I JUST WATCHED THIS BEFORE MSM 112 EXAM & CLEARED | TOP 10 INTEGRATION EXAM QUESTIONS 2025

Añadido: 2026-05-20

[00:00:03]Okay. So, welcome intellectuals to siding institute. So, in our today's lesson, we are going to focus on integration.

[00:00:11]Since we've just done differentiation, so it's not going to be a very big deal for us to understand integration.

[00:00:18]So, just a quick introduction that I can give you is that uh integration is just the opposite of differentiation. That's why it's sometimes called ant differentiation. So just the opposite of what we are doing is what you'll be doing.

[00:00:35]Okay. So uh looking at the question that I have here, I've got about 10 questions which came in the last academic year exam. And you're going to see that these questions carried about almost 40 marks. So that makes it very important for you to focus on integration too much as you are preparing for mathematics. The reason being that during exams or I may say during tests you uh for both test one and test two you are not assessed on integration because it is the last topic. So you expect a lot of questions to come during an exam. So that's why I'm going to uh first focus on integration then after that you'll be preparing again with um these other courses. So it's very important for you guys who are not registered with our tions to register now because this is the only lesson that you are going to access for free but this other lesson starting from where we started in all the courses will only be accessed by those who are going to register. So without wasting much of our time uh let's take a look on the first question that we have here. So evaluating the integral we have been given to x /x then we has to integrate to find the value using the uh using the limits that you have been given here. When you have these numbers, these are just limits that you have been given which are the value of x. The value which is on top is the upper variable of x and the one on the lowest the lower variable that means it's the it is usually the lesser number. So what you're going to do here when you have been given the question like this for us to integrate we are going to use u substitution. Okay, it is the method that we are going to be using in most of the question. But there are some of the question that use substitution can't be used as we are going to apply other concepts like integration by parts. So what you're going to do here, you're going to find the value of u. When you look at two hex / x you can see that we have two functions here. We have two in hex. If you have to forget about the constant because whenever we are integrating we ignore the constant. Okay. So we have two functions here. We have the first function is lin x and the other function is the one in the denominator which is just x. So which one can we use as x so that we can just be able to differentiate it then we integrate it very easily. So we don't need to choose by randomly or by favoring one function in preference to the to other. So what we need to do we need to follow the L which we use which we call L I A T. These are just letters. These are initials but you can remember maybe by calling it light.

[00:03:56]So L here stand for logarithm.

[00:04:06]Then I stand for I stand for invest trigonometry.

[00:04:14]So inverse trig functions.

[00:04:17]So if you have something like uh like sin inverse of x cos inverse of x sign a tan those you need to choose it first. Okay. Then here is for algebra.

[00:04:34]All the function that are going to be in the form of algebra. For example, this x is in the form of algebra. Then when you look at lin x is in the form of logarithm. So you need to say that anything that is that is in the form of logarithm logarithm the chances of choosing it first in preference to the other is high. Okay. Then t is just for for trigonometric function or trig functions or trig functions.

[00:05:06]Then the last which we go for is an exponential function.

[00:05:12]So you need to understand this exponential functions and I'm sure this is going to make a lot of sense as we be solving uh more questions because as for now it may look so confusing.

[00:05:29]So looking at the function that we have we have x and we have x this is in form of logarithm and this is an algebra. So which is coming first here? You can see that L is coming first than A. So we are going to use U to be X not X. So we're going to say U is going to be equal to X. After that now we are going to differentiate U which we are going to call DU. differentiate you in terms of x that is dx dud dx because most of you used to get confused whenever I said dud dx just means differentiate u in terms of x so the the derivative of reinhex we all know by now that is just 1 /x okay after that now the next thing that we are going to do we are going to make gu the subject of the formula or gx let's just be making dx the subject of the formula. So if you if you have to cross multiply here we're going to say that dx is going to be multiplied by one and you're going to have dx is going to be then this x is going to be multiplied by u by du then we're going to have x du okay so now with that in our mind we can go now and press the given values so I may recommend first I may recommend first you integrate then you focus on the on the variables that we have there.

[00:07:00]So let's just focus on integrating two lining x / x. Okay. So you're going to appreciate that what we are going to have here.

[00:07:12]We know that this lin x is just the value for u. So where there is lin x we're going to put you. So we're going to have two. Instead of lighting x we're going to have two u. Then we say over x.

[00:07:25]Then why there is dx? We're going to put x du because dx is equal to x du. So we say the integral of x du. You can see that x and x are going to cancel. Then our integral becomes 2 u du. So this be makes a lot of sense.

[00:07:47]Whenever you are using u substitution, make sure you remain with only u in the expression because you can't integrate two v more than one variable. Okay? If we have x and y if x is not canceling just know that you substitution has failed. So with this information in our mind we can differentiate now that. So what are we going to do?

[00:08:13]We are going to we're going to just say uh when we were integrate uh when we were differentiation we used to multiply uh 1 to the power then we subtract we used to multiply if you have for example 2x^ 2 we used to multiply this to this we have four then we subtract one to the power we used to have 4x but here it's just the opposite we are going to be um add one to the power then we divide by the power by the new power. So what you're going to do you're going to say you're going to have two u since this u has got the any number any variable or I may say any number has an um has just a silent power of one. So this u has got a power of one. So we need to add another one. Then we divide by the new power here which is 1 + 1.

[00:09:16]Okay. So with that in our mind, we're going to say our our integral is going to become 2 u ^2 / 2 of which this and this are going to cancel. That means our integral just become u ^2. But we need to remember that they asking us to evaluate we need to include this actually uh these variables that we have there. So now the most key point here which you shouldn't miss is that if you have used u substitution that means to say you even need to convert okay you need to convert the limits you have to change the limits which were in terms of x now in terms of u. So in this case we have u we have u which is equal to x from our initial assumption which we made. Then we are going to say if we use now one we replace y there is x what will be the value of one? So we say for for x = 1 u = l where there is x we put one and this just going to give us zero. So that now is is now going to become our upper upper value is going to be zero. Let's now look at the lower value.

[00:10:42]So for x= the natural log u there the value of is going to be the same x now why there is x we replaced by e and we know that in e is just one. So with that in our mind we are going out to just uh get done with this integral. We can just we can just um simplify that. So we have zero we have one then we have u 2. So this can be written now as u ^ 2 then 0 then 1. So what we are going to do we are going to say we're going to replace while there is going to to put zero. Okay. So we're going to put zero. While there is u there we put zero. This is 0 squared.

[00:11:35]Then we are going to subtract. The same way that we did with zero we do with one. Now we subtract. So where there is u you're going to put one we squared it since u is squared.

[00:11:47]Then you're going to have the answer which is going here is going to give us zero and here is just going to give us one. So our answer is just1 and here you can see that we don't have the answer but actually the answer supposed to be1. Here there was just a typo and it was corrected.

[00:12:06]So this supposed to be the correct answer. So that's what you basically need to know about differentiation. It's just the opposite of I mean integration is just the opposite of differentiation.

[00:12:19]And always whenever you are using you substitution remember the the initials li to stand for logarithm investig functions algebra trig function then e for exponential functions.

[00:12:35]After that another emphasis whenever you're using u substitution don't forget to change the limits the assumption that you made that u is equal to x. Now replace the value of x by the values that you have there. So that you may have the new values for you since the derivative that we have at the end is not in terms of x like we used to see here. It is in terms of u. So we need new values. So that's the only concept that you need to get and with the next question I'll try to be a little bit faster since you've gotten most of the important concept that you need to know.

[00:13:17]So we have now this question here question 10 evaluate the the value of the integral of x sin x dx. Okay. So in this case what we need to know first we are going to forget first about these values. We want to try if you substitution is going to help us. Okay.

[00:13:43]So from you substitution you have the integral of x sin x.

[00:13:52]So the function that we have here we have x and sin x which one can we use as u. So just use you write apply it and you can see that we have algebra here and we have trig algebra is coming before trig. So we go for algebra. So what you going to going to say we are going to let u to be equal to x that means d u dx the derivative of this u is just going to be equal to 1 then we can make dx the subject of the formula if you have to cross multiply you're just going to get dx to be equal to du.

[00:14:32]So replacing now in the integral that we have we are going to have the integral now is now going to become where there is x we put u then sin x then where the dx we're going to put du since dx is equal to du we're going to do this so from this you can tell that there is no way we can cancel this x and we have also so this can't work that means you substitution can't work so If you use substitution fails, if you substitution fails, you go for another method which we are calling integration by parts. Okay. So this is the other method that we are going to use to differentiate uh this function which u substitution has failed us.

[00:15:24]So you have the function shouldn't forget we have the function x sin x we looking for the integral. So according to differentiation by parts or integration by parts I mean so we going to say that the derivative this is the formula that you need to know by hat the derivative I mean the in the integral of u dv is equal to u * v minus the integral of v du.

[00:15:59]This is the formula that you need to know. Now the first important thing we need to identify the value of U, the value of V, what is DV and what is DU.

[00:16:13]This is the next thing now which we're going to use from light. The same light that we are using.

[00:16:21]We said U is X and that remains true. So U is X.

[00:16:29]If this is u, this u stand for x that means the remaining part is dv here. So that means this is our dv. So you say dv is equal to sin x here sin x dx.

[00:16:52]And this makes a lot of sense because this is what we are looking for. We have u and we have dv. This is u and dv. So now in the formula we need to have u.

[00:17:04]Yes, we have it. We need to have v. How do we get v? We are going to get v from dv. Okay. And this is found by just integrating this part because uh mathematically we are going to introduce okay let me write here to make it more clear. So we say sin x dx.

[00:17:30]So I'm going to introduce the integral here. You also introduce the integral here. So now the thing that you need to remember here is that the integral of dx is just x.

[00:17:45]Okay. So the same applies the integral of dv is just v. So you say v is equal the integral of sin x dx. What is the integral of sin x? Okay.

[00:18:04]From what we started from on differentiation we went to say that the derivative of sin x was positive x positive cosine x.

[00:18:17]Okay. And the derivative for cos x was sin x. Now because integration is just the opposite that means sin x will now be giving us cos x and cos x will be giving us the positive of sin x when we are integrating which is just the opposite of differentiation from what we read.

[00:18:44]So after that what we are going to do next since we have v and we have dv and we have u so we can just come and replace all the values that we have. So we have u is x and we have dv is sin x dx and you can see that basically we are just rewriting this uh question that we are looking for. After that you can see u is equal x. So you say x then v is = to cos x. So cos x we are subtracting by the integral of of v.

[00:19:32]Here the integral of v is cos x. So you say cos x then what is du?

[00:19:48]Okay so we have to find du. Okay so now here is what we are going to do. We have u we have u. So we are going to find dudx. Okay. So I'm going to say can use this simple parts. I'm going to say since we have u which is = x then du dx is just the derivative of x which is just one. We can cross multiply.

[00:20:22]It's just 1 / one. We can cross multiply. You're going to find that uh gx is going to be equal to du. That means to say whenever you've seen du there replace dx. So I'm going to put here. Here we had v which is just this one. And where there is du we are putting uh dx.

[00:20:50]The reason we are doing that we want uh the the letters to be matching because we can't have x here and have du here.

[00:20:59]that is not going to to be to make sense. Okay. So after that you can see that this is just the question that we are looking for. So we can focus on u on this part or maybe we go by writing.

[00:21:16]So you can multiply out this one. You're going to get the answer is just going to be equal to x cos x because we have a negative there. And you know we have this negative and this negative are going to give us positive. So we're going to have positive the integral of cos x then dx.

[00:21:40]Okay you can go ahead and say x sin x dx is going to be equal to -x cos x. We just rewrite the integral of cos x.

[00:21:54]The derivative of cos x was sin x. Now the integral is going to be the opposite. Instead of us having the negative, we're going to be having the positive. So you're going to have positive sin x. That is the integral of cosine x. So you're going to say positive you're going to have positive sin x.

[00:22:23]So this is how far now which we have.

[00:22:26]But let's remember that we have this um these values of x which are the limits that we need to have here. We didn't use substitution. You substitution failed us. So no need to change the limits. We are just going to take them like the way they have. So I'm going to say uh to to now since we are find the derivative of this to now find to now include now the effect of pi there to the power pi and zero there.

[00:23:09]So we have pi and we have zero over of this. So we are just going to write this as you say x cossine x + sin x then you say this should be pi zero. So how do we do this? Just the same way. So we're going to start by replacing x everywhere.

[00:23:41]Where there is x we put pi. So we're going to say pi.

[00:23:47]Then cosine where there is x we put pi plus sin where there is x we put pi.

[00:23:59]Then now we are subtracting by this one with zero. So it's a negative.

[00:24:06]Why there is x here? We put zero. So I'm going to have 0 cossine 0 plus sin here you have zero.

[00:24:20]Okay. So you can now just simplify this<unk> cos pi. Pi you know when you convert it into larger just give us 180. You know by now this. So it's more like they're saying cos 180 is just -9 cos 180 is just I mean -1 according to special angles. So we're going to have what is cos pi we're going to replace with1. So we have pi there and multiply by1 then plus sin pi it's more like sin 180 which is just zero. Okay, that is just zero. Sin 18 is 0. Just have to take note of that. So adding plus 0 there.

[00:25:09]Then we are subtracting 0 m* this just going to give us 0 and sin 0 is just 0.

[00:25:16]So it's more like we just looking with the value which is<unk> *1 just going to give us pi. And going back to check for our answer going to that we have our answer which is correct there which is just B. So that's how we go about these questions. So integration is very interesting. You just need to to abide to the principles that are behind solving any kind of question that you're going to meet.

[00:25:48]So after that we have now this question.

[00:25:52]Okay.

[00:25:53]We have now this question which is question 14. Which of the following answers equal to the integral? So we're looking for the answer which is equal to the integral of 1 / 7 x + e to the power pi.

[00:26:12]So looking at this in the brackets we have two functions. Okay. The first function is um the first function we can let the first function to be a which is just 1 / 7x and the other function to be b which is just e to the power pi. So if you are differentiating you are integrating two functions we just use uh the simple formula or the identity. So you say the integral this is not even an identity just a so if you are integrating two function a plus b it's just the same that you can just integrate separately you can separate a gx plus the integral of b gx okay so that means to say this is now going to be written as the integral we separate this 1 / 7x x you say dx then plus also the integral of e to the power pi then dx.

[00:27:26]So I'm going to start with this one.

[00:27:29]Okay. So, so we need to just understand that the integral, let me even write it here in a box so that you be able to remember the integral of k dx is just going to be equal to k where k is a constant. If k is constant the integral of that is just going to be equal to the same constant now multiplied by x.

[00:28:03]Okay. So what we are going to do in this case we are going to factor out 1 / 7 in this because this 1 / 7 x can also be as 1 / 7 then 1 / x because when we multiply out this we just going to go back to this. So it's nice we can even write it like this just one and the same. So I'm going to write like this. I'm going to say 1 / 7. I'm going to put it outside the integral. Then I'm going to say the integral of 1 /x because 1 / 7 is just the constant then dx.

[00:28:46]Then when you look at this e to the power pi okay the integral e to the power e to the power pi is just a a constant. It's different from e to the power x. Okay. because x becomes a variable but pi is just a number which we already know. Okay, it can be 180 in two legs or it can be 3.142 if you are using a calculator. That is the constant. These are numbers. So this is a constant. E to the power pi is a constant. That means the derivative of this is just going to be equal e to the power pi. then you multiply it by x because I said that whenever you are integrating a constant you say a constant multiplied by x so I'm multiplying it by x so I can put x here so whenever you are integrating for us to know that you are integrating you have to put it to some constant okay so you're going by saying what is now the integral of 1 /x okay so if you have to recall we we used to say that the dy dx of 1 /x is used to give us lin x. Now because integration is the opposite. Okay, integration is the opposite. That means the integral. Okay, let me do this. It's supposed to be this. So we we say the the dwy dx of rein x used to give us 1 /x that means the integral of 1 /x now is going to be equal to rein x because this is just the opposite. So you say the integral of this is just going to give us lin x. So whenever you see 1 /x they're asking for the integral just say lin x then you have x e to the^ x + c.

[00:30:51]So multiplying 1* x you have x / 7 and + x e to the power this is not x but pi and plus c. So you can even look for the answer that we have the answer which looks like this and it is no other than C.

[00:31:13]So C is our answer there. So integration is very simple.

[00:31:19]Just need to be smart.

[00:31:23]So we have another question. The area of the region bounded by X = 1, y = 2 and x =1 and y = 3x^2 + 4 is given by so yeah we need to find the area this is application of integration so we can apply integration in finding the area so the area bounded by the c so we need to understand that the value of x these are just values x the bigger one is just the upper variable and the and the smaller one here is the our variable. So our main work here these are the same values which used to see for example the bigger one is on top and the smaller one is below there. So the only consideration that we need to have here is on the value of the function. So we have two function we have y = 2 and we have y = 3x^2 + 4.

[00:32:31]So what you need to understand here if you have been given two functions you have f of x and uh g of x from the fact that if f of x is greater than g of x the area is going to be equal to the integral the integral of f ofxus - g of x.

[00:33:05]Okay, that's how you find the area. But if if g of x is the one which is greater than f of x, that means the integral is the area is just going to be just the same way. We just change we going to start now with d of x minus f of x.

[00:33:34]So now looking at this we want to see which one is bigger between this one and this one so that we may find the area by subtracting the smaller one from the bigger one. So we going to say why we already have this two. So for us to estimate if this one is bigger we are going to put any number because we have the number which is running the values of x are running from negative from one all the way to -1. So we can put any number it can be one or negative one because these two values are included that's why they are bounded by uh by that curve. Okay.

[00:34:15]So for us to estimate we can take any number can either take -1 or 1. As we can see that if we square this three let me use -1 3 -1 2 + 4 you're going to get a number because this will give us positive pos1* by this 3 + 7. Even if we use positive 1 we still going to have seven. So we're going to say that this is greater than this one. Since the area is going to be equal to the integral of this is going to be is going to be the first one to write 3x^2 + 4. Now we subtract the smaller one which is two.

[00:35:00]Always test to see which one is greater.

[00:35:02]Don't just look because sometimes they may trick you.

[00:35:08]Okay. So after that you can just simplify now because we know that we have like terms there like 4 - 2 can give us 2. So we're going to have 3x^ 2 - 2 then gx.

[00:35:27]So from that now we can find the our area which is just going to be equal our area you know we have to put b here and a where b is just the bigger variable which is just one so I'm going to put one here and one here for 3x^ 2 - 2 then gx that means to say we just have to start by integrating how do we integrating we had the one to the power Then we divide by the new power. So we say how can we integrate 3x²? We just add one here. We're going to have 3x to the^ 3 then divide by the new power which is just three. Then minus how do we integrate a constant? Any constant is integrated by just adding hex to it.

[00:36:23]Okay. That's why I was saying that when you integrating k we just say k multiply by x. So I'm going to have 2x there. But let's not forget that these are bounded in the x-axis by one the one.

[00:36:42]So we can deal with this here. You know 3 and 3 can go into this.

[00:36:48]So our expression remains to be x ^ 3 - 2x then you have 1 then -1.

[00:36:59]So let's try to simplify that. Now we see if we are going to get it right. So we have - 2x then we have one.

[00:37:13]So what can we do?

[00:37:16]We going to say we start with the upper variable. So we say 1 to the^ 3 - 2 where there's x we put one.

[00:37:33]Then we subtract by the next variable which is x now to the power -1 ^ 3 - 2 the -1 and we simplify you're going to have 1.

[00:37:49]This is going to give us -2us uh -1 to the^ 3 will always give us a negative. So we're going to have -1 and this and this are going to give us a positive. So we're going to have positive2.

[00:38:11]So you can see that here going to have -1. Then minus here going to have pos1.

[00:38:19]So it's1 you're going to have positive one. Of course uh let me see. Let me see something here.

[00:38:39]So we need to we need to have -1 here then minus one.

[00:38:46]Okay, let me try to go back to the question.

[00:38:51]So after we had one here, we have 3 x ^ 3 then over 3. Okay.

[00:39:00]So we are saying minus.

[00:39:16]I'm trying to understand where this became a minus. So we supposed to have a positive because 4 - 2 supposed to give us positive2 not - 2. So we're supposed to have positive2 there. So positive2x positive 2x. So this is where the mistake is. So we supposed to have positive here and supposed to have supposed to have so expression remains to be x ^ 3 now 2x.

[00:39:48]So now we can simplify then 1 -1.

[00:39:52]So we can just replace now the values.

[00:39:55]So if you have to put one here we start with one. you're going to have 1 to the^ 3 is just one. Then two here we put one you're going to have positive2.

[00:40:04]Okay. Then we subtract by the one with the negative. So putting a negative there going to have -1.

[00:40:11]Then putting a negative there we're going to have -2 because we're going to have -2. So we're going to have our answer to be three. The negative what is in the brackets there is -3.

[00:40:25]So I have three and positive3 here because negative and negative is going to give us positive3 then we're going to add we're going to get six as our answer.

[00:40:36]So looking at the option that we have been given you can see that g is the correct answer there.

[00:40:47]So this question you have already even been told to say that you can use substitution method or any other method which is just you are okay with it. So to find the value of that integral which is just e to the power so we need to find I'm going to forget about this. So I'll say the integral of e to the power sin x then we have cossine x then dx.

[00:41:13]Okay. So we have this sin x and sin x or adjust uh adjust.

[00:41:22]So using light it can just confuse confuse us. So what you do if you have bought a trig for you to choose the value of you just use the one from the function which is complicated. So we have this is more complicated than cos x. So say the value of x is going to be sin x. Don't take the the or expression here. So say just just say the value of u is going to be sin x. That means du dx is going to be the derivative of sin x which is just cosine x.

[00:41:56]Okay. So how can we make now gx the subject of the formula? You can cross multiply here. You're going to say that you're going to have dx then cossine x.

[00:42:07]If you have to multiply then it's going to be= x du can divide by cos x everywhere by cos x everywhere. So our dx is going to be = x du over cossine x.

[00:42:29]That's how you That's how you actually found it. Okay, that's how you get in fact do we need to have here x? We just needed to have one not x. So there's no need for this x. So it is du over cos x. Okay, because our integral was just cos x. I supposed to just write 1 cos x over one.

[00:42:57]So after that so we have our dx to be d cos x. So we can go and replace here. So I'm going to say e where there is sin x you put u because sin x is equal to u.

[00:43:11]Then you say cossine x but x is given by u over cos x of which cos x and cos x can.

[00:43:25]That means e to the power u becomes uh just du there and v makes a lot of sense with you substitution because we only have one variable which is just you. How can we integrate um e to the power u? Okay, how can we integrate that? So you can see that the integral you need to remember that the integral I mentioned this of e to the^ x is just e to the power x.

[00:44:02]The reason is just very simple. Okay.

[00:44:05]This one is just very simple. Whenever you are integrating you can multiply this by x and you divide by x you still get to the same value.

[00:44:18]So that's just the fact here. So even here the integral of e to the power u is just going to be equal to e to the power u.

[00:44:29]But what we need to remember with you substitution we need to change the limits that we have been given. Okay, before we apply the units, we need to change them. If you go direct, you get the wrong answer. Okay, you almost get this one.

[00:44:48]So, what you need to do, you're going to say, you've already made an assumption to say that u is equal to sin x. So, what is going to be the value of u when x = pi? This one.

[00:45:04]So when x= pi the value of u is going to be equal to sin pi which I say that pi is just 180. So sin 180 is just going to be equal to zero. What happens now when x is equal to zero. What will be the value of you?

[00:45:20]Say sign s° just zero. So in this case they're just telling us to say that now deal with this e to the power u. Now put the limits 0 and 0.

[00:45:37]So if you have to say they telling us to say that e to the power 0 - e to the power 0 just give us zero. So the answer here is d.

[00:45:51]Okay. So now let's try to test one of the question which looks uh to be more complicated but very simple. If you can just remember few concept which we which we area. So this question is integration using partial fraction. So we are has to express first the partial fraction of this expression and after that we need to evaluate this integral.

[00:46:23]So what we are going to do here we have been given to say 5x - 4 over 2x^ 2 + x - 1. So the first thing that you can just do here we need to actually factorialize this. This is uh that can be factorialized and it's no longer my work to teach you how to factorialize. So you say + x - 1. So you can say the product is this multiply by this is -2 and the sum is the coefficient of x which is just one. So what is going to be the factors those two number that we can add to have one and we multiply to have -2 is just 2 and -1.

[00:47:12]Okay. So use this now to uh to factor out this. you're going to get 5x - 4. It's going to be equal to 2x - 1. Those are the steps that I've skipped + one which you just need to do it. So when you look now at this, how can we break them into partial fraction 5x - 4 over 2x -1 then x + 1.

[00:47:45]These are just we have to so we have to the denominator is made um is is a product of two functions. So we just need to break this into two partial fractions. So we're going to have a over this. So it is 2x - 1 then + b over x + 1 this one.

[00:48:08]So what can we do next? We are going to multiply now everything by the denominator. So we are going to multiply this by the denominator which is 2x -1 then x + 1 this 2x + 1 then x + 1 then this 2x + 1 then x + 1 you multiply everything. So we are going to say that if you have to multiply this and this are going to cancel we're going to remain with 5 x - 4. Then this and this are going to cancel. We're going to remain with a multip by this which you can even factor out a m* x going to have a x and then a multiplied by one you're going to have just a. Then you can see that this and this are going to cancel with this. So b * 2 2 2x you're going to have 2 bx then b * 1 you're going to get b. So what you do you start by correcting those that have x we have a x then we have 2 bx x is common so we can cancel it out.

[00:49:29]That means our first equation becomes 5 = a + 2 b. This is our equation one.

[00:49:38]How can we find our equation two? We going to take now constants. We have -4.

[00:49:43]The other constant that we have which doesn't have a variable is a and the other one is b. So it's plus b there as our equation two.

[00:49:54]So it's very simple for us to actually find the value uh the value for the value for a or b. So what we can do here we can just make um a the subject of the formula or be the subject of the formula. So what we are going to do here we are going to say uh we are going to say here we have this is -2 it's not positive so this is negative this is1 so it's supposed to have we're supposed to have negative here And you're supposed to have negative here. So if you have to if you have to expand that you're going to have negative here b minus this just give us this then multiply by1 supposed to give us b. So it's supposed to have b here.

[00:51:02]So I want to use this to make b the subject of the formula. What are we going to do?

[00:51:07]We are going to say we can take B to the other side is going to be positive and we've remained with a and this can go to the other side is going to be pos4.

[00:51:21]This is going to be our equation three.

[00:51:23]Okay. So we're going to put this equation three into uh into this expression. Let's just use this one. So we're going to say 5 = a + 2 * b. But our value of b is a + 4.

[00:51:46]So we're going to 5 = a. 2 * this is 2 a. Then 2 * 4 is 8. Taking it to the other side is going to be 5 - 8. Then this is going to give us 3 a. So 3 a is just going to give us 5 - 8 is just -3 / 3 / 3. Our a is going to be equal to -1.

[00:52:11]But we can recall that b is = a + 4. So have to put -1 here which is the value of x. We're going to find that -1 + 4. Our b is just going to be equal to 3.

[00:52:26]So the partial fraction which we made is now going to become so we're going to say the partial fraction for 5x - 4 it's going to be where here we had 2x² n + x - 1 is just going to be equal to we had a and a we had a here and we had b there so our a is going to be1 1 / 2x + 1.

[00:53:03]It's actually 2x -1, not 2x + 1. 2x - 1.

[00:53:07]This is the same mistake you even made.

[00:53:09]Don't need to repeat it. 2x - one.

[00:53:14]So, so it's 2x - one. Then we are adding by the value of b which was 3. Then over we had x + one in the denominator x + one. So that's how you you write this into partial fraction. This question had eight marks.

[00:53:43]If you go through past papers I'm going to confirm that. So after that now we need now to find the integral.

[00:53:52]This one we need to find the integral of 5x - 4 / 2x^2 + x -1.

[00:54:03]We need to find the integral which is just very simple because when you look at this is just the same as this. So it is okay for us to actually say this can also be written as we can where there is this we put the partial fractions which we actually found. So we're going to say our partial fraction going to be -1 / 2x - 1 then + 3 then x + one there then dx from what I told you you can remember that I said if you are different if you are integrating two functions a plus b it's just the same with we can integrate just uh by separating.

[00:54:58]So we're going to separate that. We're going to find the integral uh specifically for for -1 then 2x - 1 dx.

[00:55:14]Then we going to by the integral of um 3x + 1.

[00:55:27]So finding the integrals are just one and uh the same way. So I'm going to use the other page which is likely to be the last one.

[00:55:39]So you're going to say we have look we looking for the integral of -1 / 2x -1.

[00:55:48]So you can let u to be 2x -1.

[00:55:55]That means how can we differentiate u we said u dx is going to be equal the derivative of 2x is just two. Then you can appreciate that we can make now dx the subject of the formula. So making dx the subject of the formula this multiply by this we're going to have 2 dx then du can divide by two can divide by two that means this and this are going to this only this and this are going to cancel.

[00:56:27]So that means dx is going to be equal to du / 2.

[00:56:33]Okay. and now replace. Okay, let's replace. What do we have? We have -1.

[00:56:40]Then we have here we are going to say u because is just equal to that. Then why there is dx? We're going to put du / 2. This of course can be simplified because we are going to we can put our two here can multiply this. We're going to have 2 u. Okay.

[00:57:07]Multiply by that.

[00:57:09]Then after that you can factor out because f can be factored out. We don't need a constant. So going to remain with 1 / u because if you have to multiply this and this we still get back to this.

[00:57:25]Then we have du.

[00:57:29]What do we know about the derivative of 1 /x is just in x.

[00:57:37]That means even the derivative of I mean the the integration or the integral of 1 / u is just going to be equal to alu. So you say in U but we need to remember that U. So you just have to include that but to gain a lot of marks u = 2x - 1. So I'm going to say the answer is going to be equal to half then lean 2x - 1. Then you know we never ended from there. We add also another integral which was actually 3 then x + 1. We need to integrate that.

[00:58:23]So 3 x + 1 then dx.

[00:58:29]What do we need to know? We say x = to u= that then dx is going to be equal to one.

[00:58:39]dx just the derivative of this just one.

[00:58:43]This is zero. We can make dx the subject of the formula. So this multiply by this. So dx is just going to be equal to du.

[00:58:58]So with that now we can just replace 3 over y there's x + 1 we put just u.

[00:59:07]Then y there is dx we put du because dx is equal to du. So we said du.

[00:59:14]Same applies to what we did. We don't need this three. So we can take it out.

[00:59:18]are going to remain with 1 / u because if you have to multiply this and this we still get back to this. So we don't change anything then du the same integral the same integral which we had here. So can say that we going to have the the integral of 1 / u is just u but in this case our u is = x + 1. So we're going to get three L where now there is say X + one.

[00:59:55]If you are that students um who like recording things you can remember that if you have a uh Lin X is just the same as Lin X to the power this A.

[01:00:13]So you can maybe write it as lin x + 1 to the^ 3 and this can also be written as lin 2x - 1 le to the power if you want to further your your answering but since we were just hading here so our final answer is going to be the this 2x - 1 to the^ then plus this which is lin x + 1 to ^ 3.

[01:00:54]So that's how you find the integral that's how you can get the 8 marks. So 8 marks to for partial u for partial fractions which we did in semester one and four marks for integrating.

[01:01:11]So we can now do the last question.

[01:01:14]We have to do the last question.

[01:01:17]So this question also was uh came to section B and it has also eight marks. If it's not eight marks, I'm sure it's supposed to be maybe seven marks.

[01:01:32]It can't go less than that. Should be within that length. So we need to integrate that. And because these guys are very much okay they they want you to pass. They even tell you to say that let you to be what let you to be 4 - cos^² x.

[01:01:52]So what you can write for you to make the uh the differentiation to be very simple you're going to this can also be as cos x now squared because of this.

[01:02:06]So how can we find dx?

[01:02:10]The derivative of a constant is just zero. So now this part we use chain rule. So chain rule will tell us to multiply this by this. You're going to have -2.

[01:02:21]Then we subtract one to this. We're going to have now this cos x. We have subtracted already one.

[01:02:33]So it has remaining with one. So you can't even put one here. It have no no any meaning. So now after that you multiply by the derivative of this. This is what I was telling you. Whenever you are multiplying by by chain rule, if you have um if you have for example u to the power x, just multiply x there u then subtract x by one then uh find the derivative of u.

[01:03:12]So du dx that's that's those simple. So you we now need to multiply by the derivative of cos x which is just going to be sin x.

[01:03:25]So du dx is going to be equal to because this is negative and this is negative.

[01:03:31]We can say 2 cos x * by this we're going to have two and start with sin x then plus oh I mean cos 2x there.

[01:03:46]So after looking at the question you can see you have in the in the numerator there sin 2x.

[01:03:54]So I can think of writing this as sin 2x because from where we started with trigonometry you can remember that sin 2x was just an identity for 2 sin x cos x that means this our dx it's just going to be equal to sin 2x because this is equal to sin 2x sin 2x.

[01:04:23]So we need to make now dx the subject of the formula. So you can cross multiply.

[01:04:28]You're going to have dx then sin 2x is equal to du can multiply by sin 2x then by sin 2x.

[01:04:42]Okay. So this are going to cancel out.

[01:04:45]you're going to have dx which is just going to be equal to d u over sin 2x.

[01:04:54]So we after we just have the value of u and uh the value for uh the value for gx then we are free now to actually convert okay just actually to replace so we're going to have the integral I'm going first to forget about this but don't forget that since we are using u substitution we need to change these limits in terms of u. So we're going to say the integral now becomes 9 sin 2x.

[01:05:30]Now where there is this we replace by u because that's what we have been told and that's what we were using. While there is dx, you're going to put this value which is du over sin 2x of which sin 2x and sin 2x are going to cancel.

[01:05:53]The integral becomes 9 / u then du which is making a lot of sense. We can factor out 9. We don't need it is a constant.

[01:06:02]So I'm going to have 1 / u then du.

[01:06:09]How do how do we differentiate one over u? Something we've been doing. So we're going to have 9 lu.

[01:06:22]Okay. So after that we shouldn't forget that we have these values for x.

[01:06:30]So we're going to find we know that u is equal the simplified one we were using 4 minus cos x then squared. So what is going to be the value of u at x =<unk> / 2 because you know by now that / 2 is no other than 90°.

[01:06:51]You can convert that into legs. So we're going to have what is be the value of u.

[01:06:57]So the value of u is going to be equal to 4 minus cos 90 there. Here we put cos 90 squared. We know that cos 90 is zero. So 0 square is just going to be z. So at this value x u is just going to be equal to 4. Okay. So this has now been changed to four.

[01:07:26]Then we can now actually check the value for u at x =0.

[01:07:36]So we can say u is going to be equal to we have 4 minus cos 0°.

[01:07:47]So cos 0 is just one. You know this.

[01:07:52]So it is 4 - 1. it is just going to give us three. So this now becomes three. So that means what they want you to do, they just need you to actually now put this uh we can put uh this. Let me try to ignore.

[01:08:16]So, I'm going to ignore actually the I'm going to ignore this nine because it's a constant. I'm going to put the brackets here and put on top we have four and down we have three. Okay. So, what can we do here? We can just replace now.

[01:08:38]Think I can do here.

[01:08:41]This is just the final answer we are looking for. We've done everything.

[01:08:45]So we have nine then we have lin lu in those variables.

[01:09:00]So we going to say where there is we put three sorry it's r four. We start with the upper one. Then we are subtracting by the lower which is L three.

[01:09:16]If you can remember what we did in semester one, you can remember that when you have rein A minus L B, our answer when you're subtracting logarithm, you're going to have lin then a over b. If you're subtracting, we used to if you adding we used to multiply. So this can also be written as 9 then le 4 over 3.

[01:09:48]So that's how you go about these questions. So we've covered almost everything which you need to know about integration.

[01:09:56]It's one that we only answered the one question on application but you're going to provide more.

[01:10:03]So uh this is just a call to register for for tition so that we may revise together as you are as you about to close soon. You know most of you are complaining about your CA which is just less than maybe even 10 can't decide your final exams your final results. So if we just prepare smartly, you can make things very very easy for you can pass. So we are doing we are doing a promotion uh which is the only promotion which you can just even try to miss because we are going to do revision in everything starting from test that you wrote stream one and stream two in all courses.

[01:10:52]I say in all courses until you write your exam and all the courses are just going at 100 quarter that means to say each course cost is just 25 quarter but you don't need to pay in units you just have to pay the the total of 100 quarter and be able to revise with us until you write. So I've got physics starting from where you started with law up to uh electrical electrical energy and capacitance.

[01:11:26]We have chemistry starting from chemical kinetics.

[01:11:30]So we understand those topics that's usually come in an exam in the all the way up to organic chemistry.

[01:11:40]I know most of you started so it's a must. There are some question which will come on organic chemistry. Then we have biology. Some of you started with DNA replication and you hand all the way up to genetics up to genetics.

[01:12:03]Then if it is in mathematics you know mathematics must we started with trigonometry.

[01:12:11]And you have a lot of complicated topics in semester 2 all the way up to um up to integration and I've been emphasizing on integration stud more on integration because there will be a lot of questions which will come here and we also have what course am I missing here I think I've talked about trick DNA chemical genetics that means just physics I mentioned starting from slow all the way up to electrical inent capacitances.

[01:12:45]So all these we are doing the revision but our main focus we first starts with exam questions those test that you wrote we get done with them. So because there's a possibility that some questions are going to be repeated but because they are not going to repeat everything that's why you need to go back to revise those concept that were not add in class or maybe even when I was teaching you you miss some concept it's a time to revise back everything so just inbox me on this number I think most of you you have my number so you can just inbox me WhatsApp me so that you just register now so that you'll be able to access all the lessons because this was just a bonus lesson that I gave you for most of you wanted to know uh how the division is being done. So thank you for watching guys. See you in the next um lesson as you register. So don't forget to register and share this video to your