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### Video Transcript

Okay, everybody. So in this video, we're gonna take a look at problem number 22 where we have a integral along the curve. Um, so now I'm gonna write down the integral here we have the integral along the curve, which is which is denoted with the vector function are, which is primer charged by t, and we're integrating the function of along the curve. Okay, so and what we're given in this problem is we're given f and were given both eggs. And why, As a function of t So let me write down your your function of here. Take a look at this. Okay, Um, so the function f is just x minus y plus three, this is the function, and we have DS, which you, which you recall, is really just the square root of X prime of square ex prime squared. Excuse me? Plus why Prime Squared multiplied by DT. Okay, so now because we're given both X and y is a function of T, I'm going to substitute everything, um, as an expression for tea. Ok, so X, I know is just, um, co sign of tea. So that's eggs. And as for why we have sine of t. Okay, so I'm going to subtract off sign of tea and then at three, that's the function f expressed as a expressions for tea. Um, now I'm gonna take the derivative of X and y with respect to t I differentiate co signed gets negative sign negative scientist e squared. Plus, I'm gonna do the same thing with why so I get close i to you squared, multiplied by DT and then as her the limit of integration. I have from zero all the way up to two pi. Okay, so I have done the majority of the work here we have to do now is to crank it out. So as you can see here we have science square plucks go sine squared, which is gonna give you one. I'm just cross it out because it's one. Okay, so I have now co Scient e minus scientific plus three integrated with DT firm zero to two pi. Okay, so now let's crank it out. We have Nick. We have If you, uh, integrate co star, you get signed. Of course. Scientist E and then you integrate negative science. You get close I and then you get three t um, evaluated at the two endpoints, which is zero and two pi. Okay, so if I evaluate sign of tea at two pi, get zero. Of course. I have zero plus co sign of two. Pi is really just one plus three times to pie minus. Sign at zero is zero plus co sign of zero. CO sign of zero is still one plus zero. Of course. Now that's Ah, I see. Evaluate this whole thing. We have one plus six pi minus one, which is six pie. Okay, so this right here is the answer for problem number 22. And we're done for this video. Thank you. Bye bye.

University of California, Berkeley

Integrals

Vectors

Vector Functions

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