 Let's talk some more about calculating antiderivatives. We've seen already how we can use the power rule for antiderivatives. That is, if we wanna integrate a function of the form x to the n dx, we end up with x to the n plus one over n plus one plus a constant. Don't forget that constant. This of course only works when n doesn't equal negative one, otherwise use the natural log in that situation. Well, what if you have more than just a power function? What if you have like a polynomial? Well, just like derivatives, there actually is a constant multiple rule and a sum and difference rule that works just like it does for derivatives. So constant multiples, if you have a function f of x and you have a constant, like a two or three or a pi in front of it, just a constant, you can actually factor this constant out of the integral. And so you get the integral, or k times the integral of f of x. This is just like it is for derivatives and limits. Also, if you take the integral of f of x plus g of x, this will equal the integral of f plus the integral of g. So the antiderivative of sum is a sum of antiderivatives. It's also true for the subtraction as well. You can take the integral of f minus g and that is the integral of f minus integral of g. Now these properties we've seen before, these two right here, this is true for derivatives, this is true for limits and it's now true for indefinite integrals or what we call antiderivatives. When something shows up over and over and over again, it kind of deserves a name and this condition right here is known as the linearity property that integrals are linear, antiderivatives are linear. So we're derivatives and limits. And in fact, linearity shows up a lot in calculus. It shows up enough that people start paying attention to it and we actually start studying it. And the study of this linearity property is what's known as linear algebra. Turns out there's a lot of linear algebra going on inside of calculus. Linear algebra is about amongst other things, systems of linear equations, matrices, vectors, a lot of cool stuff there. You should check it out sometime, but linear algebra has huge applications to calculus because of this linear property here. Basically everything in calculus is secretly linear. Calculus is just linear algebra plus limits, but we can talk about that some other time here. Let's see some examples of how we can use this linearity property to help us calculate some integrals. So take the integral two V cubed. Well, if we, by the constant multiple rule, we can take the two out, we get V cubed DV and then by the power rule, its antiderivative will be V to the fourth over four plus a constant. Which if you distribute that two through, where you get two over four, V to the fourth plus two times a constant, that simplifies just to be one half V to the fourth plus a constant. Whoa, whoa, whoa, whoa, what just happened there? We had a moment ago, two times a constant and now we just have a constant. What happened to the two times a constant? Well, one thing to remember here is that this number C is what we call an arbitrary constant. It's a constant, it's the y intercept. Well, you don't know what it is yet. It's unspecified. And so because we don't really know what C is, multiplying it by two, since we don't know what C is, we don't know what two C is either. And it's really just a placeholder for a number that we're gonna plug in later. And this arbitrary constant, it's not any more specific saying two C as opposed to C. And so you often see this abusive notation where arithmetic operations will happen to this unknown constant. And since it's unknown, it will just kind of absorb it. And I honestly like to think of this plus C as this gelatinous blob, this monstrous creature that wants to take over the city. And you have over here, it's attacking the city, it's eating everything. We have these tanks that are trying to come from the National Guard to protect everyone. And they come shoot it, but it doesn't matter what happens, it just kind of eats it, whoop, everything's gone. The blob itself will come and eat the tank, it eats the soldiers, it eats the buildings, it eats everyone, everything just gets eaten inside the blob. And this blob is none other than plus C. That whatever happens to this plus C, it's just gonna absorb it. And so the blob eats everything. That's what you wanna remember when it comes to this plus C right here, all right? So we don't have to worry about times in C by any constant. As another example, factor out the 12, we end up with Z to the negative fifth. Notice because it's a reciprocal, I can write this as a negative power. By the power rule, we end up with Z to negative four. We raise it by one, divide negative four plus a constant. You see, I'm not gonna distribute the 12 onto the plus C because my blob just eats it. Negative four goes into 12, negative three times. So you get negative three Z to negative four plus a constant. Since the original function was written using a reciprocal, I might write the answer that way as well. Negative three over Z to the fourth plus a constant. There's our answer right there. How about a polynomial? What do you do with a polynomial like this? Well, because of the sum rule, it's kind of like the sum pig rule right there. Because of the sum rule, you can break this up into three separate integrals. We get the integral of three Z squared, DZ, don't forget the differential, minus the integral of four Z, DZ. And then plus the integral of five DZ. So because of the sum and difference rule, you can break up into three integrals. Because of the constant multiple rule, you can pull out those multiples three times the integral of Z squared, DZ. You get minus four times the integral of Z, DZ. And then you get plus five times the integral of DZ. And then using the power rule, we end up with three Z cubed over three plus a constant. We'll call it C one. Next, we then get, I guess it's a minus, minus four times, that was a really big four, four times Z squared over two plus another constant C two. And then we're gonna add to this five Z plus another constant C three, like so. So we can do each of these integrals separately. And I'm gonna rewrite things just a little bit because after all three goes into three right here, two goes into four, that leaves two behind. That's the simplification we can get. And so then we're gonna end up with a Z cubed minus two Z squared plus five Z. And then we have these constants C one plus C two plus C three. What do we do with them? Well, these constants are, again, our gelatinous blobs. We have over here plus our C one. We have then next another blob plus C two. And then we have a final blob plus C three right here. Well, what happens when all of these blobs come together? It makes a mega blog. This is sort of like the power rangers of gelatinous blobs. They come all together because we take an unknown plus an unknown plus an unknown. Guess what you got? I don't know. It's an unknown. And so we can actually bring these all together as just a single constant plus C. And so this is how you're gonna see the answer for these type of integrals. You'll see this function Z cubed minus two Z squared plus five Z, but then you'll see this plus C at the end of it. You don't have to keep track of all of the constants. Just a single plus C at the end takes care of everything. I got a little bit crowded right here. When you see an integral like this one, this integral of X squared plus one over the square root of X, we might be tempted to be like, oh, I'll do the quotient rule here, but we don't have a quotient rule yet, right? And because we don't have a quotient rule, we have to try to simplify this thing algebraically before we can take its anti-derivative. And if we were to do that, algebraically what we can do is we can break it up as two fractions. We get X squared over X to the one half, DX. And then we get plus the integral of one over this X to the one half. Power is always more preferable in calculus than radicals. We get like this. So for the first one, X squared over X to the one half, if you subtract those powers, you will end up with X to the three halves power, DX. And then this one over here is the integral of X to the negative one half power. And so taking the anti-derivatives here by the power rule, we're gonna get X, add up one to three halves. That gives you five halves. And you're gonna divide by five halves. Then next you're gonna get plus X. Well, the negative one half power, if you add one to it, you're gonna get positive one. And so you divide by one half plus a constant. And when you divide by those fractions, multiply instead by the reciprocals, you get two fifths X to the five halves, plus two, the reciprocal of one half, X to the one half plus a constant. And this gives us our anti-derivative. All right, we'll do some more examples of this in the next video. So look for that in the links below. See you then.