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Welcome back to recitation.

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In this video, I'd like us to
do the following problem.

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I want us to show that the
function, if I integrate x to

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the n, e to the minus x from
1 to infinity, that that

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actually converges for any value
of n, and I want us to

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show this without using
integration by parts n times.

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So I'd like you to figure out
a way to show that this

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integral converges.

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And again, what we mean by
converges, is for a fixed

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value here, say, some b, if I
let, as the limit is b goes to

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infinity, that that sequence
of values converges to a

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finite number.

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That's what we mean,
again, when we

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say an integral converges.

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Ultimately, we want to
show this is finite.

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So show this is finite, without
using integration by

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parts n times.

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So I'll give you a little while
to work on it, and then

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I'll be back, and I will
show you how I did it.

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OK.

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Welcome back.

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Well, I want to show you how we
can show that this integral

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actually convergences
for any n.

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And I don't want to have to use
integration by parts and

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kill off powers of x in
order to do that.

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So again, the integral is from 1
to infinity of x to the n, e

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to the minus x dx.

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And if you recall, Professor
Jerison was showing, in the

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lecture video, that if you can
show even from, not 1 to

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infinity, but from very far
out to infinity, that that

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integral converges, from 1 to
whatever the far out value is,

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this integral is finite.

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OK?

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Doesn't blow up.

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It's going to be potentially a
very big number, but it is

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going to be a finite number.

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There are no places
where we run into

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trouble with this function.

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It's a continuous function
from 1 to infinity.

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So we can go very far out and
say, OK, from very far out to

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infinity, the integral
converges.

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And then that's going
to be enough.

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So we did see that kind
of technique earlier.

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But I just want to remind you,
that's what we're going to do.

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Now, you might have thought
about this problem and said,

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well, I know that x to the n is
much smaller than e to the

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x for values of x very large.

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So you might have said--

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I think you used this notation
also in the lecture. x to the

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n is much smaller than e
to the x for large x.

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So what if we tried
to do a comparison

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with those two functions?

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We're going to see, that's
not quite enough.

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But let's say x is very large,
and let's look at a

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comparison.

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If I say, the integral from,
say, some very large r to

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infinity of x to the n, e to the
minus x dx, it's certainly

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going to be much smaller than
the integral from big r to

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infinity of e to the x times
e to the minus x dx.

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And you think, well,
you know, that's a

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pretty good first step.

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I'm doing all right.

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But what happens here?

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What's e to the x times
e to the minus x?

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It's e to the x plus negative
x, so it's e to

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the 0, so it's 1.

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So this is integrating
the constant 1.

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Well, the constant 1 from r to
infinity, think of that.

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It's the line y equals
1 from r to infinity.

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It's an arbitrarily
long rectangle.

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That's got a lot of area.

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It's got infinite area.

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So this integral diverges.

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That doesn't mean this
one diverges, right?

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Because this one is smaller
than that one.

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So this one here could still
converge, even though this

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integral diverged.

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Again, let me remind you.

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Why does this integral
diverge?

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Because this is actually
equal to 1.

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If I integrate 1 from
r to infinity, I

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get something infinite.

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So you might have started with
that, but that's not quite

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good enough.

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Right?

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What is going to be good enough,
is if I pick any

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constant in front of this n--

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I can put any constant
in front of this n I

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want, bigger than 1.

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And that's going
to help us out.

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So I'm going to pick the
constant 2, because it's going

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to be easy, and it's a
nice fixed number.

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If I--

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so this is how you do it
correctly, or one strategy to

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do it correctly.

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If I take x to the 2n, instead
of just x to the n, for any n,

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there's some r big enough so
that x to the 2n is much less

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than e to the x for all x bigger
than sum, bigger than

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or equal to sum r.

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So if I go far enough out in
x-values, x to the 2n is much

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smaller than e to the x.

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So let's say that value
is capital R, and

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then we're much smaller.

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This is not very formal, but
it's getting closer to a

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formal kind of thing.

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Then that means x to the n
is much smaller than e

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to the x over 2.

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Right?

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And that's going
to be the key.

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Let's anticipate why that is.

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The problem with the
substitution of e to the x was

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that e to the x times e to
the minus x gave you 1.

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But if I use e to the x over
2, I'm going to end up with

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some function, e to the minus
something, and that's going to

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be good, because that's
going to converge.

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So if you tried e to the x
first, and you saw you didn't

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get a good function, you didn't
have to stop there.

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You could say, well,
I was close.

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The problem is, I cancelled
off all the e to the minus

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power, which is what I
want to keep around.

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If I want to keep some of that
around, then I have to have a

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little less power of
e to the x there.

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I have to have, I can't
just have e to the x.

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I should have something like
square root of e to the x.

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So e to the x over 2.

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That's going to help us out.

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So this is, this is kind of, as
you're working on this type

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of problem, this is some of the
thought process you want

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to go through.

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So what do we see here?

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We have x to the n is much less
than e to the x over 2

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for x bigger than
or equal to r.

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So now let's do a comparison
with our

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new comparative function.

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So I'm going to come over, and
this will be our last line to

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finish this off.

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So now we're integrating from r
to infinity x to the n, e to

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the minus x dx.

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And we know that's going to be
much less than integral from r

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to infinity, e to the x over
2, e to the minus x dx.

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And now let's figure
out what this is.

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e to the minus x plus x over 2
is e to the minus x over 2.

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And the good news is, this is,
we know this converges.

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I'll check, and I'll show you,
remind you that it converges.

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But we know this converges.

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And the reason is, because e to
the minus x is a function

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that decays so fast
as it goes to 0.

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That's really why you
get the convergence.

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Actually, you could even compare
this to x to the minus

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2 right away.

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And you could get something
like, you know this decays

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faster than x to the minus
2, and we know x

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to the minus 2 converges.

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So you could even compare
it to that.

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You could do a second
comparison in here.

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But I'll actually calculate
this, just to remind

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us how we do this.

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So e to the minus x over 2.

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If I want to find an
antiderivative, I'm going to

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guess and check.

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I know I'm going to have an e
to the minus x over 2 again,

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and then I need to be able to
kill off a negative 1/2.

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So I should put a negative
2 in front.

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Let's double check.

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This is basically a substitution
problem.

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An easy substitution.

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So e to the minus x over 2.

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Its derivative is negative 1/2
x, and then itself again.

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The negative 1/2 times negative
2 gives me a 1.

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So again, I'm just, it's an easy
substitution problem, but

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I always want to check.

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So I evaluate that from
r to infinity.

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Well, the point is, e to
the minus infinity--

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this is 0.

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As x goes to infinity, this
quantity goes to 0.

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So I get 0 minus a negative
2, so plus 2, e to the

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minus r over 2.

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For some fixed, big r.

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Well, that's finite.

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That's a finite number.

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So we come back here.

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This integral converges.

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That integral was
this integral.

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And this integral, then, is
bigger than this one.

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So this one converges.

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Now, this had a lot of pieces to
it, so I'm going to remind

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us sort of what was happening.

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So let's go back to the original
function, and I'll

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just take us back through
one more time.

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OK.

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So the original problem was,
show that this integral from 1

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to infinity of x to the n e to
the minus x dx converges.

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And I reminded you that you knew
from lecture that if I

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could show that it converged
for some very large number

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down here to infinity, that was
sufficient, because this

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function is continuous
from 1 to infinity.

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I don't have to worry about
places where I might get

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infinite area in a
finite interval.

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I'm always going to
have finite area

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and a finite interval.

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So if I start at some big R to
infinity and that converges,

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then I'm good.

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Because from 1 to r,
that'll be finite.

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So then the point is,
you want to compare.

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And I mentioned, a comparison
that doesn't quite work, but

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is a good first test. Because
you know x to the n is much

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less than e to the x for
a sufficiently large x.

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You might think to compare it to
that, but the problem again

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was when we do that
substitution, we actually get

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an integral that diverges.

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But a divergent integral bigger
than something doesn't

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mean this one diverges.

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If the divergence was on, the
inequality was the other way,

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then you could show
this one diverged.

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But we actually show convergence
this way.

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Or we're trying to show
convergence in this direction,

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so we need to say, if
this one converges,

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then that one converges.

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This diverging doesn't tell
us anything about this.

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So then we say, OK.

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This one didn't work, but
it almost worked.

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So what if I figure out a way
to compare x to the n to a

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slightly smaller thing
than e to the x?

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And that slightly smaller thing
is e to the x over 2.

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It's not really slightly
smaller.

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But the smaller function
is e to the x over 2.

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OK?

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And this is a way to think
about how that works.

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Is that for any power of x to
the 2n, I can still get it

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smaller than e to
the x for some

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sufficiently large number.

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OK?

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And you don't even have to
think about these R's

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as being the same.

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I can change them, I can
make this bigger.

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This doesn't compare to
this problem, also.

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So don't be confused
by those two R's.

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OK.

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So then we found something
we wanted to

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compare x to the n with.

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Then we come back over here,
and we actually see that we

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get a good comparison, because
we're able to see that the

248
00:10:21,590 --> 00:10:24,010
integral on the right
hand side converges.

249
00:10:24,010 --> 00:10:27,200
This integral is bigger
than this integral.

250
00:10:27,200 --> 00:10:28,670
So this integral converges,
so we know

251
00:10:28,670 --> 00:10:30,260
this integral converges.

252
00:10:30,260 --> 00:10:34,450
And then, the integral from 1 to
r of this is finite, so the

253
00:10:34,450 --> 00:10:37,580
integral from 1 to infinity
of this converges.

254
00:10:37,580 --> 00:10:40,370
And that's sort of the strategy
for doing these types

255
00:10:40,370 --> 00:10:40,440
of

256
00:10:40,440 --> 00:10:41,760
problems. OK.

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00:10:41,760 --> 00:10:43,400
That's where I'll stop.