IGCSE ADDITIONAL MATHEMATICS PAST PAPERS MAY/JUNE 2020 | CAIE
Dec 18, 2020
Welcome back once again. It's been a while. Schedule has been tight. Okay. Now, this is one of the reasons additional method May June 2020 question. I find that it's a bit interesting to work with. So here I'm going to present the solutions to this. Welcome, once again, my name is misses Chang. If you are interested with what you're looking at, don't forget to subscribe. And then press the notification bell. So that you can receive the next video that I will be uploading in the future. Okay. Now you can see that this is one of the IGC edition method question. You can look at the title below. And then the topics is on integration. All right, let's get started. All right, now before we get started, the main objective for this question is we are given two functions here. FX and GX. One is linear equation. The other one is rational function. We need to evaluate the area bounded by the curve, the blue curve. And then bounded by the straight line expression, and at the same time bounded by the X axis. And the vertical line X equals two. So what we are looking at here is we are looking at the area under the straight line. And at the same time, we are also going to find this area here. So we are looking at the total area between this region and this region. All right, since we are clear of the objective, let's get started. Okay. Now how are we going to approach this questions? Typically, is quite direct. I'm going to apply the dividing conquest strategy. Here we're going to work out or evaluate the area under these straight line. Basically, if you look at it from the geometry perspective, we are looking at area for triangle, let me label this. This is area or triangle, let me label this as a B and O here. Since this is the origin, so all we need to do is work out the area for this triangle a or B since this triangle is triangle triangle. Okay. Now, to do this, first of all, we have to figure the Y interception and then the X interception here. So the X and Y intersects, we can obtain it directly from this equation here. X plus 1.2. You can see that. From there, we can identify that the Y in the set is given as 1.2 here. And then we ask the X in the set is given by negative 1.2. So I hope that you guys are clear how to work out the X intercept and then Y intercept here. So just to recap to call the X and Y intercepts, the steps here, just to walk through one more time. We will let for X in the set. So for X in a set, we will let the Y equal to zero. And then in this case, the Y that I'm referring to is the function GX, which is Q and S plus one one by two. So violating your Y equals zero, you want X plus 1.2 to zero. That would produce X equals negative 1.2. That's how we get these accents. Now, to get the Y intercept. We will let X equals zero in this case. So once we let X equal zero, our Y, which is given by GX function here, is actually equal to 1.2. So that is our Y intercept. So now we are ready to obtain the area for triangle. So area okay, let's work this out. A real triangle. A or B in this case is given as our label this as a one is easier to refill it on. So the width is 1.2 and then the height so I'm going to. So that would be like, in this case, you get 1.44. So zero point 7 two. You need screen. That's the aerial for the triangle a or B no, next we are going to obtain the area under the curve. From zero to two here. Okay, so I hope you guys are okay with getting the area for triangle that shouldn't be a problem. All right, now just as a reference, I will write my result here to give it as a reference. So the area for triangle is given as 0.72 units square here. Let's look at how do we work out the area under the cuff. From zero to two. Now, to do that, we have to apply integration. By using a low boundary of zero, and then the upper boundary of two. So basically area. Okay, let me write this down. Area under the curve. In this case, I'm interested to get the area under the curve. From X equals zero to X equal to two for this particular function of function. Three X squared plus 13 X plus ten. The X. Now, the first glance, this function is pretty intense. And then packed with flavor though. Hopefully it's good for digestion. Now, in order to work this out, we are going to look at the functions and then see how we are going to simplify the integration. So there we can work with the integration more easily. Just by looking at this whole function is point difficult to integrate it directly. So one way we can do this would be we can break it down using partial fraction on this one method. We can actually break it down using partial friction. Okay, let's give that a try. So the method that we are using is partial fraction method. To do that, we will let phi X plus two X squared plus 13 X plus ten. Now, first of all, we have to check whether we can factorize the denominator in this case. So we notice that this is quite easy to factorize, I would say. So there is basically three X plus ten. And then X plus one. Okay, just let me quickly check that to see whether I got the right. Okay. So there you have it. There is the factorization of this function. Effects. All right. Well, next, the partial fraction. So we will let this fight X plus 12. Equal to its partial fractions form. And then from here, we again this software a and B so that we can get his partial fractions for me. One way we can do this would be. Okay. That would be equal to high X buster. By comparison. Now the easiest way to obtain a and B is by is to use substitution. So we can substitute X equal to negative one in this case. So once you substitute X equal negative one, we indirect get rid of the variable a so in this case, we only have B negative one here, so we have a 7. And then in this case, there is negative one also. So there is a 7 that will you be equal to a one. Now the same thing next, we can choose another value where we can substitute into this equation here, equation number one. So we can substitute usually for me, I prefer to use a zero. It's easier. So from here, we have a sub zero zero into equation number one. We have a and then three X becomes zero. So we have ten, but anyway, we knew that B is one. So that must be a ten. X is zero and the right hand side on the left will therefore be two. And then from here, we conclude that the partial fraction for 5 X plus two over three X plus ten X plus one is equal to two over three X plus ten. Plus one over X plus one. Of course, you can do a very quick check to see whether we obtain the rational functions here. So just let me quickly check through that. In case I made any mistake. So two X plus two days. So I have 5 X plus, okay. So after checking, yes, I'm quite happy with the result. Now we can check on this. This is the partial fraction. So I'm going to copy down this result at the site here so that we can continue on with the integration. So from here, we follow the 5 X plus 12 three X plus ten. Okay, that is X plus one. Is equal to two. Three X plus ten. Plus one over X plus one thing. Now there is our partial fraction. I'm going to take this out. At this moment. Okay. Now we're going to continue on with this integration. So by integrating this rational function, we are saying that we can actually integrate this that. So we can use a divide and conquer strategy to integrate this rational function by using its partial fractions for. Now this recap this note here, I'll just write it down in there. I'm going to erase that off. Just a piece of knot here. We knew that integrating one over X with you long X and then therefore integrating one over X plus one would you long X plus one here that is the result. Here I did not put C here because our focus is mainly on the integration of the function. And then the toughest part would be integrating this part here. Two three X plus ten. So we have to be a bit careful with this. The best way to do this would be to maybe from my perspective, I prefer to bring this number out here. So therefore, I will have one of X plus ten over three. The X. So if I'm here, it's easy to see that this format looked quite similar to this one here. So from the pasta integration result, we can conclude that this is the same as X plus ten over three. So using this result result number 100 result number E and B, we can continue our work from step number two here. I hope that's okay. Okay, so take a moment to just capture this result. And we're going to use, all right, so you've got ID. I'm going to fill this off. So we have here, we have after integrating, we have two over three. Long X plus ten over three. Because this is definitely integration. Substitute this value in. And then we know that this is non X plus one. Let's see what the okay. And then down after is basically just a numerical computation. Just let me give that a try here. Like the word tree, so we have the subsidy the two first into here. We will have a long, let me see there will be 16 tree. Minus of lawn ten over three. That's my first part. And then the second one is basically just treat. So from here, I can simplify further. I would have a long, let me see here. I will have to add to it to 5. So that's all I have. Long tree. So if you want, you can just leave it as it is. If you have calculator by all means, you can achieve this on into your copy. Or you may want to write this in a different form for instance, so you may like this. 1.6. Plus long tree. Now, this will be the area underneath the curve. The blue region. So this one will be write this down. You will need square. All right, so I hope that is of help to your integration technique. But anyway, don't get too excited with the integration. We still have to find the total area bounded by the straight line bounded by the curve and the X axis together with the vertical line X equal to two. So let's do this. So the total area, I would just right here for the area. That we need to answer in this question would be a given S just labels a T here is basically a one plus 8. So we have our a one here. I'll just write down 0.72 plus or whatever result that we are looking at. One 1.6 plus long trip. You and me. Squared. Okay. There you go. That is the. Integration technique that we use to evaluate the area under the curve and the straight line thing. All right, I hope that this has been useful to you. Thank you very much for watching this video. 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