Evaluate $$\int_C P(x,y)dx+Q(x,y)dy$$ where $P(x,y) = y^2 $, $Q(x,y) = x$, and $C$ is the part of the graph $x = y^3$ from $(-1,-1)$ to $(1,1)$.
I was trying the parametrization: $$x = t $$ $$y = \sqrt[3] t$$ $$dx = 1$$ $$dy = \frac{1}{3}t^{-2/3}$$ from which: $$\int y^2dx+xdy=\int_{-1}^1 \left(t^{2/3}+ \frac{1}{3}t^{1/3}\right) dt= \frac{17}{10} $$ which is not the correct answer, correct answer is $\dfrac{6}{5}$.
With your non-smooth parametrization $(t,\sqrt[3] t)$ for $t\in[-1,1]$ (it is not differentiable at $t=0$): $$\int_C y^2dx+xdy=\int_{-1}^1 \left(t^{2/3}+ \frac{1}{3}t^{1/3}\right) dt= 2\int_{0}^1 t^{2/3}\,dt=\frac{2}{2/3+1}=\frac{6}{5}$$ where we applied the fact that $t^{2/3}$ is an even function and $t^{1/3}$ is an odd one.
A smooth parametrization of $C$ is $(t^3,t)$ for $t\in[-1,1]$. Hence $$\int_C y^2dx+xdy=\int_{-1}^1 \left(t^2\cdot 3t^2+ t\cdot 1\right) dt= 6\int_{0}^1 t^4\, dt=\frac{6}{4+1}=\frac{6}{5}$$ (note that $t^4$ is an even function and $t$ is an odd one).