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Numerade Educator

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Problem 13 Easy Difficulty

Evaluate the line integral, where $ C $ is the given curve.

$ \displaystyle \int_C xye^{yz} \, dy $,
$ C: x = t $, $ y = t^2 $, $ z = t^3 $, $ 0 \leqslant t \leqslant 1 $

Answer

$$
\frac{2}{5}(e-1)
$$

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Video Transcript

Okay. So what you want to do in this problem is you want to change everything to T. So you're gonna put tea in place of X. T. Squared for whitey cube for Z. And then dy since why is T squared Dy will be to T. T. T. All right so you get T. T. Squared. Eat the T. Square times T. Cubed times to T. D. T. And he's going from 0 to 1. Okay so you have to integral de Erdogan E. T. To the fifth Times T. T. Square T. T. to the 4th DT. Yeah some substitute here. I'm gonna let you be T to the fifth then d'you is five T. To the fourth D. T. And if T. is zero U. is zero to the 5th which is zero If T. is one You is 1 to the 5th which is one. Okay so this turns into two. Now I'm doing Uz Erdogan eat to the U. Oh but forgot I need five T. To the fourth. And I just have T to the fourth. So I'm gonna put five in there and 1/5 out the front. So I have two fists. E. To the U. D. U. The integral will be to the U. E. To the U. From 0 to 1. So 2/5 it's the one minus E. To the zero Or to 5th E -1