I'm reading this as

with

.
The value of the integral will be independent of the path if we can find a function

that satisfies the gradient equation above.
You have

Integrate

with respect to

. You get


Differentiate with respect to

. You get
![\dfrac{\partial f}{\partial y}=\dfrac{\partial}{\partial y}[x^2e^{-y}+g(y)]](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%3D%5Cdfrac%7B%5Cpartial%7D%7B%5Cpartial%20y%7D%5Bx%5E2e%5E%7B-y%7D%2Bg%28y%29%5D)


Integrate both sides with respect to

to arrive at



So you have

The gradient is continuous for all

, so the fundamental theorem of calculus applies, and so the value of the integral, regardless of the path taken, is
Answer:
Step-by-step explanation:
y + 5 = -2(x + 3)
y + 5 = -2x - 6
y = -2x - 11
Answer:
Step-by-step explanation:
Let x represent the number of 3-lb bags purchased. Then the total purchase was ...
$2(8 -x) +$5.50(x) = $37
16 +3.50x = 37 . . . . . . . . . divide by $, collect terms
3.50x = 21 . . . . . . . . . . . . . subtract 16
21/3.50 = x = 6 . . . . . . . . divide by the coefficient of x
You bought 6 3-lb bags of peanuts and 2 1-lb bags.