Answer
$\frac{1}{3}(10\sqrt{5}-2)$
Work Step by Step
We set up the integral
$$\int_{C} f(x, y, z) d s=\int_{a}^{b} f(g(t), h(t), k(t))|\mathbf{v}(t)| d t$$
Where
$$f=\frac{x^3}{y}$$ and $$\mathbf{r}(t)=t\mathbf{i}+\frac{t^2}{2}\mathbf{j}$$ and the $t$ range is $$0 \leq t \leq 2$$
Thus, we have:
$$\int_{0}^{2} \frac{t^3}{t^2/2} |\sqrt{1^2+(2t/2)^2}| dt = \int_{0}^{2} 2\cdot t \cdot \sqrt{1+t^2} dt = \int_{1}^{5} u^{1/2}=\frac{1}{3}(10\sqrt{5}-2)$$
where $u=1+t^2$ and $du=2tdt$
