WEBVTT mathematics/ap-calculus-ab/hovasapian
00:00:00.000 --> 00:00:03.700
Hello, and welcome back to www.educator.com, welcome back to AP Calculus.
00:00:03.700 --> 00:00:07.100
Today, we are going to be discussing the areas between curves.
00:00:07.100 --> 00:00:10.900
Let us jump right on in.
00:00:10.900 --> 00:00:21.100
Let us look at the following, let me go ahead and work in blue today.
00:00:21.100 --> 00:00:28.100
Let us look at the following situation.
00:00:28.100 --> 00:00:34.100
We have ourselves a little bit of graph here.
00:00:34.100 --> 00:00:38.800
We have one graph, let us go ahead and call this one f(x).
00:00:38.800 --> 00:00:43.700
And then, we have another one let us say something like that.
00:00:43.700 --> 00:00:46.900
We will just go ahead and call this one g(x).
00:00:46.900 --> 00:00:50.900
Let us pick an interval from a to b.
00:00:50.900 --> 00:00:57.600
We have got this and that.
00:00:57.600 --> 00:01:15.000
The question is how can I find the area between f and g?
00:01:15.000 --> 00:01:18.800
How can I find this area right here?
00:01:18.800 --> 00:01:26.300
It is exactly what you think.
00:01:26.300 --> 00:01:33.000
The area under f(x) is just this, it is just the integral.
00:01:33.000 --> 00:01:37.700
As you know, from a to b of f(x) d(x).
00:01:37.700 --> 00:01:39.500
Let me go ahead and do that in red.
00:01:39.500 --> 00:01:46.700
That takes care of everything underneath f(x).
00:01:46.700 --> 00:01:48.500
Let me go back to blue.
00:01:48.500 --> 00:01:59.500
The area under g(x), that is just the integral from a to b of g(x) dx.
00:01:59.500 --> 00:02:05.900
I will go ahead and do this one in black, that is this region right here, underneath g(x).
00:02:05.900 --> 00:02:12.900
You know already for years and years and years now, this area is going to be the area of f - the area of g.
00:02:12.900 --> 00:02:19.900
And that will give me the area that I'm interested in, this portion right here.
00:02:19.900 --> 00:02:21.900
That is it, it is nice and simple.
00:02:21.900 --> 00:02:27.700
The area above - the function of the area below, that is all.
00:02:27.700 --> 00:02:35.500
Let me go back blue.
00:02:35.500 --> 00:02:47.700
Since f(x) is greater than or equal to g(x) on all of the interval ab,
00:02:47.700 --> 00:03:10.300
the area between is the integral from a to b of f(x) d(x) - the integral from a to b of g(x) dx.
00:03:10.300 --> 00:03:15.000
Nice and straightforward.
00:03:15.000 --> 00:03:18.800
That is going to be equal to, I will go ahead and put those integrals together.
00:03:18.800 --> 00:03:26.900
It is going to be the integral from a to b of f - g dx, that is it.
00:03:26.900 --> 00:03:30.100
Generally, you are going to express it like this, f - the g.
00:03:30.100 --> 00:03:38.800
And then, as a practical matter, when you solve the integral, you are going to separate it out as integral of f - the integral of g.
00:03:38.800 --> 00:03:44.600
This is when over the entire interval f is above g.
00:03:44.600 --> 00:03:50.200
Upper function - the lower function, that is essentially how it goes.
00:03:50.200 --> 00:04:02.900
Area equals the integral from a to b, we will write upper, the upper function - the lower function d(x)
00:04:02.900 --> 00:04:07.800
or whatever variable you happen to be integrating with respect to.
00:04:07.800 --> 00:04:17.100
What if we have the following situation?
00:04:17.100 --> 00:04:23.300
Where we are over a given interval that actually cross.
00:04:23.300 --> 00:04:33.500
What if we have the following situation?
00:04:33.500 --> 00:04:41.600
We have got, we will call this one f(x) and then we can do something like this.
00:04:41.600 --> 00:04:46.200
We will call this one g(x).
00:04:46.200 --> 00:04:54.300
Let us go ahead and call this a, we have got that one.
00:04:54.300 --> 00:04:58.400
We will just put b over here, something like that.
00:04:58.400 --> 00:05:06.600
Now where they meet, I’m going to go ahead and call this c.
00:05:06.600 --> 00:05:19.000
The area between the curves is going to be this area right here, this area and that area.
00:05:19.000 --> 00:05:28.600
However, from a to c, it is f that is on top and it is g that is lower.
00:05:28.600 --> 00:05:35.900
But from c all the way to b, now I have g is on top and f is the one that is lower,
00:05:35.900 --> 00:05:40.200
where we find the area by taking the upper function - the lower function.
00:05:40.200 --> 00:05:42.600
You have to split this up into two integrals.
00:05:42.600 --> 00:05:47.500
You have to integrate from a to c doing f – g.
00:05:47.500 --> 00:05:52.700
You have to do the integral from c to b of g – f.
00:05:52.700 --> 00:05:54.400
That is all, it is that simple.
00:05:54.400 --> 00:05:57.000
You just add the integrals together, you just have to separate them.
00:05:57.000 --> 00:06:01.100
You have to find out where they meet, to find that x value.
00:06:01.100 --> 00:06:07.800
How do you do that, you set the two functions equal to each other and you solve for x.
00:06:07.800 --> 00:06:10.700
Let us go ahead and write this out.
00:06:10.700 --> 00:06:19.100
On the interval ac, f is greater than or equal to g.
00:06:19.100 --> 00:06:29.500
The area is actually equal to the integral from a to c of f - g dx.
00:06:29.500 --> 00:06:40.300
Of course from c to b, here to here, in this particular case, it is g that is greater than or equal to f.
00:06:40.300 --> 00:06:52.700
The area = the area from c to b of g – f dx, that is it upper – lower.
00:06:52.700 --> 00:07:02.900
That is all you really have to know.
00:07:02.900 --> 00:07:36.300
Whichever graph is higher up on a given interval is the first entry.
00:07:36.300 --> 00:07:47.900
The first entry is the upper graph, entry in the integrand.
00:07:47.900 --> 00:07:53.700
The integrand is the thing that is underneath the integral sign.
00:07:53.700 --> 00:08:10.200
Because we want the integrand to be positive, because we are dealing with areas.
00:08:10.200 --> 00:08:17.800
That is it, nice and simple.
00:08:17.800 --> 00:08:22.100
A little bit of foray into some notation.
00:08:22.100 --> 00:08:45.800
Is it possible to write the integral from a to c of – g dx + the integral from c to b of g - f dx.
00:08:45.800 --> 00:08:47.900
The integral we just took, we split it up.
00:08:47.900 --> 00:08:56.600
Is it possible to write that as a single symbol, as a single integral?
00:08:56.600 --> 00:09:06.700
The answer is yes, the symbol for it is the following.
00:09:06.700 --> 00:09:25.300
For f(x) and g(x) both continuous on ab, both continuous on the interval ab,
00:09:25.300 --> 00:09:43.300
the area between the curves on of the interval is, the symbolism we use is the absolute value symbolism.
00:09:43.300 --> 00:09:49.400
a to b absolute value of f - g dx.
00:09:49.400 --> 00:09:50.800
The truth is you can actually do it in either way.
00:09:50.800 --> 00:09:52.500
You can do f – g, g – f.
00:09:52.500 --> 00:09:58.300
The absolute value sign, this is this.
00:09:58.300 --> 00:10:00.900
I will show you why in just a minute.
00:10:00.900 --> 00:10:07.300
This is the actual statement of how we find the area between two curves.
00:10:07.300 --> 00:10:12.000
If you are given the curve f, given the curve g, you take the integral from a to b,
00:10:12.000 --> 00:10:16.000
whatever interval you are dealing with of the absolute value of f(g).
00:10:16.000 --> 00:10:20.700
The absolute value of f(g) is actually telling you to do something.
00:10:20.700 --> 00:10:26.200
When we solve these, we do not use this, the symbolism asks us to do something.
00:10:26.200 --> 00:10:30.800
What the symbol is telling us is to actually separate it out, here is how.
00:10:30.800 --> 00:10:36.300
Let us revisit absolute value.
00:10:36.300 --> 00:10:45.600
I find that kids, that absolute value is one of the things that kids know how to do
00:10:45.600 --> 00:10:51.400
but they do cannot really wrap their minds around what an absolute value is saying.
00:10:51.400 --> 00:10:55.800
Let us revisit it, it is always good to revisit it a couple of times.
00:10:55.800 --> 00:11:03.000
For some odd reason, absolute value always is a little, people are not quite sure how exactly to go about it.
00:11:03.000 --> 00:11:07.700
We will discuss it now, let us revisit absolute value.
00:11:07.700 --> 00:11:14.000
The definition of absolute value is the following.
00:11:14.000 --> 00:11:23.300
The absolute value of a, whatever is between that absolute value symbols is equal to the following.
00:11:23.300 --> 00:11:33.700
It is equal to just a, if a, the thing in between the absolute value size is bigger than 0.
00:11:33.700 --> 00:11:42.400
But it is equal to –a, if a, what is in between the absolute value sign is less than 0.
00:11:42.400 --> 00:11:45.900
If whatever is in between the absolute value signs is bigger than 0,
00:11:45.900 --> 00:11:49.400
it just means drop the absolute value signs, take the number as is.
00:11:49.400 --> 00:11:54.100
In other words, what is the absolute value of 5, it is just 5.
00:11:54.100 --> 00:12:06.300
If what is in between the absolute value signs is negative, then the value of the absolute value sign is negative of a.
00:12:06.300 --> 00:12:17.600
In other words, if I had the absolute value of -5, the definition says take - -5, that is my answer which I know is 5.
00:12:17.600 --> 00:12:19.000
You do it automatically.
00:12:19.000 --> 00:12:23.200
But when you see it in the context of something like this in integral, for numbers it is fine.
00:12:23.200 --> 00:12:27.000
You know that the absolute value of -5 is 5.
00:12:27.000 --> 00:12:31.600
What is the absolute value of f – g?
00:12:31.600 --> 00:12:40.400
Let us look at f – g.
00:12:40.400 --> 00:12:45.900
Once again, this is what is important, this definition right here.
00:12:45.900 --> 00:12:48.200
If the thing between the absolute value sign, the whole thing,
00:12:48.200 --> 00:12:56.600
whether it is a number, a letter, or an expression, if it is bigger than 0, then we just take expression as is.
00:12:56.600 --> 00:13:09.400
If it is less than 0 then we take the negative of the expression as is.
00:13:09.400 --> 00:13:18.100
Now we have f – g, just like our definition.
00:13:18.100 --> 00:13:21.000
I will actually do this in reverse.
00:13:21.000 --> 00:13:30.600
If f – g, f – g, what is the absolute value of f – g?
00:13:30.600 --> 00:13:42.500
If what is in between the absolute value signs, if f - g is less than 0 which is equivalent to saying f is less than g,
00:13:42.500 --> 00:13:53.800
then the absolute value of f - g is just plain old f – g.
00:13:53.800 --> 00:14:07.700
If f - g is less than 0 which is equivalent to saying, sorry I have this backwards,
00:14:07.700 --> 00:14:12.300
if f - g is greater than 0, the thing underneath the absolute value signs
00:14:12.300 --> 00:14:16.700
which is equivalent to saying f is greater than g, I just move this g over here.
00:14:16.700 --> 00:14:22.200
If f is greater than g and the absolute of f - g is just f – g.
00:14:22.200 --> 00:14:32.100
If f - g is less than 0 which is equivalent to saying that f is less than g or g is bigger than f,
00:14:32.100 --> 00:14:38.800
then the absolute value of f - g is - f – g.
00:14:38.800 --> 00:14:42.600
What is f - f – g, it is g – f, that is it.
00:14:42.600 --> 00:14:44.400
That is all the absolute value symbol is saying.
00:14:44.400 --> 00:14:47.800
In the case of an expression, you are going to negate.
00:14:47.800 --> 00:14:53.000
If that expression is less than 0 then you negate the entire expression.
00:14:53.000 --> 00:14:57.400
When you negate a difference, the term is flipped and you are getting that.
00:14:57.400 --> 00:15:10.400
This is just a symbolic way of representing what it is it that we did, in terms of two separate integrals.
00:15:10.400 --> 00:15:24.600
The absolute value symbol accounts for all cases.
00:15:24.600 --> 00:15:38.500
It is just a shorthand notation, all cases all on the interval ab.
00:15:38.500 --> 00:15:54.000
When we actually do the integration, in practice, we still just separate
00:15:54.000 --> 00:16:10.800
the area of calculation into two or more areas, depending on how many times it crosses.
00:16:10.800 --> 00:16:23.900
In each case, we always take the upper – lower, upper – lower, upper – lower.
00:16:23.900 --> 00:16:38.900
The symbol, the integral from a to b, the absolute of f – g dx, it just gives us a compact notation.
00:16:38.900 --> 00:16:47.000
It actually tells us that if f - g ever drops below 0, I have to switch those.
00:16:47.000 --> 00:16:49.900
That is what the absolute value symbol is telling me.
00:16:49.900 --> 00:17:07.500
It just gives us a compact notation, there we go.
00:17:07.500 --> 00:17:34.200
The area of a region between two curves is the integral from a to b of the upper function - the lower function.
00:17:34.200 --> 00:17:41.400
It is probably the best way to think about it, dx, that is all.
00:17:41.400 --> 00:17:46.000
What if we are given functions not in terms of x but in terms of y?
00:17:46.000 --> 00:17:54.100
Now instead of x being the independent variable, what if we are given something like this?
00:17:54.100 --> 00:18:04.000
What if we are given functions of y?
00:18:04.000 --> 00:18:21.700
For example, x = y², let us say the other function is x = y – 2/ 2.
00:18:21.700 --> 00:18:26.900
They are going to ask, what is the area between these curves?
00:18:26.900 --> 00:18:29.800
We are accustomed to seeing y in terms of x.
00:18:29.800 --> 00:18:32.700
Here we have x, in terms of y.
00:18:32.700 --> 00:18:34.800
Now y is the independent variable.
00:18:34.800 --> 00:18:40.600
Whatever y happens to be, we do something to it and we spit out an x.
00:18:40.600 --> 00:18:50.700
Let us graph these two and see what we are dealing with.
00:18:50.700 --> 00:18:57.900
x = y², whenever you flip x and y, the role of x and y, what you have done is actually take the inverse function.
00:18:57.900 --> 00:19:05.500
If I know that my normal x, x², y = x² is my parabola that looks like that.
00:19:05.500 --> 00:19:12.100
My x = y² is my parabola that looks like this.
00:19:12.100 --> 00:19:15.900
It is just moving along the x axis, instead of the y axis.
00:19:15.900 --> 00:19:20.300
Let me erase these little arrows, it is confusing.
00:19:20.300 --> 00:19:26.900
x = y – 2/ 2, let us go ahead and put in the form that we are actually used to seeing it, as far as lines are concerned.
00:19:26.900 --> 00:19:40.200
I'm going to multiply by 2 and you are going to end up getting y is equal to 2x.
00:19:40.200 --> 00:19:44.500
I’m sorry, this should be +2 – 2.
00:19:44.500 --> 00:19:49.200
I multiply by 2, I’m going to move that 2 over, and I get y = 2x – 2.
00:19:49.200 --> 00:19:50.300
It is okay, we can do that.
00:19:50.300 --> 00:19:53.500
We can flip it around, in order to help us graph it.
00:19:53.500 --> 00:19:55.900
I can do the same thing here, if I wanted to.
00:19:55.900 --> 00:20:03.900
This is going to be y is equal to + or -√x which I know is this curve and is this curve.
00:20:03.900 --> 00:20:08.200
That is fine, you can go ahead and do that, if I need to graph it.
00:20:08.200 --> 00:20:13.400
Now y = 2x – 2, let me come down and mark -2.
00:20:13.400 --> 00:20:21.200
It is up 2/ 1, up 2/ 1, I’m going to get basically a line that looks like that.
00:20:21.200 --> 00:20:29.300
The area that I'm interested in is this area.
00:20:29.300 --> 00:20:40.300
Notice what we have here, we have an upper function, we have a lower function.
00:20:40.300 --> 00:20:41.500
There is a bit of an issue here.
00:20:41.500 --> 00:20:50.200
It is like from 0 to whatever this point happens to be, this is the upper function and this is the lower function.
00:20:50.200 --> 00:21:00.100
But from here to this x value, this is my upper function, my line is my lower function.
00:21:00.100 --> 00:21:08.500
If I were to integrate this along x, in other words, make a little rectangle like that and add this way,
00:21:08.500 --> 00:21:17.500
from your perspective, this way, moving in this direction, I have to break this up into two integrals.
00:21:17.500 --> 00:21:21.800
In this case, it is actually better to integrate along y.
00:21:21.800 --> 00:21:26.100
In other words, along the axis of the independent variable.
00:21:26.100 --> 00:21:28.500
Here the independent variable is y.
00:21:28.500 --> 00:21:31.400
It is best to integrate along y.
00:21:31.400 --> 00:21:36.600
When you are given a function of x, it is best to integrate along x.
00:21:36.600 --> 00:21:41.800
What happens here is the following.
00:21:41.800 --> 00:22:01.100
Whenever you are given functions in terms of y, the formula becomes,
00:22:01.100 --> 00:22:04.800
the area is equal to the integral from a to b.
00:22:04.800 --> 00:22:13.600
This time it is going to be the right function - the left function.
00:22:13.600 --> 00:22:17.300
Before we have upper – lower, now we have right – left.
00:22:17.300 --> 00:22:23.400
And of course, we are going to be integrating along the y axis, so it is dy.
00:22:23.400 --> 00:22:26.000
But what are a and b?
00:22:26.000 --> 00:22:34.200
If we are integrating along y, they are the points on the y axis.
00:22:34.200 --> 00:22:36.500
Let us go ahead and write it out.
00:22:36.500 --> 00:22:47.800
But what are a and b?
00:22:47.800 --> 00:23:07.300
We want the area between the curves.
00:23:07.300 --> 00:23:22.900
a and b are just the y values of the points where the two graphs meet.
00:23:22.900 --> 00:23:32.400
Points are just the y values of the points where f(y) = g(y).
00:23:32.400 --> 00:23:35.600
In other words, you do what the same thing that you do any other time.
00:23:35.600 --> 00:23:38.800
You set the two graphs equal to each other, you see where they meet.
00:23:38.800 --> 00:23:45.500
But now instead of taking the x values, you take the y values because we are integrating with respect to y.
00:23:45.500 --> 00:23:54.800
Let us go ahead and do this problem.
00:23:54.800 --> 00:23:57.300
Let us do this problem.
00:23:57.300 --> 00:24:17.100
We had this graph, let me check something real quickly here.
00:24:17.100 --> 00:24:29.000
y = 2x – 2, that is fine.
00:24:29.000 --> 00:24:36.000
We hade this graph where we have this and we have this line.
00:24:36.000 --> 00:24:47.300
This was our x = y² and this one we had y = x – 2.
00:24:47.300 --> 00:24:56.900
When you set them equal to each other, you can do it, you got x = y².
00:24:56.900 --> 00:25:01.800
Let me do this in red.
00:25:01.800 --> 00:25:08.800
x = y², and then we have this other version of it, in order to make it easier for us to actually graph.
00:25:08.800 --> 00:25:18.900
y = x – 2, you can go ahead and put the x - 2 in here.
00:25:18.900 --> 00:25:21.800
I should do it this way.
00:25:21.800 --> 00:25:30.300
I have x = y² and this becomes y + 2.
00:25:30.300 --> 00:25:38.100
I think I’m getting ahead of myself, let me go back to blue.
00:25:38.100 --> 00:25:40.700
Let me rewrite down my functions properly.
00:25:40.700 --> 00:25:42.500
This is y = 2x – 2.
00:25:42.500 --> 00:25:45.100
This function = x = y².
00:25:45.100 --> 00:25:50.900
We are looking for the area that is contained here.
00:25:50.900 --> 00:25:56.100
What I'm going to do is I'm going to find that point and that point.
00:25:56.100 --> 00:26:00.400
I’m going to find the y values of that point which are here and here.
00:26:00.400 --> 00:26:02.800
That is going to be my a and that is going to be my b.
00:26:02.800 --> 00:26:07.500
That is what is going on here because we are going to be integrating along the y axis now,
00:26:07.500 --> 00:26:12.400
taking the right function which is this one.
00:26:12.400 --> 00:26:17.400
This is the left function.
00:26:17.400 --> 00:26:20.000
Let us go ahead and see what we were dealing with.
00:26:20.000 --> 00:26:24.900
We have got x = y².
00:26:24.900 --> 00:26:32.700
I have y = 2x – 2, I got y + 2 = 2x.
00:26:32.700 --> 00:26:38.300
I have got x = y + 2/ 2.
00:26:38.300 --> 00:26:42.900
x = y², x = y + 2/ 2.
00:26:42.900 --> 00:26:48.500
I got y² = y + 2/ 2.
00:26:48.500 --> 00:26:51.900
When I solve this, I’m going to get two values of y.
00:26:51.900 --> 00:26:54.500
Move this over, turn it into a quadratic.
00:26:54.500 --> 00:26:57.700
I’m going to get two values for y.
00:26:57.700 --> 00:27:02.000
The y values that I get, those are my a and b.
00:27:02.000 --> 00:27:04.400
That is exactly what is happening here.
00:27:04.400 --> 00:27:13.600
Let us take a look at this, I went ahead and I use mathematical software to go ahead and graph this for me.
00:27:13.600 --> 00:27:19.200
You can use your calculator, any kind of online software that you want.
00:27:19.200 --> 00:27:24.900
In this particular case, I use something called www.desmos.com.
00:27:24.900 --> 00:27:29.600
It is available the minute you pull it up, you click this big red button that says launch the calculator.
00:27:29.600 --> 00:27:33.900
This screen comes up and you can actually do your graphs.
00:27:33.900 --> 00:27:36.800
That is what I use for all of the pictures that I generate here.
00:27:36.800 --> 00:27:41.400
When I do this, x = y² and I just wrote it as y = 2x – 2.
00:27:41.400 --> 00:27:49.000
When I graph this, I end up finding this point and this point.
00:27:49.000 --> 00:27:50.500
Let me go ahead and go back to blue.
00:27:50.500 --> 00:28:02.200
My y values are 1.281 and -0.781, that is here and here.
00:28:02.200 --> 00:28:17.800
The area equals -0.781, the integral from -0.781 negative to 1.281 of the right function
00:28:17.800 --> 00:28:22.200
- the left function, expressed in terms of y.
00:28:22.200 --> 00:28:30.900
That was going to be the right function, this one, in terms of y.
00:28:30.900 --> 00:28:33.500
Here we have y = 2x – 2.
00:28:33.500 --> 00:28:39.000
It is going to be y + 2/ 2 is equal to x.
00:28:39.000 --> 00:28:58.500
It is going to be y + 2/ 2 - the left function.
00:28:58.500 --> 00:29:08.100
2 - y² dy, that is it.
00:29:08.100 --> 00:29:17.900
Because I'm integrating vertically like this, taking a little horizontal strips that way,
00:29:17.900 --> 00:29:28.100
it is going to be integral from this point to this point, that is my 0.781 negative to 1.281 positive of the right function,
00:29:28.100 --> 00:29:40.300
expressed in terms of y which is y + 2/ 2 - the left function, expressed in terms of y y² dy.
00:29:40.300 --> 00:29:43.500
In the problems that we are going to do which is going to be the next lesson,
00:29:43.500 --> 00:29:56.700
when you are given a set of functions, you are just going to be given functions randomly.
00:29:56.700 --> 00:30:02.800
Sometimes they are going to be in terms of x, sometimes they are going to be in terms of y, you do not know.
00:30:02.800 --> 00:30:26.600
When you are given a set of functions and ask to find areas between regions,
00:30:26.600 --> 00:30:30.100
you will have to decide what is going to be the best integration.
00:30:30.100 --> 00:30:36.200
I’m going to integrate this along the x axis and I’m going to integrate along the y axis.
00:30:36.200 --> 00:30:37.200
What is going to be the easiest?
00:30:37.200 --> 00:30:40.700
Sometimes you can do both, but one of them is longer than the other.
00:30:40.700 --> 00:30:46.800
Sometimes it is best only to do one, either along y or along x, you get to decide.
00:30:46.800 --> 00:30:52.600
When is it not necessarily the form of the function, the only thing we have done is say that,
00:30:52.600 --> 00:30:59.500
if you are going to be integrating along x, you are going to be taking the upper function - the lower function and integrate it.
00:30:59.500 --> 00:31:04.800
If you are going to be integrating along y, you are going to be taking the right function - the left function.
00:31:04.800 --> 00:31:12.300
You, yourself, have to decide which one is best and decide how to manipulate the situation, according to what is best.
00:31:12.300 --> 00:31:15.800
It is not necessarily some algorithm or recipe that you want to follow.
00:31:15.800 --> 00:31:19.000
You want to take a look at the situation and decide what is best.
00:31:19.000 --> 00:31:26.000
In this particular case, it was best to just go this way because you have a left and a right function, and a left function.
00:31:26.000 --> 00:31:30.100
You have a series of rectangles that touch both functions.
00:31:30.100 --> 00:31:37.400
If you were to decide to do this with respect to x, which you can, you have to break it up here.
00:31:37.400 --> 00:31:43.200
You have to integrate from here to here, this being your upper function, this being your lower.
00:31:43.200 --> 00:31:46.600
And then, you have to integrate from here to here.
00:31:46.600 --> 00:31:50.400
This being your upper function, this being your lower function.
00:31:50.400 --> 00:31:52.200
We are going to do that in just a second.
00:31:52.200 --> 00:31:58.900
You will decide what is best.
00:31:58.900 --> 00:32:15.100
You will decide the best way to integrate.
00:32:15.100 --> 00:32:17.500
Let us go ahead and actually do it the other way.
00:32:17.500 --> 00:32:23.600
What will this integral look like, if we decided to integrate along the x axis?
00:32:23.600 --> 00:32:31.700
First, now we are going to integrate along the x axis.
00:32:31.700 --> 00:32:37.800
It is going to be dx, if it is going to be dx, we need the functions to be expressed in terms of x.
00:32:37.800 --> 00:32:46.300
We have got y = √x, y = -√x.
00:32:46.300 --> 00:32:48.000
We already have this one, in terms of y.
00:32:48.000 --> 00:32:51.200
y = 2x – 2.
00:32:51.200 --> 00:32:56.400
However, we have to break it up.
00:32:56.400 --> 00:33:02.500
Our first integral, from here to here, we are going to have rectangles.
00:33:02.500 --> 00:33:06.000
This is going to be one representative rectangle for that area.
00:33:06.000 --> 00:33:09.200
This is going to be a representative rectangle for that area.
00:33:09.200 --> 00:33:12.700
We are going to need the top function - the bottom function.
00:33:12.700 --> 00:33:22.800
The area is going to be the integral from 0 to 0.61 because that is the x value of where they meet.
00:33:22.800 --> 00:33:26.900
From here to here, upper – lower.
00:33:26.900 --> 00:33:35.900
It is going to be √x - - √x dx.
00:33:35.900 --> 00:33:38.800
I’m going to add the second area.
00:33:38.800 --> 00:34:00.300
It is going to be 0.61 to 1.64, upper function is √x - the lower function which is 2x – 2.
00:34:00.300 --> 00:34:05.800
This one is only slightly longer; not more complicated, it is just slightly longer.
00:34:05.800 --> 00:34:09.300
Again, you can do it both ways.
00:34:09.300 --> 00:34:11.000
Ultimately, it comes down to a personal choice.
00:34:11.000 --> 00:34:17.400
You have noticed with calculus that as the problems become more complicated, there are more ways of approaching it.
00:34:17.400 --> 00:34:20.300
You get to decide what is the best integration.
00:34:20.300 --> 00:34:26.100
Do not feel like you have to do one or the other, it is whatever you feel comfortable with, whatever your eye sees.
00:34:26.100 --> 00:34:32.700
If you prefer to stick with dx and it is not too complicated, great, go ahead and stick with dx.
00:34:32.700 --> 00:34:35.500
But sometimes you are not going to be able to do it with respect to x.
00:34:35.500 --> 00:34:37.500
We will get into more of those problems later on.
00:34:37.500 --> 00:34:45.300
Sometimes, you have no choice but to do it along the y axis, because the x integration is just going to be too complicated.
00:34:45.300 --> 00:34:49.800
That is all, thank you so much for joining us here at www.educator.com.
00:34:49.800 --> 00:34:54.500
The next lesson is going to be example problems for areas between curves.
00:34:54.500 --> 00:34:56.000
Take care, see you next time, bye.