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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

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below converges 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 that

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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,

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the finite number
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

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to infinity of cosine x dx.

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

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x divided by x to the 1/2 dx.

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So that's just square
root x, down there.

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

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of 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

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bring 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

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to know if the following
intervals intervals integrals

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converged or diverged.

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And then I wanted us to actually
compute them if they converged.

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So again, we're going to
start with the integral

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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

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of 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

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the antiderivative of 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

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that we're supposed to
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 sine x

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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.

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Right?

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

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so now it's really
the function sine b,

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b is now the variable, if
we think about it that way--

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as b goes to infinity,
sine is going

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to oscillate as it 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

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and stay arbitrarily close
to that value as be goes off

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to infinity.

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So this limit does not exist.

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And maybe what's informative is
to think about how could this

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happen as an integral?

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If we know that the
integral we're looking for

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is really the signed
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

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and explain 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 cosine x from 0

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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,

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I pick up the same amount
of area, but it's negative.

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So once I get to here
my integral, it's 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,

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and it kills it off.

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So I start off, I have
some positive value,

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and then it 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,

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the area is oscillating.

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

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is at that point.

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The area from under
the curve from 0

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to any value for cosine x is
the value of sine 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,

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then 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

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the integral doesn't exist.

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But it actually, it
doesn't exists there.

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So hopefully that
makes sense, and even

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though it's a little 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 ln x 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

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to see why is this improper.

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Well at 1, we don't of a problem
because 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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The natural log of 0
actually doesn't exist.

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The limit as x goes
to 0 of natural log

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x is minus infinity.

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

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

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So we have something
going to minus infinity

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in the numerator and
0 in the denominator.

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And we're trying to
integrate that function

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as it goes 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 antiderivative for this.

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OK?

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And we don't worry
about the constant,

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remember, in the
antiderivative because we're

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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

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and 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

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or to take derivatives of them.

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

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to integrate is x to the
minus 1, because its integral,

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because its antiderivative
is natural log instead

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of, like, another power of x.

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But, this one is not
that case so it's

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looking good for an
integration by parts.

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So again we said
for an integration

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by parts, natural log you
love to take its derivative.

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

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and u prime be 1 over x.

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And then v 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,

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so v prime is 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

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2 is 1-- x 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 u*v. So it's going
to be 2 x to the 1/2 ln x--

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

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evaluated from 0 to 1, minus
the integral of v u 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 over 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 antiderivative

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once, so we know what we get.

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

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because notice
that as x goes to 0

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we have to figure out
if this has 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.

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So I'll 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 2 x to the 1/2.

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

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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 evaluate it

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I'll just get 4 and the
minus sign in front.

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So this part is just negative
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

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is the limit as x goes
to 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

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as b goes to 0 of 2 b 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,

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I'm going to rewrite this.

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This is actually
equal to the limit

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as b goes to 0-- I'm going
to keep the natural log be

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up here, and I'm going to write
this as b to the minus 1/2

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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 the minus
1/2 in the denominator,

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it's still equal
to the same thing.

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Right?

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b to the minus 1/2
in the denominator

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is equal to b to 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,

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

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so I end up with it just here.

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That might have been confusing.

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The real point is this
quantity, one over this quantity

246
00:10:54,840 --> 00:10:56,980
is equal to that quality.

247
00:10:56,980 --> 00:10:58,290
And now what's the point?

248
00:10:58,290 --> 00:10:59,870
Why did I bother to do that?

249
00:10:59,870 --> 00:11:02,320
As b goes to 0, this
goes to negative infinity

250
00:11:02,320 --> 00:11:05,080
and this goes to infinity.

251
00:11:05,080 --> 00:11:08,820
So now I have something where
I can apply L'Hopital's rule.

252
00:11:08,820 --> 00:11:14,110
So the limit as b goes to 0,
derivative of natural log of b

253
00:11:14,110 --> 00:11:16,120
is 1 over b.

254
00:11:16,120 --> 00:11:17,995
So I get 2 over b
in the numerator.

255
00:11:17,995 --> 00:11:23,020
The derivative of b to the
minus 1/2 is negative 1/2 b

256
00:11:23,020 --> 00:11:25,400
to the minus 3/2.

257
00:11:25,400 --> 00:11:30,250
There's a lot of denominators in
the numerator and denominator,

258
00:11:30,250 --> 00:11:32,580
so let's simplify this.

259
00:11:32,580 --> 00:11:36,270
That's the limit as b goes to 0.

260
00:11:36,270 --> 00:11:37,270
OK, what do we get here?

261
00:11:37,270 --> 00:11:41,650
We have 2 times minus 1/2,
or 2 divided by minus 1/2.

262
00:11:41,650 --> 00:11:44,450
So I'm going to get a
negative 4 from this part,

263
00:11:44,450 --> 00:11:46,370
the coefficient.

264
00:11:46,370 --> 00:11:50,940
Then I have a b to the minus
1 up here divided by a b

265
00:11:50,940 --> 00:11:52,890
to the minus 3/2.

266
00:11:52,890 --> 00:11:56,070
That's actually going to be
a b to the 3/2 divided by b.

267
00:11:56,070 --> 00:11:58,260
That's going to be b to the 1/2.

268
00:11:58,260 --> 00:11:58,970
That's algebra.

269
00:11:58,970 --> 00:12:00,640
You can check it if you need to.

270
00:12:00,640 --> 00:12:02,140
I mean, you should
have gotten this.

271
00:12:02,140 --> 00:12:04,420
But if you didn't get
this, check again.

272
00:12:04,420 --> 00:12:08,290
Let me make sure I get it again,
b to the 3/2 divided by b, b

273
00:12:08,290 --> 00:12:09,690
to the 1/2.

274
00:12:09,690 --> 00:12:11,980
And that equals 0, because
it was a limit as b

275
00:12:11,980 --> 00:12:13,770
goes to 0 of this quantity.

276
00:12:13,770 --> 00:12:15,700
Now I've just got a
continuous function.

277
00:12:15,700 --> 00:12:18,649
I am, notice, I I
didn't actually--

278
00:12:18,649 --> 00:12:21,065
I kind of cheated a little
bit, because I just wrote limit

279
00:12:21,065 --> 00:12:21,950
as b goes to 0.

280
00:12:21,950 --> 00:12:25,760
But I'm always doing as b goes
to 0 from the right-hand side.

281
00:12:25,760 --> 00:12:27,210
I'm starting at 1.

282
00:12:27,210 --> 00:12:29,900
So you may have seen
this, this 0 plus

283
00:12:29,900 --> 00:12:33,750
means I'm only interested
as b goes to 0 from above 0.

284
00:12:33,750 --> 00:12:34,250
OK?

285
00:12:34,250 --> 00:12:36,890
I didn't write that in,
but notice our integral

286
00:12:36,890 --> 00:12:37,895
was between 0 and 1.

287
00:12:37,895 --> 00:12:41,020
So it only mattered
values to the right of 0.

288
00:12:41,020 --> 00:12:43,730
So this function is
defined to the right of 0.

289
00:12:43,730 --> 00:12:47,195
So I can just evaluate there.

290
00:12:47,195 --> 00:12:48,220
I can just plug it in.

291
00:12:48,220 --> 00:12:52,450
It's continuous, and so I can
just say at 0 it equals 0.

292
00:12:52,450 --> 00:12:54,061
So now let's go back
to where we were.

293
00:12:54,061 --> 00:12:55,060
What were we doing here?

294
00:12:55,060 --> 00:12:59,210
We were taking this whole
piece, was to come back in here

295
00:12:59,210 --> 00:13:01,840
and figure out what
this value was.

296
00:13:01,840 --> 00:13:03,940
We knew at 1, we got 0.

297
00:13:03,940 --> 00:13:06,480
And now we know at
0, we also get 0.

298
00:13:06,480 --> 00:13:10,650
So that question mark
I can replace by a 0,

299
00:13:10,650 --> 00:13:12,250
and I get 0 minus 4.

300
00:13:12,250 --> 00:13:17,030
So the actual answer
is negative 4.

301
00:13:17,030 --> 00:13:19,740
OK, we have one more.

302
00:13:19,740 --> 00:13:21,040
What's the last one?

303
00:13:21,040 --> 00:13:25,950
The last one-- let me write
it down over here again--

304
00:13:25,950 --> 00:13:32,550
is the integral from minus 1
to 1 of x to the minus 2/3 dx.

305
00:13:32,550 --> 00:13:35,090
Now this is interesting,
because this,

306
00:13:35,090 --> 00:13:39,030
you actually saw an example
kind of like this in the lecture

307
00:13:39,030 --> 00:13:42,520
that at this
endpoint, it's fine.

308
00:13:42,520 --> 00:13:44,920
You can evaluate the
function at that endpoint.

309
00:13:44,920 --> 00:13:46,071
At this endpoint it's fine.

310
00:13:46,071 --> 00:13:47,570
You can evaluate
the function there.

311
00:13:47,570 --> 00:13:49,420
So it looks good
at the endpoints.

312
00:13:49,420 --> 00:13:50,950
But the point is
that at x equals

313
00:13:50,950 --> 00:13:53,570
0, because this minus
power is putting

314
00:13:53,570 --> 00:13:56,987
your x in the denominator,
you actually, your function

315
00:13:56,987 --> 00:13:57,570
is blowing up.

316
00:13:57,570 --> 00:14:00,360
It has a vertical
asymptote at x equals 0.

317
00:14:00,360 --> 00:14:03,280
So what we want to do
is we have this strategy

318
00:14:03,280 --> 00:14:07,040
for these problems, is to
split it up into two parts,

319
00:14:07,040 --> 00:14:13,570
minus 1 to 0 of x to the
minus 2/3 dx plus the integral

320
00:14:13,570 --> 00:14:18,960
from 0 to 1 of x to
the minus 2/3 dx.

321
00:14:18,960 --> 00:14:21,014
Now the point is that
now the only place

322
00:14:21,014 --> 00:14:22,430
where I have a
vertical asymptote,

323
00:14:22,430 --> 00:14:25,010
I see it as an endpoint
on both of these.

324
00:14:25,010 --> 00:14:26,950
And this is again,
just this good thing

325
00:14:26,950 --> 00:14:29,410
we have an additive property
for these integrals,

326
00:14:29,410 --> 00:14:31,690
that the sum of
these two integrals

327
00:14:31,690 --> 00:14:34,250
is going to equal this one here.

328
00:14:34,250 --> 00:14:36,949
As long as, you know, as
long as these are converging,

329
00:14:36,949 --> 00:14:38,490
then I can say that
if this converges

330
00:14:38,490 --> 00:14:41,830
and this converges, then their
sum converges to this one here.

331
00:14:41,830 --> 00:14:46,160
OK, so that's really what
I'm trying to exploit here.

332
00:14:46,160 --> 00:14:49,340
Now what is the antiderivative
here for x to the minus 2/3?

333
00:14:51,237 --> 00:14:53,820
This is going to be-- again, I'm
going to write something down

334
00:14:53,820 --> 00:14:55,140
and then I'm going to check it.

335
00:14:55,140 --> 00:15:00,000
I think it should be
5/3, x to the 5/3.

336
00:15:00,000 --> 00:15:02,117
And then I have to
multiply by 3/5.

337
00:15:02,117 --> 00:15:04,200
Let's double check, because
this is where I always

338
00:15:04,200 --> 00:15:05,630
might make mistakes.

339
00:15:05,630 --> 00:15:09,190
If I take the derivative of
5/3 times 3/5 it gives me a 1.

340
00:15:09,190 --> 00:15:13,537
5/3 minus 1 is 5/3 minus
3/3, and that's 2/3,

341
00:15:13,537 --> 00:15:14,245
and that's wrong.

342
00:15:14,245 --> 00:15:14,745
Right?

343
00:15:16,812 --> 00:15:17,770
Because I went too big.

344
00:15:17,770 --> 00:15:18,810
It's supposed to be 1/3.

345
00:15:18,810 --> 00:15:20,351
I knew I was going
to make a mistake.

346
00:15:20,351 --> 00:15:21,610
Right?

347
00:15:21,610 --> 00:15:23,654
I'm supposed to
add 1 to minus 2/3.

348
00:15:23,654 --> 00:15:24,320
That's just 1/3.

349
00:15:24,320 --> 00:15:25,850
And so I should
have a 3 in front.

350
00:15:25,850 --> 00:15:28,600
So for those of you who were
saying she's making a mistake,

351
00:15:28,600 --> 00:15:29,510
I was.

352
00:15:29,510 --> 00:15:32,410
OK, x to the 1/3.

353
00:15:32,410 --> 00:15:35,035
So I take the derivative
times 3, I get a 1 there.

354
00:15:35,035 --> 00:15:37,024
I subtract 1 and
I get minus 2/3.

355
00:15:37,024 --> 00:15:37,982
So now I'm in business.

356
00:15:37,982 --> 00:15:39,760
OK.

357
00:15:39,760 --> 00:15:43,230
And now I just need to evaluate
that here, here if I can,

358
00:15:43,230 --> 00:15:47,940
then another one
here from 0 to 1.

359
00:15:47,940 --> 00:15:50,460
And the good news
is x to the 1/3

360
00:15:50,460 --> 00:15:53,470
is continuous at
all these points.

361
00:15:53,470 --> 00:15:56,660
It's continuous across
negative 1, 0, 0, and 1.

362
00:15:56,660 --> 00:15:58,530
So I can actually just evaluate.

363
00:15:58,530 --> 00:15:59,670
I don't have to worry.

364
00:15:59,670 --> 00:16:02,012
There is a value for
the function there.

365
00:16:02,012 --> 00:16:03,720
It's continuous through
all these points.

366
00:16:03,720 --> 00:16:05,340
So I can just evaluate them.

367
00:16:05,340 --> 00:16:06,700
So let's see what I get.

368
00:16:06,700 --> 00:16:09,540
I get 3 times 0.

369
00:16:09,540 --> 00:16:13,700
And then I get minus
3 times negative 1,

370
00:16:13,700 --> 00:16:18,200
so I get a negative
3 plus 3 times

371
00:16:18,200 --> 00:16:23,540
1, and then minus 3 times 0.

372
00:16:23,540 --> 00:16:25,880
OK, so 3 times 0
minus negative 3.

373
00:16:25,880 --> 00:16:29,259
So that's a 3 plus
3, and I get 6.

374
00:16:29,259 --> 00:16:30,550
And what's the picture of this?

375
00:16:30,550 --> 00:16:33,660
Well you should think
about what the picture of x

376
00:16:33,660 --> 00:16:35,050
to the minus 2/3 looks like.

377
00:16:35,050 --> 00:16:37,842
And you should notice
that across 0, it's even.

378
00:16:37,842 --> 00:16:38,660
Right?

379
00:16:38,660 --> 00:16:41,217
So it's actually
going to have-- it's

380
00:16:41,217 --> 00:16:43,050
going to look the same
in the left-hand side

381
00:16:43,050 --> 00:16:44,130
and the right-hand side.

382
00:16:44,130 --> 00:16:47,480
So if I had wanted to,
I could have just found

383
00:16:47,480 --> 00:16:49,237
the value of say,
this integral-- I

384
00:16:49,237 --> 00:16:51,320
like positive numbers
better-- the integral from 0

385
00:16:51,320 --> 00:16:54,220
to 1 of this function,
multiplied it by 2,

386
00:16:54,220 --> 00:16:57,140
and it would have given
me the actual value.

387
00:16:57,140 --> 00:16:59,530
Because it's exactly the same
function to the right of 0

388
00:16:59,530 --> 00:17:00,404
and to the left of 0.

389
00:17:00,404 --> 00:17:01,980
It's a reflection
across the y-axis,

390
00:17:01,980 --> 00:17:04,080
so you get the same value there.

391
00:17:04,080 --> 00:17:06,690
So you actually would've gotten
3, for this, multiplied by 2

392
00:17:06,690 --> 00:17:07,760
and you get the value.

393
00:17:07,760 --> 00:17:09,854
If this one had
diverged, that one also

394
00:17:09,854 --> 00:17:11,770
would have had to diverge,
and the whole thing

395
00:17:11,770 --> 00:17:12,600
would have diverged.

396
00:17:12,600 --> 00:17:14,016
And that's because
of the symmetry

397
00:17:14,016 --> 00:17:15,900
of the function over 0.

398
00:17:15,900 --> 00:17:18,720
So I'm going to just
briefly review what we did.

399
00:17:18,720 --> 00:17:19,920
And then we'll be done.

400
00:17:19,920 --> 00:17:22,820
So we come back over here.

401
00:17:22,820 --> 00:17:25,230
I gave you three integrals.

402
00:17:25,230 --> 00:17:28,360
We wanted to see if they
converged or diverged, right,

403
00:17:28,360 --> 00:17:31,130
and if they converged
find what they were.

404
00:17:31,130 --> 00:17:33,500
So the first one was cosine x.

405
00:17:33,500 --> 00:17:36,020
And we found that diverged
for a different reason

406
00:17:36,020 --> 00:17:37,590
than what we've seen before.

407
00:17:37,590 --> 00:17:40,540
Because as x goes to infinity,
the problem is the areas

408
00:17:40,540 --> 00:17:44,180
are varying up and down
and they're bounded always.

409
00:17:44,180 --> 00:17:46,680
But the area under the
curve is constantly changing

410
00:17:46,680 --> 00:17:49,560
and it's not converging
to a fixed value.

411
00:17:49,560 --> 00:17:53,260
It's varying between minus
1 and 1 over and over again.

412
00:17:53,260 --> 00:17:56,060
And then (b), we had
this integral from 0

413
00:17:56,060 --> 00:17:59,440
to 1 natural log
x over root x dx.

414
00:17:59,440 --> 00:18:02,150
And the point here was a
little more complicated,

415
00:18:02,150 --> 00:18:04,690
so let me come to that one.

416
00:18:04,690 --> 00:18:07,130
We used an integration by parts.

417
00:18:07,130 --> 00:18:09,330
We had an integral
they converged easily,

418
00:18:09,330 --> 00:18:14,350
because this was an easy
improper integral to determine.

419
00:18:14,350 --> 00:18:16,190
And then we had this
other thing that we now

420
00:18:16,190 --> 00:18:17,519
have to evaluate it.

421
00:18:17,519 --> 00:18:19,310
You wound up having to
use L'Hopital's rule

422
00:18:19,310 --> 00:18:21,037
to evaluate it.

423
00:18:21,037 --> 00:18:23,370
But then you're able to show
that this using L'Hopital's

424
00:18:23,370 --> 00:18:26,490
rule, this actually has
a value at each endpoint,

425
00:18:26,490 --> 00:18:28,210
a finite value at each endpoint.

426
00:18:28,210 --> 00:18:33,720
It converges then to 0 as you,
when you put in these bounds,

427
00:18:33,720 --> 00:18:36,484
and then you have a
fixed value for this one

428
00:18:36,484 --> 00:18:37,650
when you take that integral.

429
00:18:37,650 --> 00:18:39,983
So you ended up with another
integral that was improper.

430
00:18:39,983 --> 00:18:41,490
But you could show it converged.

431
00:18:41,490 --> 00:18:43,310
And then you could
use L'Hopital's rule

432
00:18:43,310 --> 00:18:45,018
to find what happened
at these endpoints.

433
00:18:45,018 --> 00:18:46,940
And you get a value of minus 4.

434
00:18:46,940 --> 00:18:50,816
Then the third one was this
integral minus 1 to 1 x

435
00:18:50,816 --> 00:18:52,141
to the minus 2/3 x dx.

436
00:18:52,141 --> 00:18:53,640
And the point I
wanted to make there

437
00:18:53,640 --> 00:18:55,340
is that just because
the function is

438
00:18:55,340 --> 00:18:58,415
well-behaved near the endpoints
doesn't mean that it's still,

439
00:18:58,415 --> 00:19:00,212
you know, that
everything is hunky-dory.

440
00:19:00,212 --> 00:19:01,670
You have to be
careful and you have

441
00:19:01,670 --> 00:19:04,640
to check at places where
the value of the function

442
00:19:04,640 --> 00:19:06,160
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 --> 00:19:18,740
And so you split it up
into it's two pieces

448
00:19:18,740 --> 00:19:20,960
so that you have the endpoints.

449
00:19:20,960 --> 00:19:22,790
Now these two new
integrals represent

450
00:19:22,790 --> 00:19:24,670
where the problem
might happen, and then

451
00:19:24,670 --> 00:19:28,240
you do your, find
your antiderivative,

452
00:19:28,240 --> 00:19:31,270
and evaluate, and see if you can
get something that converges.

453
00:19:31,270 --> 00:19:34,740
So that's the idea of
these types of problems.

454
00:19:34,740 --> 00:19:36,900
So hopefully that
was helpful, and I

455
00:19:36,900 --> 00:19:38,546
think that's where I'll stop.