-
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.