Integration Powers E93776

Question: Let's say you've got the integral of a x power n dx. So you can see this. If they've given you this and you're able to see the symbol here, it means you get this coefficient that you're able to see. You write the way it is and then you write the x the variable which you are respecting to the power you add a one. So this is n it will be n + 1 over n + a 1 just like that. So the moment you just add a 1 to the power get it and then write it down here as a denominator you come and say plus c. For this one we always say plus c because there are no limits here. For this one we don't write the plus c. So that's it. Let's try to answer question a. So for question a we're going to say equal to okay x the variable that you're able to see here and then we say 4 + a 1. So this 4 + a 1 over 4 + 1 plus here it's a 3x^ 2 + a 1 over 2 + 1. I hope you're able to see what is happening. We are we are adding a one to each power. So this power it's a 4. 4 + 1 and then get this power. You write it here as a denominator. There is a three here which is right over here. You add a 1 to the power. So it will be 2 + 1 and then we get this 2 + 1. We write it down here as a denominator. Thereafter we shall say minus a 1. This one x. Okay. So minus 1 x. Whatever number is here without a letter you put this letter which is right here. So it will be 1 x. So now from this stage we can clean up we shall say x and then the power there it will be power five that's 4 + 1 will give us a 5 over 5 + a 3 x^ 3 over a 3. Okay. And then here min - x. Finally we can say equal to x^ 5 over a 5 + x^ 3 - x + a c. So this is our answer. The moment we see that we've simplified to the fullest we can now introduce a plus c. What this plus c means? It means you're dealing with an indefinite integral and there's a constant which you don't know. We always put a plus C at the end. So that's how it is with integration. Let's go to the next question which is question B and see what we're supposed to do on question B. So for question B, question B, we've got the integral and the limit here. it's a one. That's a lower limit. The upper limit it's a two. And then here we've got x plus a 2 x^ -2 dx. Okay. So this is what you need to do. You can see that we've got a letter here which is x^2. So now to integrate the way it is here, it being a letter down here, it cannot work out. So we're supposed to use the laws of indices. Look at this. for indices when you've got u a one over a power n. So to to move this up here using the laws of indices it will be it will be a power n and this is the law we are applying we just move this denominator become a numerator and its power will have a negative sign. This is what we've done right here and this is how it's looking. So from this stage now we can integrate. So we shall say equal to x power here it's a power 1 + a 1. There is a power one already here. So it will be 1 + 1 over 1 + 1. Then here it will be + a 2 x^ -2 + a 1 over 2 + 1. We simplify it will be x^ 2 over 2 and then here it will be + a 2x^ -1 / -1. Okay. So we simplify further. Let me clean up here. We simplify further from here we shall say x^ 2 over a 2 - 2 x^ -1 cuz this negative and positive will give us a negative which is equal to here we shall say x^ 2 / 2 - 2 / x and then we put the limit limits here. The lower limit will be one. Upper limit is a is a two. So we substitute the limits. It will be 2^ 2. We get this two. Put it where x is over two. And then -2 over a two. We put a two where we're able to see x here and here. The upper limit. And then when we do that, we shall say minus. We now substitute the lower limit. So one here and also one down here. So this is how it will be. And when we simplify further here, it will be uh 4 over a two. So when we say 2^ 2 it will give us a 4 over this two minus a one. This is a one. Okay. Or let me just write it like this so that we can understand well. And then here we shall say minus I'm running out of space. So I'll write this down here. It will be 1 / 2 minus 1 / I mean 2 over 1 for this one. Okay. So from here since I'm running out of space, let me just pick it up from here. We've got this two into two. it's a one into four it's a two. So it it will be a 2 - one because 2 into 2 1 into two it's a one and then here we shall say minus here we've got 1 / 2 - a 2. So we can clean up this is one minus here it will Transcript: Let's say you've got the integral of a x power n dx. So you can see this. If they've given you this and you're able to see the symbol here, it means you get this coefficient that you're able to see. You write the way it is and then you write the x the variable which you are respecting to the power you add a one. So this is n it will be n + 1 over n + a 1 just like that. So the moment you just add a 1 to the power get it and then write it down here as a denominator you come and say plus c. For this one we always say plus c because there are no limits here. For this one we don't write the plus c. So that's it. Let's try to answer question a. So for question a we're going to say equal to okay x the variable that you're able to see here and then we say 4 + a 1. So this 4 + a 1 over 4 + 1 plus here it's a 3x^ 2 + a 1 over 2 + 1. I hope you're able to see what is happening. We are we are adding a one to each power. So this power it's a 4. 4 + 1 and then get this power. You write it here as a denominator. There is a three here which is right over here. You add a 1 to the power. So it will be 2 + 1 and then we get this 2 + 1. We write it down here as a denominator. Thereafter we shall say minus a 1. This one x. Okay. So minus 1 x. Whatever number is here without a letter you put this letter which is right here. So it will be 1 x. So now from this stage we can clean up we shall say x and then the power there it will be power five that's 4 + 1 will give us a 5 over 5 + a 3 x^ 3 over a 3. Okay. And then here min - x. Finally we can say equal to x^ 5 over a 5 + x^ 3 - x + a c. So this is our answer. The moment we see that we've simplified to the fullest we can now introduce a plus c. What this plus c means? It means you're dealing with an indefinite integral and there's a constant which you don't know. We always put a plus C at the end. So that's how it is with integration. Let's go to the next question which is question B and see what we're supposed to do on question B. So for question B, question B, we've got the integral and the limit here. it's a one. That's a lower limit. The upper limit it's a two. And then here we've got x plus a 2 x^ -2 dx. Okay. So this is what you need to do. You can see that we've got a letter here which is x^2. So now to integrate the way it is here, it being a letter down here, it cannot work out. So we're supposed to use the laws of indices. Look at this. for indices when you've got u a one over a power n. So to to move this up here using the laws of indices it will be it will be a power n and this is the law we are applying we just move this denominator become a numerator and its power will have a negative sign. This is what we've done right here and this is how it's looking. So from this stage now we can integrate. So we shall say equal to x power here it's a power 1 + a 1. There is a power one already here. So it will be 1 + 1 over 1 + 1. Then here it will be + a 2 x^ -2 + a 1 over 2 + 1. We simplify it will be x^ 2 over 2 and then here it will be + a 2x^ -1 / -1. Okay. So we simplify further. Let me clean up here. We simplify further from here we shall say x^ 2 over a 2 - 2 x^ -1 cuz this negative and positive will give us a negative which is equal to here we shall say x^ 2 / 2 - 2 / x and then we put the limit limits here. The lower limit will be one. Upper limit is a is a two. So we substitute the limits. It will be 2^ 2. We get this two. Put it where x is over two. And then -2 over a two. We put a two where we're able to see x here and here. The upper limit. And then when we do that, we shall say minus. We now substitute the lower limit. So one here and also one down here. So this is how it will be. And when we simplify further here, it will be uh 4 over a two. So when we say 2^ 2 it will give us a 4 over this two minus a one. This is a one. Okay. Or let me just write it like this so that we can understand well. And then here we shall say minus I'm running out of space. So I'll write this down here. It will be 1 / 2 minus 1 / I mean 2 over 1 for this one. Okay. So from here since I'm running out of space, let me just pick it up from here. We've got this two into two. it's a one into four it's a two. So it it will be a 2 - one because 2 into 2 1 into two it's a one and then here we shall say minus here we've got 1 / 2 - a 2. So we can clean up this is one minus here it will