# How to evaluate this line integral over a plane curve?

How do I integrate $f$ over the given curve?

$$f(x,y)= \frac{x^3}{y};\quad C: y=\frac{x^2}{2} \quad \text{for}\; 0 \leq x \leq 2.$$

I can't figure this out... can anyone show me how to solve it?

The answer is supposed to be $\displaystyle \frac{10\sqrt{5}-2}{3}$.

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Parametrize the curve $C$ by $\gamma : [0,2] \rightarrow \mathbb{R}^2$ with $\gamma(t) = (t, \frac{t^2}{2}).$ Then, $$\int_{\gamma} f(x,y) \mathrm{ds} = \int_0^2 f(\gamma(t)) ||\dot{\gamma}(t)|| dt = \int_0^2 \frac{t^3}{\frac{t^2}{2}} ||(1, 2t)|| dt = \int_0^2 2t \sqrt{1 + 4t^2} dt.$$ Solve this integral by substitution.