Parameterize the curve by
[tex]\mathbf r(t)=\begin{cases}x(t)=t-5\\y(t)=2t^{3/2}\end{cases}[/tex]
Then
[tex]\mathbf r'(t)=\langle1,3t^{1/2}\implies\|\mathbf r'(t)\|=\sqrt{1+9t}[/tex]
The length of the curve [tex]\mathcal C[/tex] is then
[tex]\displaystyle\int_{\mathcal C}\mathrm dS=\int_{t=5}^{t=6}\left\|\mathbf r'(t)\right\|\,\mathrm dt=\int_5^6\sqrt{1+9t}\,\mathrm dt[/tex]
[tex]=\displaystyle\frac19\int_5^6\sqrt{1+9t}\,\mathrm d(1+9t)=\dfrac2{27}(55^{3/2}-46^{3/2})\approx7.10398[/tex]
