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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

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

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that that actually converges
for any value of n.

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And I want us to show this
without using integration

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by parts n times.

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So I'd like you to
figure out a way

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to show that this
integral converges.

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

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here, say, some b,
if I let-- the limit

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as b goes to infinity, that
that sequence of values

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converges to a finite number.

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That's what we mean, again, when
we 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

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by parts n times.

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

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and then 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

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show that this integral
actually convergences for any n.

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

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and 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

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was showing, in
the lecture video,

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that if you can show even
from, not 1 to infinity,

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but from very far
out to infinity,

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that that integral
converges, from 1

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to whatever the far-out value
is, 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,

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but it is going to
be a finite number.

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There are no places
where we run into trouble

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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,

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from very far out to 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

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and said, well, I
know that x to the n

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is much smaller than e to the
x for values of x very large.

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So you might have said-- I think
you used this notation also

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in the lecture. x to
the n is much smaller

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than e to the x for large x.

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So what if we tried to do
a comparison with those two

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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 comparison.

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If I say, the
integral from, say,

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some very large R to infinity of
x to the n e to the minus x dx,

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it's certainly going to be
much smaller than the integral

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from big R to infinity of e
to the x times e to the minus

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x dx.

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

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a 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 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,

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even though this
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,

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I get something infinite.

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So you might have
started with that,

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but that's not
quite good enough.

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Right?

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What is going to
be good enough, is

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if I pick any constant
in front of this n--

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I can put any constant in
front of this n I want,

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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,

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

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If I-- so this is how you do
it correctly, or one strategy

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to 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

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is much less than e
to the x for all x

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bigger than or equal to some R.

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

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is much smaller than e to the x.

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

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and then we're much smaller.

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

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to a formal kind of thing.

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Then that means x to
the n is much smaller

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than e 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

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was 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

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going to end up
with some function,

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e to the minus
something, and that's

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going to be good, because
that's going to converge.

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So if you tried
e to the x first,

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and you saw you didn't
get a good function,

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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 power,

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which is what I
want to keep around.

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If I want to keep
some of that around,

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then I have to have a 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

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working on this type
of problem, this

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is some of the thought process
you want 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

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our new comparative function.

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

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to finish this off.

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

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e to the minus x dx.

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

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from R 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 that

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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 2

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right away.

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And you could get
something like, you

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know this decays faster
than x to the minus 2,

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and we know x 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,

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just to remind 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,

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I'm going to 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,

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but 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-- 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

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to the minus R over 2.

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For some fixed, big R.
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,

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so I'm going to remind us
sort of what was happening.

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So let's go back to
the original function,

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and I'll just take us back
through one more time.

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OK.

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So the original
problem was, show

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that this integral from 1
to infinity of x to the n e

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to the minus x dx converges.

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And I reminded you that
you knew from lecture

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that if I could show
that it converged

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for some very large number
down here to infinity,

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that was sufficient,
because this function is

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continuous from 1 to infinity.

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I don't have to
worry about places

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where I might get infinite
area in a finite interval.

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I'm always going to have finite
area in 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,

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but is a good first test.

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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,

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but the problem again was when
we do that substitution, we

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actually get an
integral that diverges.

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But a divergent integral
bigger than something

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00:09:13,410 --> 00:09:15,270
doesn't mean this one diverges.

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If the divergence
was-- the inequality

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was the other way, 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, then

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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

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to a 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,

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I can still get
it smaller than e

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to the x for some
sufficiently large number.

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OK?

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00:10:01,911 --> 00:10:03,910
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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00:10:12,260 --> 00:10:14,890
So then we found something we
wanted to compare x to the n

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00:10:14,890 --> 00:10:16,170
with.

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Then we come back over
here, and we actually

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see that we get a good
comparison, because we're

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able to see that the integral on
the right-hand side converges.

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00:10:24,010 --> 00:10:27,166
This integral is bigger
than this integral.

246
00:10:27,166 --> 00:10:28,540
So this integral
converges, so we

247
00:10:28,540 --> 00:10:30,260
know this integral converges.

248
00:10:30,260 --> 00:10:33,160
And then, the integral
from 1 to R of this

249
00:10:33,160 --> 00:10:34,900
is finite, so the
integral from 1

250
00:10:34,900 --> 00:10:37,580
to infinity of this converges.

251
00:10:37,580 --> 00:10:39,990
And that's sort of
the strategy for doing

252
00:10:39,990 --> 00:10:41,260
these types of problems.

253
00:10:41,260 --> 00:10:41,760
OK.

254
00:10:41,760 --> 00:10:43,400
That's where I'll stop.