PROFESSOR: Welcome
back to recitation.
In this video I want us to
work on some problems looking
at integrals that may
converge or diverge.
So these are improper integrals.
And we want to know if
each of the integrals
below converges or diverges.
And if they converge, I
want you to compute them.
I want you to actually
evaluate it, find a number that
is the area under the curve.
So that's really, remember,
what the integral is.
So we should be able
to find, actually,
the finite number
that represents that.
So there are three of them.
The first one is
the integral from 0
to infinity of cosine x dx.
The second one is the integral
from 0 to 1 of natural log
x divided by x to the 1/2 dx.
So that's just square
root x, down there.
And the third one is the
integral from minus 1 to 1
of x to the minus 2/3 dx.
So why don't you take a
while to work on this.
Pause the video.
When you're feeling
good about your answers
bring the video back up.
I'll be back then to
show you how I do them.
OK, welcome back.
Well hopefully you were able
to answer these questions.
The question again
was that we wanted
to know if the following
intervals intervals integrals
converged or diverged.
And then I wanted us to actually
compute them if they converged.
So again, we're going to
start with the integral
of cosine x dx.
Then we'll look at the
integral of natural log
x over root x dx.
And then we'll look at the
integral from minus 1 to 1
of x to the minus 2/3 dx.
So let's look at the first one.
And I'll rewrite it
up here so we have it.
OK.
And this is kind of
interesting because maybe this
is a little bit different than
what you have seen previously
in the way of saying this
is potentially improper.
Because cosine x certainly
doesn't blow up anywhere.
It's bounded between
minus 1 and 1.
So the function itself
is not blowing up.
But let's look at what we get
when we try and evaluate this.
So if we take this
integral, we know
the antiderivative of cosine is
negative-- no it's just sine.
Sorry, it's just sine x.
Right?
The derivative of
sine is cosine.
So we get sine x.
And what we're
supposed to do, is
that we're supposed to
take-- let me rewrite that x.
That's horrible.
We're supposed to take the limit
as b goes to infinity of sine x
evaluated from 0 to b.
Right?
And so this is pretty
straightforward.
Again now it's the limit at
b goes to infinity of sine b,
because sine of 0 is 0.
Now here's where we run into
trouble, because this limit
doesn't exist.
Right?
because as b goes to infinity,
sine b, the function sine x--
so now it's really
the function sine b,
b is now the variable, if
we think about it that way--
as b goes to infinity,
sine is going
to oscillate as it always
does between minus 1 and 1.
And it's going to
continue to do that.
It's not going to
approach a certain value
and stay arbitrarily close
to that value as be goes off
to infinity.
So this limit does not exist.
And maybe what's informative is
to think about how could this
happen as an integral?
If we know that the
integral we're looking for
is really the signed
area under the curve.
So let me explain
briefly what's happening.
Let me just draw a
quick picture of cosine
and explain briefly
what's happening.
So cosine starts off like this.
Right?
So what happens if I wanted
to integrate cosine x from 0
to infinity.
I first pick up this much area.
Sorry this graph is a
little sloped, I realize.
I pick up this much area
and that's all positive.
And then as I keep
moving over here to here,
I pick up the same amount
of area, but it's negative.
So once I get to here
my integral, it's 0.
The area above and the
area below are equal.
So then I start the process
again with the value 0.
And so I accumulate
some negative area.
Then over here it's the same
amount of positive area,
and it kills it off.
So I start off, I have
some positive value,
and then it becomes less
positive and goes to 0.
Then I get some negative
value accumulated.
Then it comes up
and goes to 0 again.
So the point is as I'm
moving off to infinity,
the area is oscillating.
And the area, remember, is
actually what the value of sine
is at that point.
The area from under
the curve from 0
to any value for cosine x is
the value of sine at that point.
That's what we're seeing here.
So the point is that this
integral, even though cosine x
is a bounded function,
the area is accumulating,
then disappearing, then becoming
negative, then disappearing,
then accumulating,
then disappearing.
So there's no value
that it's approaching.
It's varying between all these
values over and over again.
So this is a weird
one, maybe, where
the integral doesn't exist.
But it actually, it
doesn't exists there.
So hopefully that
makes sense, and even
though it's a little different,
you understand the idea.
So now I'm going to go to (b).
So what is (b)?
It's the integral from 0 to
1 of ln x over x to 1/2 dx.
OK.
And what I want to point
is we probably want
to see why is this improper.
Well at 1, we don't of a problem
because natural log of 1 is 0.
And we can put a 1 down here
and nothing bad happens.
At 0 we have a problem.
The natural log of 0
actually doesn't exist.
The limit as x goes
to 0 of natural log
x is minus infinity.
And then natural log--
or sorry, natural log?
Zero, evaluating square
root x at 0, gives you 0.
So we have something
going to minus infinity
in the numerator and
0 in the denominator.
And we're trying to
integrate that function
as it goes towards that value.
So we have to figure out
kind of what's going on here.
So let's not worry about
the bounds at the moment.
Those are obviously going
to be important at the end.
But let's figure out what we get
as an antiderivative for this.
OK?
And we don't worry
about the constant,
remember, in the
antiderivative because we're
going to evaluate.
So what's the best way
to attack this one?
Well probably you should see
that you got a natural log
and you've got a power of x.
So this is really set up to
do an integration by parts.
Because remember,
you like to take
derivatives of natural log.
And powers of x are
happy to be integrated
or to take derivatives of them.
The only power of x that's
maybe a little bit annoying
to integrate is x to the
minus 1, because its integral,
because its antiderivative
is natural log instead
of, like, another power of x.
But, this one is not
that case so it's
looking good for an
integration by parts.
So again we said
for an integration
by parts, natural log you
love to take its derivative.
So we're going to let
u be natural log of x
and u prime be 1 over x.
And then v prime is
x to the minus 1/2.
Right?
This is x to the 1/2
in the denominator,
so v prime is actually
x to the minus 1/2.
So v is going to be x to the
1/2 with a correction factor.
So it's going to need a
2 in front, I believe.
Let me double check.
Derivative of this
is-- 1/2 times
2 is 1-- x to the minus 1/2.
So I'm doing OK.
So now what do I get when I
want to take the integral?
I take u*v. So it's going
to be 2 x to the 1/2 ln x--
and then now I'm going
to put in the bounds--
evaluated from 0 to 1, minus
the integral of v u prime.
So v is x to the 1/2, and u
prime is x to the minus 1,
really it's 1 over x.
And so I'm going to keep the 2
in front, integral from 0 to 1,
x to the 1/2 over x
is x to the minus 1/2.
This will be nice because we've
already taken an antiderivative
once, so we know what we get.
All right, now this one
we'll have to look at,
because notice
that as x goes to 0
we have to figure out
if this has a limit, OK?
We have to figure out
of this has a limit.
And we'll probably need to use
L'Hopital's rule to do that.
So let's finish
this part up first.
So I'll just keep this here.
0 to 1, that's a
0, and then minus.
OK, x to the minus
1/2, its antiderivative
was 2 x to the 1/2.
So I have another 2, so I get
a 4 c to the 1/2 evaluated
from 0 to 1.
So this is easy because
here, I just get 4.
Right?
This part right here,
when I evaluate it
I'll just get 4 and the
minus sign in front.
So this part is just negative
4, because at x equals 1 I get 1
and at x equals 0 I get 0.
So that's fine.
This part is fine.
This part, when I put in 1
for x, notice what happens.
I get a 1 here.
Natural log of 1 is
0, and so I get 0.
And so the question
really is, what
is the limit as x goes
to 0 of this quantity?
I know right here-- again,
I have a minus 4 here.
And then the question
is what do I have here?
it?
It could blow up still.
It could diverge.
We're going to see what happens.
So let me just put a line.
It's the same problem, but I
want to distinguish for us.
So we're actually
wanting to find the limit
as b goes to 0 of 2 b to
the 1/2 natural log b.
Right?
That's what we're interested in.
That's the only part we
don't yet understand.
So let's see what happens.
Well, if I want to make this
into a L'Hopital's rule thing,
right now I have 0
times negative infinity.
So to put it into
a form I recognize,
I'm going to rewrite this.
This is actually
equal to the limit
as b goes to 0-- I'm going
to keep the natural log be
up here, and I'm going to write
this as b to the minus 1/2
in the denominator.
Let's make sure we understand
what just happened.
I had a b to the 1/2
in the numerator.
If I put it to the minus
1/2 in the denominator,
it's still equal
to the same thing.
Right?
b to the minus 1/2
in the denominator
is equal to b to the
1/2 in the numerator.
OK?
Really I'm taking 1 over this,
and then I'm dividing by it.
That's the way you
want to think about it.
You want to think about
saying I'm taking 1 over this,
and I'm taking it in the
numerator and the denominator.
so I end up with it just here.
That might have been confusing.
The real point is this
quantity, one over this quantity
is equal to that quality.
And now what's the point?
Why did I bother to do that?
As b goes to 0, this
goes to negative infinity
and this goes to infinity.
So now I have something where
I can apply L'Hopital's rule.
So the limit as b goes to 0,
derivative of natural log of b
is 1 over b.
So I get 2 over b
in the numerator.
The derivative of b to the
minus 1/2 is negative 1/2 b
to the minus 3/2.
There's a lot of denominators in
the numerator and denominator,
so let's simplify this.
That's the limit as b goes to 0.
OK, what do we get here?
We have 2 times minus 1/2,
or 2 divided by minus 1/2.
So I'm going to get a
negative 4 from this part,
the coefficient.
Then I have a b to the minus
1 up here divided by a b
to the minus 3/2.
That's actually going to be
a b to the 3/2 divided by b.
That's going to be b to the 1/2.
That's algebra.
You can check it if you need to.
I mean, you should
have gotten this.
But if you didn't get
this, check again.
Let me make sure I get it again,
b to the 3/2 divided by b, b
to the 1/2.
And that equals 0, because
it was a limit as b
goes to 0 of this quantity.
Now I've just got a
continuous function.
I am, notice, I I
didn't actually--
I kind of cheated a little
bit, because I just wrote limit
as b goes to 0.
But I'm always doing as b goes
to 0 from the right-hand side.
I'm starting at 1.
So you may have seen
this, this 0 plus
means I'm only interested
as b goes to 0 from above 0.
OK?
I didn't write that in,
but notice our integral
was between 0 and 1.
So it only mattered
values to the right of 0.
So this function is
defined to the right of 0.
So I can just evaluate there.
I can just plug it in.
It's continuous, and so I can
just say at 0 it equals 0.
So now let's go back
to where we were.
What were we doing here?
We were taking this whole
piece, was to come back in here
and figure out what
this value was.
We knew at 1, we got 0.
And now we know at
0, we also get 0.
So that question mark
I can replace by a 0,
and I get 0 minus 4.
So the actual answer
is negative 4.
OK, we have one more.
What's the last one?
The last one-- let me write
it down over here again--
is the integral from minus 1
to 1 of x to the minus 2/3 dx.
Now this is interesting,
because this,
you actually saw an example
kind of like this in the lecture
that at this
endpoint, it's fine.
You can evaluate the
function at that endpoint.
At this endpoint it's fine.
You can evaluate
the function there.
So it looks good
at the endpoints.
But the point is
that at x equals
0, because this minus
power is putting
your x in the denominator,
you actually, your function
is blowing up.
It has a vertical
asymptote at x equals 0.
So what we want to do
is we have this strategy
for these problems, is to
split it up into two parts,
minus 1 to 0 of x to the
minus 2/3 dx plus the integral
from 0 to 1 of x to
the minus 2/3 dx.
Now the point is that
now the only place
where I have a
vertical asymptote,
I see it as an endpoint
on both of these.
And this is again,
just this good thing
we have an additive property
for these integrals,
that the sum of
these two integrals
is going to equal this one here.
As long as, you know, as
long as these are converging,
then I can say that
if this converges
and this converges, then their
sum converges to this one here.
OK, so that's really what
I'm trying to exploit here.
Now what is the antiderivative
here for x to the minus 2/3?
This is going to be-- again, I'm
going to write something down
and then I'm going to check it.
I think it should be
5/3, x to the 5/3.
And then I have to
multiply by 3/5.
Let's double check, because
this is where I always
might make mistakes.
If I take the derivative of
5/3 times 3/5 it gives me a 1.
5/3 minus 1 is 5/3 minus
3/3, and that's 2/3,
and that's wrong.
Right?
Because I went too big.
It's supposed to be 1/3.
I knew I was going
to make a mistake.
Right?
I'm supposed to
add 1 to minus 2/3.
That's just 1/3.
And so I should
have a 3 in front.
So for those of you who were
saying she's making a mistake,
I was.
OK, x to the 1/3.
So I take the derivative
times 3, I get a 1 there.
I subtract 1 and
I get minus 2/3.
So now I'm in business.
OK.
And now I just need to evaluate
that here, here if I can,
then another one
here from 0 to 1.
And the good news
is x to the 1/3
is continuous at
all these points.
It's continuous across
negative 1, 0, 0, and 1.
So I can actually just evaluate.
I don't have to worry.
There is a value for
the function there.
It's continuous through
all these points.
So I can just evaluate them.
So let's see what I get.
I get 3 times 0.
And then I get minus
3 times negative 1,
so I get a negative
3 plus 3 times
1, and then minus 3 times 0.
OK, so 3 times 0
minus negative 3.
So that's a 3 plus
3, and I get 6.
And what's the picture of this?
Well you should think
about what the picture of x
to the minus 2/3 looks like.
And you should notice
that across 0, it's even.
Right?
So it's actually
going to have-- it's
going to look the same
in the left-hand side
and the right-hand side.
So if I had wanted to,
I could have just found
the value of say,
this integral-- I
like positive numbers
better-- the integral from 0
to 1 of this function,
multiplied it by 2,
and it would have given
me the actual value.
Because it's exactly the same
function to the right of 0
and to the left of 0.
It's a reflection
across the y-axis,
so you get the same value there.
So you actually would've gotten
3, for this, multiplied by 2
and you get the value.
If this one had
diverged, that one also
would have had to diverge,
and the whole thing
would have diverged.
And that's because
of the symmetry
of the function over 0.
So I'm going to just
briefly review what we did.
And then we'll be done.
So we come back over here.
I gave you three integrals.
We wanted to see if they
converged or diverged, right,
and if they converged
find what they were.
So the first one was cosine x.
And we found that diverged
for a different reason
than what we've seen before.
Because as x goes to infinity,
the problem is the areas
are varying up and down
and they're bounded always.
But the area under the
curve is constantly changing
and it's not converging
to a fixed value.
It's varying between minus
1 and 1 over and over again.
And then (b), we had
this integral from 0
to 1 natural log
x over root x dx.
And the point here was a
little more complicated,
so let me come to that one.
We used an integration by parts.
We had an integral
they converged easily,
because this was an easy
improper integral to determine.
And then we had this
other thing that we now
have to evaluate it.
You wound up having to
use L'Hopital's rule
to evaluate it.
But then you're able to show
that this using L'Hopital's
rule, this actually has
a value at each endpoint,
a finite value at each endpoint.
It converges then to 0 as you,
when you put in these bounds,
and then you have a
fixed value for this one
when you take that integral.
So you ended up with another
integral that was improper.
But you could show it converged.
And then you could
use L'Hopital's rule
to find what happened
at these endpoints.
And you get a value of minus 4.
Then the third one was this
integral minus 1 to 1 x
to the minus 2/3 x dx.
And the point I
wanted to make there
is that just because
the function is
well-behaved near the endpoints
doesn't mean that it's still,
you know, that
everything is hunky-dory.
You have to be
careful and you have
to check at places where
the value of the function
is going off to infinity.
So in the second case, the
left endpoint posed a problem,
and everything else was fine.
In this case, the left and
right endpoints are both fine,
but in the middle
there's a problem.
And so you split it up
into it's two pieces
so that you have the endpoints.
Now these two new
integrals represent
where the problem
might happen, and then
you do your, find
your antiderivative,
and evaluate, and see if you can
get something that converges.
So that's the idea of
these types of problems.
So hopefully that
was helpful, and I
think that's where I'll stop.
