Start | Next week will be exam review, which I know will make you very happy. As I said I've almost got the final exam in
place. I have to check a few things on it. Over the weekend I will make up some practice
problems for everybody to work on that
will help you with the final, okay?
That's good and then next week we'll do a lot of review and then you have the final on the 13th I forget what time, 11 AM or something? 11:30 okay I'll post |
0:30 | the location this weekend so for the
many many of you who are not here, you'll
have some information. Hopefully for
those of you who are here you will be
rewarded just a little bit for doing me
the courtesy of showing off on a regular
basis. All right, so let's practice some
more u-substitutions because I can
assure you you'll need to know these
for the final.
Let's give you guys a few of these. I'll take them from the book. I think they'll come out better. |
2:35 | Ok, there's 4 to get you started.The first one, you're gonna wanna do that trick where you make u=x-5 and x=u+5.
I'll give you guys some clues to help you along. u=x-5 |
3:02 | here you want u=2x^4-1 here you want u = 12x^2+4x+3 here you can want u=ln|x| |
3:34 | Alright, let's start trying these problems.
So, first one we let u=x-5. How do we know what to do with a problem like this? Well, remember, you're always looking to see if the integrand contains a function and its derivative. Problem is, the derivative of x-5 is 1. That's not x, and the derivative of x is 1 then it's not x-5 so when you |
4:00 | get that type you do this little trick.
Ok? Because the derivative of u, du, is dx. So you do have the function and it's derivative the problem is what you do with x. So if you let u+5=x, then you can rewrite this integrand. This becomes u+5, this becomes u, and this just becomes du. |
4:32 | Didn't really change, all you really did was use a little trick to turn this from a
fraction that you can't break up into a
fraction that you can make u/u +5/u
You don't even really need to do u substitution, there's
another trick to this but I won't
confuse you. One trick per-type
one technique is usually all you need. Ok? So u/u =1, alright.
|
5:06 | The anti-derivative of 1 is u, and the anti-derivative of 5/u is 5ln|u|.
Ok? Because the derivative of u is 1 so the integral of one is u and the derivative of log is 1/u so the antiderivative of the integral of 5/u is 5 log(u) and then you just substitute back. It becomes (x-5)+ln|x-5|+C. |
5:41 | Ok? Much easier when I do it. Part of it is I've
done many many many integrals in my life.
Far too many. It's not whether it's thousands it's in the tens of thousands. You know, it's hard to be fooled on these anymore. So as I said, the clue is that this is the same power |
6:05 | these x's are the same power, which means
that derivative of 1 isn't going to help
you find the derivative the other so you
simply have to find a way to rewrite this.
Where instead of being unable to play with that integral you're able to play with the integral. Ok? Boom. All right this is the straightforward cut. Here if I let u =2x^4-1, the derivative is 8x^3dx |
6:37 | So I look at my integrand and I say, I have to substitute for everything. So (2x-1)^8 is going to become immune to the 8th. But then I have to substitute for the rest. I have x^3 dx and this gave me 8x^3 dx so if I divide by 8, |
7:00 | I'll have x^3dx, and that's the remaining part to substitute for. So this
x^3dx becomes 1 du, so put the 1/8 on the outside, so this is (1/8)u^9/9.
Which is u^9/72, and then substitute back. |
7:41 | So remember what the U substitution does is it helps you figure out the chain rule backwards.
Because when you take the derivative of this, you get the 9/72 which is the 1/8, (2x^4-1)^8, which is that, times the 8x^3, so another 8 cancels |
8:05 | and you get the x^3. So you can always
check if you're doing these correctly by
differentiating your answer and seeing
if you get back to your integral. You
will have lots of time on the final, or
you should have lots of time for the final.
We get two and a half hours of something crazy for the final, or 2 hours and 15 minutes I expected most of you will be done in about an hour and a quarter. |
8:30 | Maybe a little longer so you can take longer, you shouldn't need more
than an hour to an hour a quarter for
the exam, so you can use the rest of the
time to check your answers. So you'll
have time to integrate and then check by going back and differentiating.
You should have time, ok? I'm not trying to make the exam last the whole period of time. Some of you of course will stay till the very last minute that's ok. So far so good? Alright, let's do integral number 3. So again, this is a relatively |
9:04 | straightforward one. I'm looking to have
a higher power of X in the bottom than
on top. It's one power difference so I
know which one to make the derivative
and which one to make u, so u = 12x^2+4x+3, and du=24x+4 dx.
Now looking integrand and I see this will be u, that's not a problem and |
9:30 | this is 6x+1dx, so to turn this
into 6x+1, I just divide by 4.
Ok, substitute it. So the bottom becomes u, the top becomes 1/4du |
10:01 | that's 1/4 ln|u|+c, well you don't really need the +C but whatever, 1/4ln|12x^2+4x+3|+C.
12x^2+4x+3 is always positive, you don't actually need the absolute values but whatever. If you're not sure, leave them in. Even if you're writing absolute value of 3. |
10:31 | You're silly, but it's not really wrong, ok? So far so good? Ok, last one.
So here, I see dx/x. that's 1/x, right? log(x^2) so what do I do with that squared? Well first of all just use your rules of logs, change that into 2log(x) |
11:04 | Remember the rules of logs. If you have something to a power, you can bring the power in front. Maybe you forgot that rule. Well now, so this is the same thing as saying this is a 2 in the denominator so I can pull that out, and I can say this is 1/x, times dx/log(x) if I really wanted to break it up. So if u = ln(x), |
11:37 | then du = 1/x dx.
Does everybody see what I did? Where are we lost? How did I get the 1/2?So this is 1/2, so I can pull that out, 1/x, because you can imagine there's a 1 |
12:04 | in here. So 1/2 becomes here, 1/x becomes this, and then I'm left with dx/log(x). Ok? You don't have to
break it up like that but that's so you
can see what's going on. All right, so now
we can substitute. So ln(x) becomes u,
and you want to write dx as du.
|
12:30 | So this is 1/2ln(u), which is 1/2 ln(ln(x)).
Ln(ln) shows up in things like chemistry, physics. I can't think of where it's showing up in business. You use a log graph to smooth things, are you watching the Stock Market? So the Dow Jones broke 24,000 yesterday, but if you took the stock market you look on a |
13:00 | logarithmic scale it's much more normally behaved. So the stock market right now if you look back way back go back say 1929 stock market just this looks like this. Ok? With ups and downs in there but if you put it on a smooth log scale so if you look at what that means is in other words going from 23,000 to 24,000 is an increase of 1,000 out of 23,000. Going from 10,000 to 11,000 an increase |
13:30 | of 1,000 out of 10,000, ok? So it's a much bigger increase going from 10,000 11,000 is 10% going from 23,000 to 24 is 4.something percent, so if you adjust for that then the stock market really kind of looks more like that. Ok? So in other words if you think about inflation, if you think about the increase from 23 to 24 is not the same as an increase from 3 to 4 then it'sstill good, it's still going up, but it's |
14:00 | not as dramatic an increase. We're still, it's still been quite a run since Donald Trump was elected a little more than a year ago. It's gone from 18,000ish to 24,000ish that's a 33 percent increase in the stock market. This is a huge huge increase. So you're really getting a bit of a blip right there, ok? if this tax bill doesn't pass it's gonna come down in a hurry, but the good news is according to my Twitter feed, stock market it's gonna I mean it |
14:34 | the tax cut bill is going to go through and you will all suffer for it and I won't because I'm old and you're young. There you go. But we won't get into a discussion of its merits as a tax bill. You guys are all business suits you should be paying a great deal of attention to what's going on in Congress and pull these things on a regular basis on basis because you'll interview for jobs and |
15:00 | they'll ask you what do you think of the tax code. What do you think of this. Next week we
get the economic indicators for the rest
of the year, you're gonna be expected to
know that stuff. But you don't have to know
it today, and it's not on the final.
Boom. So for example, we have a business I don't really care what's called and |
15:36 | marginal demand is D'(x)= -2000x/((sqrt)(25-x^2)). Ok? Where D(x) is the number of units sold at x dollars per unit.
|
16:19 | Ok? So this is D'(x), this tells you how the demand is changing where D is the number we sell at a price of whatever x is, ok? So find the demand function |
16:40 | Not the marginal demand, the demand, given that D(3)=13,000.
So what topic have we you been doing for the last three days? You know exactly what you need to do so, |
17:05 | this is the kind of thing that might
show up in in your job as I keep
emphasizing you have to be able to do
these things because you have to know if
somebody else is giving you nonsense.
It's not really, generally, I don't think most of you this is gonna be well how you're gonna spend your day. Well, I'll tell you my first job at the Chicago Board of Trade we had to figure out and try to come up with us a function that would track the way the |
17:33 | bond future, the forward future contract,
was behaving, so we didn't, we weren't
very successful, it was too complicated at
the time, given the math in the computer
power that we had but we gave it a shot.
So I did get to use some high-level math early on but generally you have to be able to understand what's going on so you need to understand the phrase marginal demand. Marginal is it means derivatives. It means how is the demand changing so technically it means that |
18:00 | the last units you sell, it really means,
how is the demand changing it changes
according to whatever this function is that they made up and put in the
book, ok? So if we want to find the
original function where they gave us the
derivative, when we go backwards, that's
an integral. You're gonna want to
integrate -2000x/((sqrt)(25-x^2))dx.
|
18:36 | So you look at this you go, I'm not scared, I know my calculus let's let u = 25-x^2
not the (sqrt) of 25-x^2, just 25-x^2.
Because remember we can integrate the square root of u. Then du is -2xdx. So I look and I say well, I kinda have -2xdx. I have -2000. |
19:06 | So I multiply this by a thousand I have -2000x du.
Now I've got everything I need to substitute. I look here and I say ok, that's the square root of u, and -2000xdx, is |
19:34 | 1000du. So they're the same integral, but the second integral is much easier to do so 1/(sqrt)u, it's the same as saying u^(-1/2). Why would they give it to you as D(3)? Why not D(4)? It's so stupid, whatever, I didn't |
20:06 | make up the problem. Ok, so the integral
of this is u^(3/2/(3/2), bless you.
Which flip and make that 1000* (2/3)u^(3/2) which is a thousand which is 2,000/3(25-x^2)^(3/2)+C. |
20:48 | Maybe you say ok, I'm done. Except we're not quite because we also told you D(3) is 13000.
There's one last step, I would have chosen a number you can take |
21:04 | the square root of but whatever. So this is
the function, but that's generalized. So
now we can find it exactly because we
know I plug in 3 and we get 13,000.
So 13,000= 2000/3(25-3^2)^(3/2)+C, I take back what I said. |
21:34 | All right, 25-9=16, square root of 16 is 4, 4^3 is 64. You get 13,000=2000/3(64)+C. If I were to give you a problem that
looked like this on the final, getting
to about here is good enough, ok? I'd give
you easier numbers so let's see you get
39=128+C, so C=-89.
|
22:13 | So function becomes (2000/3)(25-x^2)^3/2-89.
And that's your demand, and that's fun. So how many of you got all the way to the end? Yes? |
22:43 | oh it's supposed to be u^1/2, I apologize, you guys should have said something sooner. It doesn't fundamentally change it that much.
|
23:06 | Let's see, 25, 9, sqrt 16, is 4,
you 13/8 = C, so you get (2000/3)(25-x^2)^(1/2)+13/8. Did you get there?
|
23:37 | All right, no you need to get all the way
there to get the full credit, let's give
you another one
Ok, this time we'll give you marginal profit.
|
24:04 | the marginal profit is
There's (9000-3000x)/(x^2-6x+10)^2, where P is profit.
|
24:40 | Ok, find the profit function given that P(3)=1500.
|
25:02 | So it's the same kind of problem, so we want to integrate.
So we want to integrate (9000-3000x)/(x^2-6x+10)^2 dx. So you look at this and you say, well, let's let u =x^2-6x+10. |
25:32 | And then du = 2x-6dx. Did I copy that right? No I did. I'm being dumb, I'm just reading this backwards.
Alright so I can multiply this by 1500 and I would get 3000x-9000dx. |
26:07 | Which is essentially what you have up there with a minus sign so this would then become, -1500du/u^2. or, -1500 integral u^-2du.
|
26:33 | It's always going to do something like this. Yes?
I think, yes, you could do a minus sign either way. You also could go up to here in the beginning and factor out 1500, there are a couple different ways to do it. So the integral of this is -1500*u^-1/-1du, which is the same as 1500/u, which is 1500/(x^2-6x+10)+C. |
27:21 | You end up in the same spot. Ok, then it says P(3)=1500, so 1500=(1500/(3^2-6(3)+10))+C.
|
27:43 | Look at that. 9-18+10=1, so c=0. So your profit equation is just 1500/(x^2-6x+10) |
28:06 | How do we like that? Alright, that's enough for today,
enjoy your weekend I'll, put the paper homework up later this afternoon, have fun with it.
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