Welcome to the presentation on the indefinite integral
or the antiderivative.
So let's begin with a bit of a review of the
actual derivative.
So if I were to take the derivative d/dx.
It's just the derivative operator.
If I were to take the derivative of the expression
x squared-- this is an easy one if you remember the
derivative presentation.
Well, this is pretty straightforward.
You just take the exponent.
That becomes the new coefficient, right.
You actually multiply it times the old coefficient, but in
this case the old coefficient is 1, so 2 times 1 is 2.
And you take the variable 2x.
And then the new exponent will be one less than
the old exponent.
So it'll be 2x to the 1, or just 2x.
So that was easy.
If I had y equals x squared we now know that the slope at any
point on that curve, it would be 2x.
So what if we wanted to go the other way?
Let's say if we wanted to start with 2x, and I wanted to say
2x is the derivative of what.
Well, we know the answer this question, right?
Because we just took the derivative of x squared
and we figured out 2x.
But let's say we didn't know this already.
You could probably figure it out intuitively, how you can
kind of do this operation that we did here, how
you can do it backwards.
So in this case the notation-- well we know it's x squared--
but the notation for trying to figure out 2x is the derivative
of what, we could say that-- let's say 2x is the
derivative of y.
So 2x is the derivative of y.
Let's get rid of this of what.
Then we can say this.
We can say that y is equal to-- and I'm going to throw some
very fancy notation at you and actually I'll explain why we
use this notation in a couple presentations down the road.
But you just have to know at this point what the notation
means or what it tells you to really do, which really is
just the antiderivative or the indefinite integral.
So we could say that y is equal to the indefinite
integral 2x dx.
And I'm going to explain what this squiggly line here is and
dx, but all you have to know is when you see the squiggly line
and this dx and then something in between, all they're asking
is they want you to figure out what the antiderivative
of this expression is.
And I'll explain later why this is called the
indefinite integral.
And actually this notation will make a lot more sense
when I show you what a definite integral is.
But let's just take it for granted right now that an
indefinite integral-- which I just drew here, it's kind of
like a little squirrely thing-- is just the antiderivative.
So y is equal to the antiderivative essentially,
or the indefinite integral of the expression 2x.
So what is y equal to?
Well y is obviously equal to x squared.
Let me ask you a question.
Is y just equal to x squared?
Because we took the derivative, and clearly the derivative
of x squared is 2x.
But what's the derivative of x squared-- what's the
derivative x squared plus 1?
Well, the derivative of x squared is still 2x.
What's the derivative of 1?
Right, derivative of 1 is 0, so it's 2x plus
0, or still just 2x.
Similarly, what's the derivative of x squared plus 2?
Well the derivative of x squared plus 2 once
again is 2x plus 0.
So notice the derivative of x squared plus
any constant is 2x.
So really y could be x squared plus any constant.
And for any constant we put a big c there.
So x squared plus c.
And you'll meet many calculus teachers that will mark this
problem wrong if you forget to put the plus c when you do
an indefinite integral.
So you're saying Sal, OK, you've showed me some notation,
you've reminded me that the derivative of any constant
number is 0, but this really doesn't help you solve
an indefinite integral.
Well let's think about a way-- a systematic way if I didn't do
it for you already-- that we could solve an
indefinite integral.
Let me clear this.
A bolder color I think would make this more interesting.
Let's say we said y is equal to the indefinite integral of--
let me throw something interesting in there.
Let's say the indefinite integral of x cubed dx.
So we want to figure out some function whose derivative
is x to the third.
Well how can we figure that out?
Well just from your intuition, you probably think, well it's
probably something times x to the something, right?
So let's say that y is equal to a x to the n.
So then what is dy/dx, or the derivative of y is n.
Well we learned this in the derivative module.
You take the exponent, multiply it by the coefficient.
So it's a times n.
And then it's x to the n minus 1.
Well in this situation we're saying that x to the third is
this expression, it's the derivative of y.
This is equal to x to the third.
So if this is equal to x to third, what's a and what's n.
Well, n is easy to figure out.
n minus 1 is equal to 3.
So that means that n is equal to 4.
And then what is a equal to?
Well a times n is equal to 1, right, because we just have a 1
in this coefficient, this has a starting coefficient of 1.
So a times n is 1.
If n is 4, than a must be 1/4.
So just using this definition of a derivative, I think we now
figured out what y is equal to.
y is equal to 1/4 x to the fourth.
I think you might start seeing a pattern here.
Well how did we get from x to the third to
1/4 x to the fourth?
Well, we increased the exponent by 1, and whatever the new
exponent is, we multiply it times 1 over that new exponent.
So let's think if we can do a generalized rule here.
Oh, and of course, plus c.
I would have failed this exam.
So let's make a general rule that if I have the integral
of-- well, since we already used a, let's say-- b
times x to the n dx.
What is this integral?
This is an integral sign.
Well my new rule is, I raise the exponent on x by 1, so it's
going to be x to the n plus 1.
And then I multiply x times the inverse of this number.
So times 1 over n plus 1.
And of course I had that b there all the time.
And one day I'll do a more vigorous-- more rigorous proof
and maybe it will be vigorous as well-- as to why this b
just stays multiplying.
Actually I don't have to do too rigorous of a proof if you just
remember how a derivative is done, you just multiply this
times the exponent minus 1.
So here we multiply the coefficient times 1 over
the exponent plus 1.
It's just the inverse operation.
So let's do a couple of examples like this really fast.
I have a little time left.
I think the examples, at least for me, really
hit the point home.
So let's say I wanted to figure out the integral
of 5 x to the seventh dx.
Well, I take the exponent, increase it by one.
So I get x to the eighth, and then I multiply the coefficient
times 1 over the new exponent.
So it's 5/8 x to the eighth.
And if you don't trust me, take the derivative of this.
Take the derivative d/dx of 5/8 x to the eighth.
Well you multiply 8 times 5/8.
Well that equals 5 x to the-- and now the new exponent will
be 8 minus 1-- 5 x to the seventh.
Oh, and of course, plus c.
Don't want to forget the plus c.
So I think you have a sense of how this works.
In the next presentation I'm going to do a bunch more
examples, and I'll also show you how to kind of
reverse the chain rule.
And then we'll learn integration by parts, which is
essentially just reversing the product rule.
See you in the next presentation.