Question: [![enter image description here](https://i.stack.imgur.com/xT3KN.jpg)](https://i.stack.imgur.com/xT3KN.jpg)

This is my work and explaining why I am confusd in this specific problem

Answer: 
We could use $x=t, y=t^{3/2}$ to parametrize the curve, but let's use $x=t^2, y=t^3$ instead.

Then $r^{\prime}(t)=\langle x^{\prime}(t), y^{\prime}(t)\rangle=\langle2t, 3t^2\rangle$, so $\big|r^{\prime}(t)\big|=\sqrt{(2t)^2+(3t^2)^2}=\sqrt{4t^2+9t^4}$.

Then $\displaystyle L=\int\_{C}ds=\int\_{C}\big|r^{\prime}(t)\big|dt=\int\_0^{\sqrt{5}}\sqrt{4t^2+9t^4}dt=\int\_0^{\sqrt{5}}t\sqrt{4+9t^2}dt$

and $\displaystyle\int\_{C}f(x,y)ds=\int\_{0}^{\sqrt{5}}\sqrt{4+9t^2}\sqrt{4t^2+9t^4}dt=\int\_0^{\sqrt{5}}t(4+9t^2)dt=\int\_0^{\sqrt{5}}(4t+9t^3)dt$

and, as you indicated, you want to calculate $\displaystyle \frac{1}{L}\int\_{C}f(x,y)ds$.
The curve is parametrized: $r(x) = (x, x^{3/2})$ is a perfectly acceptable parametrization. Note though, that $|r'| \ne |y'|$, so your later work is not correct.