| Start | Alright, so suppose we wanted to combine these two fractions.
Say, you remember how to do this, you don't really need to take a note but we need a common denominator. So you would multiply the left fraction by (x+1)/(x+1) and you'd multiply the right fraction by (x-3)/(x-3). And then you would get |
| 0:32 | I'm gonna sort of skip a step.
You would get (5x+5+8x-24)/(x-3)(x+1). And that simplifies into (13x -19)/(x-3)(x+1). Ok, and you guys, you know that from 9th grade algebra, or 10th grade. Whenever you did these. |
| 1:03 | They were very annoying because the teachers immediately did hard versions of them.
So here's something we're gonna do called partial fractions. The method of partial fractions also known as partial fraction decomposition abbreviated PFD which sounds vaguely obscene but it's not. |
| 1:30 | 13x minus, so now, here's the thing.
Suppose we started with this. And you wanted to figure out what were the two fractions that we originally had? The two rational expressions? So if I started with this how do I get backwards to that? Ok, it looks scary but it's really not that hard. You'll master it because you're in 126. |
| 2:00 | So what you do is you say well, this must've been some A over (x-3)
plus some B over (x+1).
Then what do we do? You say well, if I took this and I got a common denominator I would know that it's A times x+1 |
| 2:30 | plus B times x-3
and each of them is over (x-3)(x+1).
Alright? Remember, pretend I don't know what the answers are. So I get a common denominator. Multiply this by x+1 and multiply this by x-3. Well that means that, now this is equal to 13x-19/(x-3)(x+1). |
| 3:07 | So that means that this must be the same as that, cause the common denominator.
In other words 13x-19 has to equal A(x+1)+B(x-3). Make sense? Remember you're just working in the other direction. |
| 3:32 | Distribute.
Cause they all have the same denominator so I can ignore the denominator. The numerators must be equal. So I distribute and I get Ax+A+Bx-3B equals that. So far so good? Alright, now group the two x terms and the two constant terms. So that's 13x-19 is Ax+Bx+A-3B, or if I factor out the x |
| 4:17 | got that, right?
Take your time. So again, what we did is we wrote this as A/(x-3) + B/(x+1), and I got a common denominator by multiplying the top and bottom by x+1 |
| 4:35 | and by x-3. And then I said since the numerator of this equals the numerator of this, then 13x-19 has to equal A(x+1) + B(x-3).
Then I distribute the A, distribute the B, and then I rearrange. I put the two x terms together and the two constant terms together. And pull out the x. |
| 5:00 | I'm waiting until people are caught up.
Ok, so then, here's the thing that a lot of people have trouble with. Then A+B must be 13, and A-3B must be -19. Why is that true? Well this is something x. And this is the something x. So they have to be the same. This is the constant and this is the constant, so they have to be the same. Ok? |
| 5:30 | There's no x in these terms, so they have nothing to do with the 13x.
That make sense? Again, this is (A+B)x + (A-3B) so since this is times x, its gotta equal 13. Since this is not times x, it has to equal -19 cause the two things are the same. Ok? Now I just have to solve this equation. Which you remember with pain from algebra class. |
| 6:01 | So let's see, if I subtract the bottom equation from the top equation, B - (-3B), I get 4B equals 32.
B=8. Which is kind of what we expected. And then A=5. That means then that I now know, which I knew in advance, that this 13x-19/(x-3)(x+1) is the same as 5/(x-3) + 8/(x+1). |
| 6:39 | Ok, so I can work backwards through a lot more steps.
And I can, when I start with a messy expression like this, I can figure out what the original expression must've been. You say that's nice, why would I care? Well cause I'm gonna give you an integral like this, ok? I didn't invent the subject. |
| 7:00 | If you could go back in time and you could find Isaac Newton what would you do?
Exactly. He wouldn't last very long. Unfortunately Leibniz also invented calculus so gotta take them both out. [inaudible] So in other words if I had the integral of 13x-19/(x-3)(x+1)dx, I now know that I could rewrite it as the integral of |
| 7:33 | 5/(x-3) plus the integral of 8/(x+1).
Ok. Well this integral is just 5 ln|x-3| and the other one is 8 ln|x+1| +c. |
| 8:06 | In fact, to remind you, if you have integral of dx/(Ax+B) that's equal to (ln|Ax+B|)/A.
Or you could write 1/A out front. Doesn't matter. Ok? |
| 8:32 | Ok, so we're gonna do about, we're gonna do 3 or 4 of these so you guys get the hang of it.
Ok, then we'll do the hard ones. I hate to tell you, no, I don't think we're gonna be able to get enough time in to do this all in one hour and 20 minute session, but I'll try. Ok? So we'll do another one as a team, make sure everybody understands it. |
| 9:02 | So, let's pick a good example.
That looks like a fun one. Ok. |
| 9:32 | Ok, so let's suppose I have |
| 10:00 | Alright, I have the integral of 5x+1 on top, 2x^2-x-1 on the bottom.
So how will I know that I can do this with partial fractions? Well, what if I tried u substitution? If I let u=2x^2-x, du would be 4x-1, that's not gonna-I'm not gonna be able to turn that into 4x-1. So I can't do u substitution. I don't know what I would do with integration by parts. I could try sticking some trig stuff in there, that would probably be very entertaining. |
| 10:32 | Right, or I could say aha, I remember this is uh, partial fractions.
That should be your first instinct if you see something with x's on top, and something with one more power of x on the bottom. Ok, or maybe two more powers. So first of all, since we love to factor that factors into (2x+1) and (x-1). I know you guys can all do that in your heads. |
| 11:02 | So that means that, forget the integral symbol for a minute, 5x+1/(2x+1)(x-1)
is some A/(2x+1) + some B/(x-1).
|
| 11:34 | Ok. So remember my technique. I need to get a common denominator.
I need to multiply A by x-1 and multiply B by 2x+1. So it's gonna be A(x-1)/(2x+1)(x-1) +B(2x+1)/(2x+1)(x-1). |
| 12:24 | And then we can get rid of the denominators cause they're all the same.
And we just know that 5x+1=A(x-1)+B(2x+1). |
| 12:43 | In fact, when you're at this step, the other way to think about it is to clear the denominator, multiply through by 2x+1 times x-1.
And 5x+1 would equal A times this one, plus B times that. So they sort of criss cross. |
| 13:00 | Ok, they cross multiply.
So you can sort of save yourself a step if you want. So you get to 5x+1 is A times what was under B, plus B times what was under A. Ok. Distribute. Ax-A+2Bx+B. Got that step? |
| 13:31 | Distribute the A, distribute the B. What happens on April 13th?
Saint John's 1PM. What day of the week is that? Wednesday. You don't like Hofstra either. We hate Albany and Binghamton the most. I hope that's on video. How I feel about Albany and Binghamton. |
| 14:01 | You'll be there?
Holding down at basketball. Yeah, I wasn't sad to see that. Alright, so group the x terms, group the non x terms. Ok? Yes. Uhh, in other words which one is A and which one is B? |
| 14:32 | I mean you could put A over x-1 and B over 2x+1.
Sure you can do that. Yeah well at some point, you'll end up in the same spot. But, you'll get there in a roundabout way. Ok, there's no right way to divide it up, it doesn't matter. Ok, then factor out the x. |
| 15:01 | Ok. So now we know that A+2B
equals 5. And -A+B equals 1.
Cause again, 5x, A+2Bx. +1, +(-A+B). So, the 5 has to equal the A+2B and the 1 has to equal the -A+B. |
| 15:30 | Ok.
Alright, now I could solve this very easily. Just add those two equations together, the A's will drop out. And you get 3B=6 or B=2. And once B=2, A=1. This step seems to bother a lot of you. You get to here just fine and then you can't figure out what A and B are. |
| 16:01 | Ok, you're gonna have to practice that.
So that means that that integral, integral of 5x+1/(2x^2-x-1) dx is the integral of 1/(2x+1) dx minus 2/(x-1)dx. |
| 16:37 | Everybody can see that?
So the first integral is the natural log of 2x+1 all over 2. And the second one is natural log of x-1 over 1. Plus c. So here, the messy part is doing the partial fraction decomposition. The integration's easy. |
| 17:02 | Ok, it's breaking the integral up that's hard. Hi Paige.
Hey. I have a question. Go ahead. Why is it minus? Why is it minus? It shouldn't be, it should be plus. Well, thank you to all 180 of you who didn't notice. Ok, it should be plus. Other questions? Ok you get to come to the front and clean the erasers. Yes. |
| 17:35 | Well, you take this equation, you add this equation.
A-A gives you 0. 2B+B gives you 3B. And you're adding, 5+1 is 6. You guys ready for one? Why is it divided by 2? |
| 18:08 | You guys ready for one? I'll give you one to do on your own.
|
| 18:31 | Alright, I'll give you guys 5 minutes.
|
| 27:27 | Alright that's long enough.
For those of you who are watching this at home in the video, 8 minutes to place. |
| 27:39 | Alright.
Cause these are the easy ones. Then we're gonna do the hard ones. Ready? So first let's factor this. |
| 28:01 | Right?
Ok, so this is A/(x-2) + B/(x-3). So we know now that x-4/(x-2)(x-3) is A multiply top and bottom by x-3 |
| 28:37 | plus B times x-2
(x-2)(x-3). Ok?
Yes. Doesn't matter which one you put under A and which one you put under B. Ok? Feel free to do either order, in the end you'll end up with the same answer. |
| 29:01 | Ok?
Now cancel those, and I know that x-4 is A(x-3)+B(x-2). Distribute. And you get x-4 is Ax-3A+Bx-2B. Yes. |
| 29:31 | Ok, what's your question?
I just said that. It doesn't matter which one is over x-3 and which one is over x-2, as long as you factor correctly. Ok, that's the hard part. Ok, so make that Ax+Bx+(-3A)-2B. |
| 30:05 | So you have to do things like watch your minus signs, and your grouping, and all of that stuff.
Factor out the A. I mean the x, excuse me. Plus -3A-2B And as I said before, then we get to the part where everybody has trouble. |
| 30:33 | Which is interesting.
But I guess this is something that we just don't enforce enough when we're doing it in oh, in 9th grade math. Ok, these are called systems of equations. So we have two equations. We have A+B=1. We have -3A-2B=-4. |
| 31:00 | So first of all, if -3A-2B is -4, then +3A+2B is +4.
Ok, so you can ignore, you can get rid of the minus sign. Now let's do this a different way. Some people seem to have trouble with the subtracting equations. Let's say that B is 1-A and substitute, cause many of you are more comfortable doing it that way. Ok, so this becomes 3A+2(1-A)=4. |
| 31:34 | Everybody seeing what I'm doing?
Who's lost? Once again, A+B has to be 1, -3A-2B is -4 so +3A+2B is +4. And then since A+B is 1, B is 1-A. So now I put it in for B and I get 3A+2-2A=4 or A=2. |
| 32:09 | Ok.
And since A=2, B is 1-A, so B is -1. Therefore, the integral of x-4/(x^2-5x+6) dx is |
| 32:33 | integral of 2/(x-2) plus, or minus, the integral of dx/(x-3).
Ok. So there's lots of steps. It's really not fun to grade this. So this is 2ln|x-2| - ln|x-3| + a constant. |
| 33:18 | Loving these?
Were any of you able to get this one? Wonderful. [inaudible] Once you get the hang of this, the problem is just it's a lot of messy algebra, but that's a problem with the whole course. |
| 33:33 | Ok, both differential and integral calculus is it's all the algebra that you hated back when you did that in high school, right?
Ok, now of course these are the easy ones. How do I make it harder? |
| 34:19 | Suppose I had this integral.
The problem is x^2+9 does not factor. |
| 34:32 | Or it doesn't factor with reals. It's (x+3i)(x-3i), there's no imaginary numbers in this course.
No complex numbers. So what do we do? Well we still do partial fraction decomposition, but it's slightly more complicated. So, if you have a quadratic term that can't be reduced, you have to change what's in the denominator, I'm sorry you have to change what's in the numerator. |
| 35:01 | So you still have A/(x-1) + now you have Bx+C/(x^2+9).
Why? Cause it's gonna work, ok? So, if this is x^2 you need a x term and a constant term. So you have Bx+C. If this was x^3, you would need Bx^2 + Cx+D. Ok, and it can get really brutal but we're not gonna make it that hard. Ok? |
| 35:30 | But, we do test these, ok?
So again, because we have x^2+9, we have a Bx+C. That's necessary because when you do the decomposition you could end up with an x term as well as a regular term. You'll see how this works. So let's get a common denominator. Equals, ok A is gonna be multiplied by x^2+9. |
| 36:08 | It's getting kind of boring writing these denominators over and over again.
Plus Bx+C(x-1)/(x-1)(x^2+9). Ok, so again the numerator is slightly different on one of the rational expressions. |
| 36:33 | So throw out the denominators and you get 10=A(x^2+9)+(Bx+C)(x-1).
And then we're gonna have to do some algebra again. |
| 37:03 | Ok, so distribute.
You get 10=Ax^2+9A plus, gotta FOIL, Bx^2-Bx+Cx-C. And then use some grouping. |
| 37:31 | So 10 is Ax^2+Bx^2 +Cx-Bx +9A-C.
And then factor. So you get 10 is x^2(A+B) +x(C-B) +9A-C. |
| 38:11 | You loving these?
Not so much. Ok, so what do we know? What does A+B have to equal? What does A+B have to equal? |
| 38:31 | It has to equal 0. There is no x^2 term on the other side.
So A+B has to equal 0. C-B also has to equal 0, and 9A-C has to equal 10. So let's see. This says that A=-B, this says that C=+B. |
| 39:02 | So when I substitute we get -9B-B=10, so -10B=10. B=-1.
You understand why A+B has to be 0? You understand why C-B has to be 0? Ok, you can make these hard. |
| 39:35 | I took, ok so I took this equation and I said A is -B, so I put -B in for A.
And I said C is +B so I put B in for C. And I got -9B-B=10. So B comes out -1 which means C is -1 and A is +1. We're not done folks. So now, I'm gonna erase this first line. |
| 40:13 | I now have to find the integral of A, 1/(x-1) dx, right, plus integral of -x-1/(x^2+9) dx.
|
| 40:37 | I'm not even close to done.
Ok. So this first integral is easy. This is ln |x-1|. What about these other two integrals? Break it up. Make this the integral of -x/(x^2+9) dx and the integral of -1/(x^2+9). |
| 41:11 | Alright how do we do that middle integral?
U substitution. Yeah I know. Like I said, try grading these. You have no sympathy cause you're the ones who have to do it. Ok well the integral of x/(x^2+9). Let u=x^2+9, du is 2xdx. |
| 41:46 | So 1/2 du = xdx.
And this becomes, I'll slow down in a second, -1/2 du/u. |
| 42:13 | This is -1/2 ln|x^2+9|
Doesn't really need the absolute value. So we have two of the integrals.
First one is natural log of x-1, second one is -1/2 x^2+9. |
| 42:31 | What about the integral of -1/(x^2+9)
Any ideas?
The minus isn't really very important. That's inverse tan of x/3. The 3 comes from the square root of 9. |
| 43:09 | I think that's right. I'm missing a third?
Let's see that's 1/(1+(x/3)^2) times 1/3 1/3 tan inverse |
| 43:36 | Ok, so to help you memorize that the integral of dx/(x^2+a^2) is 1/a tan^-1 x/a
plus c. Ok?
I know. I don't test these cause it's 400 steps. |
| 44:01 | But I don't know if were gonna test them or not. We'll find out.
Yes. Don't need to do u substitution on that one. What you would do is you would do a trig substitution, you would let x=3 tan theta You probably wouldn't let x=3 tan theta. But that's what you would do. Ok? Yes. x^2 + a^2, so this is 3^2. |
| 44:31 | Ok? If that was x^2 +a there would be 1 over the square root of a.
Ok? Let's practice one of these. Ok, I know this is hard. It's your first day doing it. Remember the alphabet was hard when you were in 1st grade. Eh, kindergarten. I hate these examples. Alright I have to do a little erasing. |
| 45:01 | We'll just erase this for the moment.
|
| 45:40 | Alright, let's keep you guys busy for a few minutes.
I'm gonna start erasing things. |
| 53:05 | Alright folks. I hope we're excited.
Way too much fun. It's nice to have an x term in the denominator. That's a lot less work to do. I hope I copied this correctly. I did. Ok. So I know that 2x^2-x+4/x(x^2+4) is some A term over x, plus Bx+C/(x^2+4). |
| 53:43 | And by the way it could be Ax+B/(x^2+4) and C/x. It doesn't really matter.
Ok, in the end you'll get the same answer. Alright, so I'm tired of writing that denominator but I know that 2x^2-x+4 has to equal A(x^2+4)+x(Bx+C). |
| 54:13 | Everybody agrees?
Were we all able to get that step? Ok, that's by clearing the denominator. A gets multiplied by x^2+4 Bx+C gets multiplied by x, and then the numerator has to equal the numerator. So distribute. |
| 54:33 | This has to equal Ax^2+4A+Bx^2+Cx.
And then you do a little grouping. So 2x^2-x+4 is Ax^2 + Bx^2 + Cx+4A. |
| 55:12 | And you pull out x^2.
And you get x^2(A+B) + Cx +4A. |
| 55:31 | So I know that A+B has to equal 2.
I know that C has to equal -1. And I know that 4A has to equal 4. That worked out nicely. So we could give you a messier one where the numbers work out well. You get A is 1, and then if A is 1 B is 1. So now I know I can rewrite this as the integral of 1/x plus the integral of x-1/(x^2+4). |
| 56:17 | And I could break that into two integrals so I have dx/x plus the integral of xdx/(x^2+4) minus the integral of dx/(x^2+4).
|
| 56:41 | So this integral is the log, |x|.
This integral is gonna need u substitution, so we'll do that in a second. And this integral is gonna be 1/2 tan^-1(x/2). |
| 57:02 | Ok? Cause you take the square root of 4, and it's 1 over that, tan^-1 of x over that.
Ok? That's the formula. There's x^2 + a^2, a^2 is 2. So far so good? Now we just have to do the middle integral. For the integral of xdx/(x^2+4), do u substitution. |
| 57:32 | du is 2xdx, so 1/2 du is xdx.
This is 1/2 integral of du/u which is 1/2 ln|u|. Or 1/2 ln(x^2+4), you don't need the absolute value bars because x^2+4 is always positive. |
| 58:11 | You can put them in if you want. It's just unnecessary.
How much fun was this topic? Yeah? I can give you one more, we have time. Do we have time? Who votes for time? |
| 58:31 | One. You're outnumbered, gotta tell you.
So the key to this is you have to break them up properly, ok? Then you do the algebra. So, I put some stuff up on Blackboard to help you from my book. And we'll see everybody on Wednesday. |