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Problem 30 Hard Difficulty

Scalar line integrals in $\mathbb{R}^{3}$ Convert the line integral to an ordinary integral with respect to the parameter and evaluate it.
$\int_{C} x e^{y z} d s ; C$ is $\mathbf{r}(t)=\langle t, 2 t,-4 t\rangle,$ for $1 \leq t \leq 2$

Answer

$$\int_{c} x e^{y z} d s=\frac{\sqrt{21}}{16}\left[\frac{1}{e^{8}}-\frac{1}{e^{32}}\right]$$

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Video Transcript

All right, Number 30 We have the line and girl X e to the y Z. With our tea, you could ah t comma two t come a 40 or in greeting from wanted to. All right, So this is our xsara. Why this or z? Our prime of tea. It's gonna be the driven of each of these squared. And then at some together in the square root three scared of one plus for plus 16. That's one squared plus Cities Group is foursquare medical scared of 21. You know, plugging everything back into original equation, substituting all the exes for our parametric forms. We get in to go from one to ah t t eats and negative a T squared times through 21 80. Now it's a little bit of Ah, you substitution week. New girlfriend wanted to negative 16 t Ah, it's a negative. A T squared DT simplifying one. Call this the e to *** 18 t squared. Integrating further. We gotta get 21/16 times each of negative 80 squared. So you're from 2 to 1. That gives this negative route 21 over a 16 times each and negative 32 each of negative eight medicals Route 21/16 times, one over E to the eight minus one over E to 36.

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