The arc length is given by the definite integral
[tex]\displaystyle \int_1^3 \sqrt{1 + \left(y'\right)^2} \, dx = \int_1^3 \sqrt{1+9x} \, dx[/tex]
since by the power rule for differentiation,
[tex]y = 2x^{3/2} \implies y' = \dfrac32 \cdot 2x^{3/2-1} = 3x^{1/2} \implies \left(y'\right)^2 = 9x[/tex]
To compute the integral, substitute
[tex]u = 1+9x \implies du = 9\,dx[/tex]
so that by the power rule for integration and the fundamental theorem of calculus,
[tex]\displaystyle \int_{x=1}^{x=3} \sqrt{1+9x} \, dx = \frac19 \int_{u=10}^{u=28} u^{1/2} \, du = \frac19\times\frac23 u^{1/2+1} \bigg|_{10}^{28} = \boxed{\frac2{27}\left(28^{3/2} - 10^{3/2}\right)}[/tex]
