Answer
Result: \[ I = \frac{6}{5} \]
Work Step by Step
Step 1: In our case, we have the curve: \[ x=t^{2/3}, \quad y=t, \quad (-1\leq t\leq 1) \] We have for the differentials: \[ dx = \frac{3}{2} t^{-1/3} dt, \quad dy = dt \] Therefore, for the integral: \[ I = \int_C (x^2 - y^2)dx + xdy \] We have: \[ I = \int_{-1}^{1} (t^{4/3} - t^2) \left(\frac{3}{2} t^{-1/3} dt\right) + t^{2/3} dt \] \[ I = \int_{-1}^{1} \left(\frac{3}{2} t - \frac{3}{2} t^{5/3} + t^{2/3}\right) dt \] \[ I = \left[\frac{3}{2} \cdot \frac{2}{3} t^2 - \frac{3}{2} \cdot \frac{3}{8} t^{8/3} + \frac{5}{3} t^{5/3}\right]_{-1}^{1} \] \[ I = \frac{5}{3} \left(1^{5/3} - (-1)^{5/3}\right) \] \[ I = \frac{6}{5} \] Result: \[ I = \frac{6}{5} \]
