Math 2414 Part II Section 7.8
Okay, so what do we have to do first? We have to rewrite this as a limit and then an integral
with finite bounds. So the limit as T approaches infinity from three to T. IF you want to you can pull that
two through the integral and through the limit. So if I wanted to make my life a little bit easier, I can pull
that all the way through like this? (Someone walked in late and Yosko made the class sing happy
birthday, you have been warned.) Okay, do I know this integral? Ln absolute X where X ranges from
three to T. So do you see why I write the X equals X equals stuff? It's because I would always confuse
things. So two, I rewrite the limit and evaluate my bounds and what do I get?
I don't really need the absolute value here because three is always positive. Now I can take that
limit right? Graph Ln for me in the air, Y equals Ln of X. It looks like that? So as T gets large what do the
Y values in that function do? They get very large, so this is two. So this piece goes to infinity? So it
doesn't really matter if the rest goes is constant because you're subtracting a number and multiplying
the whole thing by two? So this is what the answer is. So the actual integral is infinity, but the answer to
the question that they asked was if it was convergent or divergent or not. So this is divergent because
the infinity showed up. Oh look at that, what do I do?
Rewrite it with a limit? Is that okay? Now integrate that? Integrate that sucker now. Is it U subs?
Integration by parts, let's see. Is it integration by parts? So what would we try for U? R? Okay, then what
would Du be? Dr, that means that this piece E to the R over three Dr would have to be the Dv? Is that
something I can easily take the antiderivative of? Okay, recall the derivative with respect to X of E to the
star X is just E to the star X times star right? So the integral of E to the star X Dx is E to the star. If here I
multiply by star what am I going to do here? Divide, okay? Yeah, so this is really E and then it's one-third
times R. So up here I would have that E to the R over three as a power, what goes in front?
Variables? Okay check it, take a derivative. You would do three and E to the R over three. Then
you multiply by the coefficient of R which is one-third. What happens to three over one third? So is this
the correct antiderivative? So this would be the limit as T approaches negative infinity and do Uv. So
what's that three R E to the R over three from T to six minus the integral of three to this and I still have
the bonds of T and six. Is that okay? Uv minus the integral of V Du? Why did I not change the bounds
when I did this substitution? It's not a variable substitution. Okay, then we start plugging in stuff. So I get
eighteen E squared ... ks that okay?
Okay here I'm going to have nine because when I do the antiderivative wasn't it exactly the
same thing I did when I found Dv. I have Dv and I found V? Where did the nine come from? I had a three
right here, and then when I took the antiderivative I multiplied by the reciprocal of the coefficient of R.
Do we got it? So eighteen E squared minus three T E to the T over three. Okay minus E squared right?
Now it's going to be this plus E to the T over three. Is that okay? How many squares do I have? I have
nine E squared minus three times the limit as T approaches the limit of negative infinity of T E to the T
over three plus nine times the limit as T approaches negative infinity E to the T over three.
Graph E to the t for me? Good, it looks like this? So I can do this one really easily right? What's
that? Zero right? So that one’s zero. So I have nine E squared, this one is zero. How do I do this one?
When we first learned about the top one we just kind of plugged in and see what happens. What
happens when you plug it in? You get negative infinity and you get E to the negative infinity which is
negative infinity times? Huh? Do you remember what that is? When you have something indeterminate
like this, it was a product of a power or difference. You always had to rewrite that as a fraction.
Remember you were never allowed to use L'Hospitals unless you had a fraction.
Now we have to rewrite this as a fraction. Can you think of a convenient way to write this as a
fraction? Nope, I can only use the pieces I have. So I have T E to the T over three. That's either E to the T
over three over T to the negative one. Or that's T on top and e to the negative three on the bottom.
Which ones is easier to use? The one on bottom? What's the derivative of the bottom? Remember for
L'Hospitals, let's do it this way. So you would go like this. Rewrite the limit and do one-third E to the T
over three. This is negative T to the negative two. That got worse, do we see why? Because if I put
everything back to the top this is one-third negative.
That's more T's than what I started with. So you don't want to do it that way. So you go back
here and try L'Hospitals again. Now I would do this on top and this on bottom right? Actually this isn't
L'Hospitals, I'm just rewrite it. So we're doing L'Hospitals and I get a one and what's on the bottom? No
not three, negative one-third E to the negative T over three. Remember before you quit, you clean it up
with algebra to get it into the simplest form. So this nine E squared minus three. I'm going to move this
up to the top before I take the limit. So I should have a plus and a three here E to the T over three? Is
that right? Okay, then what is that limit as T goes to negative infinity?
So what's the answer, nine X squared. That was a good problem, it was a review of a lot. So
remember, anything you learned in math is fair game for test. How do we feel about hat one? Did we
remember L'Hospitals? Oh look at that? What would we do? Split it? Where? Split it at zero. Zero is nice
because it has properties that make it ideal. In fact things disappear when you evaluate at zero except
trig and exponential. Okay, so we're allowed to do two pieces. Let's do one at a time. Let's do the
negative one first, you can pick either one you want. So we're going to consider this guy first. Okay, so
we would rewrite this as the limit as t approaches negative infinity.
So I have from T to zero. Okay how do I integrate that? What? U-subs? U, Du, so one-half Du is
X Dx? Okay, now I still have my limit, it hasn't been touched yet. Now I’m rewriting this as one over U
and then a one-half Du. Do ya'll see it? Yeah? Probably should change my bounds. Lower was X equals
negative infinity and so I'm going to get U equals.... oh wait. I have T don't I. U doesn't make since, it's t.
So U is on plus T squared? Upper, so X equal zero and so U is one. Is that okay? Okay integrate that
sucker. Ln absolute value T, somethings wrong this should be a T. This is still a U, I just wrote this one
wrong. There's no way I could have done it like this.
Don’t ya'll see that if I integrated with respect to U and plug back into U I wouldn't get back
numbers? Anytime you have integral with bounds you're going to get a number answer? Okay, so I just
copied it wrong. This should be one plus T squared. Okay, U equals one plus T squared and U is one. I
still have one-half hanging out in the front. Okay so I have Ln one minus Ln of one plus T squared. So Ln
of one? Zero, so graph Ln of one for me in the air. Okay, then I let the operand with X in this case go to
infinity because on plus T squared goes to infinity? The T goes to negative infinity? So what's that piece
going to? Infinity?
Do ya'll see that one plus T squared is going to infinity as T goes to negative infinity. So if you
plug in infinity in Ln, it approaches infinity. What you really have is one-half times negative infinity.
Remember this is going to infinity, by you have the negative in front of it. Okay, so you know that's just
negative infinity. So this guy isn't zero, what is it? Negative infinity to zero? That diverges, so what does
the original one do? That was in my original problem, do I have to do the other side? Okay, let's just say
that I did the other side and I get a number like twelve. How would the original one behave? There's a
sum right here right? This one is running off to infinity, or negative infinity and then I'm adding twelve.
Does that make a difference?
No, so if you have to split something and you get one piece that diverges. You're done, right?
Because if it is infinity and you add or subtract something from infinity. It’s still infinity. Therefore, the
original guy diverges. Does everyone understand why there's no need to check to other one. What was
it, zero and infinity? It doesn't matter what it does, what's always fun is when one is seven or the second
one is infinity. Because if you do the infinite one first you're done right? If you get something with a
number first you have to do the other one to check. So anyway, we sort of talked about this question
already on the first page.
We talked about looking at this yellow region S and how we know the region wasn't closed in. It
went on forever, so why does this converge? That was the question, what is so interesting about this
integrand so that this would converge? So that's the reason why we're asking this question right now. Is
there something that's special about that power of X in the denominator? Could that be the reason that
this converged for the first example? Maybe for a different value of P it diverges? So that's what we're
going to do right. Are there particular values of P that make a difference? We know right that P was two
it would what? Converge, that was the first example?
So um recall, what example did we do? Was it like three to infinity? A couple pages ago did we
do this? Okay, so divergent means that it's approaching infinity and not a finite value. So what if I do this
instead, what if from three to infinity I did from one to infinity? I'm actually including more area right?
I'm actually more area right? I'm also including the area between one and three? Right? So, if this guy is
divergent and I'm including even more area. What does this guy do? It's still divergent right? Does that
make since? Now that would not be clear if you did from ten to infinity, right? Because you're using less
area there?
If I add more area obviously if I'm going to infinity and I add more area it's still divergent. Does
that make since? Now you're going to find out that this doesn't really matter, it's still divergent. But do
ya'll see that if I include less area it doesn't look as obvious? But the integral from one to infinity of two
over X Dx is actually, I can pull that constant out and write it like this? So we know this guy right here is
divergent. What happens if I don't have the two in front? What can I say about this? My answer is
infinity and I got it from doubling something. Wouldn't I have to double infinity? Something very very
large? So does this seem reasonable that this is divergent too?
Yeah? Okay, so thus when P is one the integral is divergent. So that's one P value that we looked
at. There's infinitely many P values right? So hopefully instead of checking every one of them we can
sort of classify them into this sort of group. Okay so we're going to now consider everything but equal
one. So now assume P is not equal to one. If I look at one over X to the P that's my integrand right? If
you notice earlier, what interval am I looking at? Where or what values of X am I looking at? Be even
more specific, this is what the bounds say right? This is really X is one or greater. Okay, I'm not talking
about P I'm talking about X.
Okay, so if I go back to this one over X to the P that means I'm always plugging in something one
or bigger, right? So I'm never plug in zero because that's a domain issue, but the bounds tell me that X is
one or bigger. So will this ever be negative? Huh? The whole thing. Even if P is negative, will the fraction
be negative? What does a negative P tell us? What does it do? It flips the position right? It flips the
position of whatever the P position is? Okay, so this is going to be positive. Okay, so let's look at this
further. So if I look at the integral from one to infinity one over X to the P Dx. I don't know what to do
with that infinite bound, so I need to write it down with a limit.
Nothing surprising there, but I'm going to integrate this right? I need to be able to write this in a
form that I can integrate. We already said let's not consider P equal one. If P is any number other then
one to take an integral I would have to use a power rule and rewrite it with a negative exponent? If P
equals one I would have to use an Ln right? I have to do the antiderivative, but since I throw down a P
equal one. It might be convenient to look at like this? See what I'm doing here? Just like if I integrated
one over X to the seventh, you never integrate it like this. You always clean it up with algebra and
integrate it like that.
That's all we're doing here. Check the integral? What is it? What? X to the negative P plus one
and what goes on the bottom? Negative P plus one right? Is that okay? I have bounds right? X equal one
to X equal two? Is that okay? Okay, so then we can plug in and I'll get the limit as T approaches negative
infinity will be X to the negative T plus one, oh wait. I'm plugging in for X right? So T to the negative P
plus one over negative P plus one minus what? It looks like this but what does it simplify to be? What is
that back term? It's one, what's one to any power? One, okay and then I have a denominator that's
negative P plus one.
Do you notice that I can factor out a negative P plus one on the bottom? I'm going to go back
over here. Everyone see that the numerator is one? You see I can rewrite this as... um, I can pull this
through the limit because it has no T's in it. Is that true? Okay so then I would get T and then negative P
plus one minus one right? Is that true? Is everyone okay with that? You just change the numerator and
factor out the denominator in both? So this is what we're going to look. We're going to talk about
different cases with this argument. Does everyone agree that if we do that integral that's what we get?
So let's think about a number line here.
So we've already said that these are P values. When P is one it is divergent. So now we're
considering all the other numbers. Let's look at things bigger then P equal one. Then we'll look at things
smaller. Okay, so Case One: Let P be bigger than one. So I'm testing this side of the number line. I can
put a D or a C or just split it somehow. Okay, P is bigger than one so tell me about this guy? Positive or
negative? Positive? P is bigger than one, give me a number bigger than one. Two, negative two plus one.
Is that positive or negative? Negative seven plus one, negative eleven plus one. So no matter what, that
is negative if P is bigger than one.
Is that okay? So then let's look at this piece right here. So T and a negative P plus one, we'll
negative P plus one is a negative number right? So this is like this whole thing is a negative number?
Yeah? Okay, so is this true? Can I pull out a negative out front? Is that true? So I can write this as one
over T to the P minus one? If I do that, now this power is positive or negative? It's positive? It was
negative to begin with. If I factor out a negative and then flip the position it becomes positive right? So
everyone okay with that? Yes? So now let's look at limit there as T approaches infinity I have T to some
positive power. What's it going to?
What's it going to? That looks like a T, T to a positive number. It's going to infinity right? So as T
approaches infinity. One over T minus P does what? Approaches what number? What does that do? The
bottom is getting very very large. What is the fraction doing? It's going to zero. Okay, so then I had the
integral from one to infinity one over X to the P Dx. We said that was equal to this. We just said that this
piece was the same thing. Isn't this the same as this? Factor out that negative one? So what is that going
to be? I'll have a one over negative P plus one. This piece is going to what? Zero? So I just have a
negative one in here? What now? P? Is this okay?
So I can write this as and distribute this negative one and this is like one over P minus one if I
multiply every term in the denominator by negative one. The negative on the bottom? Is that true?
Okay, what was? Some number right? So does this integral converge or diverge? It converges to this
right? Does that make since? We get something finite. So here's our number line. This is converge all
over here. It's bigger than one. So now I need to check what? Less than one, yes. I'm just going to
rewrite this top part so I can see it. That's always true right? Isn't that what we found here? Isn't that
always true?
Okay, so what's case two? P is less than one. So we're going to do the same thing we did here. If
P is less than one we're going to consider the same thing we do when P is greater than one. Do ya'll see
what happens to this and to this follows the same logic? Okay? So if P is less than one. Okay really? Then
negative P plus one is positive or negative? Positive right, so if I look at T to the negative P plus one.
That's like T to a positive number right? Yeah? As T approaches negative infinity, what does this guy do?
What? Yeah, okay it diverges. So I have a number line. I have one and here I have a big C and over here I
have a big D. What am I going to put on the left of one?
Diverges, because if I look at this integral, it runs off to negative infinity right? It doesn't really
matter, negative infinity minus one? That's still infinity right? If I multiply by a constant, this is just the
constant. So this whole thing is infinite? Okay, so this diverges when P was less than one. So here's our
summary. So you're going to see this again in future chapters (this summary will come into play when
you do series tests in chapter 11, so remember it). So when we had type one that meant that one of the
bounds was infinite, right? So it's pretty obvious when you have the type one. Okay, you're going to
have partial fraction type one question and then the type two.
The type two is a little more inconspicuous. So discontinuous integrand, what's the integrand?
The function that you're integrating right? Integral sign right? Integrand Dx, right? Okay, so if F is
continuous okay? Look at the interval, it says A to B, but A is included and B is not. This is very
important. In other words F is not continuous at B. Yes? Okay, then the integral from A to B F of X Dx
equals. Well we have a problem, here's A, and here's B. I'm good here and I have an issue there and it's
continuous everywhere in the middle right? That's what the number line looks like? I really don't have a
problem over here. I only have a problem when I get closer and closer to B from that direction.
Okay, I have to draw this every time. So then I will have the integral. Oh sorry, I'll have the limit
as T approaches B from the left of the integral from A to T F of X Dx. Why did I put from the left? That's
where the function is continuous. You have to come from a continuous side. I can't come from the right
because it's not continuous over there. I have to use the internal that I'm given and come from the
direction that's defined. Okay, so what if it was just discontinuous at A. So my number line would be A
and B, like this one is defined and this one's not. Everything in between isn't defined. What would I
write? Limit as T approaches infinity the right.
I have to come from numbers where it's continuous and approach the discontinuity. So the
integral from T to B F of X Dx. Okay, so what do we call the integral if the limit exists? Convergent, okay
so what if the limit doesn't exist? Okay, now what if I have a discontinuity somewhere in-between the
bounds? I have to split it? So if I have a discontinuity and I'm given this integral from A to B F of X Dx and
I recognize that somewhere in the middle, A is defined and B defined somewhere in the middle. The
function is not, I have to split it at that discontinuity. I have to go from the integral from A to C F of X Dx
plus the integral from C to B F of X Dx.
Then this one, we would get a limit as T approaches C from which direction? Left, then this one
is the limit as T approaches C from the right, because this is what you have. You have A and B, these are
both defined somewhere in between just like this. So then if you're going from A to C you're coming
from the left and from the right. Is that okay? I have to draw a number line every time just so I know
what to do. Should we try one? Oh, I didn't finish this thought. If F has a discontinuity and both of these
guys are convergent then this guy is convergent right? If you're adding up two values that are
convergent, that means that they're finite right?
What's a finite number plus another finite number? Another finite number, so therefore the
original one converges. What happens if this guy ran off to infinity? I'm done right? I only have to test
one and make the conclusion. So let's try one. That should be like negative one-fifth. Look there's no
discontinuity here. Is there a discontinuity right here? What's the domain of this? All real number, it's an
odd root right? So I made a typo and so we're going to rewrite this as this. Now it has a discontinuity.
Where is the discontinuity? So I need to rewrite this sucker how? I would do the integral from one to
three X minus one to the negative one-fifth Dx.
I have to rewrite it because I recognize that there is an issue, right? I rewrite it as the limit as T
approaches one from the right from T to thirty-three of this. Okay does everyone see why we put the
limit there? How do you integrate this? U-subs, this is just an integral that you've been doing right?
What's a good candidate for U? What about Du? If you want to do this in your head you can. If you want
to change it to U's you can do that. Let's go ahead and write it out, if you're good enough and you see
this you're a pro. Okay so lower bound is T, X equal T right? So we get T minus one? Upper, X equals
thirty-three and so U equals thirty-two.
I'm not touching my limit part. I'm just adjusting the bounds so that limit will be adjusted. I
don't have to change this though. So this is T minus one. Is that right? Integrate that sucker for me. It's U
to the four-fifths evaluated from three to T minus one. So I still have that limit out front. This is thirty
two to the one-fifth to the fourth power minus five-fourths T minus one to the one-fifth to the fourth
power. Okay, this is a constant so if I do the limit of this it's just the constant? You get five-fourths times
what? Sixteen over one minus five-fourths? I need the limit. So I get T minus one and that's under a root
and then I'm going to raise that to the fourth power.
So if you think about your roots, that's like the cubic root right? The fifth root is a little more S
shape. We know there is no domain restriction because if I plug in one I get zero right? It doesn't really
matter from which direction because it's an odd root. So what do I get if I plug in? Zero? So what's the
answer, twenty? How's that? Cool? What about this next one? So now what? I have a domain
restriction of Pi? Okay so let's write it out so we can see it. So Tan X is what over what? So this has a
discontinuity when Cosine is zero, right?
Obviously it keeps going, we're really looking at quadrant one and two right? So between
quadrant one and two there's only one discontinuity? It's at X equal Pi over two? Think about your
graph, there's lots of way. There's asymptotes and it's like this? So there's asymptotes at Pi over two
right? So how do I do this one? Split it, so I'm going to write this as the integral from zero to Pi over two
plus the integral from Pi over two Pi Tan X Dx. So look at your picture, zero to Pi over two. That's this
part right? Then from Pi over two to Pi right here? So it's going to look like this and this. So I need to put
limits. The zero is okay, it's the Pi over two that's the issue.
So I do zero Pi over two Pi closed closed, this is the only one that's open right? So I'm coming
from the left, right? Let's do one at a time. This is so I don't have to do so much work. If it's divergent
then I'm done, if it's convergent I have to keep going. So let's consider that guy, that's the limit. Do you
remember the integral of Tangent? Remember if you forget you can do a U sub, let U equal Cosine. We
can also do one like we did on the handout. That's still true right? You can also let U equal Secant. You
let U equal Secant, you get one over U Du? That's Ln absolute U, that's where the Secant comes from,
right? So, my bounds X equal zero and two.
Okay so I need to know what Secant does. Okay, so Pi over two. Okay Secant, do you
remember the graph of Secant? Secant is one over what? Oh I'm sorry, it has the same discontinuity as
Tangent right? Okay remember it's like when you graph Cosine you graph Pi middle low. So Cosine goes
like this, then the Secant touches like this? I'm going to do it in green, it looks like this and there's one
down here. Okay, so look at the graph of that. We're going to go Pi over two, but we're coming from the
left. So we're going this way, what is it? What's Secant of zero? One, so this is minus Ln of one which we
know that's just minus zero. So that part is just a constant.
But this guy is going to infinity right? So one of the pieces diverges, do we have to test the other
piece? Does that make since? It diverges. If you look in your book I think they do one with Secant. Okay,
look at this one. What about this? What is the domain of this? It's X can't be one-fourth. So I have a
discontinuity somewhere in the middle. So I need to rewrite this as zero to one-fourth plus one-fourth
to one, right? Let's do one piece at a time, do ya'll see that you can't say that this equals something?
Notation is everything, make sure you don't say that this is equal to one of the pieces because it's not.
One is a zero, so let's consider that guy.
So I do the limit and zero and one are allowed. The problem is somewhere in-between. How do
we feel about that? From the left? Yes? So here someone said they knew the integral of this. What's the
antiderivative? One-fourth Ln absolute value. You can do U subs if you want to. We've done enough of
these in that partial fraction section that if it's linear on the bottom you do one over something linear.
Ln of the denominator right? If you take the derivative and multiply by the coefficient of X. So now
you're going to divide by the coefficient of X? So do U subs, if you don't know what I'm taking about use
U subs. Do we see this?
Okay, so far so good? Now what? So I still have my limit hanging out in the front right? So this
is one-fourth Ln four T minus one and still and absolute value. Minus one-fourth Ln of absolute negative
one which will be zero. If you lose the absolute value, what happens? If you forget to write it down you'll
mess up. So that part is zero, so graph Ln for me in the air. Something like this? So what point is this
right here if this is Y equal Ln of X. What point is this? Okay so I have Ln of four T minus one, but it's still
going to be the same basic shape right? Okay, if I plug in one-fourth from the left. What is happening to
that operand to that log? If you plug one-fourth into here what's one-fourth times four?
One minus one to get zero? But we're approaching one-fourth right? So what we're saying is
what's happening to the graph? This operand is going to zero, so let me write this down. So four T minus
one is approaching zero as T approaches one-fourth. So looking at our picture that's like the operand
approaching zero. So it's going to negative infinity? This is because this part right here is approaching
zero. Let me be more specific, zero with a plus. That absolute value makes sure it's not negative? We
don't have anything over here. We don't know what happens to zero from the left. So it's important that
I have that absolute value. So this is infinity right? Do I have to test the other piece? It diverges right? Do
I have to test the other piece? No, so the original one does what? Diverges, perfect, now go home.