The Indefinite Integral or Anti-derivative

An introduction to indefinite integration of polynomials.

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.