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
I don't think it is. I may be wrong though
Answer:
10950
Step-by-step explanation:
1 year = 365
365 × 30
10950
Answer:
x=-4
Step-by-step explanation:
(x+4) ^ (1/3) + (2x+8) ^ (1/3) = 0
Subtract (x+4) ^ (1/3) from each side
(x+4) ^ (1/3) - (x+4) ^ (1/3)+ (2x+8) ^ (1/3) = -(x+4) ^ (1/3)
(2x+8) ^ (1/3) = -(x+4) ^ (1/3)
Cube each side
(2x+8) ^ (1/3) ^3= -(x+4) ^ (1/3)^3
2x+8 = -(x+4)
Distribute the negative sign
2x+8 = -x -4
Add x to each side
2x+8 +x =-x+x-4
3x+8 = -4
Subtract 8 from each side
3x+8-8 =-4-8
3x =-12
Divide by 3
3x/3 = -12/3
x = -4