www.mathcentre.ac.uk/.../9.7a%20Simple%20applications%20-%20areas.mp4

2 videos in this series.

Areas as summation and
integration as the reverse of

differentiation have shown that
the area under a curve.

Above the X axis and between two
ordinates. That's two values of

X can be calculated by using
integration. And what we want to

do is develop that idea and look
at some more examples of that in

this particular video.

So. Let's begin by exploring
the question. We've got. The

function Y equals X times X
minus one times X minus 2.

What's the area contained
between this curve?

On the X axis, well, the very
nature of the question suggests

hang on a minute. There might be
something a bit odd here. Let's

draw a picture. So let's start
by sketching this curve.

Well, if we put Y equals 0 then
we can see that each one of

these brackets could be 0 and
that will give us a series of

points on the curve. So why is 0
then? This could be 0 EX could

be 0. Or X minus one could be 0.
In other words, X could be equal

to 1. Or X minus two could be 0.
In other words, X could be equal

to two, so there on the X axis.
I know that this curve is going

to go through their through
there and through there.

What about when X is very large?
What's going to happen to Y when

X is very large, X minus One X
minus two are really very

similar to X, so effectively
we've got X times by X times by

X. You know as if X is very
large and positive, so most why

be so? We've got a bit of curve

there. Similarly, if X is large
but negative, again taking one

off and taking two off isn't
going to make a great deal of

difference, so we've got a large
negative number times by a large

negative number times by a large
negative number. The answer is

going to be very large, but
negative 'cause we three

negative numbers multiplied
together. So there's a bit of a

curve down here.

Now how can we join that up?
Well, assuming the curve is

continuous, it's going to go up.

Over. Down I'm back up again.

So that's the picture of our
curve and straight away we can

see what the question is asking
us. Find the area contained by

the curve and the X axis it
wants these two areas.

But they're in different
positions. One area a is above

the X axis and the other one
area B is below the X axis.

So perhaps we better be

cautious. And work out these two

areas separately. Now what we've
been told, what we know is that

if we integrate.

Between the limits. So in this
case that's not an one hour

function X times by X minus one
times by X minus two with

respect to X. We will get this
area a, so the area a is equal

to the result of carrying out
this integration and

substituting in the limits of
integration. So let's do that

first of all.

We will need to multiply out

these brackets. Well, X times by
X is X squared and times by that

ex again is X cubed.

I've got X times by minus two
gives me minus 2X.

And minus one times by X gives
me another minus X. So

altogether I've got minus 3X
coming out of these brackets and

I need times it by X, so that's
minus three X squared.

Then I've got minus one times by
minus two. That's +2 and I need

to times that by X. So that's

plus 2X. To be integrated with
respect to X so it's carry out

the integration. Remember we had
one to the index and divide by

the new index, so that's one to
the index. Makes that X to the

power 4 / 4 - 3.

To integrate X squared, we had
one to the index that's X cubed

divided by the new index. That's
3 + 2. XX is X to the power one.

We don't always right the one
there, but we know that it means

X to the power one. So we add 1
to the index that's X squared

and we divide by the new index.

These are our limits of
integration nought and one.

We can do a little bit of simply
simplifying in here. We can

cancel three with a 3 under, two

with a 2. And now we can put
in our limit of integration. The

top 1 one in.

So that's. X to the
power four. When X is one is

just one over 4 minus X cubed.
When X is one. So that's just

one plus X squared. When X is
one. So that's just one again.

Minus and now we substitute in
our lower limit. That's zero,

but when we put zero in there
that's zero and that's zero as

well, and that's zero as well,
because X is equal to 0 in each

case. So the whole of the SEC
bracket here is just zero.

So we simplify that we get a

quarter. So the area a is
1/4 of a unit of area.

What about area B? We can find
that by doing the same

integration, this time taking
the limits between one and two.

So let's look at that one.

B.

Will be the integral between one
and two of X cubed.

Minus three X squared
plus 2X with respect

to X. So let's have
a look what we're going to get

the integral of. This should not
be any different to what we had

last time. So if we integrate
the X cubed, we have X to the

4th over 4.

Minus three X cubed

over 3. +2 X
squared over 2 and

this is to be

evaluated. The limits one and
two, and again we can cancel the

threes and we can cancel the

tools. So now let's substitute
in our upper limit.

X to the power four when X
is 2, so that's 2 multiplied by

itself four times and that gives
us 16 over 4.

Minus X cubed which is 2
multiplied by itself three

times, so that's eight.

Plus X squared which is 2
multiplied by itself twice, so

that's four. Minus now we put
the one in, so that's a quarter.

Because X to the power four is
just 1 - 1 because X cubed is

just one when X is one.

Plus one. Now
let's simplify each of these

brackets separately. 4 into 16
goes four times, so we four

takeaway 8 + 4. Will that
bracket is now 0.

Minus or we have 1/4 takeaway
one add one so this bracket is

just a quarter.

So I'll answer
is minus 1/4.

Let's just have a look where B

was. Bees here
it's underneath the X

axis.

Now, if you remember, or if you
look back at the video on

integration as summation, one of
the things that we were looking

at when we had a piece of curve.

And we were looking at the area
underneath that curve.

We took small strips.

And these small strips were of
height Y and thickness Delta X,

and so the little bit of area
that they represented. Delta a

was why times by Delta X. Now
that's fine for what we've done.

But notice this Y is positive.

And so when we add up all
these strips that are above

the X axis, we get a positive
answer, IE for a the answer

we got was a quarter.

But for B.

These why values are actually
below the X axis and so this

area, in a sense is a negative
area because the wise are

negative. Hence our answer is
minus 1/4. But the question said

what is the area and clearly
physically there are two blocks

of area and so the total area.

Must be 1/4. That's the area of
that and a quarter because 1/4

is the area of that. The minus
sign merely tells us it's below

the X axis. So the answer to our
original question, what's the

area that the that's contained
between the curve and the X

axis? Is given by
the area is 1/4 +

1/4 which is 1/2.

Let's just check on something
what would happen.

If we took the integral of our
function between North and two.

Remember what our function is.
It's X cubed minus three

X squared plus 2X.

Integrated with respect to X.

So let's do that integration and
again the answer shouldn't be

any different from what we've
had before X to the 4th over 4

- 3 X cubed over 3 + 2
X squared over 2 to be evaluated

between North two. Again, you
can do the canceling the threes

there and the tools there.

Equals that substituting the

upper limit. So that's X to
the power four. When X is 2, two

multiplied by itself four times
is 16, and to be divided by 4.

Minus X cubed minus 2 multiplied
by itself three times, that's

eight. Plus X squared that's
2 multiplied by itself twice, so

that's plus 4.

Minus. And when we put zero
in that zero, that's zero and

that one zero. So the whole of
that bracket is 0.

Counseling here. This leaves us
with a four, and so we

have 0 - 0 is 0.

Which is what we would expect,
because in a sense, what this

process is done is added

together. The two
areas that we got.

With respect to their signs, 1/4
plus minus 1/4 is Nord.

What that means is that we have
to be very, very careful when

we're calculating areas.

Clearly if we just integrate it
between the given limits, what

we might do is end up with an
answer that is the value of the

integral and not the area and
what this example shows is is

that the area that is contained
by our function.

The X axis and specific values
of the ordinance I values of X

may not be the same as the value
of the integral, so bearing that

in mind, let's have a look at
some more examples.

This time. Let's take.

Function of the curve Y equals X
times by X minus three and let's

investigate what's the area
contained by the curve X access

and the audience X equals 0 and
X equals 5. Well, from what

we've learned so far, one of the
things we've got to do is a

sketch. So here's our X&Y axes.

If we set Y equals 0, then we
can see that either X is 0 or X

can be equal to three. So that
gives us two points on the curve

there at the origin and there
where X equals 3. We also know

that we're going up to X equals
5, so let's Mark that there.

When we multiply out this
bracket, we can see we've got

X times by. X gives us X
squared and it's a positive X

squared, which means that
it's going to be a U shape

coming down like that.

Now it wants the area between
the X axis, the curve and these

ordinance. So let's draw that in
there. And again we can see

we've got two lots of area and
area a which is below the X axis

in an area B which is above it.
This means we've got to be

careful, workout each area
separately so the area a is the

integral. Between North

And three of X times by X minus
three with respect to X.

Let's multiply out the brackets

first. So X times by X gives
us X squared X times Y minus

three gives us minus 3X.

We're integrating that with
respect to X.

So let's do the integration.

X squared integrates 2X cubed
over three and one to the index

and divide by that new index,
minus three X. This is X to the

power one, add 1 to the index
and divide by the new index.

That's to be evaluated between

North. And three.

Equals we put the upper limiting

first. X cubed X is equal
to three, so this is 3 cubed.

That's 27 over 3.

Minus three times by X squared
and X is 3, so we

have minus three times by 9
is 27 over 2.

Minus now we put the lower limit
in X cubed is zero when X is

zero and X squared is zero when
X is zero. That is minus 0.

Equals so now we've got 27 over
3 - 27 over 2. We can cancel

by the three here. That's nine,
so we have 9 minus and twos into

27 go 13 1/2. That will simplify
to that and so we have minus

four and a half. So that's the
area a. Let's now have a look at

the area be.

So that's the integral of X
squared minus 3X with

respect to X. This time
between 3:00 and 5:00.

Will get exactly the same as
we've got here, 'cause we're

integrating exactly the same

function. But the limits
are different there

between 3:00 and

5:00. So let's turn over the

page. Area B is equal
to. Remember that it's this that

we are evaluating.

Between 3:00 and 5:00, so we
put in our upper limit first.

X cubed when X is equal
to five, is 5 multiplied by

itself 3 times and that's
125 over 3.

Minus.

X squared when X is 5, that's
five times by 5 is 25.

25 times by three is 75
over 2.

That's our first.

Bracket minus now we put
in the three.

So that's 27 over 3.

Because if you think about it,
we've actually worked this out

before. Minus X is 3.

3 squared is 9 times by that is
27 over 2.

OK. Let's
tidy up these bits first. 125

over 3 - 27 over three.
Well, that's 98 over 3 -

5 - 75 over 2 minus
minus 27 over 2. So I've

got minus 75 + 27 and
that is going to give me

48. Over. 2.

So if you divide.

By three there threes into nine
goes three and threes into eight

goes two, and there are two
overs that's 32 and 2/3 - 2

into 48 goes 24. So the area
B is 32 and 2/3 - 24.

That's eight and 2/3, so the
actual area that I've got.

Is the area of a the actual area
of a which is four and a half.

The minus sign tells us that
it's below the X axis.

4 1/2 plus the area of B which
is 8 and 2/3 if I want to

add a half and 2/3 together then
they've really got to be in

terms of the same denominator,
so the four and the eight gives

me 12. Plus 1/2 now the
denominator is going to be 6, so

that's three sixths plus four
sixths, which is what the 2/3 is

equal to. I've got 7 sixth here,
which is one and a six. That's

13 and 1/6 is the actual value
of my area.

Let's take.
Another example, this time.

Let's have a look at the
function Y is equal to.

X squared plus

X. +4.

Now if we set Y equals 0,
then we're trying to solve

this equation X squared plus
X +4 equals not.

To find where the curve crosses
the X axis.

Doesn't look very promising.
Everything here looks sort of

positive got plus terms in it.
Let's check B squared minus four

AC. Remember the formula for
solving a quadratic equation is

X equals minus B plus or minus
square root of be squared minus

four AC all over 2A, but it's
this bit that's the important

bit. the B squared minus four
AC, 'cause If that's negative.

IE, when we take a square root,
we can't have the square root of

a negative number. What that
tells us is that this equation

has no roots. So B squared well,
there's 1X, so B is equal to 1,

so that's one squared minus four
times a while a is one 'cause

there's One X squared and C is

4. Well, that's 1 - 16 is minus
15. It's less than zero. There

are no roots for this equation,
so it doesn't meet the X axis.

Let's have a look at a picture
for that. There's The X Axis.

Why access I'd like to be able
to fix a point on the curve and

I can buy taking X is zero.
Those two terms would be 0, so

that gives me Y equals 4.

So that's a point on the curve,
but it's always going to be

above the X axis, and when
because X squared is positive.

With The X is positive or
negative, X squared is always

going to be positive, so it's
always going to be that way up

and it's going to look something

like that. So we want the
area between this curve, the X

Axis and the ordinates X equals
1 and X equals 3.

X equals 1, let's say is there,
X equals 3, let's say is there.

That's the area that we're
after. No real problems with

that area, so let's do the
calculation. The area is equal

to the integral between one and
three of X squared plus X +4

with respect to X.

Integrating each of these at one
to the index.

And divide by the new index.

Same again there and then four
is 4X. When we integrate it

and this is to be evaluated
between one and three.

So let's just turn this over.
Remember, the area that we're

wanting. Is equal to. This
is what we have to work

out X cubed over 3 plus
X squared over 2.

Plus 4X that's to be evaluated
between one and three.

So we put the upper limiting
first, so when X is equal to 3,

three cubed is equal to 27 over

3. Plus X squared when X
is equal to three, that's nine

over 2 + 4 X when X
is equal to three, that's 12.

Now we put the lower limiting.

So when X is One X cubed is
one 1 / 3 + 1, X is one

that's X squared is one and
divided by two, and when X is

one that's plus 4.

Now, some of these we can
workout quite easily, but let's

do the thirds bit first with 27
over 3 takeaway. A third is 26

over three we have 9 over 2
takeaway one over 2. So that's

plus eight over 2, and we've 12
takeaway 4, so that's +8.

Will cancel down to their.

If I do freeze into 26, eight
threes are 24 and two over slots

2/3 + 4 and 80s twelve so
altogether that's an area of 20

and two. Thirds

Now.
We can workout the area between

a curve, the X axis and some
given values of X the ordinance.

We can extend this technique
though to working out the area

contained between two curves.
Let's have a look at that

supposing. You've got the curve
Y equals X times 3 minus X,

and we've got the straight line
Y equals X.

We say what's the area that's
contained between these two

curves? Well, let's have a
look at a picture just to

see what that looks like.

If we set Y equals 0 again
and that will give us the values

of X where this curve crosses
the X axis clearly does so there

when X is zero and also when
X is equal to 3.

We've also got X times by minus.
X gives us minus X squared, so

for a quadratic it's going to be
an upside down U. It's going to

do that, just missed that point.
Let's make it bigger so it goes

through it. The line Y equals
XY. That's a straight line going

through there. And so the line
crosses the curve at these two

points. Let me call that one P
and this one is oh the origin.

And the error that I'm
interested in is this one. In

here the area core between the

two curves. Well before I can
find that area, I really need to

know what's this point P. Where
do these two curves cross? Well,

they will cross when their Y
values are equal, so they will

intersect or meet when X times
by 3 minus X is equal to X.

Multiply out the bracket
3X minus X squared is

equal to X.

Take X away from both sides. 2X
minus X squared is then zero.

This is a quadratic, and it
factorizes. There's a common

factor of X in each term, so
we can take that out.

And we're left with
two minus X equals 0.

Either the X can be 0 or the two
minus X com 0, so therefore X is

0 or X equals 2, so this is
where X is equal to 0. Here at

the origin, when X is zero, Y is
00. In there we have not times

by three is nothing. If we put
zero in there, we get nothing.

Or why is equal to? Well, This
is why equals X, so presumably

why must be equal to two?

And let's just check it over
here. If we put X equals 2 with

2 * 3 - 2, three minus two is
one times by two is 2. So it's

this here. That's the point P.

So let's just put that into
their. Now this is the area that

we want. So we can look at it
as finding the area under the

curve between North and two and
finding the area under the

straight line. And then
subtracting them in order to end

up with that.

So let's do that. Let's
first of all, find the

area under the curve.

Equals now this will
be the integral between

North and two of
our curve X times

by 3 minus X.

With respect to
X.

So let's first of all multiply
out the bracket X times by three

is 3 XX times Y minus X is
minus X squared X.

Do the integration integral of X
is X squared over 2?

And we need to multiply it by
the three that we've got there.

And then the integral of X
squared is going to be X cubed

over three that to be evaluated
between North and two.

So I put the two in first.

So we have 3.

Times 2 to 4.

Divided by two.

Minus X cubed over 3.

X is 2, so that's eight
over 3.

Minus and when we put the zero
in, X squared is zero and X

cubed is 0, so that second
bracket is 0.

Equals what we can cancel it

to there. 326 minus threes into
eight goals two and there's two

over, so that's 2/3, so the area
that we've got is 6 - 2

and 2/3, which is 3 1/3.

So that's the area under the
curve. What now we need is the

area under Y equals X.

So this will be equal to the
integral between Norton two of X

DX. Equals.

Integral of X is just X squared

over 2. Between North and
two, substituting the two 2

two to four over 2.

Minus zero in and the X squared
over 2 gives gives us 0.

And so we've just got two
choosing toward those two,

nothing there. So here we've got
the area under the curve 3 1/3

here. We've got the area under Y

equals X2. And so the area
between them.

Let's look at that

positions. We've calculated this
area here under the curve.

That's the one that is free and
the 3rd, and we've calculated

the area of this bit of.

Triangle that's here
underneath Y equals X, and the

answer that is 2 and So what
we're looking for is the

difference between the big
area and the triangle. So

that's 3 1/3 - 2 gives us 1
1/3.

So we've seen how we can extend
the technique of finding the

area under a curve.

Between the X axis on between
two given ordinance to finding

the area between two given
curves. Let's take another

example. This time, let's use
the function Y equals sign X. We

better define a range of values
of X, so I'm going to take

X to be between North and pie.

Let's say I want the area that's
cut off from that by the line

Y equals one over Route 2. So
a little sketch as always.

We're finding these areas.
It's important to know

where they are.

So why equals sign X
between North and pie?

So it's going to look.

Like that? Naspi Nazira Why is
equal to one over Route 2? Well,

that's a straight line.

Somewhere, Let's say across.

There, so this is the area that

we're after. Area between the
line and that curve. What we

need to know. What are these
values of X?

Well, why is equal to one over
Route 2 and Y is equal to sign

X, so one over Route 2 is equal
to sign X where they meet?

So the values of X will be the
solutions to this equation.

Well, one of the values we do
know is that when sign X equals

one over route 2X must be equal
to π over 4.

I working in radians, so I've
got to give the answer in

radians, but the equivalent
answer in degrees is 45.

So that's that one there and the
one there is 3 Pi over 4.

So now I can workout what it is
I've got to do I can find the

area under the sign curve
between these two values. I can

find the area of this rectangle
and take it away. That will

leave me that area there. Just
let me put these answers in.

I found their and there, so it's
first of all find the area under

this curve. So the area.

Under The
curve is equal to

the integral between pie
by four and three

Pi 4. Of
sign XDX equal so

I have to integrate

sign X. Well, the
integral of sine X is minus Cos

X. To evaluate that
between pie by four and

three Pi 4.

So we put in the upper limit
first minus the cause of three π

by 4. That's the first bracket.

Minus minus the cause of
Π by 4.

Now cause of three π by 4 is
minus one over Route 2 and that

minus sign gives me plus one
over Route 2. The cause of Π by

4 is one over Route 2, minus
minus gives me plus one over

Route 2, so that is 2 over Route
2, which is just Route 2.

So I now know this area here.

I need to do is calculate the
area of this rectangle.

Then I can take it away from
that. Well, it is a rectangle.

Its height is one over Route 2.

So area.

Equals one over Route 2. That's
the height of that rectangle.

Times by this base it's.

Width, well, it runs from Π by
4, three π by 4, so the

difference is 2π by 4. In other
words, pie by two, so the area

of the rectangle is π over 2
times Route 2.

So the total area that I want
is the difference between these

two π over 2 Route 2 and

Route 2. In the area.

Equals Route 2 -
π over 2 Route

2. Put all that over a common
denominator of two Route 2,

which makes this 4.

Minus pie.
Leave that as that. That's the

area that's contained.

In there, between the straight
line and the curve.

I want now just to
go back to the example

where we were looking at
the area that was between

these two curves.

If you remember we

had. This sketch.

The curve.

So that
was the

picture that

we had.
And we calculated the area

between the two curves here.

By finding the area underneath.

X times by 3 minus X.

Then the area underneath
the line Y equals X and

taking them away.

Another way of looking at this
might be to go back to 1st

principles if you remember when
we did that, we took little

rectangular strips which were
roughly of height Y and

thickness Delta X. Well,
couldn't we do the same here,

have little rectangular strips
for that. Not quite rectangles,

but they are thin strips and
their why bit, so to speak would

be the difference between these
two values of Y here.

So if we were to call this, why
one, let's say, and this one Y

2, then might not the area that
we're looking for the area

between these two curves Y1 and

Y2? Between these ordinates Here
Let's call them A&B.

Might not we be able to work it

out? By doing that.

In other words, take the upper
curve and call it why one type

the lower curve and call it Y 2.

Subtract them and then integrate
between the required limits.

Well, let's see if this will
give us the same answer as we

had in the previous case.

So our limits
here are not 2.

And why one is this one?

Let's multiply it out first, X
times by three, is 3X and X

times Y minus X is minus X

squared. Take away why two? So
we're taking away X and

integrating with respect to X.
So if 2X minus X, sorry,

3X minus X. That gives us
2X minus X squared to be

evaluated between Norton two
with respect to X. So let's

carry out this integration.

The integral of X is X squared
over 2, so that's two times X

squared over 2. The integral
of X squared is X cubed over

three between North and two. I
can cancel it to their and

their substituting.

2 Twos are 4.

Minus.

X cube when X is 2, that's eight
over 3 minus and when I put zero

in that second bracket is just 0

equals. 4

minus. 2 and 2/3 which
just gives me one and third

units of area, which is exactly
what we had last time. So this.

Gives us a convenient formula.

Being able to workout the area
that is caught between two

curves, we call the upper curve.
Why won the lower curve Y two

and we subtract them and then
integrate between the ordinance

of the points of intersection?