By now you have had a lot
of opportunity to practice

differentiating common
functions.

And you have learned many
techniques of Differentiation.

So for example, if I were to
give you a function, let's call

this function capital F of X.

If it's a fairly simple
function, you'll know

how to differentiate it.

And so by differentiating it,
you'll be able to calculate its

derivative, which you'll
remember we denote as DF DX. So

that's a process that you're
very familiar with already.

Now in this unit we're going to
refer to DF DX as little F of X.

So little F of X.

Is the derivative of big
F of X. Let me write

that down little F of X.

Is the derivative.

As big banks.

And as I say, that's a process
that you're very familiar with.

What we want to do now is try to
work this process in reverse.

Work it backwards. In other
words, we want to start with

little F. And try and
come back this route.

And try and find the function or
functions capital F which when

they are differentiated will
give you little Earth. So we

want to carry out this process
in reverse. We can think of this

as anti differentiation.

So we can think of this anti
differentiation as

differentiation in reverse.

So big F of X we're going to
call the anti derivative of

little F of X.

So by these two results in mind
and try not to get confused with

a capital F in the lower case F
little F is the derivative of

Big F. Because little F
is DF DX.

And big F is the anti derivative
of little laugh.

Now you may have already looked
at a previous video called

integration as summation.

And in that video, the concept
of an integral is defined in

terms of the sum of lots of
rectangular areas under a curve.

To calculate an integral, you
need to find the limit of a son.

And that's a very cumbersome
and impractical process. What

we're going to learn about in
this video is how to find

integrals, not by finding the
limit of a sum, but instead

by using antiderivatives.

Let's start off
with the example.

Suppose we start off with the
function capital F of X.

Is equal to three
X squared plus 7X.

Minus 2.

And what I'm going to do is
something that you're already

very familiar with. We're
going to differentiate it.

And if we differentiate it.

We get TF DX equals.

Turn by turn the derivative of
three X squared is going to be.

2306 X.

And the derivative of Seven X is
just going to be 7.

What about the minus two? Well,
you remember that the derivative

of a constant is 0, so when we
do this differentiation process,

the minus two disappears.

So our derivative DF DX is just
six X +7.

So I'm going to write little F
of X is 6 X +7.

And think about what will happen
when we reverse the process when

we do anti differentiation.

Ask yourself what is an anti
derivative of 6X plus 7?

Now we already have an answer to
this question and anti

derivative of 6X plus 7.

Is 3 X squared plus 7X minus
two, so the answer.

Is F of X is 3 X squared plus
7X minus two, but I'm afraid

that's not the whole story.
We've already seen that when we

differentiate it, three X
squared plus 7X minus two, the

minus two disappeared.

In other words, whatever number
had been in here, whether that

minus two had been minus 8 or
plus 10 or 0, whatever it would

still have disappeared. We've
lost some information during

this process of Differentiation,
and when we want to reverse it

when we want to start with the
six, XX have Seven, and working

backwards, we've really no idea
what that minus two might have

been. It could have been another

number. So this leads us to the
conclusion that once we found an

antiderivative like this one,
three X squared plus, 7X minus

2. Than any constant added
onto this will still be an

anti derivative.

If F of X.

Is an anti derivative.

A little F of X.

Then so too.

Is F of X the anti derivative
we've found plus any constant at

all we choose. See is what we
call an arbitrary constant. Any

value of see any constant we can
add on to the anti derivative

we've found and then we've got
another antiderivative. So in

fact there are lots and lots of
antiderivatives of 6X plus

Seven, just as another example,
we could have had over here.

Three X squared plus 7X minus 8.

Or even just three X squared
plus 7X where the constant term

was zero. So lots and lots of
different antiderivatives for a

single term over here.

Let's do another example.

Suppose this time we look at.

A cubic function that suppose F
of X is 4X cubed.

Minus Seven X squared.

Plus 12X

minus 4.

Again, you know how to
differentiate this. You've

had a lot of practice
differentiating functions

like this.

So the derivative DF
DX is going to be.

Three 412 X squared.

The derivative of minus
Seven X squared is going to

be minus 14X.

And the derivative
of 12 X is just 12.

So think of our little F of X as
being the function 12 X squared

minus 14X at 12.

Notice again the point that
the minus four in the

differentiation process
disappears.

So now ask the question, what's
an anti derivative of this

function here little F.

Well, we've got one answer. It's
on the page already. It's this

big F is an anti derivative of

Little F. But we've seen
that any constant added on

here will still yield
another antiderivative, so

we can write down lots of
other ones, for example.

If we add on a constant.

And let's suppose we add on the
constant 10 onto this one.

Then we'll get the function 4X
cubed, minus Seven X squared.

Plus 12X plus six is also an
anti derivative of Little F.

And the same goes by adding
adding on any other constant at

all we choose. If we add on
minus six to this.

Will get 4X cubed, minus Seven X
squared plus 12X and adding on

minus six or just leave us this
so that that is another

antiderivative. There's an
infinite number of

antiderivatives of Little F.

Now in all the examples we've

looked at. I've in this in a
sense, giving you the answer

right at the beginning because
we started with big F, we

differentiated it to find little
F, so we knew the answer all the

time. In practice, you won't
know the answer all the time, so

one way that we can help
ourselves by referring to a

table of antiderivatives. Now a
table of antiderivatives will

look something like this one.

There's a copy of this in the
notes accompanying the video and

a table of antiderivatives were
list lots of functions.

F.

And then the anti derivative.

Big F in the next column and
you'll see every anti

derivative will have a plus.
See attached to it where C is

an arbitrary constant. Any
constant you choose.

What I want to do now is I
want to link the concept

of an antiderivative with
integration as summation.

Let's think of a function.

Y equals F of X.

Let's suppose the graph of the
function looks like this.

I'm going to consider
functions which lie entirely

above the X axis, so I'm
going to restrict our

attention to the region above
the X axis were looking up

here.

And also I want to restrict
attention to the right of the Y

axis, so I'm looking to the
right of this line here.

Let's ask ourselves what is the
area under the graph of Y equals

F of X?

Well, surely that depends on
how far I want to move to the

right hand side. Let's
suppose I want to move.

To the place where X
has an X Coordinate X.

Then clearly, if X is a large

number. The area under this
graph here will be large,

whereas if X is a small
number, the area under the

graph will be quite small
indeed. Effects is actually 0,

so this line is actually lying
on the Y axis than the area

under the graph will be 0.

I'm going to do note the
area under the graph by a,

so A is the area.

Under

why is F of X?

And as I've just said, the area
will depend upon the value that

we choose for X. In other
words, A is a function of XA

depends upon X, so we write
that like this, a is a of X and

that shows the dependence of
the area on the value we choose

for X. As I said, LG X larger
area small acts smaller area.

Ask yourself. What is the height
of this line here?

Well, this point here.

Lies. On the curve Y equals F of

X. So the Y value at that point
is simply the function evaluated

at this X value.

So the Y value here is just F of

X. In other words, the
height of this line or the

length of this line is just
F of X. The function

evaluated at this point.

Now I'd like you to think what
happens if we just increase X by

a very small amount.

So I want to just move this
point a little bit further

to the right.

And see what happens.

I'm going to increase X by a
little bit and that little bit

that distance in here. I'm going
to call Delta X. Delta X stands

for a small change in X or we
call it a small increment in X.

By increasing axle little bit.

What we're doing is we're adding
a little bit more to this area.

This is the additional
contribution to the area. This

shaded region in here.

And that's a little bit extra
area, so I'm going to call that.

Delta A, That's an incrementing
area changing area.

I'm going to write down
an expression for Delta

a try and work it out.

Let's try and see what it is,
but I can't get it exactly. But

if I note that the height of
this line is F of X.

And the width of this column in
here is Delta X. Then I can get

an approximate value for this
area by assuming that it's a

rectangular section. In other
words, I'm going to ignore this

little bit at the top in there.

If I assume it's a rectangular
section, then the area here this

additional area Delta A.

Is F of X multiplied by Delta X.

Now, that's only approximately
true, so this is really an

approximately equal to symbol.

This additional area is
approximately equal to

F of X Times Delta X.

Let me now divide both sides by
Delta X. That's going to give me

Delta a over Delta. X is
approximately equal to F of X.

How can we make it more
accurate? Well, one way we can

make this more accurate is by
choosing this column to be even

thinner by letting Delta X be
smaller, because then this

additional contribution that
I've got in here that I didn't

count is reducing its reducing
in size. So what I want to do is

I want to let Delta X get even
smaller and smaller and in the

end I want to take the limit.

As Delta X tends to zero of
Delta over Delta X.

And when I do that, this
approximation in here will

become exact and that will give
us F of X.

Now, if you've studied the unit
on differentiation from first

principles, you realize that
this in here is the definition

of the derivative of A with
respect to X, which we write as

DADX. So we have the result that
DADX is a little F of X.

Let's explore this a
little bit further.

We have the ADX is a little
F of X.

What does this mean? Well,
it means that little laugh

is the derivative of A.

But it also means if we think
back to the discussion that we

had at the beginning of this
video, that a must be the anti

derivative of F. A is
an anti derivative.

F. In other words, the area is
F of X. The anti derivative of

little F was any constant.

That's going to be an important
result. What it's saying is that

if we have a function, why is F
of X and we want to find the

area under the graph? What we do
is we calculate an anti

derivative of Little F which is

big F. And use this expression.

To find the area.

There's one thing we don't know
in this expression at the

moment, and it's this CC.
Remember, is an arbitrary

constant, but we can get a value
for. See if we just look back

again. So the graph I drew.

And ask yourself what will be
the area under this curve when X

is chosen to be zero? Well, if
you remember, we said that if

this vertical line here had been
on the Y axis.

Then the area under the
curve would have been 0.

So this gives us a
condition. It tells us that

when X is 0, the area is 0.

When X is 0.

Area is 0. What does this mean?
While the area being 0?

X being 0.

Plus a constant and this
condition then gives us a value

for C, so C must be equal to
minus F of North.

And that value for C and go back
in this result here. So we have

the final result that the area
under the graph is given by big

F of X minus big F of note.

Let me just write that
down again.

OK.

Now let's look at this problem.
Supposing that I'm interested in

finding the area under the graph
of Y is little F of X.

Up to the point.

Where X has the value be.

So I want the area from the Y
axis, which is where we're

working from above the X axis up
to the point.

Where X equals B.

We can use this boxed
formula here.

When X is be will get a of B.
That's the area up to be.

Will be F of B.

Minus F of note.

So this expression will give you
the area up to be.

Suppose now I want.

Area up to A and let's
Suppose A is about here.

So now I'm interested.

In this area in here, which is
the area from the Y axis up to a

well, again using the same
formula, the area up to a which

is a of A.

Is F evaluated at the X value
which is a?

Subtract. Big F of note.

So I have two expressions on
the page here, one for the

area up to be.

And one for the area up to A.

Ask yourself How do I find
this area in here? That's

the area between A&B.

Well, the area between A&B we
can think of as the area up to

be. Subtract the area up to A.

So if we find the difference of
these two quantities, that will

give us the area between A&B.

So we get the area.

Under Wise F of X.

From X equals A to X equals

B. Is the area under
this graph from X is A

to X is B is given by?

Once the area up to be, subtract
the area up to A.

So if we just find the
difference of these two

quantities will have F of B
minus F of A.

Add minus F of not minus minus F
of notes. These terms will

cancel out. So In other words
that's the result we need. The

area under this graph between
A&B is just.

Big F of B minus big FA.

Let me just write that
down again.

The area and Y equals F of X
between ex is an ex is B is

given by big F of B minus big F
of a where big F remember is an

anti derivative of Little F.

Any answer derivative?

So this is a very important
result because it means that if

I give you a function little F

of X. And you can calculate an
anti derivative of its big F.

And you can find the area
under the graph merely by

evaluating that anti
derivative at B, evaluating

it A and finding the
difference of the two

quantities.

Now in the previous video on
integration by summation, the

area under under a graph was
found in a slightly different

way. What was done there?

Was that the area under the
graph between A&B was found by

dividing the area into lots of
thin rectangular strips?

Finding the area of each of the
rectangles and adding them all

up. And doing that we had this
result that the area under the

graph between A&B was given by
the limit as Delta X tends to

zero of the sum of all these
rectangular areas, which was F

of X Delta X between X is A and
X is be. If you're not familiar

with that, I would advise you go
back and have a look at that

other video on integration by
summation, but that formula.

For this area was derived
in that video.

And this formula
defines what we mean

by the definite
integral from A to B

of F of X DX.

That's how we defined
a definite integral.

And if we wanted to define a
definite integral, if we wanted

to calculate a definite integral
in that video, we had to do it

through the process of finding
the limit of a sum.

Now the the process of finding
the limit of a sum is quite

cumbersome and impractical. But
what we've learned now is that

we don't have to use the limit
of a sum to find the area. We

can use these anti derivatives.

Because we've got two
expressions for the area, we've

got this expression. As a
definite integral, and we've got

this expression in terms of
antiderivatives, and if we put

all that together will end up

with this result. The definite
integral from A to B.

F of X DX.

Is FB.

Minus F of A.

In other words, the definite
integral of little F between the

limits of A&B is found by
evaluating this expression,

where big F is an anti
derivative of Little F.

Let me write that formula
down again.

The integral from A to B.

F of X DX is given by
big FB minus big F of A.

Let me give you an example.

Suppose we're interested
in the problem of finding

the area under the graph
of Y equals X squared.

And let's suppose for the sake
of argument we want the area

under the graph. Between X is
not an ex is one, so we're

interested in this area.

In the previous unit on
integration as a summation,

this was done by dividing
this area into lots of thin

rectangular strips.

And finding the area of each of
those rectangles separately.

And then adding them all up.

That gave rise to this formula
that the area is the limit.

As Delta X tends to 0.

Of the sum from X equals not to
X equals 1.

Of X squared Delta X.

So the area was expressed as the
limit of a sum.

And in turn, that defines the
definite integral. X is not to

one of X squared DX.

Now open till now if you
wanted to work this area out

the way you would do it would
be by finding the limit of

this some, but that's
impractical and cumbersome.

Instead, we're going to use
this result using

antiderivatives.

Our little F of X in this
case is X squared.

And this formula at the top of
the page tells us that we can

evaluate this definite
integral by finding an

antiderivative capital F. So
we want to do that first.

Well, if little F is X
squared, big F of X, well we

want an anti derivative of X
squared and if you don't know

what one is, you refer back to
your table.

An anti derivative of X
squared is X cubed over 3

plus a constant.

So in order to calculate this
definite integral, what we need

to do is evaluate.

The Anti Derivative Capital F.
At B. That's the upper limit,

which in this case is one.

And then at the lower limit,
which in the in our case is

zero. Let's workout F of one.

Well, F of one is going to be 1

cubed. Over 3, which is
just a third.

Plus C.

Let's workout big F of 0.

Big F of 0.

Is going to be 0 cubed over
three, which is 0 plus. See, so

it's just see. And then we
want to find the difference.

Half of 1 minus F of note.

Will be 1/3 plus C minus C, so
the Seas will cancel and would

be just left with the third.

In other words, to calculate
this integral here.

All we have to do is find the
Anti Derivative Capital F.

Evaluate it at the upper limit
evaluated at the lower limit.

Find the difference and the
result is the third. That's the

area under this graph between
North and one.

Now let me just show you how we
would normally set this out.

We would normally set this
out like this.

We find. An anti derivative of X
squared, which we've seen is X

cubed over 3 plus C.

We don't normally write the Plus
C down, and the reason for that

is when we're finding definite
integrals, the sea will always

cancel out as we saw here, the
Seas cancelled out in here and

that will always be the case.

So we don't actually need to
write a plus C down when we

write down an anti
derivative of X squared.

It's conventional to write down
the anti derivative in square

brackets. And to transfer the
limits on the original integral,

the Norton one to the right hand
side over there like so.

What we then want to do is
evaluate the anti derivative at

the top limit that corresponded
to the FB or the F of one. So we

work this out at the top limit
which is 1 cubed over 3.

We work it out at the bottom
limit. That's the F of a or

the F of Norte.

We work this out of the lower
limits will just get zero and

then we want the difference
between the two.

Which is just going to give
us a third. So that's the

normal way we would set out a
definite integral.

So as we've seen.

Definite integration is very
closely associated with Anti

Differentiation.

Because of this, in general it's
useful to think of Anti

Differentiation as integration
and then we would often refer to

a table like this one of
antiderivatives as simply a

table of integrals.

And there will be a table of
integrals in the notes and in

later videos you'll be looking
in much more detail at how to

use a table of integrals. So
you think of the table of

integrals as your table of
antiderivatives and vice versa.

Very early on in the video, we
looked at this problem. We

started off with little F of X
being equal to four X cubed.

Minus Seven X squared.

Plus 12X minus 4.

We differentiate it.

To give 12 X squared
minus 14X plus 12.

And then we said that an anti
derivative of 12 X squared minus

14X plus 12.

Was this function
capital F plus C?

We've got a notation we
can use now because we've

linked antiderivatives to
integrals and the notation

we would use is that the
integral of 12 X squared.

Minus 14X plus 12 DX
is equal to.

4X cubed

minus Seven X squared plus 12X
plus any arbitrary constant.

And we call this an
indefinite integral.

And we say that the indefinite
integral of 12 X squared minus

14X plus 12 with respect to X is
4X cubed minus Seven X squared

plus 12X. Plus a constant
of integration.

OK, In summary, what have
we found? What we found

that a definite integral?

The integral from A to B of
little F of X DX. We found that

this is a number.

And it's obtained from the
formula F of B minus FA,

where big F is any anti
derivative of Little F.

And we've also seen the
indefinite integral of F

of X DX.

Is a function big F of X plus an
arbitrary constant? Again, where

F is any anti derivative of
Little F&C is an arbitrary

constant. So in this video we've
learned how to do

differentiation in reverse.
We've learned about

antiderivatives. Definite

integrals. And indefinite
integrals in subsequent videos.

You learn a lot more teak
techniques of integration.