# 3How do you find the lengths of the curve y=2/3(x+2)^(3/2) for 0<=x<=3?

Feb 18, 2017

$= 4 \sqrt{6} - 2 \sqrt{3}$

#### Explanation:

Length, s, is:

$s = {\int}_{0}^{3} \sqrt{1 + {\left(y '\right)}^{2}} \mathrm{dx}$

$y ' = {\left(x + 2\right)}^{\frac{1}{2}}$

$\implies s = {\int}_{0}^{3} \sqrt{3 + x} \mathrm{dx}$

$= {\left[\frac{2}{3} {\left(3 + x\right)}^{\frac{3}{2}}\right]}_{0}^{3}$

$= \frac{2}{3} \left({6}^{\frac{3}{2}} - {3}^{\frac{3}{2}}\right)$

$= 4 \sqrt{6} - 2 \sqrt{3}$