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# Evaluate $\int_{C} \frac{x^{2}}{y^{4 / 3}} d s,$ where $C$ is the curve $x=t^{2}, y=t^{3},$ for $1 \leq t \leq 2$.

## $\frac{1}{27}\left(40^{3 / 2}-13^{3 / 2}\right)$

Integrals

Vectors

Vector Functions

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

Okay, folks. So in this video, we have this integral here to evaluate. We have this Interpol along the courtesy of X squared over. Expert over. Why? To the power of for 1/3 I'm going to substitute an expression for X here. We have an expression for X, which is t squared, something to substitute that in and to do something with why. OK, so X squared is really just he squared on to the power of to over t to the three. Uh and that's for why, um that raised to the power of for third by two years. I'm not I'm not gonna write. Yes, I'm going to write the square root of X prime squared. Plus why Prime scored DT. Okay, but now if you look at this this time right here, this'll time right here. Excuse me. This is t to the power fourth over T to the power fourth, which is really just one. So I'm going toe something. So I'm going to ignore this term because that's one. And then I'm going to evaluate X prime and white prime. Um, so for X, I have t squared. That means X promise to t And then I raise that of the power of to I get 40 square. It was, um if I do the same thing with why I'm going to get three key squared to the power of to Okay, whatever that is multiplied by DT. And then I have 40 squared plus 94th DT. But now I am going to Ah, pull out. I guess you can pull out a a term of tea here. I'm the plot. A factor of tea. I have t multiplied by four plus 90 squared, multiplied by DJ. Okay, I kind of jumped a step here because I could have just I really should have told you that. Dumb that the way I did this, um, is by writing whatever's inside of the square root as t squared multiplied by four plus I e squared, OK, and then pulled this time right here, out of the square root. That's why I have this tea here. Um and then I have four plus 90 squared left inside of the square root. And then now I'm gonna do a u substitution, which I'm sure a lot of your very familiar with, um, I'm going to define you variable a new variable called you, which is defined as four plus nineties word. So now do you is 18 t g t. Okay, um, that I'm gonna substitute this back in here. I get I'm the square root of you. T t t is really just to you over 18. Okay, I have 1/18 which is a constant. So I'm gonna pull it out of the integral. Um, integrated. I mean, integrating u to the power of 1/2 do you that I can write as u to the power 3/2 over 3/2. Let's figure out what the limits of integration is. Well, originally, the limits integration for tea is between, um is between one and two. But now, when tea is one when t is equal to one, U is equal to four plans nine, which is 13. And when t is equal to u is equal to four plus nine times four, which is Ah ah, 40. Okay, um, so now I'm going to take these two numbers right here, and then put them here, which is for evaluation purposes. Um, 13 and 40. Now I'm gonna have 1/18 times to third. I I just took this time right here on the denominator. I pulled it out of the denominator, um, multiplied by 40 to the power three. Half minus 13 to the power of three. Okay, whatever that is. So now I have I'm gonna simplify. There's a little bit more. I have. No, I hear one over, um 27. Multiply by 40 to the three. Half minus 13 to the 3/2. That is the answer for problem number 23. And we're done for this video. Thank you for watching, but I

University of California, Berkeley

#### Topics

Integrals

Vectors

Vector Functions

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp