Answer
Result: \[I = -\frac{39}{2}\]
Work Step by Step
We have the path of integration: \[C: \text{line segment from } (3,4) \text{ to } (2,1).\] We can parametrize the curve above as: \[x = 3-t, \quad y = 4-3t, \quad 0 \leq t \leq 1.\] Then the integral: \[I = \int_C (y - x)dx + xydy\] can be calculated as: \[\begin{align*} I &= \int_0^1 (4-3t - 3 + t)(-tdt) + (3-t)(4-3t)(-3tdt) \\ &= \int_0^1 (-1 + 2t - 3t + t^2 - 36t + 39t^2 - 9t^3)dt \\ &= \int_0^1 (-1 + 2t - 36 + 39t - 9t^2)dt \\ &= \left(-t + \frac{t^2}{2} - 36t + \frac{39t^2}{2} - \frac{3t^3}{3}\right)\Bigg|_0^1 \\ &= [- (1-0) + (1-0) - 36(1-0) + \frac{39}{2}(1-0) - 3(1-0)] \\ &= [-1 + 1 - 36 + \frac{39}{2} - 3] \\ &= [-39/2 - 3] \\ &= -\frac{39}{2} \end{align*}\] Result: \[I = -\frac{39}{2}\]
