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.
Best Answer
$f(x,y) = x^2e^{-y} + y^2$ is a valid potential. Let's use it: $$ x^2e^{-y} + y^2 \Bigr|_{(1,0)}^{(2,1)} = (4e^{-1} + 1) - (1+0) = \frac{4}{e}. $$