🤔 Find out what you don't know with free Quizzes 🤔Start Quiz Now!

# Integrate $f$ over the given curve.$f(x, y)=\left(x+y^{2}\right) / \sqrt{1+x^{2}}, \quad C: \quad y=x^{2} / 2$ from $(1,1 / 2)$ to (0,0)

Integrals

Vectors

Vector Functions

### Discussion

You must be signed in to discuss.

Lectures

Join Bootcamp

### Video Transcript

Okay, folks. So in this video, we're gonna take a look at problem number 28 which is a line integral problem. So let's go ahead and start doing the London to girl. We have integration along, see of f effects by DS. And you know the drill DSS could be rewritten, has has an expression in terms of X, um, and f for f I'm going to plug in. Why? As a function of X and so let's ah, after you plugging. Why? Well, let's just do it step by step we have for for the function if we have for the numerator we have X plus y squared over one plus x squared here. So that's, um that's the function f multiplied by the S, which I'm going to rewrite it. Well, if you remember, the S is simply an infinite asthma line segment, which you can write as square root of the X squared plus y squared. Now, if you do a little bit of algebra here, you're going to arrive in in this problem. You're going to arrive at one plus X squared DX. Okay, um, so let's block in the S. We have one square root of one plus x squared DX. Now, have you noticed we have the same thing here and here, So they're going to cancel out. We're going to arrive and X plus, why squared the X? But I'm not going to write. Why squared? I'm going to plug in. Why? The function of X? We're going to arrive at X to the fourth over four. And there's also a factor of of negative one here that I should add. Um, so now I'm going toe. Now, you see, that thing is really just, you know, to into girls. Here was Aiken. Sublet. Split this integral into two parts. We have from 0 to 1 x dx, um, plus 1/4 from 0 to 1 x of the power. Fourth, the X, and this is very simple. We have one have here plus 1/4 times, 1/5, and this is gonna This is going to add up to 11/20 which is the answer for problem number 28. And we're done for this video. Thank you very much.

University of California, Berkeley

Integrals

Vectors

Vector Functions

Lectures

Join Bootcamp