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
with
The value of the integral will be independent of the path if we can find a function
You have
Integrate
Differentiate with respect to
Integrate both sides with respect to
So you have
The gradient is continuous for all
