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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 $

$$

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

$$

Vector Calculus

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Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Boston College

so for us to evaluate this line, integral. One thing that we should notice is that here we have a d y as opposed to a D s. So what we would want to do is essentially first find what is D Y by D? T. So, like, this is equal to something. And then we could just kind of multiply the DT over and we get de y busy, but whatever. So that's essentially what we're going to do in this case, eso was find that so if we take the derivative of this with respect to t, we end up with two t multiply that over, we get d Y is equal to t d t. So we're just going to replace D Y with that. And then we can make our other substitution for X and Z as well. Yeah, we get it. Just integrate from 01 So let's do all that here. So our new balance is going to be from 0 to 1. X is going to be t y is ah t squared e is eso e r e e to the y z eso again. Why is t squared? Z is t cubed and then d wise was to be two times t Then we have DT actually, just grow simplify the stone a little So zero toe one looks like to t to the fourth times e raised to the to t to the fifth and we can go ahead and use a substitution of you is opportunity to fifth, which would give us the differential. Both this here eso We need a five here, So let's just multiply this by five that weaken divide by five as well. And now that turns into d you. So if I come over here maybe 1/5 integral of so if I plug zero in that zero plug in 11 um and then we'd have And actually let me just pull that two out as well. So it would be 2/5 and then the e to the you, do you and we'll evaluating that would just give us eat the youth. So you have to fifth c e to the u evaluated from 01 eso we plug in one, which is just gonna be e plug in zero. That's once a minus one. And so this here should be what we get for evaluating that line. Integral

University of North Texas

Vector Calculus