Introduction to Integration: Rules & Curve Equations | Class 29 | AS Level P1 9709

Generated automatically from the lecture captions, so spoken maths may be transcribed imperfectly.

Okay. So the first class for integration of as and the last topic for as for P1. So hopefully we will be completing this in three four classes maximum four classes with the passers and uh okay first of all integration another name of this integration is also called it's also an anti-ifferiation because as I told you in the last previous class as well that differentiation integration they are they come in hand to hand they are a part of calculus overall we call them calculus uh but they are very much related to each other so We have initially let's suppose if we have the y function or the fx function

when we differentiate this we get the first derivative the fdash or the dy by dx this is what we get when we differentiate this but if you want to get back to our original function again if you want to reverse this effect we integrate it. So this is what is actually integration. So, so the first application that you now understand is simply the first application is getting your original function getting your function back function back from dy by dx or from the double derivatives as well or the double derivative as a d² y upon dx² by integrating it twice. If it is your first derivative, we would integrate it

once and we will get the original function. And if it is double derivative, we would if we would have to integrate it once, we will we will get the first derivative. Then we would integrate again, we will get the original function back again. So this is the first thing that you will learn. Then there is there are another uh more things as well to this. So it's not very big but we will see this. But this is the first thing that we will do. But even before inputting this using this uh the application we have to learn how to integrate just like we did how to differentiate. So over here what do we

have the formula or how do we integrate the first one first of all this is the sign of integral you will see something like this there's not a bracket a bracket is something like this it goes in and then goes the other way. So this is the uh sign or the notation used that that will tell you that you have to integrate that we are trying to integrate this and whatever the function is it will be written over here and then dx. So this means that you are going to integrate the inside this bracket whatever is inside the bracket with respect to x. So integrating with respect to x this is what it it actually

means. And now [clears throat] the first formula is that if you have the integral x power n the same way we had over there as well differentiation x power n derivative of x power n over there what we did was that we brought the power down the power came down and multiplied and power was reduced by one but now over here it is absolutely opposite first of all we add the power n + one first we add the power and then we divide the new power n + 1. And then another thing is we also put in whenever you do the integration, we also put in the value the the constant C. This constant is called the constant of

integration. Integral and this is you have to write this down. Whenever you integrate this, you put in the plus C as well. This is you can say part of the formula. After you integrate all of the terms in the in the equation, you put in plus C as well. I will tell you why do you put this plus C later on. But right now you have to understand this is something that you need to show. So for example, if you have the integral of x^ 2 dx or if you have the integral of square rootx or if you have the integral of - 6x - 4. So how would you integrate this? First of all, simple. What do we do? The x power adds by 1.

2 + 1 divided by the new power 2 + 1 plus c. So this becomes x cq upon 3 + c. This is it simple. Over here this is x power square root square root x means x power.5 and then this would become x^.5 + 1 divided by.5 + 1 + c and then what would you get x^ 1.5 upon 1.5 + c. If you want to simplify this thing further, 1 divide by 1.5 will give you 2 upon 3. So you can even write it down 2 upon 3 x^ 1.5 + c. But this this is the integrant answer. This is just further simplification. This is your integral and the rest of it was simplification. Then the last one min - x just like in differentiation as

well the number the coefficient was just there. You just multiply the power if it comes down and divide over here as well. This minus 6 you can even bring this out. You can write this coefficient outside the integral - 6 x^ - 4 dx and then this will become - 6 still outside x^ -4 + 1 / - 4 + 1. So this will become - 6 x^ - 3 upon - 3 - - cancel out 3 * 1 3 * 2. So this becomes 2 x^ - 3 + c. This is your answer. This is the basic thing that you have to learn. First of all, how to integrate and then we would move on to how to further apply this integration. Another formula or thing you need to learn is if let's suppose there's an

integral of a constant integral of constant a it can be anything a can be 100 or minus4 or 6 or million any any constant so in differentiation what what happened was when you differentiate a constant the constant would disappear it would become zero over here it won't disappear. It will get the variable. So this will become a x plus c because this is the opposite in the differentiation your constant was getting removed. It was getting eliminated became zero. Over here the integral of a constant is that now the the constant will get a variable as well. It would become ax. So let's suppose now if we have

example integral of 4x q - 6 x - 4 dx. So this would become 4 is the coefficient. Four would remain as it is. X power 3 + 1 would become 4 / the new power 4 - 6 X power 1 + 1 will be 2 / 2 and then minus this is a constant I told you when you integrate a constant it would get the variable so - 4x and then plus c if you simplify this thing so x^ 4 - 3x x² - 4x + c. This is your answer. This does at the integral. Sometimes you may also need to simplify your equation before you start integrating. Okay? Sometimes your equation won't be as simple. Let's suppose it is something like this. integral of

x² - 7 upon under rootx. Now in this one you cannot directly start integrating you won't be able to. So first you will have to simplify use a bit of indices and separate the terms before you get it in this in this format. Separate separate terms and then that would become easier for you to integrate. So in this case what it would become in order to simplify this thing you can do x² upon under rootx - 7 upon under rootx you you separated the fraction so this would become x^ 2 -.5 and use indices you know how to do this 2 -.5 will be 1.5 - 7 x^.5 dx and now you going to start integrating. So x^ 1.5 will become add + 1 ^ 1.5 will

become 2.5 divided by 2.5 then - 7 x^.5 + 1 will be positive.5 divided by positive.5 + c and if you want to simplify this thing further you will get 1 upon 2.5 will be 2 upon 5 x^ 2.5 5 -4 x^.5 plus c. This is your answer. Okay. Any questions? Anyone? >> Nope sir. >> Later on when when they ask you to find the area that is another application. >> Okay. Okay. Okay. >> Last fraction. >> Last fraction. Sometimes I'm telling you that the equation won't be as simple. It can be something like this as well or even more complicated than this. So before you start integrating you use simple indices. You move things around

and you separate the terms before you start integrating because the rule of integration look it applies on a separate terms is x power something. So if it is in this format in a bracket from something like this and you have a denominator as well. So you cannot directly apply x^ 3 upon 3 on this thing and this would become 7x and you this would become x^ 1.5 upon 1.5. You cannot do this like this. So you will you what do you do? You first use a bit of indices you separate the fractions you simplify the powers and then you start integrating. So what did I do? So I separate this x² upon this this under rootx. It is in the

denominator of both of these. So first of all x² upon under rootx I separated this thing and minus and 7 upon under rootx you can see both of the denominators are same so I can go back to the same step again back the denominators are same so I can combine the fraction so if the fractions are fraction is the same I can separate into the same denominators as well and then I use indices this x^.5 x^2 when they're dividing we know the power subtract so 2 -.5 will become 1.5 and this is x^.5 if I move it upwards it would become.5 so 7 x^.5 and now you can see these are in separate terms now x to the power of n

something like that so the power adds by one and then divided by the new power so 1.5 + 1 will be 2.5 and divided by this new power 2.5 minus.5 + 1 will be.5 5 / the new power.5 and then this and then I have just simplified it. This is your integral. If you want to simplify, you can simplify this further. But this is what they had asked you to find the integral. Okay. Okay. One more example. There's a third formula as well for integration. But before moving on, let's clear this. If you have integral 2x cq - 5x + 1 dx. Now this is also the basic one. Not very difficult. The two is a coefficient. It

remains there. X power adds by 1. 3 becomes 4 and divided by the new power 4 - 5 x^ 1 1 + 1 it's 2 upon 2 new power plus 1 is the 1 is a constant so it will get the variable x so 1 this this would become 1 x then plus c if you want to simplify this would become x 4 upon 2 - 5 upon 2 sorry 5 2x² + x + c. This is your answer. And now the last formula to learn for integration. If you have integral of a bracket raised to the power n in fact it can be anything inside the bracket. Whatever is inside the bracket doesn't matter. you will do it the same way. So if the bracket is raised to the power, it can the bracket in the inside

of the bracket is a function of x. Okay, it's something with x. So how do you integrate this? What happens? First of all, no changes inside the bracket. Whatever is inside the bracket will remain inside the bracket. Same as it is. The power of the bracket increases by one + one divided by the new power and plus one. And one more thing is that you will also write the derivative of the inside of the bracket in the denominator and then plus c. So you will differentiate the inside of the bracket and you will put it in the denominator. Now for example you have integral of 4x + 3^ 5. So look at the formula which I've

written down and I'm just going to follow the steps from the formula. So no changes to the bracket whatsoever. 4x + 3 doesn't change. 4x + 3 is there. And the power of the bracket increases by one. So this would become six. Five becomes six divided by the new power 6. And now the derivative of the inside of the bracket, the derivative of 4x + 3. The derivative of 4x + 3 is simply four. That also comes in the denominator. And then plus c. This is your answer. If you want to simplify 4x + 3^ 6 upon 24 plus c. This is your answer. This the integral. Let's do another example. If you have let's suppose

under square <unk>2x + 1. So first of all two [clears throat] the square root means power.5 and now no changes to the bracket to the inside of the bracket. The power of the bracket adds by 1. So.5 becomes becomes 1.5 divided by 1.5 and the derivative of the inside of the bracket 2x + 1 that is derivative of 2x + 1 is 2. So two comes in the denominator plus C. This is your answer. Let's suppose you have 1 upon 5x + 3² dx. They're asking you to integrate this thing. So first you would move the bracket above. So this will become 5 x 5 x + 3 power - 2 dx. And now the exact same thing. No changes to the inside of the bracket. 4 5 x + 3

5 x + 3. Same - 2 + 1 will become minus one divided by minus 1. And the derivative of the inside 5 x + 3 will be 5 + c. This is your last and final integration rule that you need to learn. Okay. Now the first application of using integration. Find the equation of the curve through 1A 2 for which dy by dx is given to you under <unk>2x + 1. simple question and a very very powerful you will get very similar questions the past as well. So they have given you the dy by dx sorry it is 2x + 1 power.5 and they have given you a point through which this curve is passing through why have they given you a point of so that

we can find the value of c to find c this is the condition They have given us given it to us. Whenever you do the integration right now if you would integrate this you you want to get the original function back. I told you if you want to go back to the original function you would integrate this. So when you integrate you will get y is equals to the integral of this 2x + 1 power.5. So you would do it the usual way. What would happen? The bracket will not change. This would become 1.5 / 1.5 multiplied by the derivative of the inside of the bracket which is 2 plus C. Right? Now because you don't know the

value of C, this equation is incomplete. When you just find the integral till here, you call it the general solution. This is the general solution not the specific solution. Okay, not the particular solution because this is still incomplete. The value of C will change the equation will change the graph. If if your C value is four, it will be a different curve. If your C value is seven, it will be a different curve. It will be know about transformation, right? It will go up and down because of this value of C. So this graph is still not the exact graph that they are talking about. Now using this point we know we would we

would substitute this point into this thing and then we will find the value of C and then we will get the particular graph the particular equation that they are asking you to find. So now if if X is we know if X is 1 your Y is Y is 2 and if you substitute we will get the value of C find the value of C. So this would become Y is 2 X is 1. So 2 * 1 + 1 power 1.5 upon 1.5 * 2 is 3 + C. You can use your calculator. This becomes 2 + 1 is 3. 3 power 1.5 upon 3. So under root 3. So your C comes out as 2 minus under root 3. This is the answer of your C and now you would replace this in the original function in the integration

that you have found 2x + 1^ 1.5 upon 3 + 2 - <unk>3 this you have replaced the value of c this is the exact solution this is the particular solution You will call this the particular solution and this is a very fast replica question and this is all how always you will do this integrate it and then use the point that they have given you to find the value of C and then replace the value of C in its Yes. Okay. Look at this. The part A, B, A, C and D they are not of a concern. That is differentation. The second part part B is the thing we need to do. They have given you look they have given you the double derivative

the double derivative is 6 x² and if you move the x cube above this will become 4 x^ minus 3 and they've given you that this is a stationary point and you know a stationary point is where your single derivative value the dy by dx it is equals to zero. From differentiation if you can recall at the stationary point at the turning point at the minimum maximum point your derivative first derivative is equals to zero. In the first part they're just asking you to find the nature of stationary point. So you would just substitute the value of x the x value of the stationary point in the double derivative and see

if it is positive negative. If it is positive you will say it is minimum. If it is negative you will say it is maximum. So that part the first part is also of differentation. The second part is of concern right now. Find the equation of the curve. So they have we have the double derivative. So this means we will have to integrate it twice. So if I would integrate it once I will get the single derivative. I'm integrating it once. 6 x² - 4x - 3 and dx sorry uh dx I will get the first derivative. So integrate the usual way this would become 6 x^ 3 upon 3 power adds by 1id by new power - 4x^ -2 upon -2 plus c.

This is your single derivative. If you simplify this will become 2x cq + 2x - 2 + c. How will you get the value of c? this value of C before we try to integrate again you will find you will have to find the C anyone how will you get the C substitute the stationary points in X and Hi >> stationary point you have the value of yes there we are very close but slightly wrong we know good good that's one that's good the value of the derivative it will be zero when your x value is you can see minus one we we don't have the y coordinate over here to put in we know that the value of the first derivative it will be zero

when a value of x will be the turning point x which is minus one and that is how you can find the value of this first c. We know that dy by dx the first derivative is zero when x is minus1. So we can use this condition and find the value of c. So this would become uh minus2 and + 2. So c is zero. So that is easier. So the first derivative we have found it is 2x cub + 2x - 2 + 0. And now you'll have to integrate it again. If you will integrate it now again you will get the original function back and you will do the integration the usual way. 2 x 4 upon 4 + 2 x -1 upon -1 + c. Simplify x4 upon 2 - 2 upon x + c. And now to

find the value of C, this value of C, this let's suppose this is C1. This is your C2. Whenever you would integrate your C, your plus C would come in. Okay? Whenever you would integrate, so don't forget this. Now you have the X value, we know that X value is minus one on this curve and the value of Y is 9 upon 2. And now we can get the value of C2 as well. So 9 upon 2 is equ= to -1 power 4 upon 2 - 2 upon -1 + c2 so 9 upon 2 1 upon 2 + 2 + c2 [clears throat] 9 upon 2 - 1 upon 2 8 upon 2 4 - 2 is 2 so c2 is 2 this means that your original function back it is X 4 upon 2 - 2 upon X + 2. This is the answer.

Five marks. Look. Okay, let's find another question similar. Question number eight. Our curve is uh such that dy by dx is given to you 8. And if I bring the bracket above this would become 3x + 2 power minus 2. The curve passes through point 2 comma 5 whole 2 upon 3. Okay. So look they have given you a point as well through the curve so that you can get the value of c. So if you integrate this so y will be equal to 8 bracket 3x + 2 power min - 2 integral so this will become 8 is a coefficient so 8 will still be 8 the inside of the bracket doesn't change 3x + 2 remains 3x + 2 power adds by 1 so -2 becomes minus1

divided by minus1 and multiplied by the derivative of the 3x + 2 which is 3 only and then plus c. So now you simplify 8 - 8 upon 3 3x + 2 power - 1 + c. And now the point that is given to you to find value of c 5 upon 3 is equ= to - 8 upon 3 and 3 * 2 + 2 power - 1 + c. So 5 whole 3 upon 2 upon 3 will become 17 upon 3 plus 8 upon 3 and this would become 6 + 2 is 8 power - 1 is equals to C. You can use your calculator to simplify make your life easier. 17 upon 3 + 8 upon 3 * 8 power - 1 6 and now you would replace it. So y = - 8 upon 3 3x + 2^ -1 + 6. This is your answer final. This is it. Four marks. Easy. Four marks.

Okay, I'm stopping the recording. You can copy this down.