 Welcome back, everyone, to our lecture series, Math 1210, Calculus I for students at Southern Utah University. As always, I'm your professor today, Dr. Andrew Misseldine. We are in our final lecture for the series. Woo, woo, woo, woo, or I mean, this is sad. Oh no, I want to learn more calculus. Well, that opportunity, of course, is presented in front of you, right? As my Calculus I viewers who have been watching these last couple of videos, you'll actually notice me say some comments about how some of these videos, which are at the end of Calculus I are actually the first couple of videos of Calculus II. Calculus II actually picks up and continues the discussion of anti-derivatives for a good long while. And so things like the fundamental theorem of calculus and this U substitution that we talked about in the previous video and in this video as well are something like the first moves that one wants to learn when playing the game called Calculus II. So if you are dying to see more calculus, feel free to click into the playlist I have for Calculus II videos. Even if you're not a Calculus II student, take a look at those. Those are gonna be pretty awesome as well. But for those who are trying to finish up Calculus I, what I wanna do is provide some more examples of this technique, which I call U substitution. And so this is the substitution rule from 5.5 in James Stewart's Calculus textbook here. So we've seen some examples in the past where and in the past videos for we wanted to identify a U and that U we choose is typically a function inside of one which parentheses often do that but there are other delimiters that tell us exactly who's inside who. So like for example, you see this integral here the integral of X times the square root of one minus X right there. And so you do see that there's this one minus X inside of the square root. And so the square root often acts like these parentheses that there's a function inside of the square root. So far to see that I would be inclined to set my U to be one minus X right there because it's the inside function. But we have to calculate the inner derivative D U which in this case would just be negative DX which I don't have the negative sign there but we can always correct that just by timing a double negative, negative on the inside, negative on the outside and which case then your negative one that's right here combines with the DX together to give you this D U. And we can write our integral now as negative the integral of the square root of U D U but we still got this sort of like this elephant in the room right here. What do we do with the X? Now I've cautioned you in the past that if you can't make the substitution work then you're gonna have to try something else right. And that could be by trying a different substitution sometimes your first U substitution doesn't work. And if it doesn't work that doesn't mean give up just means maybe try something else. And after all we can massage this problem a little bit cause when you look at this right here U equals one minus X. We often see that as an assignment. I'm deciding that U is equal to one minus X. That's fine but it's also an equation. And as an equation it can be manipulated. For example we could subtract one from both sides and we end up with U minus one equals negative X. And if we times both sides by negative one we end up with X is equal to one minus U. And this equation right here we can substitute in above for the original expression X there. And so when you do that you end up with negative the integral you're gonna get a one minus U times U to the one half aka the square root of U du. And so does this function right here put us in a simpler situation than we started with? The one of the guiding principles always about U substitution is that the value that U you choose should always make the function simpler than it was. We were trying to integrate X times the square root of minus X. Now we're trying to integrate one minus U times the square root of U. Does that make things a little bit better at all? Anything better? And I would say that we're in a better situation than we were because beforehand we had a one minus X inside of the square root. But now the one minus U is outside the square root and we have just a monomial inside of the square root. We're now in a situation where this square root of U we could distribute onto these pieces. And if we distribute them, we end up with negative the integral of U to the one half minus U to the three halves du like so. And by the power rule and the linearity properties of antiderivatives, we can calculate that antiderivative absolutely. So I would say our U substitution is a success because it put us in a situation that was easier to calculate than we had before. And so using the power rule for U to the one half, the power will raise to three halves, you add one to it, divide by three halves. For the next one, we raise the power by one so it becomes five halves, then divide by five halves. And don't forget your plus C. And so then if you divide by fractions, of course, you just multiply by the reciprocal. So you're gonna get negative two thirds U to the three halves plus, I'm distributing the negative sign throughout there, plus two fifths U to the five halves plus a constant. And then remove the U and go back to the original variable X there. You're going to get negative two thirds times the times one minus X raised to the three halves power and then add to that two fifths one minus X raised to the five halves power plus a constant. And so this gives us our antiderivative right there. U substitution should always be used to simplify the integral into something that's easier to take an antiderivative of. But after all, this assignment of U is also an equation. If you can manipulate the equation, you potentially could use that to your advantage to what makes, to make the U substitution work. Let me give you another example of this one. This is a little bit more involved here, but same basic idea. Because we have a one plus X squared inside of the square root, I would be inclined to set one plus X squared as my inner function U. And then your inner derivative will be two X DX right there. And so if we partition this a little bit, we're going to need a two over two for the coefficient. Moving things around a little bit, we end up with a one half that sits in front and X to the fourth that we don't know what to do with yet. We get a square root of one plus X squared. And then we have this two X DX. So this portion right here becomes your U and this portion right here becomes your DU. But what do you do with the X to the fourth, right? Well, the idea is again, your assignment for U can actually be, it's an equation that you can manipulate. Like if you subtract one from both sides, notice you'll get X squared equals U minus one. And then making the recognition, we'll have to get X to the fourth. If I square both sides, X to the fourth will equal U minus one squared. Make that substitution in above. And when you do that, your integral becomes one half the integral of the X to the fourth became a U minus one squared. The square root of one plus X squared becomes a U to the one half. And then the two X DX becomes a DU. Is this a function for which we have a better chance at integrating? Well, when you look at that, even if it's not easy, the question is doable, right? U minus one squared, I could foil that out. I could foil that. And if you did that, you would end up with a U squared minus two U plus one times that by U to the one half DU. Well, that U to the one half I could distribute, just like the last example, you could distribute that through. And then you're gonna get this polynomial like expression, one half in front, you're gonna end up with, so when you distribute the U to the one half, you're adding one half to each of the powers. So you're gonna get U to the five halves minus two times U to the three halves plus just U to the one half DU like so. And those are things I can anti-differentiate because I just have a combination of powers of U. So using the anti-derivative rules here for the power rule, we'll increase each power by one. So five halves becomes seven halves divide by seven halves. The next one, you're gonna get two U, you're gonna raise the power to be the five halves divide by five halves. And then lastly, your U's power will raise to three halves divide by three halves and add a constant here. And every time you divide that fraction, you really could have multiplied by a reciprocal. And here I'm also gonna distribute the one half through on all of these things. So we're gonna end up with one half times two sevenths times U to the seven halves. Then we're gonna get negative one half times two fifths. And then there's another, of course, two U to the five halves. And then lastly, we're gonna get plus one half times two thirds times U to the three halves plus a constant right here. And simplify where we can, you'll notice that the one half cancels with a lot of the twos that are in the numerator. So you get all that cancellation right there. And then substitute back in the original expression for U, which it fell off the screen right here. Remember that was X squared plus one. So in the end, we have a one seventh, one plus X squared to the seven halves power. Next, we're gonna get negative two fifths, one plus X squared to the five halves power. And then finally, we are gonna get a plus one third times one plus X squared to the three halves power plus an arbitrary constant. And this of course gives us the anti-derivative we were looking for. A little bit more involved, but that U assignment we did earlier can be massaged as an equation to pick up the pieces that we couldn't do beforehand. So we'll talk some more about anti-derivatives using U substitution in the subsequent videos. Click on the link to see those. And I'll see you next time, everyone.