?

254
00:10:59,870 --> 00:11:02,610
As b goes to 0, this goes to
negative infinity and this

255
00:11:02,610 --> 00:11:05,080
goes to infinity.

256
00:11:05,080 --> 00:11:08,820
So now I have something where I
can apply L'Hopital's rule.

257
00:11:08,820 --> 00:11:14,110
So the limit as b goes to 0,
derivative of natural log of b

258
00:11:14,110 --> 00:11:16,120
is 1 over b.

259
00:11:16,120 --> 00:11:17,995
So I get 2 over b in
the numerator.

260
00:11:17,995 --> 00:11:23,130
The derivative of b to the minus
1/2 is negative 1/2 b to

261
00:11:23,130 --> 00:11:25,400
the minus 3/2.

262
00:11:25,400 --> 00:11:29,570
There's a lot of denominators
in the numerator and

263
00:11:29,570 --> 00:11:32,580
denominator, so let's
simplify this.

264
00:11:32,580 --> 00:11:36,390
That's the limit
as b goes to 0.

265
00:11:36,390 --> 00:11:37,270
OK, what o we get here?

266
00:11:37,270 --> 00:11:41,650
We have 2 times minus 1/2, or
2 divided by minus 1/2.

267
00:11:41,650 --> 00:11:44,770
So I'm going to get a negative
4 from this part, the

268
00:11:44,770 --> 00:11:46,370
coefficient.

269
00:11:46,370 --> 00:11:50,940
Then I have a b to the minus
1 up here divided by a b

270
00:11:50,940 --> 00:11:52,890
to the minus 3/2.

271
00:11:52,890 --> 00:11:56,070
That's actually going to be a
b to the 3/2 divided by b.

272
00:11:56,070 --> 00:11:58,260
That's going to be
b to the 1/2.

273
00:11:58,260 --> 00:11:58,970
That's algebra.

274
00:11:58,970 --> 00:12:01,010
You can check it
if you need to.

275
00:12:01,010 --> 00:12:02,140
I mean, you should
have gotten this.

276
00:12:02,140 --> 00:12:04,420
But if you didn't get
this, check again.

277
00:12:04,420 --> 00:12:07,460
Let me make sure I get it again,
b to the 3/2 divided by

278
00:12:07,460 --> 00:12:09,690
b, b to the 1/2.

279
00:12:09,690 --> 00:12:12,560
And that equals 0, because it
was a limit as b goes to 0 of

280
00:12:12,560 --> 00:12:13,770
this quantity.

281
00:12:13,770 --> 00:12:15,430
Now I've just got a continuous
function.

282
00:12:15,430 --> 00:12:18,840

283
00:12:18,840 --> 00:12:19,730
I am, notice, I didn't actually,
I kind of cheated a

284
00:12:19,730 --> 00:12:21,950
little bit, because I just wrote
limit as b goes to 0.

285
00:12:21,950 --> 00:12:25,760
But I'm always doing as b goes
to 0 from the right-hand side.

286
00:12:25,760 --> 00:12:27,210
I'm starting at 1.

287
00:12:27,210 --> 00:12:30,470
So you may have seen this, this
0 plus means I'm only

288
00:12:30,470 --> 00:12:33,950
interested as b goes
to 0 from above 0.

289
00:12:33,950 --> 00:12:34,250
OK?

290
00:12:34,250 --> 00:12:37,060
I didn't write that in, but
notice our integral was

291
00:12:37,060 --> 00:12:37,895
between 0 and 1.

292
00:12:37,895 --> 00:12:41,020
So it only mattered values
to the right of 0.

293
00:12:41,020 --> 00:12:43,730
So this function is defined
to the right of 0.

294
00:12:43,730 --> 00:12:47,195
So I can just evaluate there.

295
00:12:47,195 --> 00:12:48,220
I can just plug it in.

296
00:12:48,220 --> 00:12:52,450
It's continuous, and so I can
just say at 0 it equals 0.

297
00:12:52,450 --> 00:12:54,250
So now let's go back
to where we were.

298
00:12:54,250 --> 00:12:55,060
What were we doing here?

299
00:12:55,060 --> 00:12:57,970
We were taking this whole piece,
was to come back in

300
00:12:57,970 --> 00:13:01,840
here and figure out what
this value was.

301
00:13:01,840 --> 00:13:03,940
We knew at 1, we got 0.

302
00:13:03,940 --> 00:13:06,480
And now we know at
0, we also get 0.

303
00:13:06,480 --> 00:13:10,820
So that question mark I can
replace by a 0, and

304
00:13:10,820 --> 00:13:12,250
I get 0 minus 4.

305
00:13:12,250 --> 00:13:17,030
So the actual answer
is negative 4.

306
00:13:17,030 --> 00:13:19,740
OK, we have one more.

307
00:13:19,740 --> 00:13:21,040
What's the last one?

308
00:13:21,040 --> 00:13:25,950
The last one-- let me write
it down over here again--

309
00:13:25,950 --> 00:13:32,550
is the integral from minus 1 to
1 of x to the minus 2/3 dx.

310
00:13:32,550 --> 00:13:35,850
Now this is interesting, because
this, you actually saw

311
00:13:35,850 --> 00:13:40,820
an example kind of like this
in the lecture that at this

312
00:13:40,820 --> 00:13:42,520
endpoint, it's fine.

313
00:13:42,520 --> 00:13:44,920
You can evaluate the function
at that endpoint.

314
00:13:44,920 --> 00:13:46,250
At this endpoint it's fine.

315
00:13:46,250 --> 00:13:47,570
You can evaluate the
function there.

316
00:13:47,570 --> 00:13:49,420
So it looks good at
the endpoints.

317
00:13:49,420 --> 00:13:52,770
But the point is that at x
equals 0, because this minus

318
00:13:52,770 --> 00:13:56,390
power is putting your x in the
denominator, you actually,

319
00:13:56,390 --> 00:13:57,570
your function is blowing up.

320
00:13:57,570 --> 00:14:00,360
It has a vertical asymptote
at x equals 0.

321
00:14:00,360 --> 00:14:03,600
So what we want to do is we have
this strategy for these

322
00:14:03,600 --> 00:14:08,120
problems, is to split it up into
two parts, minus 1 to 0

323
00:14:08,120 --> 00:14:15,550
of x to the minus 2/3 dx plus
the integral from 0 to 1 of x

324
00:14:15,550 --> 00:14:18,960
to the minus 2/3 dx.

325
00:14:18,960 --> 00:14:21,600
Now the point is that now the
only place where I have a

326
00:14:21,600 --> 00:14:23,200
vertical asymptote,
I see it as an

327
00:14:23,200 --> 00:14:25,010
endpoint on both of these.

328
00:14:25,010 --> 00:14:27,690
And this is again, just this
good thing we have an additive

329
00:14:27,690 --> 00:14:30,690
property for these integrals,
that the sum of these two

330
00:14:30,690 --> 00:14:34,250
integrals is going to
equal this one here.

331
00:14:34,250 --> 00:14:37,360
As long as, you know, as long as
these are converging, then

332
00:14:37,360 --> 00:14:39,450
I can say that if this converges
and this converges,

333
00:14:39,450 --> 00:14:41,830
then there's some convergence
to this one here.

334
00:14:41,830 --> 00:14:46,160
OK, so that's really what I'm
trying to exploit here.

335
00:14:46,160 --> 00:14:49,340
Now what is the antiderivative
here for x to the minus 2/3?

336
00:14:49,340 --> 00:14:52,250

337
00:14:52,250 --> 00:14:53,540
This is going to be-- again, I'm
going to write something

338
00:14:53,540 --> 00:14:55,140
down and then I'm going
to check it.

339
00:14:55,140 --> 00:15:00,000
I think it should be
5/3, x to the 5/3.

340
00:15:00,000 --> 00:15:02,650
And then I have to
multiply by 3/5.

341
00:15:02,650 --> 00:15:04,430
Let's double check, because this
is where I always might

342
00:15:04,430 --> 00:15:05,630
make mistakes.

343
00:15:05,630 --> 00:15:09,190
If I take the derivative of 5/3
times 3/5 it gives me a 1.

344
00:15:09,190 --> 00:15:13,300
5/3 minus 1 is 5/3 minus
3/3, and that's

345
00:15:13,300 --> 00:15:14,245
2/3, and that's wrong.

346
00:15:14,245 --> 00:15:15,495
Right?

347
00:15:15,495 --> 00:15:17,100

348
00:15:17,100 --> 00:15:17,770
Because I went too big.

349
00:15:17,770 --> 00:15:18,810
It's supposed to be 1/3.

350
00:15:18,810 --> 00:15:19,920
I knew I was going to
make a mistake.

351
00:15:19,920 --> 00:15:21,610
Right?

352
00:15:21,610 --> 00:15:23,760
I'm supposed to add
1 to minus 2/3.

353
00:15:23,760 --> 00:15:24,320
That's just 1/3.

354
00:15:24,320 --> 00:15:25,850
And so I should have
a 3 in front.

355
00:15:25,850 --> 00:15:27,740
So for those of you who were
saying she's making

356
00:15:27,740 --> 00:15:29,510
a mistake, I was.

357
00:15:29,510 --> 00:15:32,410
OK, x to the 1/3.

358
00:15:32,410 --> 00:15:35,035
So I take the derivative times
3, I get a 1 there.

359
00:15:35,035 --> 00:15:37,240
I subtract 1 and I
get minus 2/3.

360
00:15:37,240 --> 00:15:37,982
So now I'm in business.

361
00:15:37,982 --> 00:15:39,760
OK.

362
00:15:39,760 --> 00:15:44,020
And now I just need to a value
that here, here if I can, then

363
00:15:44,020 --> 00:15:47,940
another one here from 0 to 1.

364
00:15:47,940 --> 00:15:52,510
And the good news is x to
the 1/3 is continuous

365
00:15:52,510 --> 00:15:53,470
at all these points.

366
00:15:53,470 --> 00:15:56,660
It's continuous across negative
1, 0, 0, and 1.

367
00:15:56,660 --> 00:15:58,530
So I can actually
just evaluate.

368
00:15:58,530 --> 00:15:59,670
I don't have to worry.

369
00:15:59,670 --> 00:16:02,020
There is a value for
the function there.

370
00:16:02,020 --> 00:16:03,720
It's continuous through
all these points.

371
00:16:03,720 --> 00:16:05,340
So I can just evaluate them.

372
00:16:05,340 --> 00:16:06,700
So let's see what I get.

373
00:16:06,700 --> 00:16:09,540
I get 3 times 0.

374
00:16:09,540 --> 00:16:13,930
And then I get minus 3 times
negative 1, so I get a

375
00:16:13,930 --> 00:16:23,540
negative 3 plus 3 times 1,
and then minus 3 times 0.

376
00:16:23,540 --> 00:16:25,880
OK, so 3 times 0 minus
negative 3.

377
00:16:25,880 --> 00:16:29,260
So that's a 3 plus
3, and I get 6.

378
00:16:29,260 --> 00:16:30,550
And what's the picture
of this?

379
00:16:30,550 --> 00:16:33,920
Well you should think about what
the picture of x to the

380
00:16:33,920 --> 00:16:35,050
minus 2/3 looks like.

381
00:16:35,050 --> 00:16:37,842
And you should notice that
across 0, it's even.

382
00:16:37,842 --> 00:16:38,660
Right?

383
00:16:38,660 --> 00:16:42,160
So it's actually going to have,
itt's going to look the

384
00:16:42,160 --> 00:16:44,130
same in the left-hand side
and the right-hand side.

385
00:16:44,130 --> 00:16:48,270
So if I had wanted to, I could
have just found the value of

386
00:16:48,270 --> 00:16:49,360
say, this integral--

387
00:16:49,360 --> 00:16:51,320
I like positive numbers better--
the integral from 0

388
00:16:51,320 --> 00:16:54,420
to 1 of this function,
multiplied it by 2, and it

389
00:16:54,420 --> 00:16:57,140
would have given me
the actual value.

390
00:16:57,140 --> 00:16:59,580
Because it's exactly the same
function to the right of 0 and

391
00:16:59,580 --> 00:17:00,200
to the left of 0.

392
00:17:00,200 --> 00:17:03,030
It's a reflection across the
y-axis, so you get the same

393
00:17:03,030 --> 00:17:04,080
value there.

394
00:17:04,080 --> 00:17:06,220
So you actually would've gotten
3 for this multiplied

395
00:17:06,220 --> 00:17:07,760
by 2 and you get the value.

396
00:17:07,760 --> 00:17:10,590
If this one had diverged, that
one also would have had to

397
00:17:10,590 --> 00:17:12,600
diverge, and the whole thing
would have diverged.

398
00:17:12,600 --> 00:17:13,920
And that's because of
the symmetry of the

399
00:17:13,920 --> 00:17:15,900
function over 0.

400
00:17:15,900 --> 00:17:18,720
So I'm going to just briefly
review what we did.

401
00:17:18,720 --> 00:17:19,920
And then we'll be done.

402
00:17:19,920 --> 00:17:22,820
So we come back over here.

403
00:17:22,820 --> 00:17:25,230
I gave you 3 integrals.

404
00:17:25,230 --> 00:17:28,360
We wanted to see if they
converged or diverged, right,

405
00:17:28,360 --> 00:17:31,130
and if they converged
find what they were.

406
00:17:31,130 --> 00:17:33,500
So the first one was cosine x.

407
00:17:33,500 --> 00:17:36,230
And we found that diverged for
a different reason than what

408
00:17:36,230 --> 00:17:37,590
we've seen before.

409
00:17:37,590 --> 00:17:40,840
Because as x goes to infinity,
the problem is the areas are

410
00:17:40,840 --> 00:17:44,180
varying up and down and they're
bounded always.

411
00:17:44,180 --> 00:17:46,960
But the area under the curve
is constantly changing and

412
00:17:46,960 --> 00:17:49,560
it's not converting
to a fixed value.

413
00:17:49,560 --> 00:17:53,260
It's varying between minus 1
and 1 over and over again.

414
00:17:53,260 --> 00:17:56,760
And then b, we had this integral
from 0 to 1 natural

415
00:17:56,760 --> 00:17:59,440
log x over root x dx.

416
00:17:59,440 --> 00:18:02,260
And the point here was a little
more complicated, so

417
00:18:02,260 --> 00:18:04,690
let me come to that one.

418
00:18:04,690 --> 00:18:07,130
We used an integration
by parts.

419
00:18:07,130 --> 00:18:10,220
We had an integral they
converged easily, because this

420
00:18:10,220 --> 00:18:14,350
was an easy improper integral
to determine.

421
00:18:14,350 --> 00:18:16,400
And then we had this other
thing that we now have to

422
00:18:16,400 --> 00:18:17,700
evaluate it.

423
00:18:17,700 --> 00:18:19,120
You wound up having
to use L'Hopital's

424
00:18:19,120 --> 00:18:21,120
rule to evaluate it.

425
00:18:21,120 --> 00:18:23,370
But then you're able to show
that this using L'Hopital's

426
00:18:23,370 --> 00:18:26,530
rule, this actually has a value
at each endpoint, a

427
00:18:26,530 --> 00:18:28,210
finite value at each endpoint.

428
00:18:28,210 --> 00:18:32,590
It converges then to 0 as you,
as when you put in these

429
00:18:32,590 --> 00:18:36,650
bounds, and then you have a
fixed value for this one when

430
00:18:36,650 --> 00:18:37,650
you take that integral.

431
00:18:37,650 --> 00:18:39,740
So you ended up with another
integral that was improper.

432
00:18:39,740 --> 00:18:41,490
But you could show
it converged.

433
00:18:41,490 --> 00:18:43,710
And then you could use
L'Hopital's rule to find what

434
00:18:43,710 --> 00:18:44,640
happened at these endpoints.

435
00:18:44,640 --> 00:18:46,940
And you get a value
of minus 4.

436
00:18:46,940 --> 00:18:51,070
Then the third one was this
integral minus 1 to 1 x to the

437
00:18:51,070 --> 00:18:52,340
minus 2/3 x dx.

438
00:18:52,340 --> 00:18:54,830
And the point I wanted to make
there is that just because the

439
00:18:54,830 --> 00:18:57,850
function is well-behaved near
the endpoints doesn't mean

440
00:18:57,850 --> 00:19:00,550
that it's still, you know, that
everything is hunky-dory.

441
00:19:00,550 --> 00:19:03,820
You have to be careful and you
have to check at places where

442
00:19:03,820 --> 00:19:06,160
the value of the function is
going off to infinity.

443
00:19:06,160 --> 00:19:10,640
So in the second case, the left
endpoint posed a problem,

444
00:19:10,640 --> 00:19:11,860
and everything else was fine.

445
00:19:11,860 --> 00:19:14,660
In this case, the left and right
endpoints are both fine,

446
00:19:14,660 --> 00:19:16,600
but in the middle there's
a problem.

447
00:19:16,600 --> 00:19:19,160
And so you split it up into it's
two pieces so that you

448
00:19:19,160 --> 00:19:20,960
have the endpoints.

449
00:19:20,960 --> 00:19:23,430
Now these two new integrals
represent where the problem

450
00:19:23,430 --> 00:19:27,350
might happen, and then you
do your, find your

451
00:19:27,350 --> 00:19:29,800
antiderivative, and evaluate,
and see if you can get

452
00:19:29,800 --> 00:19:31,270
something that converges.

453
00:19:31,270 --> 00:19:35,820
So that's the idea of these
types of problems. So

454
00:19:35,820 --> 00:19:37,080
hopefully that was helpful,
and I think

455
00:19:37,080 --> 00:19:38,690
that's where I'll stop.

456
00:19:38,690 --> 00:19:39,046