Find the line integral of $f(x, y)=y e^{x^{2}}$ along the curve $\mathbf{r}(t)=4 t \mathbf{i}-3 t \mathbf{j},-1 \leq t \leq 2$.

$\frac{15}{32}\left(e^{16}-e^{64}\right)$

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Oregon State University

Harvey Mudd College

Baylor University

Idaho State University

Okay, folks. So in this video, we're gonna be talking about problem number 21 on your book. We have a line integral to solve. We have the 1st 1 we have the function, which is a function of x and y. And that function is, um, why multiplied by e to the power of X squared. That's weird. Um, and then we're given the curve, which is parameter prized by the parameter t. Um, and that curve is right here from this point all the way down to this point. And of course, I you know, I I did a little bit of work beforehand. And you know what your gash really given in the book is this we have are as a function that he is for t I a minus three t j. Okay. And then I took this thing and I rewrote it like this because that's that. That makes it a little bit simpler toe, um, to, uh, toe to analyze. Because if you run it in this form, you can see that that this is really just the vector. If you Ah, first right out the vector for negative three, which is which is somewhere here for negative three and then you let the TV you let the variable t Very. You're going to get a curve. A line segment, to be precise. That looks like that Looks exactly like this. OK, so we're going to evaluate the function f along the curve. We're gonna integrate it alone the curve. So let's run it out. We have integral along the curve. Are, um why e x squared multiplied by the Yes, of course. But DS, we're going to rewrite Yes, in another way, which is the usual way of rewriting. Thean Vent has a malign segment. Um, we have X as a function of t is 40. So ex prime is for and four score to 16. So we have 16 plus. Why prime while? Why? As a function of t is negative three t why prime this negative three? Okay, so why Prints Square to give you nine more by by DT. Okay, so I'm winner. He raised those, and then I have a square root of 25 multiplied by the tea. Um, but that is a constant, so I can pull it out of the integral we have. We have five because that's what scared of 25 is, um, multiplied by why? Which is a negative three t multiplied by E to the power of X squared by, um, X is a function of t and that function this 14. So x squared gives you 16 t squared, multiplied by DT. Now, we can try to figure out what the limits of integration is. Well, we started off from X equals for tea, which we started. We started off from this point, which is when four. Which is when tea Kool some. Your problem is it is a problem in 21 when to use negative one. Um, this is our starting point. This is our starting position. Um, X equals 40. Um, and let's see here. I'm having a little bit of it. Problem here? Oh, yeah. Never mind. We have where we have the variable t ranging from negative 1 to 2. Okay, um, that's our limited integration from negative one or two. I thought we were gonna have to figure out the limits of integration ourselves, but turns out, I over thought I thought the problem was harder than it really is. Um, so let's Let's do this. We have five multiplied by 93 which is a negative 15. I kind of pulled out of the I pulled the constant out of integral again have negative 15 times negative one to t multiplied by E to the power of 16 2 squared D T. Now I'm gonna do a little bit of a U substitution here. I'm going to find a new variable called you, which is defined as 16 t squared so you can see that the EU is 32 t d t um, which I'm going to substitute back into this expression. I have negative 15 times You The power of you T d t is the same thing as do you over 32. Now, as for the limits of integration when t is negative one you is 16 When t is to you is 16 times for whatever that is 16 times for. Uhm, I'm going to pull this constant out of the integral so I have negative 15 over 32 into grow you Do you? I have this thing again multiplied by eu because the integral doesn't change you, um, from evaluated from from 16 to 64 um So now I have this constant again, Um, multiplied by whatever is inside of the bracket E to the power of 16 minus e to the power of excuse me E to the power of 64th minus E to the power of 16. All right. I think, uh, I think that's it. That's the the ultimate answer that we were looking for. It doesn't look very pretty, but But that's that you if you don't like this, you can pull, pull out your calculator and plugged the whole thing, this whole chunk into your calculator in and get a some kind of a decimal expression out of this. But I'm not gonna do that in this video. I'm gonna leave leaving, like, leave it like this. Um, so this is a problem. 21 ar answers right here. Thank you very much for watching

University of California, Berkeley