work done in a vector field

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,

$\displaystyle \frac{W}{q}=-\int _{\text{P1}}^{\text{P2}}E\circ dl$

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,

$\displaystyle \frac{W}{q}=-\left[\int _{-2}^5x \text{dx}-\int _0^{-1}2y \text{dy}\right]$= -19/2 = -9.5

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