What is the line integral of...

∫C xydx+(x-y)dy; C is from (0,0) to (2,0) and (2,0) to (3,2).

I look at this and I don't even know where to start.

Break C into two segments:

(i) (0, 0) to (2, 0) via x = 2t, y = 0 for t in [0, 1].

So, ∫ (xy dx + (x-y) dy)
= ∫(t = 0 to 1) (2t * 0 * 2 + (2t - 0) * 0) dt = 0

(ii) (2, 0) to (3, 2) via x = 2 + t, y = 2t for t in [0, 1].

So, ∫ (xy dx + (x-y) dy)
= ∫(t = 0 to 1) [(2 + t)(2t) * 1 + ((2+t) - 2t) * 2] dt
= ∫(t = 0 to 1) (4 + 2t + 2t^2) dt
= (4t + t^2 + 2t^3/3) {for t = 0 to 1}
= 17/3.
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By (i) and (ii),
∫c (xy dx + (x-y) dy) = 0 + 17/3 = 17/3.

I hope this helps!