Having trouble finding the area between this parametric curve and the y axis can someone go over my work please. Thanks in advance!

$\displaystyle x= t^2-5t, y= t^\frac{1}{2}$

Interescts the y-axis when x=0 or when t=0,5

So, A= $\displaystyle \int_0^5 \! x(t)*y' \, dt$

= $\displaystyle \int_0^5 \! (t^2-5t)(\frac{1}{2t^\frac{1}{2}}) \, dt$

= $\displaystyle \frac{1}{2}\int_0^5 \! t^\frac{3}{2}-5t^\frac{1}{2} \, dt$

= $\displaystyle \frac{1}{2}[\frac{2}{5}t^\frac{5}{2}-5*\frac{2}{3}t^\frac{3}{2} \,\bigg|_0^5$

= $\displaystyle \frac{1}{5}t^\frac{5}{2}-\frac{5}{3}t^\frac{3}{2} \,\bigg|_0^5$

= $\displaystyle 1-(\frac{25}{3})^\frac{3}{2}$