Compute $\int_{\gamma}z^{2}\bar{z} \, dz$ where $\gamma$ is the path parameterized by $\gamma (t) = t+it^{2}$ for $t\in [-1,2]$.
The integrand is continuous on $\gamma$ (it is the product of continuous functions), $\gamma$ itself is a smooth surve, and $\gamma '(t) = 1 + i2t$ is nonzero so we evaluate using $$ \int_{\gamma}f(z)\,dz = \int_{a}^{b}f(\gamma (t)) \gamma '(t) \, dt$$ In our case this is $$ \int_{\gamma}z^{2}\bar{z}\,dz = \int_{-1}^{2} (t+it^{2})^{2}(t-it^{2})(1+i2t)\,dt = \frac{-141}{2} + i\frac{2628}{35}$$
The answer is so ugly I really don't want it to be right. Am I misunderstanding completely how to do this?