\(\displaystyle \int{(2x-y^{2})dx+(y^{3}-2xy)dy}\)
(1,1) to (3,2)
I can't see all your picture, but you have:
\(\displaystyle x=1+2t, \;\ y=1+t, \;\ 0\leq{t}\leq{1}\)
\(\displaystyle \int\left[(2x-y^{2})i+(y^{3}-2xy)j\right]\cdot(dxi+dyj)\)
\(\displaystyle \int_{0}^{1}(t-1)(t^{2}-2t-1)dt=\frac{3}{4}\)
