A Level Maths Made Easy - Integration AS

welcome to another AEV maths Made Easy video this time we're doing the year 12 integration again a topic that can be kind of tricky the actual integrating normally isn't too bad but the applied questions can be a little bit tricky so you're probably wondering what is integration if you haven't seen it before essentially in maths every operation so timesing dividing Etc has an inverse operation so with timesing the invers is dividing with adding the inverse is subtracting with squaring their Square rooting and so on that is a fundamental principle of maths essentially the idea is if I do something to a number I should be able to undo it and get back to my original value so for example if I do 2 * 3 I get six I should be able to do something to six to get back to the number two as an example integration is the inverse of differentiation so its job is to undo differentiation now if we remember when we differentiate so if I was going to differentiate let's say x cubed what we do d by DX what we do is we bring power to the front power to front which when I say that it does mean multiply essentially and then we subtract one from the power subtract one from power and when we do this actually I put a little subtract sign just to make it super clear when we do this if we're doing integration we want to be able to do the exact inverse now inverse doesn't mean the opposite so we can work out how to integrate something without actually knowing how to integrate something by just doing these steps backwards and doing the opposite so the common mistake I see when students do this is they they do it the opposite they'll bring power to the front or they'll leave it there or whatever or they divide by the power so instead of multiplying by the power they divide by the power then they subtract one but that's not the same that's doing the opposite not the inverse when you invert something you also work backwards so you're starting from step two and doing the opposite so you're not dividing by the power and then adding one instead you're adding one to the power so that would be x^ 3 + 1 so 4 and then what's the opposite of multiplying by the new Power you divide by the new power and that's it you can test this out so I'd recommend if you differentiate xq and then you want to integrate it you should get back to X Cub so when you differentiate you get 3x^2 integrate you get 3x^ 3 / by 3 x cubed so um there is one thing though there's one problem do remember what happens if we have like a number plus 4 over here right well when we differentiate it it goes away doesn't it it becomes 3x^2 and then it goes away so when I integrate I need to have the ability to bring back the constant the problem is there's no way to work out what that constant is straight away so instead what we do is we introduce a letter remember when we don't know what a number is in math we just use a letter to represent it so plus C that is called my constant of integration because it's there just in case there was a constant when we uh in the original function that we differentiated now bear in mind this C can be any natural number it could be sorry not natural number any real number it could be zero so for example in the case of just X cubed it would be zero but if it's X Cub + 4 then we'd have a plus c which we' then have to work out or use some other piece of information to work out what that value is but the key Point here is is when you differentiate something you do lose information you lose what the constant was if there was one because I have no way of knowing if there was one I have to introduce a plus c that's just the law that's the rules okay but that's all you really have to do so if you want me to give you a general form just like I did before um that's not how it looked x to the N DX then all we do is we're going to write x to the n + 1 we're going to put a at the front and then divide by whatever the new n +1 is and then we are going to have our plus C every time you integrate Bar None So let's get to it with some other questions so jumping into something with indices just because it I want to make it a bit trickier because again the topic isn't too bad normally you'll get a question where it says integrate this giving each term in its simplest form so the first thing I would do is well I'll do what I said right so first thing we do is add one to the power so that's 4X theus1 and then we divide by the new Power which is min-1 do the same thing here we get plus 3x to the power of a half I do notice a few year 12s by the way making mistakes when you're adding or subtracting one from fractions which can be a bit it's an easy way to make a mistake right over half and then of course you have the plus C simplifying we have 4 over -1 so it's -4 then we have 3 ID half again use a calculator if you're unsure right 3 ID a half which does give you 6 x half plus C now that would be it so you get a mark for the actual integration and a mark for the simplification that's what you have to do so coming on to a slightly trickier question in my opinion uh we have this question so p and Q uh integrated gives you 2x + 10 X plus C find the value of p and Q a really common style of question not just in integration but in pretty much all topics in a level maths is the idea that they tell you to do an operation they tell you it equals something else and you need to work out the missing variables basically uh not variables con um in this case just uh algebra so we have P 2x^2 the first thing you do is You' put this into a form you can integrate so it' be p /2 xus 2 plus PQ and then DX and all you'd get is so again add one to the power gives me in fact I could write it like this minus one over since it's already over two I'm also going to have a minus one there so that does just make minus two which is something vaguely interesting I like to think uh plus PQ X and then of course plus C so again the plus C and the plus C are already the same and they're telling me that this equals 2x + 10 x + C so in other words I know that P and again if you want me to write this in another way another way you could put it is you could put the X on the bottom here cuz x to the minus one but I know that the PQ is equal to 10 how do I know this well again this is a really common again I've said this in the previous videos as well a really common uh strategy is called comparing coefficients where you look at the coefficient of the X term and since we know that these two things are equal we know that they should simplify to each other so PQ should give me 10 and everything in front of uh in this case uh the X so the two should be equal to P over minus 2 just like that so now of course we can solve so we get p is equal to-4 subing that into here we get - 4 Q = 10 so Q is equal to - 5 over two whoops that one and that one for five marks I think that is remarkably generous but again it's the comparing coefficients and it's doing the integration property that really matters so integration can be really really useful when we are given the first derivative on of an equation or a curve and we need to find it again so again remember when we differentiate something we'll get the F Prime uh you can also call this Dy by DX if you'd like it means the same thing and what we're going to do is now work backwards to get the actual equation of the curve instead and you may notice they've also given me an additional piece of information I if you can kind of guesstimate where that is going to come in handy so the first thing we need to do in order to kind of undo the differentiation we need to integrate but this thing here is pretty hard to integrate so what we're going to do is we're first going to simplify it so I'm going to write x^2 - 2 over x to the half and then all we're going to do is simplify that one more time because we don't know how to integrate fractions yet uh X2 / sorry X2 / X2 will be X x^ 3 / 2 CU 2 - half and then there would be - 2 x - half DX now integrating this we get so add one to the power and then bring it and then divide by it so I'm going to show you that it's going to be 25ths x ^ 5/2 and the reason for that is we get x ^ 5/ 2 divided 5/2 remember when you divide by a fraction you need to use keep flip chain so Times by 2 fths because we flip the second fraction so just keep that in mind then here we're going to add one we get to the power of a half divide by the new power so 2 ID by a half will give me 4 so - 4 x the half and then remember r + C now you can probably see where this is heading right it passed through here we need to find the equation for C this is not an equation for C yet because we have an unknown which is plus c but they've given me a coordinate now you're probably seeing what we need to do so when x equals 4 the uh function should equal 5 so we get 5 = 25s 4 to^ 5 / 2 - 4 to the power sorry * 4^ a half um am I doing that right yeah cool had a little brain fart there then I use my calculator just to work out all that stuff because I'm lazy and we have 2 FS * 4 power of uh 5 over 2 so 2.5 let's just call it - 4 4 to^ of 0.5 and that's a half gives me 24 over 5 so 5 = 24 5 + C so then minusing 24 over 5 from both sides we get our value for C we can sub it in and we're all done so it should be it's going to be 1 I'm pretty sure five minus answer at that too easy uh yeah let's leave it as a fraction actually because the rest of my things in it as a fraction so then we have f ofx is equal to um it was 2 fths x to the power 5/ 2 - 4 x^ a half half +2 as our final answer and again these questions can be four five marks really really nice questions if you know what you're doing and that is again why they give you that piece of information remember they never give you information you do not need this is the last bit of simple uh content we need to do before we go on two questions that are actually pretty tricky so so far we've been doing something called an indefinite integral and the reason it's called an indefinite integral is because it doesn't give us a numerical answer however we can do this which is a definite integral and it is definitely more interesting not really so you might have noticed that there's these two values right one and N what does that mean well what it means is we're going to evaluate our integral between the values of 1 and 9 I'm going to show you exactly what that means in the next question and exactly what this actually calculates in the next question as well but it is in my opinion pretty interesting so what does this mean well to be honest it kind of means the same thing so your first step actually stays exactly the same we're going to write it out in a form that we can actually integrate and then we're going to do the integration just like normal I know very interesting so far um I'm going to put square brackets you'll see why in a second but we get 2x the^ 2 / 2 gives me x^2 here we're going to have - 3x^ 3 / 2 so we're going to have over 3 over2 which is going to give 2 so - 2 x^ 3 / 2 and then of course we're going to have our plus C and we still haven't used our limits here 9 and one what all we do is we substitute both of those values in and we subtract them from each other so we start with the top limit so we're going to do 9^ 2 minus 2 9^ 3 2 plus C and then we're going to subtract 1^ 2 - 2 1^ 3 over 2 + C now you might notice that the plus C's cancel and just as a bit of a spoiler warning they will always cancel because when you sub in values it's never going to affect a constant so that that constant is always going to disappear so if you want you don't even need to write it but I wrote it just to make super clear that that is a thing now we're actually going to get a numerical answer so what we're going to do is we're going to whip out our calculator one more time and we're going to do 9^ 2 - 2 9 1.5 o okay I thought I pressed another button there fine I'll do it this way there we are subtract uh 1 2 - 2 1 28 we get a number crazy now of course the question is what on Earth does that number represent so when we differentiate we get a gradient what happens when we integrate and for that we're going to go on to the next question and believe me it's going to make a lot more sense here's the question these can be six seven marks really nice questions but we need to first start by sketching the curve and honestly this is pretty pretty quick so if you go back to the uh kind of sketching graph um videos that might make a b more sense but as you can see we're going to have three x intercepts so that's going to be x = 0 + 1 and - 3 so let's label those we have minus three over here uh we obviously have zero right through the middle here and then plus one over here and because it's a positive cubic it's going to come up down and then up again oh beautiful beautiful now hopefully you didn't read the rest of the question while I was doing that when you integrate it gives you the area between the curve and the x-axis now there is a common way of saying this that is wrong some people say that it is the area under the graph no so for example in this case from -3 to 0 that is the area under the graph but it's also this area and that that area is Not underneath it it's above it so this is the technical way of saying it it's the area of the region Bound by the curve in the x-axis I prefer to say the area between the curve in the x-axis because I'm not pretentious uh not always anyway and that's it that's what integration gives you so if I integrate this curve between Min -3 and 0 it will give me the area this first red area here interesting right it's pretty cool it's pretty cool and if I integrate it between Zer and one I will get the area over here now here's where it gets complicated a little bit complicated areas in the positive xaxis will always be positive but can you see that this area is actually underneath the X AIS that means the grade sorry the area is negative so if we want now you might be thinking how on Earth can you have a negative area and you're 100% right you can't which is why when we integrate between 0 and one we'll get let's say -4 but we're going to say the area is four not -4 the problem with this question is I cannot integrate between minus 3 and 1 because this area is positive and this area is negative which means they will cancel each other out if you have a perfectly symmetric graph they actually cancel out perfectly and you get zero which is obviously really wrong so what we need to do is we need to do two separate integrals between minus 3 and 0 we'll get a positive number let's say 28 then between 0 and 1 we'll get a negative number let's say 12 but we make it positive and then add the two values together okay this leads me to tell you what integration actually is if I ask you personally to calculate the area under the graph before you did a level maths what you would probably do is break it up into lots of little thin strips and then find out the area of each of those strips and then add them together okay valid integration makes those strips infinitely thin so it makes it per perfectly fit to this graph here just a perfect way of doing it um the way differentiation integration were kind of invented so to speak is very interesting in my opinion so again you know where we had the tangents before with differentiation but what if you made the tangent infinitely small which means that they perfectly conform to the curve of a graph it's the same thing with the areas the common way of working out the area of the UN under the graph is by splitting it up into shapes that you can actually work out the area of nor trapeziums okay so for example if you had like a curve like this You' split it up into a bunch of trapeziums which kind of fit the graph but as you can see there's there's overlap here right there's empty spaces but then they said well what if you made those trapeziums really really really thin well the thinner they are the more accurate your value so what if we made them infinitely thin well that means we'll be infinitely accurate that's the idea behind it and that's where they kind of clo equalized these kind of rules that we've used here so that's the idea behind this question we're literally just going to be integrating between minus 3 and 0 and 0 and 1 and then making them both positive and adding them together we cannot do minus 3 and 1 because the negative area will cancel out a bit of the positive so again imagine this is like 32 and this is12 that means it give me an answer of 20 instead of uh what I actually want which is 42 44 sorry so bear that in mind so let's get to it we can't integrate bracket yet so we need to expand the uh brackets so what we're going to do is we're going to have y is equal to X expanding that we get x^2 - x + 3x that be + 2x and then - 3 and expanding that one more time x cubed 2x^2 - 3x and we're going to integrate this twice once between minus 3 and 0 going do the exact same thing again between 0 and one yeah 0o and one good o spoilers anyways 0o and one and it's going to be the exact same thing again there is good news here you don't have to do the same thing again because once you've integrated it once you can just kind of move it across that might be a good way to do it so here we're going to have A4 x 4 uh plus 2 3 x Cub - 3 2 x^2 because again remember I've added one to the power and I've divided by the new Power that's all I've done then of course you have the plus C first things first you may notice if you sub in the value of zero we're going to get zero in this whole thing right we're going to get 0 plus C so I'm not even going to bother subbing in the zero but in the other one I'm going to do a quarter - 3 4 2/3 - 3 Cub - 3 2 - 3^ 2 and then again the plus C so then the C's will cancel so we're going to do 0 minus all of that can I scroll down with that right you're going to see a bit of the next question oh well that's fine uh we're going to have zero minus and then we're going to have all of this um stuff three still squared this time and yeah that's it 45 over 4 so again as I said it would be positive now again we've already done the integration so we're literally just subbing in between 1 and zero now again the good news here is we don't need to do the zero cuz it's zero right it's going to give you zero thir 1 cubed - 3 2 1^ 2 plus C and then obviously minus it's going to be 0 + C so again C's cancel out as they always will so all I'm going to do is I'm going to be kind of clever I'm going to go through I'm going to delete the Zero from the front and I'm just going to change all of these to ones okay it's all about Saving Time - 7/ 2 12 as I said is going to be a negative number so what we're going to do for the area is we're going to say it equals 45 over4 + 72 because again areas can't be negative okay that's just a A Fault In the calculation so to speak um that is because mathematically speaking you can have negative areas it's just in real life you can't have negative areas uh 71 over 6 perfect and if you want to give it a units you'd have to just say unit squar because I don't know if it's centimeters or meters or whatever that's it now some of you may have seen on your calculator that there is this handy dandy integration button you can use it but some questions will say uh Solutions based entirely on calculation calculated technology cannot be accepted so you do need to do the substitutions and things like that otherwise will give you nothing um to bear that in mind let's move on to the last example which is where things can get a bit hard but I'm going to show you it's really not that bad so here's a kind of questions which can will probably appear in the end of uh year 12 exams and obviously they can appear in year 13 exams as well where they get slightly trickier and it's trickier because this area looks whack right it looks crazy the thing is and the great thing about integration is you don't need to do that you know those end of GCSE questions right where they would have like a pentagon inside a circle inside of a rectangle you had to work at the it doesn't matter you still just are going to integrate that's the great thing about it it's integrate right because integration does the work for you you do not need to memorize any formulas for area because you can always integrate to get the area but let's start with part A and there's a special reason why they've asked you to do this for part A so find the Quant to the points where the line and the curve intersect I mean that's going to be 0 0 but whatever so to do that we're going to equate the two things so 10 x - x^2 = 2x um move everything to one side to minus um yeah I'm going to minus 2x on both sides so we get 8x - x^2 = 0 factorize to solve we're going to have X and we're going to have 8 - x = 0 so either x = z well it's not either it's and x = 8 and that does make sense cuz obviously this is our zero and that's our eight value makes sense now calculate the area of the Shaded region a lot of people here will get a bit worried scared um plenty of other adjectives I can use to describe them but remember what integration gives you it gives you the area between the curve and the x-axis so if I integrate the curve at the beginning not the straight line so this one it will give me this it will give me the area of this entire region including the white part at the bottom if I integrate the line it will give me the area of this triangular shape could you use the area of a triangle formula absolutely the problem there is you would have to sub in x = 8 to find out what the height of the triangle is but you could do it I'm not going to say you can't because that would be a lie you could absolutely do it so again you could use the area of a triangle um but why would you right um actually you know I should actually remove a bit of this because technically speaking it would only go up to here right that's why I said it's triangular shape by the way okay so if I just literally integrate the red part between 0 and 8 and then subtract the integral of 2x between 0 and 8 I'm done and it doesn't matter how complicated the area is it's the same thing so if they gave you like two curves intersecting each other and you had to work out the area that looks complicated right but it's not you're just integrating one curve between three and four and the other curve between three and four and subtracting it that's it that's why these questions when I mean eight marks is a huge amount of marks can be worth up to N9 or 10 marks you're always just going to do integration and it's so good it's really really worth the investment so let's start shall we we're going to have the integral between 0 and 8 of 10 x - x^2 I will say by the way that because this question doesn't say you know you can't use a calculator you could just use a calculator for the whole thing um but in the spirit of things I've noticed more and more recently that they have that uh disclaimer you can't use a calculator just do it properly but that's my answer well as soon as I work out anyway so again let's do this that would be 10 x^ 2 / 2 so that's 5 x^2 then you're going to have x^ 3 / 3 so minus I like to put it as a thir X Cub but it really doesn't make a difference I'm not going to write the plus C I'm I'm hoping you're okay with that now then again uh that's 2x Sid 2 so just X squ between 0 and 8 and again remember with zeros the answer is always going to be zero so I actually only need to sub in the eight for both of these right so 5 a^ 2 - 3 a cubed and again I'm technically minusing zero because if I sub in that so I'm not even going to do it and minus the green one a s oops and again remember since none of my values are sorry none of the curves are underneath the x-axis I don't need to make it positive I can literally put this in my calculator now get the marks six marks for three lines of working out that's crazy crazy so when people say these questions are difficult it's honestly practice this more than anything because it's so nice it really is anyways enough of that again I'd recommend just remembering that zero always will equal zero so you don't usually have to write it unless the the Zer is on top because then you going to have to have 0 minus something um whereas in this one it's just it's going to be the answer minus Z so I'm not going to bother putting it um get rid of that part of the calculator and yeah just sub this in so 5 8^ squ minus thir 8 cubed take away 8 [Music] squared 256 over 3 six marks eight in total right uh integration I'm not going to lie year 13 is really hard it's obnoxiously hard I'd say but this is all you really need for year 12 if you can do this is is the hardest start of integration question you can get remember the principles they have to give you enough information to do it and with the the areas just remember the core things if it's beneath the x-axis it's going to be negative so you need to make it positive yourself you have to manually do that it's one of those few times in mats where you have to do it manually um and yeah just really when you're practicing this take it slow a lot of a level students they a over complicate things uh which is not good keep it simple that's best piece of device I can give you and the second thing is take it slow they've G me 6 minutes to do this even when I was explaining it to you it didn't take me 6 minutes right so take your time as if you're explaining it to someone else do it in your head say oh okay so this is what you would do you would do between Z and eight cuz that's where the intersection points are you would do the curve minus the straight line because the curve is on top of the straight line so we have that that's what you need to do right that's what you need to do in your head is talk through it but I hope to see you in the next one where we go on to some of the stats and mechanic stuff off