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In Exercises $27 - 30 ,$ integrate $f$ over the given curve.

$$f ( x , y ) = x ^ { 3 } / y , \quad C : \quad y = x ^ { 2 } / 2 , \quad 0 \leq x \leq 2$$

$\frac{10 \sqrt{5}-2}{3}$

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okay. What we wanted to is we want to integrate the function ager, Um, the given curve. And so our function in X and Y is equal to X cubed over Why? And our curve is why equal to X squared over to an X goes between zero and two inclusive. Okay, so the first thing we need to do is, um, we need to define, um r of t. Um And so we know that, um t I plus, um and so X is going to be equal to tea. And therefore, why is he called T squared over two? So this becomes 1/2 a T squared? Um j And then, um we also know since, um, X is equal to ti. Umm, tea will go between zero and two inclusive as well. And so now what we want to do is take the derivative of our of tea, which will give us we have tea. So this will be, um I plus, um, t j. And so the magnitude of e of t is equal to the square root of one squared plus t squared, which is gonna be this Croteau one plus t squared. And so d s is that magnitude? Um, Times T t. Okay. And so what we need to do now is we have the integral from 0 to 2, um, and so of our function. And so we would have t cubed over t squared over too, because that's what y iss. And so and then this is gonna be times the square root of one plus t squared t t. Um, and so this is gonna be equal to Comes out to the integral from 0 to 2 of two tea. Um, time to square root a one plus t squared TT. Okay. And so it looks like where you have to do you use up. Um, so let's go ahead. We have the integral from 0 to 2 of two tea times the square root of one plus t squared TT. And so if we let you be one plus t squared, then de Hue is to t t t. Which is exactly what we have right here. And so, um, we have the integral and we're gonna go ahead and change or lower and upper limits. And so we're gonna let if t if t zero then us one if t is too, then you is a five. And this is gonna be you to the 1/2 do you to this because 2/3 you to the three halves and we're getting evaluated at five. And then at one. And so this becomes, um, 2/3 times, um five, route five, minus one, and so we can go ahead and, um, put it over that denominator. So we have, um 10 route five minus two, all over three.

University of Central Arkansas