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Use a calculator or CAS to evaluate the line integral correct to four decimal places.

$$\begin{array}{l}{\int_{C} z e^{-x y} d s, \text { where } C \text { has parametric equations } x=t} \\ {y=t^{2}, z=e^{-t}, 0 \leqslant t \leqslant 1}\end{array}$$

$\int_{C} z e^{-x y} d s=\int_{0}^{1} e^{-t^{3}-t} \sqrt{1+4 t^{2}+e^{-2 t}} d t \approx 0.8208$

Vector Calculus

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So in this video are asked determined lining to grow of Z eat negative X y along the curve C and were given the excessive quantity wise he quantity swear it's easy to either. The negativity and T varies from 01 Well, the first thing we're gonna have to do is we're gonna try to convert everything in Santa Cruz in terms of tea. So see, comparing in terms of tea so can exit only issues with Yes, But then we know that yes, and be written us sweat rolling off the exp i t Square was divided by DT Swear was easy, I swear. Um, here then and then what? The x by tt's derivative of TV respected to use one you're a bit of t squared with respected to use to t and the derivative of e to the negative t with respected t is negative to now If we square all these in the ad together, what we get is one plus 40 square close to the negative duty and then we take the square root of that of the plane. So now we can plug everything back unwto our original line integral. So we get in to grow from zero. So now everything is in terms of t, t and E t so that we have the integral from 01 of e to the negative tee times e to the negative t times the square times the square root of one plus 40 squared plus e to the negative, too. Key announced. We worked that out. What we get is the integral from zero to a lot of e to the negative, minus two cubed times square of one plus 40 square plus speed to the negative ET. And in the question they tell us that we should plug this into a software or into a calculator and determine it using software, and the answer that we get is 0.82