Learn about The Indefinite Integral or Anti-derivative Video

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Learn about The Indefinite Integral or Anti-derivative

Learn about the Indefinite Integral or Anti-derivative Welcome to the presentation on the Indefinite Integral or the Anti-derivative. So let’s begin with a bit of our review of the actual derivative. So if I were to take the derivative, d/dx it’s just the derivative operator. If I want to take the derivative of the expression x² this is an easy one if you remember the derivative presentation. Well this pretty straight forward you just take the exponent that becomes the new coefficient right the actual multiply times the old coefficient but in this case the old coefficient is one so 2 x 1 = 2 and you take the variable 2x and then the new exponent will be one last and the old exponent so it will be 2x^1 or just 2x. So that was easy if I had y = x² we now know that the slope at any point on that curve it would be 2x. So what if we want to go to the other way? Let’s say if we want to start with 2x and I want to say 2x is the derivative of what? Well we know the answer to this question right because we just took the derivative of x² and we figure out 2x. But let’s say we didn’t know this all ready, you could probably figure it out intuitively on how you can kind of do this operation that we did here, how you can do it backwards right. So in this case, the notation—well we know its x² but the notation for trying to figure out 2x is the derivative of what we can say that—let’s say 2x is the derivative of y. So 2x is the derivative of y. Then we can say this. We can say that y is equal to—I’m going to throw some very fancy notation at you and I’ll explain it more either—actually I’ll explain why we use this notation in a couple of presentation down the road but you just have to know it at this point what the notation means or what it tells you to really do it should just the entire derivative or the indefinite integral. So we could say that y = the indefinite integral of 2x dx and I’m going to explain what this quick line here and dx but all you have to know is when you see this quickly line in this dx and in something in between that you really all their asking is they want you to figure out what the anti-derivative of this expression is and I’ll explain later why is this called the indefinite integral and actually this notation will make a lot more sense when I’d showed you what a definite integral is. But let’s just take if for granted right now that indefinite integral which I just drew here is kind of like a little squarely thing is just the anti-derivative. So y = anti-derivative essentially or the indefinite integral of the expression 2x. So what is y equal too? Well y is obviously is equal x². Let me ask you a question. Is y just equal to x² because we took the derivative and I clearly the derivative of x² is 2x but what is the derivative of x² + 1? Well the derivative of x² is still 2x and what’s the derivative of one? Right derivative of one is zero so its 2x + 0 or still just 2x. Similarly, what’s the derivative of x² + 2? Well the derivative of x² + 2 once again is 2x + 0. So now the derivative of x² plus any constant is 2x so really y could be x² plus any constant and for that any constant we put a big C there. So x² + c and you’ll meet many calculus teachers that will mark this problem wrong if you forget to put the plus C when you do an indefinite integral. So you’re saying so—okay you’ve showed me some notation. You’ve reminded me that the derivative of any constant number is zero but this really doesn’t help you solve indefinite integral. Well let’s think about away that a systematic way if I didn’t do it for you all ready that we could solve an indefinite integral So let’s say we want to say—let say we said y is equal to the indefinite integral of let’s say—let me throw something interesting in there. Let say this indefinite integral of x³ dx. So we want to figure out some function whose derivative is x³. Well how can we figure that out? Well just from your intuition you probably think well that’s probably something time x to the something right. Well let me write. So let say that y = Ax to the n right? So then what is dy/dx. dy dx or the derivative of y is then what we learn in this derivative module you take the exponent multiply it by the coefficient so it’s dy/dx = (A n)x to the n^-1 Well in this situation we are saying that x³ is this expression is the derivative of y right. This is equal to x³. So if this is equal to the x³ what’s n? What’s A and well here what’s a and what’s n? Well, n is easy to figure out right. N - 1 = 3 so that means that n = 4 right and then what is A =? Well A times n = 1 right because we just have a one in this coefficient this is a starting coefficient of 1. So A times n is one. If n is four then a must be one-fourth right. So just using this definition of a derivative I think we now figured out what y is equal to? Y = 1/4x^4 I think you might start seeing a pattern here. Well how do we get from x³ to ¼ x^4. Well we increase the exponent by one and whatever the new exponent is we multiply it times one over that new exponent. So let’s think if we can do a generalized rule here that the indefinite integral and of course plus c. I would have failed this exam. So let’s make a general rule that if I have the integral of well since we already use A let’ say B x^n dx right, what is this integral. This is an integral sign. Well my rule is, as I raise the exponent on x right by one so it’s going to be x to the n plus one and then I multiply x times the inverse of this number so 1/n+1 of course I have that B there all the time and one day I’ll do more of more rigorous proof and maybe will vigorous as well as to why this B just stays multiplying but actually if you—I mean I don’t have to do rigorous or a proof if you just remember how derivative is done, you just multiply this times the exponent minus one right. So here we multiply the coefficient times the one over the exponent plus one. It’s just the inverse operation. So let’s do a couple of examples like this really fast. I think when the examples at least for me really hit the point home. So let’s say I want to figure out the integral of 5x^7dx, well I take the exponent increase it by one so I get x^8 and then I multiply the coefficient times one over the new exponent so its 5/8x^8 and if you don’t trust me take the derivative of this. Take the derivative dtx of 5/8x^8. Well you multiply 8 times 5/8 well that equals five x^—and then the new exponent will be 8-1 = 5x^7 + c. I don’t want to forget the plus c. So I think you have a sense of how this works. In the next presentation, I’m going to do a bunch more examples and I’ll also show you how to kind of reverse the chain rule and then we’ll learn integration by part which is essentially just reversing the product rule. See you on the next presentation.

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Learn about the Indefinite Integral or Anti-derivative

Welcome to the presentation on the Indefinite Integral or the Anti-derivative. So let’s begin with a bit of our review of the actual derivative. So if I were to take the derivative, d/dx it’s just the derivative operator. If I want to take the derivative of the expression x² this is an easy one if you remember the derivative presentation. Well this pretty straight forward you just take the exponent that becomes the new coefficient right the actual multiply times the old coefficient but in this case the old coefficient is one so 2 x 1 = 2 and you take the variable 2x and then the new exponent will be one last and the old exponent so it will be 2x^1 or just 2x. So that was easy if I had y = x² we now know that the slope at any point on that curve it would be 2x. So what if we want to go to the other way? Let’s say if we want to start with 2x and I want to say 2x is the derivative of what? Well we know the answer to this question right because we just took the derivative of x² and we figure out 2x. But let’s say we didn’t know this all ready, you could probably figure it out intuitively on how you can kind of do this operation that we did here, how you can do it backwards right.

So in this case, the notation—well we know its x² but the notation for trying to figure out 2x is the derivative of what we can say that—let’s say 2x is the derivative of y. So 2x is the derivative of y. Then we can say this. We can say that y is equal to—I’m going to throw some very fancy notation at you and I’ll explain it more either—actually I’ll explain why we use this notation in a couple of presentation down the road but you just have to know it at this point what the notation means or what it tells you to really do it should just the entire derivative or the indefinite integral. So we could say that y = the indefinite integral of 2x dx and I’m going to explain what this quick line here and dx but all you have to know is when you see this quickly line in this dx and in something in between that you really all their asking is they want you to figure out what the anti-derivative of this expression is and I’ll explain later why is this called the indefinite integral and actually this notation will make a lot more sense when I’d showed you what a definite integral is.

But let’s just take if for granted right now that indefinite integral which I just drew here is kind of like a little squarely thing is just the anti-derivative. So y = anti-derivative essentially or the indefinite integral of the expression 2x. So what is y equal too? Well y is obviously is equal x². Let me ask you a question. Is y just equal to x² because we took the derivative and I clearly the derivative of x² is 2x but what is the derivative of x² + 1? Well the derivative of x² is still 2x and what’s the derivative of one? Right derivative of one is zero so its 2x + 0 or still just 2x. Similarly, what’s the derivative of x² + 2? Well the derivative of x² + 2 once again is 2x + 0.

So now the derivative of x² plus any constant is 2x so really y could be x² plus any constant and for that any constant we put a big C there. So x² + c and you’ll meet many calculus teachers that will mark this problem wrong if you forget to put the plus C when you do an indefinite integral. So you’re saying so—okay you’ve showed me some notation. You’ve reminded me that the derivative of any constant number is zero but this really doesn’t help you solve indefinite integral. Well let’s think about away that a systematic way if I didn’t do it for you all ready that we could solve an indefinite integral

So let’s say we want to say—let say we said y is equal to the indefinite integral of let’s say—let me throw something interesting in there. Let say this indefinite integral of x³ dx. So we want to figure out some function whose derivative is x³. Well how can we figure that out? Well just from your intuition you probably think well that’s probably something time x to the something right. Well let me write. So let say that y = Ax to the n right? So then what is dy/dx. dy dx or the derivative of y is then what we learn in this derivative module you take the exponent multiply it by the coefficient so it’s dy/dx = (A n)x to the n^-1 Well in this situation we are saying that x³ is this expression is the derivative of y right. This is equal to x³. So if this is equal to the x³ what’s n? What’s A and well here what’s a and what’s n? Well, n is easy to figure out right. N - 1 = 3 so that means that n = 4 right and then what is A =? Well A times n = 1 right because we just have a one in this coefficient this is a starting coefficient of 1. So A times n is one. If n is four then a must be one-fourth right.

So just using this definition of a derivative I think we now figured out what y is equal to? Y = 1/4x^4 I think you might start seeing a pattern here. Well how do we get from x³ to ¼ x^4. Well we increase the exponent by one and whatever the new exponent is we multiply it times one over that new exponent. So let’s think if we can do a generalized rule here that the indefinite integral and of course plus c. I would have failed this exam.

So let’s make a general rule that if I have the integral of well since we already use A let’ say B x^n dx right, what is this integral. This is an integral sign. Well my rule is, as I raise the exponent on x right by one so it’s going to be x to the n plus one and then I multiply x times the inverse of this number so 1/n+1 of course I have that B there all the time and one day I’ll do more of more rigorous proof and maybe will vigorous as well as to why this B just stays multiplying but actually if you—I mean I don’t have to do rigorous or a proof if you just remember how derivative is done, you just multiply this times the exponent minus one right. So here we multiply the coefficient times the one over the exponent plus one. It’s just the inverse operation. So let’s do a couple of examples like this really fast. I think when the examples at least for me really hit the point home.

So let’s say I want to figure out the integral of 5x^7dx, well I take the exponent increase it by one so I get x^8 and then I multiply the coefficient times one over the new exponent so its 5/8x^8 and if you don’t trust me take the derivative of this. Take the derivative dtx of 5/8x^8. Well you multiply 8 times 5/8 well that equals five x^—and then the new exponent will be 8-1 = 5x^7 + c. I don’t want to forget the plus c.

So I think you have a sense of how this works. In the next presentation, I’m going to do a bunch more examples and I’ll also show you how to kind of reverse the chain rule and then we’ll learn integration by part which is essentially just reversing the product rule. See you on the next presentation.

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