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
<span>paper and pencil to record work</span>
It will be 5 cm and use 3.14 to multiply
in the first half he would have swam 1 km. there are 1000m in a km so 1000/200=5 so he swam the first 200 fly, then a 200 breast, then a 200 fly again, then another 200 breast the finish the first 1 km with a 200 fly. so we switched strokes 4 times.
PS i wouldn't recommend doing a 2 km swim fly and breast