The arc length of [tex]y=2x^{\frac32}[/tex] over [0, 1] is given by the integral,
[tex]L=\displaystyle\int_0^1\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx[/tex]
Differentiate y with respect to x using the power rule:
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac32\times2 x^{\frac32-1}=3x^{\frac12}[/tex]
Then the integral becomes
[tex]L=\displaystyle\int_0^1\sqrt{1+\left(3x^{\frac12}\right)^2}\,\mathrm dx=\int_0^1\sqrt{1+9x}\,\mathrm dx[/tex]
Substitute u = 1 + 9x and du = 9 dx :
[tex]L=\displaystyle\int_0^1\sqrt{1+9x}\,\mathrm dx=\frac19\int_1^{10}\sqrt u\,\mathrm du[/tex]
[tex]L=\displaystyle\frac19\left(\frac23u^{\frac32}\right)\bigg|_1^{10}[/tex]
[tex]L=\dfrac2{27}\left(10^{\frac32}-1^{\frac32}\right)[/tex]
[tex]L=\boxed{\dfrac2{27}\left(10^{\frac32}-1\right)}[/tex]
