If `vec(F)=yhati+xhatj` then find out the work done in moving the particle from position (2, 3) to (5, 6)

Explanation

`dw=vecF.dvecs`
`dw=(yhati+xhatj).(dxhati+dyhatj)`
`dw=ydx+xdy `
Now `ydx + xdy = d(xy)` (perfect differential equation)
`implies dw = d(xy)`
for total work done we integrate both side
`intdw=intd(xy)`
Put xy = k
then at (2,3) `k_i=2xx3=6`
at `(5,6)k_(f)=5xx6=30`
then `w=int_(6)^(30)dk=[k]_(6)^(30)`
`impliesw=(30-6)=24` Joule