| Start | We're starting a new topic, there's a bunch of new stuff to do. We have about a month worth of class and then you never have to do calculus again.
And I know that puts a smile on many of your faces. Um, alright. So let's see. So we talked about functions. And we talked about derivatives, and we talked about exponential and logarithmic functions, now we're going to go back to sort of a derivative, but going the other direction. So we learned if you have a function you can take the derivative of it and that tells you how the function is changing and increasing and decreasing and rates and all that. |
| 0:36 | Now, we're going to go backwards and do anti-derivatives.
So anti-derivatives give you different information about a function, and they're related to something called the fundamental theorem of calculus. Which we may or may not get into, but certainly not today. So anti-derivatives are as I said, working backwards, they are more difficult to do, by far. It is very easy, in general, to take a derivative. You learned how to take derivatives, they're really not that hard. |
| 1:02 | You can make mistakes, but the mechanics of it is not that difficult.
However, backwards it's much harder. It's like if you smash something, but it's very hard to put it back together again. So, cause many times you take the derivative of something simple and you end up with something long and complicated, like when we have quotient rule. So if you started with the quotient rule, to go back and figure out what the original function was is not so easy. So, anti-derivatives, as I said the other direction will be more difficult, but, I have faith that you are a bunch of smart men and women and you will be able to do it. |
| 1:37 | So, ok. The symbol for anti-derivative that we're going to use is the integral symbol.
means integral of f(x). |
| 2:04 | An integral and the anti-derivative as far as we're concerned for now we sort of use interchangeably.
So we could also say the anti-derivative. So now let's figure some of these out. So say I gave you something very simple. And ok, by the way, what does that dx mean? |
| 2:32 | Well, dx is part of the anti-differentiation. Just as the way you had dy/dx when you took a derivative, you're going to have something dx when you do an anti-derivative.
First, it says we're doing this with respect to x, and second is, it has a meaning, which will be clearer when we've talked about more stuff next week. So for now, it's always in there, when you start with the integral, and goes away when you finish the integral. So for example, if I said, what's the integral of 5? |
| 3:01 | So you say to yourself, we have an answer, 5. So what was the original function?
5x, right. So you must have started with 5x. Because the derivative of 5x is 5. So the anti-derivative of 5 is 5x, going the other direction. However, we have a small problem. It could also be 5x+1. Because the derivative of 5x+1 is also 5. |
| 3:31 | It could also be 5x+2, it could be 5x+pi, 5x+100000000
That constant, when you take the derivative of that constant, you get 0.
So there could be any constant attached to 5x. So we write +C. Where C is the constant. And unless I give you more information, you have to leave it that way. This is called an indefinite integral. |
| 4:03 | It's indefinite because we don't know what C is. I may give you more information, so lets say I tell you that,
The integral of 5dx=? and I tell you that f(0)=4.
Ok? And I say this is f(x). So we say, alright. That equals 5x+c, |
| 4:36 | And since f(0)=4, I now know 4=5(0)+c.
5(0) is 0, so c=4. Therefore the equation is 5x+4. So if I give you a little extra information, you can often figure out what the constant is. Ok? How do we feel about these so far? Pretty easy, right? Of course I'm going to give you something harder than 5. |
| 5:11 | What if I gave you 5x?
You say, well, what could it be? It must be something like 5x^2. Right? Because when we take the derivative of x^2, we get 2x. So if we have 5x^2, lets try 5x^2. |
| 5:33 | Problem is, what's the derivative of 5x^2? It's 10x.
So 10x, right. So it's 10x, not 5x. So we have to divide this by 2. Then 10x/2 = 5x. That'll get us back where we started, plus some constant. What if I gave you the integral of 7x? |
| 6:02 | Well, 7x^2 would be 14x, so again we have to divide by 2. So that 2 seems to be important.
You see why we have the 2 in there? Because now if you take the derivative back, you end up with twice as much. Because when you took the derivative of x^2 you get 2x. Here we only really want to have x. What if I had the integral of 7x^3dx? |
| 6:34 | Ok, well lets see what happens when we take the derivative of 7x^4.
And if I take the derivative of that, 4 goes in the front, so I have 28x^3, but I only want 7, so I divide by 4. So it looks like I'm dividing by whatever that power is. And we are. And that's the first integral rule that we're gonna want. So rule number 1. |
| 7:04 | Well, we actually have 2 rules.
So we had 1 where we just had a constant, and that became the constant *x +c. Ok? So when we had the integral of 5, it just became 5x. Now, what if I had Kx^n(dx). Well, the k doesn't seem to matter, |
| 7:33 | so the constant, just as with derivatives, the constant just sort of has to stay there. This becomes,
You raise the power of x by 1, because when you take the derivative you go back, and you divide by whatever that power is.
So in a derivative, you bring the power in the front, multiplying it, and subtract 1 from the exponent. So the anti-derivative, you return the power, adding 1, and divide, that way they undo each other. |
| 8:07 | Cause, lets take the derivative of x^(n+1).
f(x) = x^(n+1)/(n+1) + c. So that's what I have there, except for the k part. And the derivative, bring the power in front, and reduce the power by 1. |
| 8:36 | And the derivative of the constant is 0, so these cancel and you get x^n.
Which is what we're doing over there. So if I had the integral of x^8, |
| 9:00 | You add 1 to the 8, you get 9, and you divide by 8, and add a constant. Remember you always have to have that +C.
Don't forget it. Because the derivative of any constant is 0. And in fact, the integral of 3x^8 is 3x^9/9 + C. |
| 9:30 | So many people, including me, when you see an integral with a constant in it, so if you saw, say the integral of 6x^5,
They put the 6 outside.
That's very useful for when you do messier integrals. Cause what will happen is the thing on the inside called the integrand, can get very complicated. So pulling the constant out can often just make it a little easier to work with. They're the same thing. Now what would that be? Well that would be 6(x^6/6)+C. |
| 10:11 | That would just be x^6+C. Ok?
How do we feel about these so far? Not very hard right? You literally just add 1 to the power and divide by the new power. |
| 10:31 | Of course, we could make a chain of these.
So if we want to find that integral. It should be just as straight forward as differentiating, just going the other way. So this should just be 3x^5/5 -5x^3/3 + 2x^2/2 +3x+C. |
| 11:12 | So each time, take 4, add 1 to the power, put it over the new power. Squared, add 1 to the power, put it over the new power,
2x, that's x^1, so it becomes x^2/2, those 2 can cancel.
The integral of a constant, remember is the constant *x, and then + C. |
| 11:35 | How do we feel about these so far?
We're ok? Nothing too challenging? Ok. It does get trickier, integration can get very complicated, but we only do very straight forward versions of it in this class. We only have 2 or 3 types, because I don't know how many integrals you're going to run into in your life. At some point it just becomes an exercise in [?]. |
| 12:07 | What's the integral of the (sqrt) of x?
Well, (sqrt)x = x^(1/2). So you can rewrite that as x^1/2. And then the anti-derivative would be x^(3/2)/(3/2) + C. |
| 12:39 | Now, dividing by a fraction, you flip it and multiply.
You get 2/3 x^(3/2)+C. So far so good? Ok. What if I gave you 1/x? |
| 13:01 | That's x^-1 isn't it? Well when you add 1 to -1, what do you get? And then you have to divide by 0. That doesn't make any sense right?
Cause you get x^0/0. So what is it? Ah you seem to have forgotten something. What's the derivative of the natural log? The derivative of the natural log in 1/x. So the anti-derivative/integral of 1/x is ln(x) |
| 13:32 | But we make it the absolute mag because it's possible this is a negative number, but you cant take the log of a negative number.
So, just in case, you throw an absolute value. Again, that's the kind of thing you're going to have to just sort of know. But the odds we do something with that are very small. It's not just ln(x) its ln|x|. |
| 14:03 | So let's make sure [?].
Alright? Take a minute, see if you can figure it out. Alright, that's long enough. |
| 14:31 | So, some of these are easier than others, so x^3 goes to x^4/4,
And 11x^2 goes to 11x^3/3,
And for 8x, becomes 8x^2/2, which you can simplify to 4x^2.
+ 3x. And then -4ln|x|. |
| 15:04 | And what are we going to do with 6/(sqrt)x?
Well thats 6x^(-1/2). So we would make this 6x^(1/2)/(1/2) + C. Notice, we don't write + C for each of the terms, we can just do 1 big + C at the end. |
| 15:31 | Since you haven't figured out what C actually is, it doesn't matter until you actually calculate it.
This could be simplified, you could make that x^4/4 + 11x^3/3 + 4x^2 + 3x - 4ln|x| +12(sqrt)x + C. |
| 16:04 | And by the way, since remember your log rules, you could make this log(x)^4, if you wanted, however it's usually more convenient to not have the power.
So far so good? How'd we do on that one? We're we able to do that? Yes? No? Maybe? Menza-menz? You had trouble with the last term the square root term. Right, let's practice a couple more of those just to be sure. You have to be comfortable with powers. |
| 16:38 | Now suppose we had, the integral of (1/x^3)+(1/(cube-root)x)dx.
If we wanted to do that, say well, before we do the integral, let's rewrite this, just so we can use our power rule. (1/x^3) is x^-3, |
| 17:07 | And cube-root of x is x^1/3, so 1/(cuberoot)x is x^(-1/3) dx.
And now you just have to practice adding 1. So the integral of x^-3 is x^-2/-2. |
| 17:38 | And x^(-1/3) is x^(2/3)/(2/3).
Remember when you were in 5th grade and you said you qould never use fractions? There you go, now you're finally using that stuff. It all paid off. +C. |
| 18:01 | And then if you wanted to rewrite this, there's no need to, but if you wanted to,
this is x^2, so this becomes (1/-2x^2)
because the x^-2 is 1/x^2.
And then + (3/2)x^(2/3) + C. Now why would I want to do that? Just so it's easier to look at? If I need to work with it or plug in numbers, or something like that. |
| 18:35 | Otherwise just leave it alone. There's nothing special about one way vs. another.
And as opposed to derivatives, we don't really integrate and integrate again, there are a couple applications where that shows up, mostly in physics, but we really just integrate once and you're done. Let's do another just to make sure you get the idea. Just to make sure you're comfortable playing with powers. |
| 19:03 | Got it? Many of you are nodding that's a good sign. Alright, I would rewrite this,
as x^-5 - x^(-1/2) + 1/x + 4x^(-1/5)dx.
And you take the integral. The integral of x^-5 is x^-4/-4, |
| 19:33 | integral of x^(-1/2) is x^(1/2)/(1/2),
Dividing by (1/2) is multiplying by 2,
integral of 1/x is ln|x|,
and then x^(-1/5) would be x^(4/5)/(4/5), + C.
And as I said, you can rewrite that, that would be, |
| 20:01 | -1/4x^4 - 2(sqrt)x + ln|x| + 5x^(4/5) +C.
So far so good? Ok. One last type. We did polynomials, square roots, log, are we forgetting anything? |
| 20:41 | What's the integral of e^x? Well what's the derivative of e^x? e^x. So what does that mean the anti-derivative is? e^x.
That's not too hard. And if you put a constant up here, Well, when you took the derivative, you would multiply by this constant, when you take the anti-derivative, you're going to divide by this constant. |
| 21:08 | This is sort of working the chain rule backwards.
And the integral of e^5x, is e^5x/5 +c. |
| 21:48 | what is the integral of e^x-e^-x.
Well, the integral of e^x is e^x, And e^-x is e^-x/-1. |
| 22:02 | which is just -e^-x. So the 2 minuses cancel and become + e^-x.
Whenever you do an anti-derivative, and you're not sure you got it right? Take the derivative. The derivative of e^x is e^x, and the derivative of e^-x, is -e^-x which is exactly what we have here. because the derivative of the constant is 0. So the integrals of e are pretty easy. Ok, I mean I could say what's the integral of e^pi(x) dx |
| 22:35 | pi scares a lot of people because 'It must be something different', no.
It's just, e^(pix)/(pi) + C. And there's things you can do with integrals, so first of all, we didn't really talk, well I did for one little problem, about how you find C. |
| 23:06 | That, you usually have something called an initial condition.
And then what do you do with integrals You can do all sorts of things with integrals. Integrals are very useful for finding the area under a curve. Or for adding things up for doing a sum, in fact, the integral symbol, is an elongated S because it stands for sum. which we're going to get into more. So as I said, unfortunately, I know you're going to be sad but I have to do a short class today. |
| 23:34 | And I have to go grade exams so I will see all of you on Monday.
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