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# Evaluate $\int_{C}(x+\sqrt{y}) d s,$ where $C$ is given in the accompanying figure. (FIGURE CAN'T COPY)

## $\frac{1}{6}\left(5^{3 / 2}+7 \sqrt{2}-1\right)$

Integrals

Vectors

Vector Functions

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

Okay, folks. So in this video, we're going to have, um we're gonna have to evaluate this This integral here along the curve C if a function which is given by ex place, why did have 1/2? Yes, we have this integral here that we need to evaluate. Um, and we have a very interesting curve, which I would love to drive for you, but But I can't. So I'm gonna just have to describe it to you in words. We have a curve that that's basically two separate pieces of curves joined together. So for the first part, we have a function, which is why to the power, why equals X squared. And for the second part, we have y equals x. Okay, so we're integrating. So basically see is divided up into two parts. The first part and the second part. So we're going to separate the curve. See, I'm gonna call the first part C one, and then I'm to call the second part C two and then I'm gonna do the integration separately. Okay, So see one. Let's take care of a few things first. You know, the infinite decimal line segment? Yes. Um, at least for a two D plane. We have DS equals using the prep Pythagorean theorem. We have the X squared plus y squared. OK, Um however, this could be written as ah d x as the x squared one plus. Do you? Why the X squared? Okay, so all I did was I I kind of pulled out a factor of the x squared. If you put them back in here, you will get exactly this term right here, because this is really just d X squared. Plus the wise words. Okay, so now I'm gonna pull out the d X in front. I have the X multiplied by one plus. Why? Prime Squared. Okay, So all I did here was I I took the infant asthma lying segment DS, and then I rewrote it in a different way. Okay, because the reason I'm doing this because it's gonna make our lives over the easier, you know, is gonna make the math of the more convenient because we have a function. Um, we have we have Why is a function of X and we can easily evaluate, um Why, prime? And And that that holds true for both. The first part of the curve and the second part of the curve. Even though even though, uh, thes two parts of the curves have different functions, Um, for why we can we can always evaluate White Point. That's why I'm doing. That's why I'm rewriting this, um, ds here. Okay, so let's ah, start doing start doing the integration. We have, um x plus y to the power of 1/2. Multiply that the s and then the same thing for the second curve. We have explains why the power 1/2 move apart by DS. Um, Now I'm going to plug in for the C one. I'm gonna plug in, um, the appropriate expressions for both x and y. By that, I mean X is still X. But why is a different thing? Why, if you relate, why with X? Why is the same thing as X squared? Because that's what it is that's given in the graph, which I cannot copy here for some reason, um, to the power of 1/2 multiplied by the S by yes, is the same thing as the square root of one. Plus why Prime squared multiplied by the X. But why Prime is two X. That squared is to expert. OK, so that's for C one. What about C two c to you for C two? We have we have y equals X. So I have ah X plus X to the power of 1/2 multiplied by one plus why Prime Squared y primates ones or 11 prime once where it is still one multiplied by the X. Okay, so now we have to figure out the limits of integration. So for the first curve, as you can see on the graph, I'm assuming you're looking at the graph right now. Um, as you can see on the graph, the limits of integration for X is between X equals zero and 16. And as for the second curve, I'm gonna go between X equals one and want X equals zero. Some like going in a full loop here. Okay, So, uh, so let's crank this out. We have from 0 to 1 picks place. Thanks. So that's for the 1st 1 multiplied by one plus four x squared DX. Okay, that's for this. The first part of the curve +10 x plus root x ah route to DX. Okay, so now I'm going to write X plus access to X, because that's what it is. And then I'm gonna do a use institution here. I'm going to find you as four x squared plus one, which is exactly what's under the square root. Um, so now I have the U equals eight x of the X right, eight x t x. So now I have ah u to the power of one have multiplied by two times x t x by x dx is really just you over eight. So that's for the 1st 1 Plus the square root of Well, this one is enough is easy enough to value explains. That's the power of 1/2 the X between zero. I mean between 10 Okay, so now I'm gonna pull out the constant term 1/4, um, U to the power of 1/2. Do you zero the one plus route to this thing here? Okay, last do it. Um, actually, when you when you do a u substitution the limits of integration there the limits of integration are going to change as well. So we cannot write zero on one anymore. We're gonna We're gonna have to figure out the limits of integration for you. Well, when x zero use one when access one, you is five. I hope that did that. Math. Right. Um, so I have one and five here. Um, let's crank this out. We have 1/4 the U to the power of 3/2 over three. Half one, right. Plus route to, um, I'm doing this one right here. X squared over two plus X to the power of three. Half over 3/2. Okay. Between 10 that's crank this one out. We have, ah, 1/6 nationally does not. Drew. Actually, that is true. Um, this is two or three. 242 times three is whatever. Six. All right, um, won over six U to the power of 3/2 uh, evaluated between one and five plus route to, um, if you plug in zero for both of these two terms here, you get zero. So I'm going to ignore that. Um, minus 1/2 Was one over 3/2. Okay, so now I'm going to Ah, tidy up. All of these numbers have won over sex five to the power of 3/2 minus one minus root too. Well over two. Plus 2/3. That's a lot of numbers. Um, one of her. Six 5/5 to the power for you have minus one minus one of ever to waas uh, to That's not true. There's not. Plus, that's minus minus two route to over three. And, um, I think this will do it for me. This is a clean enough expression for me. If you're not happy with this, you can plug this into a calculator, and then you're gonna get a decimal expression we just find. But yeah, this is the answer for problem number 25 we're done for this video. Thank you.

University of California, Berkeley

#### Topics

Integrals

Vectors

Vector Functions

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp