(Headbang) Parabolas don't intersect at numbers, they intersect at points. These intersect at (0, 0) and (1, 1)

So you mean the y= x(1- x) for c2. Well, in that case, you can take "x" itself as parameter- write, for c1, x= t, \(\displaystyle y= t(1- t)= t- t^2\) so that dx= dt, dy= (1- t)dt. Now \(\displaystyle \vec{F}(t)\cdot d\vec{s}= (t+ t(1-t))dt+ (t^2+ t(1-t))(1- t)dt\). Integrate that from t= 0 to t= 1.

On c2, y= 4x(1- x) so take x= t, \(\displaystyle y= 4t(1- t)= 4t- 4t^2\), dx= dt, dy= (4- 8t)dt. Integrate \(\displaystyle \vec{F}(t)\cdot d\vec{s}= (t+ t(1-t)dt+ (t^2+ t(1-t))(4-8t)dt\) from t= 1 to 0 (note the order) so that you complete the loop.