 Welcome back to our lecture series Math 1220, Calculus II for students at Southern Utah University. As usual, I will be your professor today, Dr. Andrew Misalign. What we're going to do in lecture nine of our lecture series series, introduce a technique of integration that's called integration by parts. It's an extremely useful technique of integration, one that we'll use a lot. Not just in Calculus I, but in further classes like Calculus III, differential equations, what have you. The reason we're going to use a lot is the integration by parts is essentially the anti-derivative version of the product rule. That's right. Let me give you some explanation on how that is the case. With the product rule, let's say we have two functions, two functions u and v. These are functions of x, of course. If we take the derivative of u times v with respect to x, the product rule tells us from Calculus I, that the derivative of u times v will be u times the derivative of v with respect to x, added to v plus the derivative of u with respect to x as well. This is the product rule. Now, if we were to multiply both sides of this equation by the value dx, this differential here, it might seem a little bit weird. Isn't dx just part of the notation of the derivative? Well, in some essence, these dx is these dv's, these dv's, these r numbers. They're infinitesimal numbers. These numbers that are super, super small, right? They're smaller than any positive number, but then themselves are still positive. You can clear these denominators as if these are fractions, because after all, derivatives are limits of difference quotients, and so limits of fractions and still behave like fractions. We can clear the denominators, and if we do that, we get the statement d of uv is equal to u dv plus v du. So you get this equation of differentials. Now, if we were to integrate these things, so we integrate the left-hand side, and we can integrate the right-hand side as well, but as there's a sum, integration properties allows us to integrate the two pieces separately. If you integrate both sides, now, if you integrate just differential, you get the function uv on the left, and then on the right, you're going to get the integral of u dv plus the integral of v du. And then if you subtract the integral of v du from both sides, you get the following formula, which we refer to as integration by parts. The integral of u dv is equal to u times v minus the integral of v du. Now, when you first see the integration by parts formula, it seems kind of funky. It's like, what does this even mean? How does one actually use this? I could see how this is related to the product rule, but how do you use it? Let me show you an example. Let's consider the antiderivative of the function x times e to the 5x. Now, I want to point your attention to the first. There is no u substitution that'll help us find the antiderivative here. If we did something like u equals 5x, du equals 5dx, we're still stuck with this x in front, which is 1 5th u. If you had done a substitution like that, you would end up with, let's see, we get 1 5th u e to the u. So you can move the 5 around, but you're still stuck with something of the form x e to the x. How do you deal with that? And so integration by parts is actually the technique we're going to utilize here because u substitution doesn't quite work. Integration by parts is often useful when the integrand is a product of two functions. You'll notice there's the factor x and there's the factor e to the 5x. As opposed to many polynomial problems you see in the past, we can't really foil it out, right? I mean, do it, foil it out. OK, it's kind of already done there. It's already multiplied out. But we can still utilize that factorization with integration by parts. So when we use integration by parts, what we'll need to do is choose one function to integrate and we choose one function to differentiate. So with the formula, I'm going to write it down from before, the integral of u dv is equal to u v minus the integral of v du. So we have to make a choice about who is u and who is dv. So we have to choose a function to differentiate and we have to choose a function to integrate dv. Now, in this situation, we have to go to options. We could integrate x, that could be our u, and we could, sorry, we could differentiate x and we could integrate e to the 5x. That's a choice we could make. And if we did that, then du, if you take the derivative of x, you would end up with a dx. And if you integrated the v, you would end up with 1 5th e to the 5x. And so that's a choice that works out really well. We could take the derivative of x, we could take the integral of e to the 5x. And by doing so, integration by parts says that this integral right here, which is u dv, because we have the u, we have the dv. This will equal by the formula over here to be u v minus the integral of v du. And so if we plug in u, that's an x, we plug in v, that's 1 5th e to the 5x. In this context, when you're doing these anti-derivatives, you don't need to have a plus c, because as plus c is this gelatinous cube, that'll just kind of blend everything together. We'll get this plus c at the very end. We don't need any intermediate plus c's along the way. And then you get the minus the integral of v, which is 1 5th e to the 5x, du, which is dx here. And so this integration by parts is trying to basically push the problem to an easier integral. And hence why we call it integration by parts. We took, we found a part of the anti-derivative, but we still have an integral to calculate. And generally speaking, this new integral should hopefully be easier to compute than the previous part. So by taking the derivative of x and taking the anti-derivative e to the 5x, we are able to take a part of the anti-derivative and we can keep on going. Well, x times 1 5th e to the 5x, I guess I could put the coefficient in front. That's just good manners. Don't chew with your mouth open, put your coefficients in front. Same thing with the integral, we can bring the 1 5th out and we integrate e to the 5x dx. Can we find this anti-derivative? I sure hope so because we've already done it. It's anti-derivative is still on the screen. And so taking the anti-derivative of e to the 5x once again, we end up with another 1 5th e to the 5x. And at this point, since there's no more integrals, now I do want to write this plus c to keep track of there is this arbitrary constant that could be attached to this thing at the end. And so then a simple line just by multiplying the two 1 5ths together, we get 1 5th x e to the 5x minus 1 25th e to the 5x plus a constant. And this is our proposed anti-derivative of the function x times e to the 5x. You could try to factor this thing if you want to. I'll leave it, I'll leave it multiplied out right here. Now we could check is this the correct anti-derivative by taking its derivative. If we took the derivative of this above function 1 5th x e to the 5x minus 1 25th e to the 5x plus a constant, we could see that this is going to equal x e to the 5x. Now the derivative of constant is always going to disappear. You can factor out the coefficients. And so this will look something like the following 1 5th, the derivative of x e to the 5x. Notice that derivative is going to require the product rule because integration by parts is the opposite. The next will make a negative 1 25th, we'll take the derivative of e to the 5x and then plus zero into the derivative of constant by the product rule, like we mentioned before, the product rule, you're going to take the derivative of x which is 1 times e to the 5x. Then you're going to take x times the derivative of e to the 5x, which is going to be 5 e to the 5x. Then we take again the derivative of e to the 5x as we saw was 5 e to the 5x, like so. Distribute those 1 5ths throughout. We end up with a 1 5th e to the 5x. We end up with an x e to the 5x and then we get a minus 1 5th e to the 5x. And so we can see that the e to the 5x cancel and we're left with the original function x e to the 5x. So we have in fact found a correct anti-derivative. Integration by parts genuinely found us the anti-derivative of this function. And it's because it reverses this integration, that is integration by parts reverses this product rule situation. Now, another comment I want to do before I end this example here is we go back to the very beginning. Another possibility we could have tried to find, we could have found maybe is we've done the other way around. What if we took u to be e to the 5x and we took dv to equal just x dx in the situation? We'll take the derivative of e to the 5x is not a big deal. You just end up with 5 e to the 5x dx. Finding the anti-derivative of x is also not such a big deal. We get x squared over two. And if you try to put those things together, you would end up with x squared e to the 5x over two minus the integral of 5 halves x squared e to the 5x dx. And so you kind of see like this choice would put us in a somewhat worse situation than we've been before. Beforehand, we can take the derivative of e to the 5x no problem, but that extra x in front kind of was an obstacle we had to get around. By taking the derivative of x, we got rid of the obstacle, but by taking the anti-derivative of x, we actually make the obstacle bigger than it was before. So how one chooses you in DV does make a difference on how the calculation will go. In the next video, I'll offer you some tips on how to best choose you in DV to simplify your calculations. Check out the next video, I'll see you then.