Okay, welcome to this video where we'll be having a look at the AS pure mathematics chapter on integration. Now, as you can see, we'll be looking at all of integration here. We're going to look at finding integrals, indefinite integrals, finding functions, definite integrals, areas under curves, areas under the x-axis, and areas between curves and lines. So, as with all of these videos, we will go through one part of each topic within the chapter. And before we do that, I'm just going to quickly show you how the video works and how to use it. So let's have a quick look at that now. So when you are on one of these videos, if you click into the description, you can see the video there where I cover all of the ASP curriculum. And if you scroll down, you can also see all of the chapters. Now these will slowly be filling up until it is filled entire ASP curriculum. So hopefully by the time that you are watching this video, all of them will be actually in the description and you can click and watch each chapter. If you click onto the video and click the link in the bottom left there, you can actually see all of the chapters that are within this video as well. So you can skip through and have a look at the pieces of that chapter that you are needing to have a look at for your revision. So that's how this video works. Hopefully that's useful and helpful. If you enjoy this video, please don't forget to like the video. Don't forget to share it with some of your friends. But with that being said, let's get started. [Music] When we looked at differentiation, we looked at this particular graph and this is the graph of y= x^2. So we had a look at this and we had a look at how to differentiate that and when we differentiate it, okay, we multiply the coefficient of x there by the power. Okay, so when we differentiate it, we wrote it like this. dy over dx. There we go. We multiply the coefficient of x there, which is one. So we multiply that by two. So we got 2x and then we reduce the power by one. So we reduce that power of two down to a power of one. And we got this here 2x. Now when we're looking at integrating, we're essentially looking at doing the complete opposite here. We're just going to reverse this process. So instead of multiplying the coefficient um and reducing the power, we're essentially going to increase the power and divide the coefficient. And when we write this out, we write it out like this. Okay? And this little symbol here, which I'm going to have better in a second, but it looks something like that. We say to do the opposite, that means integrate. We put in the bracket what we want to integrate, which in this case is 2x. And then we say dx after that just means with respect to x. So that just means obviously do the opposite. That means integrate 2x with respect to x and we're going to look at how we go about actually doing that. So obviously as we've said we're going to increase the power. So at the moment that power of two uh 2x there the the power of x is a power of one. So if we increase that it becomes 2x^2 and then rather than multiplying by that coefficient we're now going to divide by that. Sorry rather than multiplying the coefficient by that power we're going to divide the coefficient by that power. And the power is two. So we're going to divide it by two. And you'll see when we do 2x^2 / 2, we get back to that original x^2 that we had up the top there. Okay. Now, the problem with this is uh and obviously this isn't going to be the final answer here. The problem with this is that there's a quite a lot of different um uh graph equations here that are going to give us 2x when we differentiate it. So, if I have a look at another example up here, we could have y equ= and let's just imagine it's x^2 + 5. Now when we differentiate that obviously we multiply the x um squared by the coefficient by two and we get 2x. So when we differentiate it we get 2x and the five then disappears. So that gives us 2x as well. Likewise we could have something like y = x^2 - 19. When we differentiate that multiply the x^2 by 2 reduce the power you get 2x and the 19 disappears. So again we get 2x. So the problem here with the integration is that we don't actually know what that constant is at the end. What that number is at the end. And because of that we have to add something onto our answer when we're integrating. And that is that we have to put plus c at the end. And that is our called our constant of integration. Again that just obviously says that we don't actually know what the constant is. So we always have to put plus c at the end. And that's going to be absolutely essential as we move through this video. every time we integrate, we're going to have to put that plus C at the end because we just don't know what that constant was unless we have any specific values on our graph. And obviously with the x squ graph, that's quite a nice one. But let's just imagine just to sort of give you an overview of everything we're going to be looking at. If I imagine that I say, okay, well on this particular graph that we've just integrated, so this x^2 + c, we have an x and a y value. Let's imagine we say, okay, well y= 5 when x = 2. Okay, so if y= 5 when x = 2, what is that equation going to be? Well, if we sub those values in, x^2 is going to be four. So that would be four plus c and that has to equal 5. So we know in that case there c would have to equal 1. And the equation of this line in particular would be x^2 + 1 or y = x^2 + 1. Okay, so something here that looks a lot more complicated. And let's see how we go about this one. So we've got these three pieces that we're going to integrate this time. So we're going to start off with this 12x cubed. So again, we're going to increase the power. So we have 12 x ^ 4. And then we're going to divide by that new power. So divide by 4. Then we're going to add to that. We've got 6x. So that's going to go up to 6x^2. And then we're going to divide by that new power, which is two. And then we're going to take away. And this one's not as nice here because we've got a power of 2/3 there. But we have 15x. We need to figure out what that power is going to be. So we need to add one to 2/3. Well, one in terms of/ third is 3/3. So if we add 3/3 to 2/3, we get 5/3. There we go. And that's not a five. So let's rewrite that. We get 5/3. There we go. So 15 x ^ 5/3. And then we need to divide by that new power. So we're dividing by 5/3. Not very nice there. There we go. But divide by 5/3. and then at the end of that plus c. Okay, because we don't know what that constant would have been at the end. There we go. We just need to simplify this all down. See what we get. These first two look okay. We got 12 / 4 which is 3. So that's quite nice. We get 3x ^ 4 for the first bit. The next bit we've got 6 / 2 which is 3 again. So 3x^ 2. And then this bit isn't so nice here. Got 15 x ^ 5/3 / 5/3. And obviously if you've got your calculator obviously you can work this out quite nice and easy. But don't forget it's not too difficult just to do 15 over 1 divided by 5/3. So times it by the flip version 35ths and there you go. That equals 45 on the top over five which is nine. There we go. So 15 divided 5/3 is 9. Just a nice little bit of working out there. Obviously not always having to rely on your calculator if you've got some nice quick working out that you can do. So there we go. Take away that becomes 9 x to the^ of 5/3. There we go. and not forgetting plus C at the end. There we go. And there we go. That's that integrated. Okay. Not forgetting that plus C. Again, that's so important. Obviously, something that's obviously new with this, even though we're doing the reverse process of differentiating, that plus C there is so important to make sure that you add on the end. But there we go. That's how we're going to integrate. So, increase the power by one, divide by that new power, and then simplify all your terms. And just watch out for any sort of complications when you got fractions in the powers there. And just being careful with that. Okay. Okay, so this question here looks very similar to one we looked at earlier in the video. We've got dy over dx = x^2 + 5 in brackets 2 all over x^2 where x cannot equal zero. Obviously we can't divide by zero so we can't have x is zero on the bottom. And it says given that y= 13 and x= 1 find y in terms of x giving each term in its simplest form. So obviously you've got a few little bits to sort out here. You've got a bracket on the top that needs expanding. You're going to have to turn this into a single line. So dividing all of those pieces by x^ squ and getting those obviously rewritten but obviously you need to get that into a single line integrate it sub these values in find that value of c. So have a go see if you can get get this question done see how far you can get. We'll go over the answer in a sec. So pause the video there and have a go. Right. Okay. So we need to sort out what's on the top to start with them. So x^2 + 5 obviously squared means we got to do a double bracket. So we have x^2 + 5 and another x^2 + 5. There we go. So if we expand this all out, we get x ^ 4. Then we get + 5 x^2 + another 5 x^2. So that's + 10 x^2. There we go. And then 5 * 5, we get + 25. So that's what we've got on the top. And that's all being divided by x^2. So obviously we just need to then divide all of these powers by x^2. It's actually not too bad if you've obviously got to this point already. The first piece there, x ^ 4 / x^ 2. Well, I'll subtract the powers, we get x^2. There we are. The next piece, 10 x^2 / x^ 2 is just 10. So, + 10. And then the last piece there, 25 / x^ 2. Well, subtract the power from x ^ 0 gives x ^ minus2. So, we get 25x to the minus2. There we go. And now it's at a point where we can actually integrate it. So if we bring this up and integrate that, let's see what we get. So x^2 well obviously increase the power. So we get x cubed / 3. There we are. Plus 10 x increasing the power of x there. Obviously there's no x before. So we'll introduce the x. And then for the last piece there, if we increase the power, it goes to minus1. So we get 25x to the minus1 / the minus1 and then we get plus c at the end. Right? Okay. So nearly done. We obviously just need to simplify this down. So we get at the start there we can obviously write it as a third if you want. You've got one one in front of the x cubed there. So we've got 1/3 x cubed + 10 x and then we have 25 / minus 1. So it becomes - 255 x^ -1 + c. And now we can go about subbing these values in. So it says over here y = 13 at the point where x equals 1. So if we obviously write this as an equation now um let's just get rid of this because we could have put our y equals at the start of that one. Obviously have decided not to as we were integrating it at that point but now we're looking at it in terms of y equals y in terms of x. We can write that out. So we've got -3 is equal to a3 * 1 cubed. So a3 * 1 which is just a third. You could put the one in if you want. Let's just write that in anyway. 1 cubed plus then we've got 10 lots of one which is just 10 take away 25 lots of 1 to the^ of minus one. There we go. 1 to the^ of minus one you can put in the bracket and then plus c. Right. So simplifying this down then we've got minus3 at the start obviously equals and we have a third * 1 which is a third plus 10 * 1 which is 10 take away 25 and then one to the power of anything is one. So it's just take away 25 plus c. Obviously we just need to obviously simplify this little bit down and see what we get. So we got a 3 + 10 take away 25. Right? So if we work that out then we got -3 equ= 10 takeway 25 is -5 add the third to that is -4 and 2/3 there we go 14 and 2/3 obviously you can check that on the calculator but there we go -4 and 2/3 plus c. So not the nicest here. How do we get from minus4 and 2/3 to minus3? We're going to have to add in uh those extra 2/3 to get back to minus14 and then add an extra 1 which is 3/3. So there's 5/3 there that we need to add for C. So our value of C is going to be 5/3. There we go. That would have to be 5/3. There we go. Obviously you could add 14 and 2/3 to -3 there and you'll get your 5/3 or two and or sorry one and 2/3. So we go there's our value of C. And obviously to finish this off, we just need to plug it back into the original equation up there for our value of C. So for our final answer here, let's just write this up here. We get y = 1/3 x cubed + 10 x - 25 x - 1 and then plus 5/3. There we go. And obviously you could write that as a decimal as well if you wanted but obviously when it comes to thirds I tend to just leave it as a fraction rather than having any recurring decimals in there. But there we go. There's our final answer. Okay. So when having a look at definite integrals let's have a look at the process for going about that. So this particular question first of all we need to integrate 2x - 3x. Now again I'm going to write that in a slightly different way. I'm going to write the rootx as x ^ of a half. So I'm going to write this as 2x - 3x ^ of a half. Now to start with all we're going to do is integrate here. So if we integrate that we will have and let's just have a look at what we get. We get 2x^2 / 2. And then increasing the power there for the power of a half we get 3x ^ of 3 over2. and we're going to divide that by 3 over2. Now, at this point here, you are okay to simplify this. So, we'll simplify this a little bit. Now, you don't have to cuz we're going to type this into a calculator anyway when we go about this, but we can simplify that. So, the first piece there, if we divide that by 2, we get x^2. And for the second piece, when we divide that by 3 over2, we get 2x ^ of 3 /2. So, now we have integrated it. What we are going to do is do a definite integral. So for this, what we are going to do is we are going to substitute the number nine, the first number on the top of the integral. We're going to substitute the number one in and we're going to subtract them away from each other. So when we write this, we put our integrated piece into a squared bracket. And you will also notice that I haven't put the plus C on the end. Now you're going to see here why I haven't put the plus C. So we'll come back to that in just a second. So the first number we're going to use is the nine. The second number we're going to use is one. And if I go about doing that, I'll get something that looks like this. So I will put nine in to start with. So that will be 9^ 2 take away 2 * 9 to the power of 3 /2. And there we go. And then we are going to take away whatever it is when we substitute one in. So if I substitute 1 in, we'll get 1^ 2 takeway 2 * 1 ^ of 3 /2. And there we go. That is going to work out our answer. Now you'll notice that we didn't need to put the plus c in because we are going to take these away from each other when doing a definite integral. You'll notice that if I had have put a plus c in the first bracket, I would be taking away a plus c in the second bracket and therefore they would cancel out anyway. So putting that in is kind of pointless. So if we work this out and type that into a calculator, you will see that I get the answer 28. So our final answer for this is 28. Now on a later question, you are going to see what we use this for, but this was just a practice of going over how to go about doing this. So as you can see, we integrate it. We put it into the squared bracket showing the nine and the one. And then we substitute nine in. substitute one in and take them away from each other. And again, if we had different numbers there, we would sub those numbers in. But again, we would just take them away from each other. And we're looking to see what our answer is. Okay. So, in this question, we are going to see what we use definite integrals for. And this question here says, find the area of the finite region between the curve with equation y = x brackets x - 4^ 2 and the x ais. So, this is going to show you what we use definite integration for. Now if we were to draw this curve and think what it looks like, it would be a cubic graph and is a positive cubic graph as well. Now you can see here we get our solutions of x= 0 and we've also got an x= 4 solution there and that's a repeated solution. So it's going to bounce on the axis. So our graph would come up through 0, bounce on four and then go back up. So if we label that on, we've got zero here and four just here. And hopefully now you can see that finite region between the curve and the x-axis. So if I highlight that, we are looking at this region here. So when we are using definite integrals, we use it to find the region bounded between two of these points. So in order to do this, we're going to have a look at the equation of the curve. We are going to do our definite integrals between 0 and four. And that will tell us the area that is bounded between those points. Again, it does tell us the area as it is bounded towards the x-axis. So, we had would have to do something different if it was bounded between the curve and another line. But again, we will look at that as well. So, in order to integrate this, we'd probably want to expand that equation. So if we expand that equation, if we do the double bracket first, we would get y is equal to x lots of. If we expand that double bracket, we get x^2 - 8 x + 16. Now we'll multiply it by the x as well. So we're going to get y is = x cubed - 8 x^2 + 16x. Now we can go about actually integrating this. And if we integrate that and again we're going to do it between these two points. So we'll write that we're going to do it between four and zero. So when we integrate that we get x um let's have a look. In fact let's just write down on what we're going to integrate. We're going to integrate x cub - 8 x^2 + 16x. When we integrate it then we can just go ahead and write it in our squared bracket. So increasing the power then we get x ^ 4 over 4 take away 8 x ^ of 3 all over 3. Just make sure I write that three a little bit better. And then that is going to be + 16 x / 2. And of course you could write that as 8x^2. Sorry 16x^2 over 2. You could write that as 8x^2. Now that's what we're going to be using for our limits here. So we'll put that into our squared bracket and we're going to use four first and 0 second. It's quite a nice one when it does have zero. It does make the second step quite nice. So when we put the numbers in, let's have a look at what we get. Putting four in to start with. And again, let's just write it in cuz we're just going to type this in on our calculator. So we have 4 ^ 4 / 4 takeway 8 * 4 cubed all over 3. and then + 16 * 4^ 2 all over 2 and we're going to take away the second one which is substituting in 0. Now if you have a look if you substitute 0 into the first piece you get 0 over four so that's 0. The second piece you get 0 over3 and the third piece you get 0 over two. So all of that just comes out as zero. Again, you can substitute zero in if you want, but just looking at it, you can see they all come out as zero. So, if you type all of that into your calculator, we get the answer and it comes out as 64 over3. And there we go. That is our area. So, the area that we are looking at there is 64 over3. You could convert that into a mixed number if you want to. You could write that as 21 and a3, but either of those answers are fine. And we could leave our answer in either form. So there we go. That is our final answer. That's what we use this definite integration for. We find the two points that are bounded by the curve and the axis. And using this, we can work out the area underneath. Okay. So when looking at specific areas, sometimes we'll be looking at areas that are also under the x-axis. And we just have to take something into account when we do this. So if we think about what this curve would look like for this particular question, we will have um something going under the x-axis. So you will see when I expand this bracket and or if I draw a sketch of it, what it would look like. So if we draw a little sketch to the side and think about our solutions, we've got x= 0 and again we're just looking at this curve here. We have got x is equal to 1 and we have x is equal to -3. So again the question just says to find the finite region between the curve and the x-axis but you will see when I draw this we have a point at -3 we have a point at one and we have a point at zero and when I draw this curve in again just drawing a sketch of it so it's not a scale but it is going to go up through -3 down through zero and then up through one and you can see here that we have a finite region above the x-axis here and we have under the one below the x-axis here. And when we are working this out, something's going to change in our answer. We'll have a look at that when we get to it. But for now, we just need to follow very similar steps. We're going to integrate our equation, and then we are going to use our definite integration between these two points, but we're going to do them separately. So, we're first going to have a look at between -3 and 0, and then we're going to have a look between 0 and 1. So, first of all, let's expand those first two brackets. So, that would give us y is equal to x^2 - x. And we're going to multiply that by the third bracket there, or the third piece. Obviously, the first piece wasn't a bracket, but we're going to multiply that by x + 3. If we expand that out, we're going to get y is equal to x^2 * x is x cubed. Then, we get + 3x^2 - 2x^ 2. So + 2x^2 and the -x * 3 is - 3x. So that is our equation expanded. Now we're going to integrate it. So if we integrate that we get x ^ of 4 over 4. Then we get + 2x cubed over 3. And then for the final piece there we get minus 3x^2 over 2. Again, we're not going to put our plus C in there because we are going to be using our limits and therefore that plus C is going to disappear anyway. So, we'll put this in brackets. We'll put our dx at the end as we know that is what we are using. So, when we put our limits in, we're going to do this in two parts. To start with, we will have a look at the first part here and then we will have a look at the second part. So if we start by having a look at the first part, we are going to put our limits in now. So that is between 0 and -3. And if we substitute our numbers and we're doing this with respect to the y equation. So we'll put y dx here. And then we'll substitute our numbers in. So when we go about doing this just so that we don't have to rewrite that full equation, then we can put zero in to start with which if you have a look just comes out as zero. And then we're going to take away and we need to now substitute -3 in. So we have -3 to the power of 4 all over 4 plus our second fraction 2 * -3 uh cubed all over 3. and then take away 3 * -3^ squared all over 2. And there we go. If we work that out and type that into our calculator for that, we get the answer. And let's just put this up here. So for our first one, we get the answer 45 over4. So that area comes out as 45 over4, which you can leave like that for the moment. And that is our first area. Moving on to the second one. And this time we are going to integrate between the points one and zero. Let's just change that n there. So that this time is between 1 and 0 again. So we don't have to write the full equation. We'll put y dx. And this time when we substitute one in, let's have a think. We get 1 ^ of 4 over 4. So that's just a quarter plus 2 * 1 cubed is just 2. So it's plus 2 over 3 take away 3 * 1^ 2. So 3 over 2. There we go. And we are going to subtract when we sub zero in. Sub zero into all of it. We already know that comes out as zero. So that comes out as zero. And that comes out as 7 /2. Now this is the important part and this is where something changes when it's under the axis. An area cannot be negative but when it comes out in this form in integration it knows that it's under the axis and it gives us the area as a - 72. So all we have to do is we change that area to a positive number. So we write that it's positive 72 and that is literally it for this extra element when it is under the axis or under the x-axis. So when you get a negative area, you know it's under the x-axis. And as an area, we just write it as the positive version. So to finish this question off, now we know that the area is positive. We just add them together. So 45 over4 plus the positive 72 and that comes out as you can do this on the calculator. It comes out as 71 over 6. So there we go. That is our final answer. 71 / 6. Again, you could turn it into a mixed number if you want, but it's not required and it's not necessary. So, we can leave that as 71 / 6. Okay. So, looking at this question, it says the curve C has equation y = x^2 + 2. It says the line y = 6 intersects curve c at the points a and b. Find the area of the finite region bounded by the curve c and the line ab. So if we draw a sketch of this to see what it looks like, the x^2 + 2 is just an x squar graph that's been translated up by two. So it would look something like that. Now if it's been translated up by two, it does cross through the x-axis at point 2 just here. And that will also help us to draw the line y= 6 as we know it has to be above that. So if we draw that line in as well, it goes across here through the 6 and it crosses over here and here. So we can call that point A and we can call this one point B. So we are looking at the area that is the that is bounded by the curve and the line AB. So that is going to be this area just here. So again this is above the x-axis but it is bounded by that line y = 6. So this is an area between a curve and a line. Now when we go about doing this we just need to think logically about how we would work that out. If we can find the coordinates of a and b we can go down and we can imagine where the x coordinate would be just here. If we do the same with B, we know the X coordinate is going to be just there. And that is going to form a rectangle. Now, if we were to work out the area of that rectangle, and then subtract the area underneath the curve just here and over here, then that would just leave us with the shaded region that is bounded by the curve and the line. So to start with, we can take many multiple steps here, but I'm going to start with solving to find the coordinates where they cross over. And we can do that by setting these equations equal to each other. So if we take the equation of the curve x^2 + 2, set it equal to the equation of the line, which is 6. And now we can solve that. So if we solve this, we get x^2= 4. And therefore if we square root both sides x has to be equal to plus or minus2. So if x is equal to plus or minus2 we can see that the x coordinate will be two over here. It'll be -2 just over there. So now we can work out one of two things. We can either work out the area under the curve or we can work out the area of the rectangle. Probably easier for us to just get the area of the rectangle so we don't forget. So for the rectangle, there we go. That is that has a base length across here of -2 to 2. So that is a base length of four. And it has a height going up here going from the x axis up to the y= 6 line. So it has a height of six. So the area of our rectangle is going to be four along the bottom multiplied by 6, which is equal to 24. So I'm going to underline that so we don't forget about that one. Now we want to find the area under the curve. This is quite nice as it's just between -2 and 2. So we're not going to have to do this twice like on some of the other questions. So if we write this out, we are going to integrate between 2 and -2. And the piece we're going to integrate is x^2 + 2. So we just write x^2 + 2 dx. Now we can actually go about integrating that. So if we go about writing this out, we would could put it into our squared brackets and we would say and let's just integrate that now. So that would become x cubed over 3 plus and then increasing the power of two just makes it 2x. And again we'll just write the limits there which is 2 and -2. So now we just need to go about actually substituting those numbers in. So if we substitute two in let's see what we get. So for the first one here if we substitute two we get 2 cubed / 3 + 2 * 2. For the next one which we're going to take away, we've got something very similar looking and that is -2 cubed / 3 and then that is going to be + 2 * -2. There we go. And that is what we need to work out. So if we type that into our calculator, let's see what we get. We get the answer 40 / 3 or 40 over 3. Now that again could be written as a mixed number but of course you don't need to write it. We can just leave it as 40 over3. So we have the pieces that we need. We know that the area of the rectangle was 24 and the area under the curve there bounded by the x-axis is 40 over3. So we already discussed at the start that to get this area here that is between them, we just need to subtract them away from each other. So we'll take the area of the curve uh sorry the area of the rectangle which is 24 and subtract the area under the curve which was 40 over3. And if we again just typing that into our calculator we get the answer that comes out as 32 over3 or of course you could write that as a mixed number. You could write it as 10 and 2/3 but 32 over3 is fine. So 32 over3 would be the final answer for the area that's bounded by the line y = 6 and a curve. Okay, well done for making it to the end of this chapter. Don't forget to check in the description for the rest of the other chapters that we'll be looking at throughout the as pure mathematics curriculum. And until next time, I will see you on the next chapter. [Music] Over and over it. [Music]