 Welcome back everyone to our lecture series Math 1210 Calculus 1 for students at Southern Utah University. As usual, I am your professor today, Dr. Andrew Misseldine. I also have to point out that this video also works as a as actually lecture 2 for Math 1210 Calculus 2 for students at Southern Utah University. And so maybe the name, the beginning of the end doesn't make a whole lot of sense for them. For them, it's more like the beginning of the beginning, right, because they're still at the beginning of the semester, even though the Calculus 1 students are at the end of the semester. But nonetheless, whichever class you're in, the content you're going to see here is just as useful for us. And so we're in section 5.5 of Stuart's textbook. You're talking about a technique that is referred to as the substitution rule, or as I often use, I usually just call it U substitution. And that's a common term you see here because the variable that used to substitute is oftenly chosen to be U. And so before we get into the nitty gritties of this substitution technique, let me kind of mention that in previous sections, we've discussed ways of anti differentiating the power rule, the sum and difference rule, constant multiple rule for differentiation. We've talked about antiderivatives of exponentials. We've had an antiderivative be a logarithm before. Basically what we're trying to do is we're turning around derivative rules and turning them into antiderivative rules. We can talk about antiderivatives, which are trig functions, antiderivatives, which are inverse trig functions, hyperbolic functions. And so how might we go about trying to reverse the product rule or the quotient rule for derivatives? How about the chain rule? Can we turn these things into antiderivative rules? Well, it turns out that the product rule, it's a little bit more complicated for antiderivatives. And so this will show up sometime in the future in calculus two known as integration by parts. And so that's not our topic for today. And we're never actually going to actually try to come up with a rule for the quotient rule that is an antiderivative version of the quotient rule. Because it turns out as you've seen before that if we can do the product rule and the chain rule, we can do quotients. And so that's kind of how we're going to handle for antiderivatives as well. So what we have on the docket for us is what can we do about the chain rule? Can we reverse the chain rule into some type of antiderivative technique? And so let's explore a little bit of what the chain rule even tells us. So imagine we have a function f of x and for the moment we're going to call it u equals f of x. And again, as the chain rule typically has a function inside of a function, this u variable is going to act as our inner function. And so if we take the derivative of u with respect to x, this is just f prime of x. This is what we've been before. Now this statement right here, this statement right here, du over dx equals f prime of x. This is a equation. And on the left hand side, you have these differentials du over dx. If a college algebra student were to start watching this video and be like, oh, what's happening at three minutes into this video? Oh, I could watch the beginning, but we'll just have it the way it is. du over dx. Oh, I could just times both sides by dx and cancel out the denominator. And they would get this statement right here, right? And that algebra student who knows nothing about calculus would actually be completely justified in doing what they just did right there, right? Derivatives are limits of difference quotients, difference quotient, quotient. Let me emphasize that they behave like quotients and you can clear denominators and get this differential statement du equals f prime of x dx. Now, you can integrate differentials. And in particular, if you take the differential right here, the integral of du of du, this is going to be equal to because you can substitute out the du. So there's a lot of clutter here. Let me get rid of it real quick. So if you take this du right here, you can substitute it with this f prime of x dx. So du is the same thing as this portion right here, f prime of x dx. So you can make that substitution. But likewise, like we said before, u is the same thing as f of x. So if u is the variable inside of g, we could also replace that with a f of x there. So the integral of gudu is the same thing as the integral of g of f of x times f prime of x dx. So kind of get rid of this clutter a little bit here. So this gives us this pretty critical equation right here. And the way we generally use it is actually go from this direction to this direction. The right hand side looks a little bit more complicated, right? Yeah, it is. And so what happens here is if we can recognize there's an inner function, there's a function inside of g, we can simplify just to become g of u. But to do that, we have to also find the inner derivative, which is sitting right here, this f prime of x dx, the inner derivative. And so this idea of u substitution is about recognizing you have a function inside of a function and recognizing that inner derivative is there as well. And so this really is saying the same thing as the chain rule, as you can see right here. If you take the derivative of g of f of x with respect to x, you'll get g prime of f of x, your so-called outer derivative. And then you get f prime of x, your inner derivative right there. And so u substitution is trying to integrate this thing right here. And so let's see via an example how one actually uses this u substitution in practice. So consider the integral of 2x cubed plus one, and all of that's raised to the fourth power, times that by 6x squared dx. This function right here, we want to integrate it, and it is a polynomial. One could sort of foil it out the 2x cubed plus one to the fourth, distribute the 6x squared there. One could treat that as an expanded polynomial, but that's a lot of arithmetic that's completely unnecessary for this problem. What we want to recognize instead is the following. We have this function 6x cubed plus one that sits inside of the fourth power, right? We have this function inside of a function, there's a chain going on right here. And so we want to play this chain game using this composition of functions. So what we're going to say is the following, we're going to take u to be that inside function 2x cubed plus one. And we want to then compute, in order to use this u substitution, we're like, okay, we have an inner function, but we have to have the inner derivative. What's the inner derivative of du? Well, du remember is just going to be f prime of x dx. Don't forget the differential there, the dx is critical here. So take the derivative of our function f of x right here. And if you do that by the usual rules of derivatives, you're going to get 6x dx. And you're going to notice that voila, it's right here. This function has the inner function and the inner derivative. And as such, we can compute the anti-derivative using this u substitution. So what we're going to do is when we see the u, the 2x cubed plus one, we replace it with u. So we end up with a u to the fourth. And then this entire 6x squared dx, this entire thing is du. And so we're going to make that substitution that 6x squared dx becomes a du. And so then we want to integrate the expression u to the fourth du, which is a much easier integral to do by the power rule. This becomes u to the fifth over five plus a constant. Don't forget your constant there. Now, u was this variable we inserted into the problem that wasn't part of the original setup. We need to go back to the variable x right here. And as u is equal to 2x cubed plus one, we see that our contender for the anti-derivative is one fifth 2x cubed plus one to the fifth plus a constant. And so I claim right here that this is our anti-derivative. And if we have any doubts whatsoever, we could take the derivative of this thing, take the derivative with respect to x of this one fifth 2x cubed plus one to the fifth plus a constant right here. And if you take the derivative of this little derivative of the constant, we'll just go to zero. We have a chain going on here 2x cubed plus one inside of the fifth power. So maybe comes as no surprise when you take the derivative here, you're going to use the chain rule because it's the opposite, the inverse of this u substitution. By the chain rule, we can take out the one fifth here. You're going to get five times 2x cubed plus one lower the power to the fourth. But then when you times by the inner derivative, you take the derivative of 2x cubed plus one simplifying this, of course, one fifth times five, they get each other. And so we end up with this 2x cubed plus one to the fourth. And then the derivative of 2x cubed plus one that inner derivative is a 6x squared and Bob's your uncle. This gives us back the original function we had u substitution is in fact the inverse operation of the chain rule. We'll do some more examples of this in the next part of these lecture videos. So look for the link that hopefully you can see right now to check those out. And I'll see you there. Bye everyone.