Learn about Indefinite integrals - part II

Khan Academy Presents: Examples of taking the indefinite integral (or anti-derivative) of polynomials.

Welcome back. In this presentation I'm just going to do a bunch of examples of taking the anti derivative or the indefinite integral of a panel of new expressions and hopefully I'll show you that it’s a pretty straight forward thing to do. So let's get started. If I want to take the indefinite integral and you could do a web search for integral and you'll see this drawn properly. Take the indefinite integral of; let me make something a big expression. Let say I want to take indefinite integral of 3x^(-5) - 7x³ + 3 – x^9. So you might already by intimidated by what I wrote down. Well, if you saw the last first issue you probably really like well the infinite integral you know it looks like fancy math, isn’t that fancy or at least it isn’t that difficult to perform. And all you have to realized now is when we took the derivative, if we took the derivative of a polynomial which is the sum of the derivatives of each of the terms and actually it turns out the same way the other way around. The anti derivative of this entire expression is just the sum of the anti derivatives of each of the individual terms. So we can just take the anti derivative of each term and we look at the answer. So what is this equal? Well in this case 3x^(-5) power so we take the exponent we add one to the exponent so now we get x ^(-4), right. x(-4) and then we multiply the coefficient times one over the new exponents. So one over the new exponent is (- ¼), right, so three times (–¼) is (–¾) and let see here we have x³. So instead of x³ let raise it by one number so we get x^4. And then we multiply the coefficient, you know, we could just either just keep the minus and say the coefficient seven or we can just say the coefficient is (-7) we multiple the coefficient times one over the new exponent. So the new exponent is four so we multiple ¼ x (-7) so (-7)/4. And now this is interesting just three, well how do we apply this? Well is it three the same thing as three times x^0, right because x^0 is just one and let say you should view it, it shows you that this rule is actually very consistent. So what's the anti driven above three? Well, if we view three is 3x^0 we raise the exponent by one so now we’re going to have external one, right. And x^1 is just x so I'm just going to leave as an x and we multiply it the old coefficient to this three or the derivative coefficient we multiply that times one over the inverse of this new exponents. So the exponent is one so the inverse of one is one so just this three. We divided, we multiply three times 1/1 which is so just three. And then finally x^9 I think you're getting the hang of this. We raise the exponent by one x^10 and then we multiply the current coefficient, well the current coefficient is (-1), right we just didn’t write the one there. We multiply the current coefficient minus one times one over the new exponent so it’s minus one over ten. There we did that wasn’t too difficult of taking the anti derivative or, oh, once again as I always forget +c, right because we take the derivative of any positive becomes zero so it might disappeared here so + c that were, this is any constant this could be a ten, could be a million, could be a minus trillion, any constant. And just to really hit the point let's take the derivative of this and just make sure we got this expression. And this hopefully is second nature to you by now. And you know if you ever run out of practice proms on your book because we love doing math so much just make our problem that’s what in doing. All right, I do to see him when I'm not recording for this just for fun. So, let's take the derivative of this. (-4) x this coefficient (-4) x (-¾) is 3x and then we subtract one from this exponent minus five. And then (-7/4) x 4 is (-7), x to the, we take one from this exponent x³ and I promise you I'm not even looking up you, I know you might think well what's up he just looks up here. But no I'm actually in my head at least working through this. And then plus the derivative of 3x, well the derivative of 3x is three is almost second nature now but you can probably this is 3x^1 and you say 1 times 3 is 3 times x^0 and then 10 x (-1/10) Well that just minus one, x to the one last in ten, x ^9 plus what's the derivative of any constant, right zero. You could almost do this constant as some number times x^0 and if take to derivative, well you multiply the zero times c and you get zero and then you might get (-1) depending on how you doing it but that’s kind of interesting question. We get a zero here and if you simplify that that is equal 3x^(-5) – 7x³ + 3 – x^9. I think we have time for one more problem like this. I think you probably got this, this is probably one of the more straight forward things you learn on mathematics and in the future presentation I'll give you more of an intuition of what you actually why the anti derivative is useful and then you know, using, we’re loading the indefinite integral but we could learn to use the definite integral which we’ll learn a couple presentation to figure out things like the area under curve or the volume of the rotational figure. Well that will confuse you too much. Let's do one more problem, so the integral of (-½) x^(-3) + 7x ^5. Well, let's start with this term of the polynomial. We raise the exponent one so x^(-2) now right, because we added one to negative three and then we multiply one over this new exponents times the old coefficient and actually I write all these types. The old coefficient is (-1/2), right? The old coefficient is (-1/2) we’re using colors. So this is a (-2) so we multiply it times (-1/2), right? And then let me switch the colors back plus we raised the exponent by one, x^6 and we multiply the old coefficient times one over the new coefficient; times 1 over 6 it’s a multiplication to the other six. And so what's the answer? Well what's (-½) times (-½)? Well, that’s positive ¼ x^(-2) and of course +c and as you can tell this was by main source of missing points on calculus quizzes. ¼ x^(-2) + 7^6x^6 + c there you go and if you want to take the derivative (-2) x ¼ is (-2/4) which is (-½) x^(-3) and then 6 x 7^6 is 7x and then you decrease the exponent by 1 x^(5) and the derivative of a constant is zero. And then if we get our original expression. So I think, hopefully at this point you’re pretty comfortable, taking a derivative of polynomial. And then given a polynomial you can actually take the anti derivative, go the other way. And never forget to do your +c and I hope you understand why we have to put that constant there because when you take an anti derivative you don’t know whether the original thing that you took the derivative up I guess had a constant there because a constant derivative is zero. Hopefully I confuse you with that last statement. I will see you the next presentation and I'll show you how to reverse the chain rule. See you soon.