 I want to do another example of finding and computing an integral using integration by parts, but this time let's do a definite integral as opposed to an indefinite integral. So we're trying to find the area under a curve, the function y equals the natural log of x over x squared as x ranges from one to e here. How does that change things if you want to take a definite integral as opposed to an indefinite integral? Well, the indefinite integral is just the family of anti-derivatives. The definite integral is the area under the curve, but because of the fundamental thing of calculus, we could use the indefinite integral to compute these definite integrals. So really a definite integral just kind of adds one extra step to the process, having a definite integral that is. We want to first find the anti-derivative and then use that in our calculation. And I claim we want to use integration by parts on this one right here. U equals, we have to find something we're going to integrate, dv equals, we have to find something we're going to integrate. Now, how do we take the derivative here? Well, you have these two parts, right? This thing factors as x to the negative two times the natural log of x. And then of course the dx will always just go with the dv. So who do we want to take the derivative of? Who do we want to integrate? Well, x to the negative two. Notice if you take its derivative, right? If we take its derivative, its derivative by the power rule being negative two x and negative three dx. You still get a power function. So it's like, okay, that's an interesting choice. But if we did that, we'd have to set the natural log as dv, in which case then we'd have to find the anti-derivative of the natural log. In which case we had 30 seconds for the Jeopardy music. Do, do, do, do, do, do, do, right? We don't know the anti-derivative of the natural log or at least we haven't seen it yet in this course. Some of you might have seen it from some other course, what have you. But we don't know the anti-derivative of natural log yet. We will actually see what the anti-derivative of the natural log is shortly. We'll actually use integration by parts to get it. But if we don't know it, we can't use it. Also, I should mention that the anti-derivative of the natural log is actually gonna be more complicated than natural log is. It's rarely a good choice to put the natural log in for DV. Instead, let's go back in time again. That was rewinding the cassette tape in case you didn't know. Some of us might be too young for such things. But anyways, if we want to, in this situation, it's nearly always the best choice to put the natural log of X in for U. Because what's the derivative of the natural log? It's one over X, DX. The natural log is a transcendental function. The one over X is an algebraic function. Transcendental just means it's not algebraic. It kind of transcends the usual rules of algebra. We like algebraic functions over transcendental functions because we can use algebra to help simplify things. So if given the choice between an algebraic, a transcendental algebraic function, we want something that's algebraic. Taking the derivative of the natural log makes it become algebraic. That's always a smart move. And then on the other hand, if the natural log of X is U, then DV has gotta be X and negative 2 DX. For which then the anti-derivative, by the anti-derivative power rule, we're going to get X to negative one, divide by negative one, which is the same thing as multiplying by negative one. We actually see it's like, hey, wait a second. By taking the derivative of the natural log, I get an algebraic function and I'll actually have a smaller power of X there. If you write it as a positive power, right? What I'm saying is that this is sort of like a win-win situation. We give rid of the natural log and we reduce the power of X in the denominator. By integration by parts, we're gonna get negative the natural log of X over X. This is our UV part. We didn't have to subtract and do the VDU. And doing that, we're gonna subtract integral. We're gonna get a negative one over X. That's my V. And then DU is a one over X DX, which if we simplify that last part, I'll just have to copy the negative natural log of X over X. We're gonna get plus the integral of one over X squared DX. Which that's a power function. We could treat the one over X squared instead as X and negative two. And for which it's anti-derivative would follow by the power rule. I mean, maybe I've already done this on this screen, right? That's kind of convenient. We end up with negative the natural log of X over X. And then we get minus one over X plus a constant, which gives us our anti-derivative. Now, wait a second. JK, what we're after is actually not an anti-derivative. I mean, we're using it to help us compute, but we're looking for a definite integral. Remember, this is a definite integral. We have to evaluate this thing from one to E. Now, what's school of thought here is just to evaluate the whole thing. Once you just find the anti-derivative separately and then evaluate it one and E. Which is perfectly fine. But another school of thought is to actually do it along the journey, right? Because the integral that shows up in the integration by parts formula, it'll have those same limits, one and E. But then the function, UV, which is part of the anti-derivative, you have to also evaluate it at one and E. So some people like to do the evaluation of the first part while they're also doing the anti-derivative of the second part. So you can evaluate the one and E at the very end, or you can do part of it along the way. Again, this is just two different schools of thought. They're both correct. Do whatever can make sense to you. In this situation, I kind of procrastinate everything to the end. And so if we plug in the one and the E, we're gonna end up with, let's plug in E first. We get negative the natural log of E over E minus one over E. And then we subtract from that negative the natural log of one over one minus one, one over one, which is just a one there. Simplifying this as we can, the natural log of E is a zero. Sorry, the natural log of one is zero. The natural log of E is itself one. I got a little bit mixed up there. And so we end up with a negative one over E due page into the signs. We get a negative one over E and then we're subtracting, actually adding to that a one. And so putting that together, we get one minus two over E. This is the exact answer. Or if you prefer an approximation, the area of the curve would be approximately 0.2642, et cetera. It's an irrational number. So we'll just, we'll take as many decimal places as we want for accuracy there. This is our answer. In the homework nearly always, you're gonna wanna put the exact answer unless specified otherwise to round it to such and such decimal places. So doing a definite integral using integration by parts is not fundamentally different than doing it for indefinite goals. There's just this extra arithmetic that follows the end of the problem because of the fundamental theorem calculus.