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. 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?