Stony Brook MAT 126 Spring 2016
Lecture 15: Improper Integrals
March 23, 2016

Start   Nice, straightforward, improper integral.
4:16Alright, that's long enough.
So what do we do? So you look at this and you see that there's an infinity in the limit of the integral.
Ok? In the top one or the bottom, it doesn't matter.
You have to rewrite the integral without the infinity.
4:30So you pick a letter, substitute for infinity.
Let's see what's a good letter you'll use today? 30 letters.
V? I don't really wanna use v.
V's are bad.
B? As in How about "W" for Wolfie?
The limit b goes to ∞ integral from 5 to b of dx/x^3. Ok.
5:03This could be a very straightforward integral.
So first let's just integrate dx/x^3. So that's x^-3 dx.
And that's, the integral of that is x^-2/2.
Also known as -1/2x^2.
You guys all got there?
Ok. So this is now gonna be
5:30the limit as b goes to ∞ of -1/2x^2.
Evaluate from 5 to b.
You plug in the b, you plug in the 5 and then you take the limit.
So this is -1/2b^2 + 1/50
6:00And if you do the limit b goes to ∞ -1/2b^2 + 1/50 That's 0, so this is 1/50.
Ok?
Straightforward?
Nice and simple, ok.
Remember I told you, if you have dx/x^p, if p is greater than 1, it's gonna converge if it's less than or equal to 1 it's going to diverge.
6:31Ok?
Alright now, there's other types of improper integrals.
These are integrals where you have a problem since you have discontinuity.
You good on this one?
7:01Suppose you had integral from 2 to 5 dx/√(x-2) That's the one they used.
Now you look and you say well, now you have to pay attention when you're doing these integrals.
Ok, especially the definite ones.
This doesn't exist at x=2, because at x=2 you have 0 in the denominator.
7:31So you can't really do this integral with the 2 in there.
So what do you do?
Well, that's what we're gonna have to figure out if the antiderivative works.
Alright, we have to integrate it. But you can't really evaluate it from 2 to 5 cause the 2 doesn't exist.
So we do our trick again, make up a letter.
Um Use B for Bonnie.
Ok, and we do the limit as b approaches 2. Now, where are you approaching 2 from? You're approaching 2 from the positive side.
8:05Of the integral from 2, from b to 5.
Of (x-2)^(-1/2) dx.
Alright, because 1 over the square root is to the -1/2.
Why do I put the plus in there?
Well because I'm going between 2 and 5.
8:30The other one you brought up to infinity, so you don't need a plus symbol or a minus symbol.
Right, but between 2 and 5, you're then starting at 2, right, if you think of the number line.
Starting at 2 and I'm going to 5 so I want the limit from the right side.
Ok?
I don't really think the plus is that important but you should know that it should be there.
Ok?
So now, same trick. We have to integrate that, so integral of (x-2)^(-1/2) dx
9:02is (x-2)^(1/2) over 1/2 also known as 2√(x-2).
So this is gonna be limit b goes to 2+ of 2√(x-2)
9:31from 2 to 5.
So plug in the 5, plug in the 2 and see what happens. When you plug in the 5 you get 2√3 - 2√0 and then, that shouldn't be 2 I'm sorry.
Jumped a step.
Gotta plug in b. -2√(b-2).
10:03Now, when you take the limit of this is 0 so this just equals 2√3.
Make sense?
So these, these aren't crazy tricky, you just have to remember. You have to do the limit.
Then you have to do the integral first using a letter and then taking the limit.
10:30Well when you have these rational functions, remember they're not defined when the denominator is 0.
So you have to look. Whenever you have a function that you're integrating, and you have a denominator, make sure, if you're doing limits of integration, make sure you remember that it's not defined where that's 0.
We sort of ignored that for a while. Now we're figuring out what you actually do with it.
11:01What's another fun one?
Ok.
Ready for another one?
I'll let you guys do it.
How about that?
14:23So what you wanna do Give you a clue.
14:30You're gonna wanna make this Ok? That's why it's improper.
What you can't do is just say that's ln|x-2| from 0 to 3. You can't just do that.
Ok, cause you have a problem in the middle.
Why? It's discontinuous.
15:01If you graphed it you have the function So you're finding the area from here to here.
That goes up to infinity.
This goes down to minus infinity with a problem somewhere in the middle. You have to figure out how to deal with it. Yes?
15:33Oh, you don't need to know how to graph it.
How do you know it's discontinuous? Well, its equal to 0 at 2 in the denominator. Right?
So the denominator is 0. So when you see a definite integral and you see a place where it gets undefined, then you have to break the integral up.
Ok.
It's annoying, I know.
Yes.
16:01We only have 7 more weeks of math.
For some of you [inaudible] You'll see in a minute.
Gotta be patient.
Have to see who can get to it on their own first.
16:30Did you do the limit? You need to do the 2 integrals.
17:10Alright, so let's pick a letter.
How about, um should I do "T" for Trump?
No.
"B" for Bernie.
Ok so the limit goes- from b goes to 2 from below.
From the left side
17:31of 0 to b of dx/(x-2) + the limit as b goes to 2 from the other side of b to 3 dx/(x-2).
As I said, you don't need to know how to graph this.
Although you should certainly be able to graph 1/(x-2).
If you can't it's ok, I forgive you. But you should be able to.
18:00Alright.
So the integral of dx/(x-2) is just the natural log of |x-2| evaluated from 0 to b.
And I'll do the other one in a second.
So you plug in b and you plug in 2 and you're gonna get ln|b-2| minus
18:34you can just call that log 2.
Cause it's the absolute value of -2.
Now, when b goes to 2 you're gonna be doing the natural log of 0. The natural log of 0 is undefined.
Ok?
So this does not exist. So you're done, you could just say this diverges.
And since one part of the integral diverges, the whole thing diverges.
Ok so as soon as you get to that step you're done. Yes.
19:05There's no other answer you can write.
You can write diverges, you can write divergent. Yes.
No, don't put infinity. Put divergent or diverges.
Or diverges.
If you grow up in a different part of America.
Ok?
Don't put infinity.
Put diverges.
Ok?
As they say on the south shore.
19:30Diverges.
Alright, let's do one more.
And then we're gonna do something new.
Trying to find a good sample problem for everybody.
I mean these could get really, kinda fun but not really too much entertainment value.
How about we'll do
20:06Let's do that integral.
21:45Just shift it to the right.
23:51Alright, let's do this one. And then we learn some new stuff.
24:01So much fun.
You know, I could've been working on Wall Street making a million dollars, treating people badly, making the world a worse place.
But no.
Was that out loud?
Microphone was on?
Hate when that happens.
Alright, so how will we do this?
So again, you gotta do infinity.
So you do the limit as h goes to infinity.
24:31You guys can decide later what "h" stands for. It doesn't really matter.
I'm doing this, I'm being silly about this for a reason.
Some people think it always has to be the same letter or that the letter has some special meaning.
It doesn't. Ok, it's just a letter.
Alright.
So we just have to integrate e^-x.
That's e^-x. That's -e^-x.
So it's the limit h goes to infinity -e^-x from 1 to h.
25:10-e^-x is -1/e^x from 1 to h.
25:32Alright, now we're gonna plug in h and 1.
We're gonna get limit h goes to infinity -1/e^h + 1/e. That's an h if you can't tell.
And an infinity, e^∞ is ∞. Remember what e looks like, right?
e looks like that.
So when x goes to infinity, e goes to infinity. Right?
So this is -1/∞ which is 0. Plus 1/e
26:04just goes to 1/e.
So when the integral converges you will get a number.
When the integral diverges it's because you got an infinity in there somewhere, just write diverges and move on. Yes.
Remember e^-x, e^-x is 1/e^+x.
Ok?
Other questions.
26:31So believe it or not you're now done with techniques of integration.
This is the hardest part in some ways because it's all just sort of playing, you know, manipulation and playing and finding tricks to solve the kind of integral.
Now we're actually gonna learn how to use these integrals.
Ok, so if you're wondering we're now shifting from Chapter 5 to Chapter 6.
Ok?
So the reason we do technique of integration is you'll set up a problem, and you'll get some kind of integral and then you have to actually evaluate the integral.
27:02So for those of you who do go on to real higher level math and physics and engineering, you'll do a problem and you'll set something up, and you have to be able to solve it.
And these techniques are ways to solve it. People run into a lot of infinity stuff, especially when you move into higher physics.
Quantum mechanics.
Things like that.
Anyone planning to take quantum mechanics?
I'm going with no.
Maybe somebody here will.
And then you'll do lots of this kind of stuff.
Ok?
27:30[inaudible] black holes Black holes go down to infinity at some point, or up to infinity depending on your point of view.
There's really no down and up out there in space.
So you can tell this to your parents, they'll be very impressed.
And they'll say to their friends "my kid's so smart" and that's good enough.
Alright.
So now, we're gonna do more on areas.
I just like saying it that way.
Don't know why.
More. On. Areas.
28:06This is my exercise.
I'll have a very large left shoulder at some point.
So remember, the integral, an integral means a bunch of different things.
An integral is a giant sum. You can add things up using an integral.
The integral is also an antiderivative.
It's the opposite of a derivative. You go the other way.
28:30Integral is also area under a curve.
So now, we'll use integrals to find some more interesting areas.
Ok?
So, remember what I just said a second ago, an integral is a sum. The "S" stands for a giant sum, ok?
Suppose you have one curve and a different curve and you wanna find the area between the 2 curves.
Ok, so let's call that f(x) and let's call that g(x).
29:03So you imagine that you wanna find the area between these.
So, well but we did the rectangles before. So you do the rectangles again. Say suppose I have a rectangle like that.
Just imagine, it's very thin, ok?
So what would the height of that rectangle be?
Any ideas?
29:30f(x) - g(x)?
Why is that true?
Because.
That's an excellent answer. Need a little more.
Very good. Alright, so this distance is f(x).
This distance is g(x).
Pretend they're the same x.
Ok? Very good. So you have this distance is f(x), this distance is g(x) so the distance between the two of them is f(x)-g(x).
30:04And the width of that rectangle is Δx.
So the way you turn that into an integral cause now you wanna add all of them up so you would do f(x)-g(x).
And then you do the limit as Δx gets closer and closer to 0 that becomes dx.
You go from a to b and that is the area between 2 curves.
30:36For example
31:06Ok, so let's say we're gonna find the area between y=x^2 and y=2x-x^2.
So one of the unfortunate parts it you have to do these graphs.
We'll probably give you a couple of points for drawing the graph correctly.
Ok, which means if you do the problem incorrectly but you drew the graph correctly you'll get something.
Ok, the graphs won't be incredibly complicated. They'll be things like quadratics, cubics, sines and cosines.
31:34e Can't be too hard because you have to do a couple of important things with this graph. So, y=x^2 is a parabola.
And 2x-x^2, that's the same as x(2-x).
32:00So that's an upside down parabola from 0 and 2.
Here, I'll blow that up for everybody to see.
So this is x^2, that's 2x-x^2 and we wanna find that area. And this is the origin.
And when you miss, just make a big dot around it so people can't tell when you do it wrong.
Ok.
So now we have to find where they intersect.
Because we're gonna go from 0 to that spot.
32:32So where do they intersect? They intersect where 2x-x^2=x^2.
So that's 2x-2x^2=0.
Factor out a 2x(1-x)=0. So they intersect at x=0 and x=1.
33:09So then the area is simply the integral from the left side to the right side. So this is x=1.
And what are we doing? Well we're adding up all those little rectangles. So all those slices from 0 to 1.
So you go from 0 to 1 and each slice has a height of the top curve minus the bottom curve.
33:35So 2x-x^2 is on top, -x^2 dx.
You should all be able to integrate something like that. It's not a tough integral.
We'll leave that there for a second.
34:13So the integral of 2x-x^2-x^2 becomes 2x-2x^2.
Ok.
The integral of 2x is x^2.
The integral of 2x^2 is 2x^3/3.
34:36Evaluate from 0 to 1.
You plug in 1, you get 1-(2/3) and when you plug in 0 you just get 0.
So this equals 1/3.
Ok?
We understand this one?
So what you're gonna do is you're gonna integrate the top curve minus the bottom curve
35:02starting with the left side and going to the right side.
Of course we can make these very annoying because we can just ask harder areas.
Ready for another one?
I heard yes.
That person who said no, you meant yes right?
35:32I heard yes again.
Oh my gosh. I gotta get my hearing checked.
Ok.
How about from y=x to y=5x-x^2?
So practice drawing the picture.
39:28You know, when the people watch the videos and go "where does he go?"
39:31He puts the problem up and he just walks away.
Talking to himself.
You're gonna have to practice doing these graphs.
I mean, if you can't graph y=x I'm nervous.
I'm not sure I want you to do the open heart surgery when the time's up.
What does y=x look like? y=x
40:00kinda looks like that.
And 5-x^2 kinda looks like that.
It's a little exaggerated but that's the difference. So you wanna find the area of that.
That shaded region. What you're gonna do is you're gonna slice it like that. You're gonna add on all of those slices.
We use the word "slice" for a reason.
Ok?
Cause what we're gonna do next week is we're gonna skim the slices.
I know you're looking forward to that.
40:31Ok.
Now we have to find where they intersect.
So they intersect where these two equations are equal to each other.
So where x=5x-x^2.
Subtract x from both sides, you get 0=4x-x^2.
Did you put 6?
Not 6. It's 4.
Pull out an x.
So this is where x=0 or x=4.
41:04So this is x=0, x=4. So you're slicing from 0 to 4.
So we're gonna wanna evaluate from 0 to 4 top curve minus bottom curve.
So top curve is 5x-x^2.
Bottom curve is x
41:30dx Ok?
So let's do that integral.
That's the integral from 0 to 4 of 4x-x^2.
You will find that that integrand equals this integrand a lot of the time.
Ok so what's the integral of 4x?
2x^2.
42:00And x^3/3 from 0 to 4.
Now look, if you're lousy at the arithmetic, plug in the numbers and leave it.
Ok?
So if you're not so good at the arithmetic at least write 2(4^2)-4^3/3 minus 2(0^2) which is 0, so that whole right side is 0.
42:33Ok?
If you're not so great at doing 32-(64/3), that's ok.
I rather you leave it like this than do the last step of arithmetic and get it wrong and then lose the points.
Ok?
But you should get 32/3.
Ok, or 10 2/3 or for those of you who use your calculators, 10.666.
If you put 10.666 we're gonna be suspicious.
43:02Ok?
Sorry?
Uh, once in a while we catch somebody doing something that makes us nervous.
No calculators in the exam. No phones, no watches.
We're on to you guys, ok, none of that stuff.
Wearing a watch, we're gonna assume you're cheating. Yes.
You got them backwards?
43:33Ah, yeah you had it backwards.
Cause you had the, cause the top curve minus the bottom curve. You did the bottom curve minus the top curve.
Alright, you want what's on top minus what's on bottom. Now, how could this get interesting?
How do we make this more interesting? Well first of all, the top and the bottom could switch.
44:34We love trigonometry.
Right? I know you love trigonometry so much.
Ah, ok.
So what does y=sinx look like?
Sinx looks like this.
Cosx looks like that.
Alright, so at π/2... I didn't draw it exactly to scale but hang on.
45:03Redraw that so it looks prettier.
Ok, so that's the 0 to π/2 mark.
Sine curve should be at its' peak there.
I'm gonna be a little obsessive about this.
One more time.
Cause it's going on the video. Not that we edit the videos.
45:35Alright, there you go.
The sine is at its top when the cosine is at 0. So in other words this is cos(x) and this is sin(x).
But right here is switched, and now this is sin(x) and this is cos(x).
So what do you think we do?
What do you think we have to do? It does matter which one we use.
46:01Break it into two integrals.
Ok, so where does sin(x)=cos(x)?
45 degrees or π/4.
So right there it switches from 0 to π/4 cosine is on top and sine is on the bottom, from π/4 to π/2 sine is on top and cosine is on the bottom.
So you have to do the integral 0 to π/4 cos(x)-sin(x)dx
46:37plus the integral from π/4 to π/2 of sin(x)-cos(x) dx.
So much fun.
47:00Ok.
The integral of cosine is -sine, the integral of sine is cosine.
So this is -sin(x) Ah, I'll move it over here.
You get -sin(x)-cos(x) from 0 to π/4 plus
47:33-cos(x)... did I do that correctly?
That's sine, not -sine.
Right, and then you're gonna have Hang on.
Plus cosine, got it.
Alright, and then this is -cos(x)-sin(x) π/4 to π/2.
48:06Ok?
You'll know you did it wrong when you mess up the minus sign's. Look at the answer and it won't make sense. You'll get a negative answer or it'll be too small or something.
So sin(π/4) is √2/2.
Cos(π/4) is also √2/2 so this becomes √2 Right? √2/2 + √2/2 is just √2.
48:32You plug in 0, sin(0) is 0 and cos(0) is 1.
Ok.
Cos(π/2) is 0, sin(π/2) is 1.
So And π/4 is gonna be -√2/2 -√2/2 which is -√2.
49:01Made a mistake somewhere again.
The whole thing is canceling out, which it should do.
Oh, it's a minus. Sorry.
Right. Now I'm confused, let's see.
Alright, -1.
Minus minus, right so plus. Good. So you get
49:302√2.
Ok?
I'll double check that later. See, I mess up the minus signs also.
Alright, another thing that you could do is instead of integrating with respect to x you might have to integrate with respect to y.
You can check me later.
Ok, so we wanna find the area
50:17Ok, so now you wanna do between x=y^2 and x=8-y^2.
So what does that look like?
50:33x=y^2 is sideways parabola And x=8-y^2 is also a sideways parabola.
Well, if you tried to isolate for y you're gonna run into some problems.
Ok, think about your slices.
When you're slicing from here to here you get the same curve on top and on the bottom.
51:02So it'd be messy. You'd have to break it up from here to here and from there to there, or instead you slice sideways.
Like this.
And you add up all of those slices.
Looks hard, but it's not.
So it's basically the same thing as before except for turned sideways.
Everything instead of being in terms of x is in terms of y.
And instead of top minus bottom you do outside minus inside.
So this is the curve 8-y^2
51:33and this is the curve y^2.
So where do they intersect? Well 8-y^2=y^2 where 8=2y^2 4=y^2 So y is + or - 2.
Ok? So down here is at y=-2
52:01and up here is at y=2.
So same idea as before. You're gonna integrate.
You're gonna integrate from -2 to +2.
And outside curve was 8-y^2.
Inside curve is y^2.
52:40By the way, you could go from 0 to 2 and double the answer.
You notice, the bottom half should be the same as the top half. It's symmetric.
So one trick you could do to make your life easier is just integrate from 0 to 2 and double.
That make sense?
Ok?
Cause, even though I don't draw it beautifully, top half of this picture is the same as the bottom half of the picture.
53:01So one of the things you could do with integration, especially on these area problems, is if you see that sort of the left and the right side are the same and the top and the bottom are the same you could do half the integral and then double.
Lots of times doing half the integral will mean one part of the integral is evaluated to 0.
Which is easier.
Ok?
So this becomes integral from -2 to 2 (8-2y^2)dy
53:31which is 8y-2y^3/3 from -2 to 2.
You plug in 2 and you get 16-(16/3) minus -16+(16/3), so these two are gonna come out the same.
54:08So it becomes (32/3)+(32/3) or 64/3.
Ok?
And as I said before, feel free to leave it here or just plug in without doing the cubing.
Fine with me.
Ok, we're not that fussy in getting the arithmetic exactly correct. It's more important that you understand what you're doing.
54:34So let's give you guys one to practice.
55:05Ok, so find the area between x=1-y^2 and x=y^2-1.
56:25Ah, that sine and cosine problem should be 2√2 minus 2. We lost a -1 in there somewhere.
56:36For those of you who really care about that stuff.
Go home, catch it later, send me an email saying you're really confused.
That I made a mistake haha. So it's 2√2-2.
Do you agree with that?
Now you don't have to ask me that in class.
59:181-y^2 you go to 1 and then you go this way.
y^2-1 you go to -1 and you go this way.
59:31And they're gonna be symmetrical.
Where are they gonna intersect?
Or where does 1-y^2=y^2-1?
Well that's where 2=2y^2
60:01y^2=1 y= +/-1 These actually, these actually intersect on the y axis.
So the right side equals the left side. So you could integrate from 0 to 1 and double the answer.
Ok?
And as I said, that often is easier to do. So let's do that this time.
60:33So the right hand curve, this curve, is 1-y^2.
That's an arrow.
And this curve is y^2-1.
So we would be integrating from -1 to 1, the right hand curve, (1-y^2) minus the left hand curve, (y^2-1)dy.
61:11And by the way, if you did this and you got a negative number you probably just did it backwards.
You had your right and your left reversed.
Now you could make this instead 2 times the integral from 0 to 1(1-y^2)-(y^2-1)dy.
61:30Why are you allowed to do that? Because the right and the left are the same.
If you just integrate from 0 to 1 you'll get some answer. You just double it.
Ok?
So this equals 2 times the integral from 0 to 1 of 2-2y^2 dy.
With these, I always say you just have to be careful. You have to be careful with the minus signs and have to be careful with the arithmetic. Yes.
62:04Well you can do that. From here you can take the 2 out.
Absolutely.
I would do that, ok, but I don't like to confuse too many people.
But yes, you could pull a 2 out of here.
Ok, so this becomes 2(2y-2y^3/3) from 0 to 1.
62:32So you plug in 1 and you get 2-(2/3).
Right? Which is 4/3.
Plug in 0, you get 0.
So it's 8/3. Be careful by the way.
You don't always get 0 when you plug in 0 in an integral.
You don't, for example when you have cosine of x.
Cosine of 0 is not 0 it's 1.
So be careful, don't automatically assume that part is 0.
63:00Ok, everybody had enough? See you on Monday.