F(x,y) is defined by the integral:

$\displaystyle \\

\int\nolimits_{(2,\pi )}^{(x,y)} {[-2uv^{2}\sin (u^{2}v)]du+[\cos (u^{2}v)-u^{2}v\sin (u^{2}v)]dv}$

Express F(x,y) as a function of x and y, eliminating the integral sign.

_______________________

My approach:

I found the curl of the function to be zero, which means it is path independent and that a potential function exists... I'm now working towards building f from F and finally calculating f(x,y)-f(2,pi)

Which I did and I got:

$\displaystyle f(u,v)=v[\cos (u^{2}v)]$

which means the final value for the integral is:

$\displaystyle y\cos \left ({x^{2}y} \right )-\pi $

Does my answer seem correct to you?