Determine the work done by the electric field E=i*x-j*2y in moving a unit positive charge from position P1(-2,0,0) to position P2(5,-1,3).

The work is defined as the integral from p1 to p2 of E "dot" dl,

Answer from book: 9.5 J

My attempted solution

Break dl into constituent vectors: dl = i*dx+j*dy

The dot product of E dot dl = x*dx - 2y*dy

And the integration becomes,

= -19/2 = -9.5

So the negative sign was removed b/c work cannot be negative?