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The problem:

19-20 Show that the line integral is independent of path and evaluate the integral.

19. $\int_C 2xe^{-y}dx + (2y - x^2e^{-y})dy$,

C is any path from $(1,0)$ to $(2,1)$

I was able to find $f(x,y)$ and show $f(x)$ and $f(y)$ are the parts inside the integrals, but when i tried taking the integrals with (2,1) as my upper bound and (1,0) as my lower bound, I kept getting 7/e, but the answer is 4/e. Can someone show me how they get 4/e? I'm afraid this isn't a mere calculation error, but something I'm not understanding correctly.

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  • $\begingroup$ What do you think the potential $f(x,y)$ is? $\endgroup$ – Randall Nov 28 '17 at 20:33
  • $\begingroup$ $x^2e^(-y) + y^2$ I'm not sure why my power is coming up weird, its supposed to be e raised to the negative y $\endgroup$ – 2316354654 Nov 28 '17 at 20:35
  • $\begingroup$ OK, that's good. Let me check then... $\endgroup$ – Randall Nov 28 '17 at 20:36
  • $\begingroup$ It's definitely $4/e$. $\endgroup$ – Randall Nov 28 '17 at 20:36
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    $\begingroup$ You can write $e^{-y}$ to display $e^{-y}$. $\endgroup$ – David K Nov 28 '17 at 20:39
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$f(x,y) = x^2e^{-y} + y^2$ is a valid potential. Let's use it: $$ x^2e^{-y} + y^2 \Bigr|_{(1,0)}^{(2,1)} = (4e^{-1} + 1) - (1+0) = \frac{4}{e}. $$

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  • $\begingroup$ oohhhh I was evaluating the integral above with those values instead of evaluating the function f(x,y) $\endgroup$ – 2316354654 Nov 28 '17 at 20:40
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    $\begingroup$ That's the kind of mistake everyone makes once, then never forgets, and never makes it again. $\endgroup$ – Randall Nov 28 '17 at 20:41
  • $\begingroup$ ............ :) $\endgroup$ – 2316354654 Nov 28 '17 at 20:44

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