00:00Hi everybody. So in this
00:02lesson we're going to get
00:02the area between two curves.
00:05Now this looks horrible, but
00:09I'm actually going to show
00:10you that it's not as
00:11bad as one would think.
00:14First thing I want to
00:15look at, I'll show you
00:16is what the area between
00:19two curves means. Well, if
00:23this is a curve f
00:23of x and this is
00:24a curve g of x,
00:25the area between the curves
00:27is
00:28this area here is enclosed
00:31area, that orange area. So
00:32that's the area we're trying
00:33to find. Now, I've deliberately
00:35put some of it below
00:37the x -axis and some
00:38of it above the x
00:40-axis. And we're normally, I'd
00:43say, that's going to cause
00:43you lots of problems. When
00:45you're doing the area between
00:46two curves, it actually doesn't
00:48cause you any problems. You
00:50can do it quite easily
00:52using this formula. Now, the
00:54formula is not
00:56given in the formula booklet.
00:59So you got to know
01:00what it is, but it's
01:01pretty straightforward. It is just
01:03the integral from A to
01:04B, where A and B
01:05are the intersection points. So
01:07A and B are going
01:07to be, let's say, here,
01:11well, they're the x coordinates,
01:14if you like, of the
01:15intersection points, here. So this
01:19is going to be A.
01:21This would be A. And
01:24this
01:24would be B. And then
01:28you subtract the functions, f
01:30of x minus g of
01:31x. The function on top
01:33is f of x and
01:35the function below is g
01:36of x. So subtract the
01:38one below from the one
01:40above. And then it's in
01:43a modular sign because it's
01:44area, it has to be
01:45positive. Now, even if you
01:47mess up and you do
01:48g of x minus f
01:49of x, you'll actually get
01:51the same answer. It'll just
01:52be
01:52negative and then you can
01:55just make make it positive
01:57and say because area has
01:59to be positive it has
02:00to be positive and you
02:01actually still get the full
02:02marks. Okay, I'm going to
02:04do three things in this
02:04lesson. The first thing is
02:07I'm going to show you
02:09why this formula works because
02:11I think it's actually really
02:12neat how it works. And
02:14then the second thing I'm
02:15going to do is do
02:18an example without a calculator
02:19and then finally
02:20I'm going to do an
02:20example with the calculator. So
02:23firstly, to show you why
02:25this whole thing works. Okay.
02:30Imagine I get the integral
02:33of fx from the integral
02:41of fx from a to
02:42b. Well, what happens is
02:44I'm going to actually put
02:45these little sections here. I'm
02:47going to call this.
02:48A, this B, this C,
02:53so B is just this
02:54bit here. C is this
02:56bit, and D is this
02:58bit. Now, if I get
03:02the integral of FX from
03:11A to B, what happens
03:15is, so imagine these are
03:16areas, the A is an
03:17area B is an area
03:18C is an area D
03:18is an area. What'll happen
03:20is I'll get A, but
03:23it'll be negative. So the
03:25integral of the, the, the
03:26integral of the blue graph,
03:28it'll give me negative A
03:29because it'll add this section
03:32here, but it's underneath the
03:35line. So it's negative. And
03:37it'll give me C. So
03:39it's negative A plus C.
03:42And it'll give me D
03:43plus D.
03:44because these C and D
03:47are underneath the blue curve.
03:51If I integrate from A
03:52to B G of X,
03:58what I'm going to get
03:59is negative A because it's
04:07going to give me this
04:09is enclosed by the green
04:12graph and the X
04:12It's also going to give
04:14me negative B. Same thing,
04:17and these are negative because
04:18they're below the x -axis.
04:21And then it's going to
04:22give me plus D. It's
04:23going to give me a
04:24positive D. And then when
04:28I subtract, when I do
04:29f of x minus g
04:31of x, when I subtract
04:32them, I'm going to get
04:34negative a plus c plus
04:36d minus g of x,
04:39which is negative a.
04:40minus b plus d. When
04:44I subtract this, I get
04:46negative a minus minus a,
04:49so that cancels the a's
04:51cancel. I'm gonna get c,
04:54so I'm gonna get c,
04:55so the a's cancel. Negative
04:58a minus negative a cancel.
05:00c is still there, fine.
05:02Then I'm gonna get my
05:03d's cancel, I have d
05:05minus d, say cancel, and
05:08I finally get
05:08it when I have C
05:10and then I have minus
05:11negative B which is plus
05:14B. So I end up
05:15with C plus B and
05:16look, what is this orange
05:18area? Well, it's C plus
05:21B. Okay, so that's I
05:24actually think that is pretty
05:26cool and it shows you
05:27how even though this is
05:30happening underneath the x -axis
05:31and above the x -axis,
05:32this formula makes things much
05:36easier. Okay, let's go down
05:42and look at two examples.
05:45One without a calculator and
05:46one with a calculator. So
05:48the first thing is first
05:49person is find the area
05:50enclosed by the graphs y
05:51equals x squared minus 3x
05:53and y equals x. What
05:56I want to do is
05:57shade in the region that
06:01I'm looking for. So the
06:02region I'm looking for is
06:03this region
06:04here. This is the region
06:07enclosed by the graphs. Well
06:11enclosed by the two graphs.
06:12This region, fine. Now, in
06:17order to use my formula,
06:18my formula will be the
06:20integral. Let's write it out
06:22here. My formula will be
06:25the integral from A to
06:27B of f of x
06:31minus g of
06:33of x dx. Now f
06:36of x is going to
06:36be my y equals x
06:38because it's above this green
06:40area and my g of
06:41x which I subtract is
06:42going to be my x
06:43squared plus 3x because it's
06:44below the green area. What
06:48I don't have is the
06:51intersection points. So I have
06:52one of them. It's going
06:54to be zero but the
06:55other one I don't actually
06:57have it here. So let's
06:58just draw, we'll draw a
06:59straight line down. I can't
07:01And obviously find it, I
07:04can't just say, oh, well,
07:05it must be four. Although
07:06it looks like four, I
07:07need to calculate it. So
07:10let's first find intersection, intersection
07:17points. Okay, to find the
07:23intersection points, how do I
07:26find intersection points? Well, I
07:29call
07:29weight them, where they equal
07:30is where they intersect. So
07:32I'm going to equate x
07:33squared minus 3x equals x.
07:39And I'm going to solve
07:40this, x squared minus 4x
07:43equals 0. I'm going to
07:44do this quite quickly. Factorize
07:48x into x minus 4
07:50equals 0. Either x equals
07:530 or x minus 4
07:56equals 0.
07:57which is x equals 4.
08:00So my intersection points are
08:01the x coordinates of the
08:02intersection points at least are
08:050 and 4, which is
08:08good. This is what I
08:09need. These are now going
08:10to be my limits. Therefore,
08:13let's write. Therefore, a equals
08:170 and b equals 4.
08:20Now I'm going to integrate.
08:21I'm going to integrate f
08:24of x minus
08:25G of x from a
08:27to b. Now I'm going
08:28to actually leave out this
08:30modulus sign when I do
08:32the integration. I will end
08:34up with a positive answer
08:35if I choose the blue
08:40graph and subtract the red
08:43graph. If, as I said
08:45before, and sometimes you won't
08:46even know which is above
08:48which, well in that case,
08:50you just choose whichever you
08:52want and then when
08:53The final answer is negative,
08:54make a positive. But here
08:56I do know. So it's
08:57gonna be x. That's my
09:00f of x, y equals
09:00x, and I'm gonna subtract
09:02x squared minus three x.
09:06And this is all dx.
09:08But my a and my
09:09b, which I should have
09:10actually written here, my a
09:12and my b, are from
09:15zero to four. Okay. I'm
09:20gonna simplify this
09:21first before integrating it. So
09:23I haven't integrated yet. This
09:24is going to be x
09:26minus minus 3x, which is
09:28actually 4x minus x squared
09:32dx. Let's integrate this, put
09:36it in a square bracket.
09:37The integral of this will
09:38be 4x squared over 2
09:41minus x cubed over 3
09:45from 4 to 0. I
09:48think I'm just going to
09:49simplified over here to the
09:51side. It's going to be
09:532x squared minus x cubed
09:56over 3 from 4 to
10:000. Let's take out my
10:03wrong brackets. This is going
10:05to be 4 squared to
10:0616 times 2 is 32
10:08minus 4 cubed to 64.
10:12So that's 64 over 3.
10:16And then I subtract
10:17Well, I'm subtracting. I'm subbing
10:19it zero here and here.
10:20So it's just gonna be
10:21zero, which is nice. And
10:24then I just have to
10:26figure out what this fraction
10:28is. So 32 minus 64
10:30over three is 96 over
10:33three minus 64 over three,
10:37which equals 32 over three.
10:44So the
10:45Area, this green area, the
10:49area enclosed by the graphs
10:52x squared minus 3x and
10:53y equals x is 32
10:55over 3. So that's fairly,
10:57I mean, it's fairly straightforward
11:00considering how complicated the question
11:03looks. My area is equal
11:07to 32 over 3. Let's
11:09just be clear and write.
11:11Therefore, area
11:13equals 32 over 3. Okay,
11:19fine. Example 2. Find the
11:23area enclosed by the graphs
11:24this and this, right? But
11:26this time we have a
11:27calculator. The working, this is
11:30pretty easy to do with
11:31the calculator to be honest.
11:33But the working I want
11:34you to show has to
11:37be, so the area equals
11:40I want you to show
11:41the
11:41integral. It's going to be
11:43the integral of the blue
11:44one. So again, which, which
11:47actual area are we looking
11:48for? Well, it's this area
11:50here. This green area. So
11:59the area is going to
12:01be the integral. Now the
12:03limits will find them in
12:04a second using the calculator.
12:05It'll be the blue one
12:07minus the red one because
12:09the blue one is about
12:09of the area and the
12:11red one is blue here.
12:12So it'll be negative x
12:15squared plus 3x plus 4
12:18minus, I'm going to put
12:19a brackets, ex, e to
12:22the x minus 3 dx.
12:25If you want to put
12:25a big bracket here, that's
12:28fine. Now what are the
12:30limits? Well, actually I don't
12:33like the look of the
12:34big brackets. Let's put a
12:35big round bracket.
12:37What are the limits here
12:40with the limits are the
12:41intersection points? So I'm going
12:42to take out my calculator
12:43and I'm going to graph
12:45this. So here I'm going
12:48to graph, eat the x,
12:52eat the x minus three,
12:56graph it, and then press
12:58tab and we're going to
12:59graph negative x squared plus
13:04three x plus four.
13:06Okay, and that looks exactly
13:12like this. Great. Now, the
13:14intersection points are analyzed graph,
13:18intersect lower bound, upper bound.
13:22So this is 1, negative
13:241 .505, negative 1 .505,
13:28and let's do it again
13:32for the 1 above.
13:34of menu analyze graph intersection
13:38points lower bound upper bound
13:40that's two point one seven
13:43four two point one seven
13:46four and then I'm going
13:49to actually just put this
13:51whole thing into my integral
13:54so I am going to
13:56press let's do my integral
13:59sign here and I'm going
14:02going to go from negative
14:041 .505 to 2 .174
14:10and the function, I'm going
14:16to put it in exactly
14:17as I see it, negative
14:18x squared plus 3x plus
14:234. And I'm going to
14:27do minus and I'm going
14:28to open a bracket and
14:30I
14:30I'm gonna do E to
14:32the X minus three. I'm
14:36gonna close the bracket. So
14:38does that look exactly like
14:39that? Yes, it does. Press
14:41X for DX, press enter,
14:44and I get 16 .3,
14:45one, two, two. 16 .3,
14:49one, two, two. The last
14:52thing I'm gonna show you,
14:54if I go into my
14:55graph here.
14:58I do menu, analyze graph,
15:06and I click this thing
15:06called bounded area. So if
15:09I click bounded area, and
15:13the lower bound, I actually
15:15click the point. And even
15:18if I hadn't got that
15:19point, it would do that
15:20for me. I can actually
15:21pick that point into section
15:22point. And then I pick
15:24this, the next
15:26intersection point is my upper
15:27point and it will give
15:29me the there this is
15:33the answer that it actually
15:34gives me 16 .31 which
15:36is exactly what I got.
15:38So my advice is actually
15:39to do it both ways
15:41to make sure you get
15:42the correct answer but clearly
15:44that is giving me a
15:45nice area of the it's
15:48the bounded area between the
15:50two curves. Okay that was
15:54a fair
15:54early long lesson, we know
15:57how to do it with
15:57a calculator, we know how
15:59to do it without a
16:00calculator. The main thing to
16:04note though is there's a
16:05formula, this is the formula,
16:06and it is not in
16:08the formula booklet. If you
16:09did not understand anything I
16:11did here, it won't affect
16:13you at all in your
16:14exam and possibly not in
16:16your life, but I actually
16:19find this really interesting and
16:20neat how it all works
16:21out even
16:22when it's underneath the line.
16:24So it's worth at least
16:26trying to follow it. But
16:28certainly if you can do
16:28the two examples below, you
16:30should be in good shape
16:31in this topic. See you
16:32in the next lesson.