What is the arc length of the curve given by f(x)=x^(3/2) in the interval x in [0,3]?

Apr 4, 2016

$\frac{31 \sqrt{31} - 8}{27} \cong 6.0963$

Explanation:

$f \left(x\right) = {x}^{\frac{3}{2}}$
$f ' \left(x\right) = \frac{3}{2} \cdot {x}^{\frac{1}{2}}$

$L = {\int}_{a}^{b} \sqrt{1 + {\left[f ' \left(x\right)\right]}^{2}} \cdot \mathrm{dx}$

$L = {\int}_{0}^{3} \sqrt{1 + {\left[\frac{3}{2} \cdot {x}^{\frac{1}{2}}\right]}^{2}} \cdot \mathrm{dx}$
$L = {\int}_{0}^{3} \sqrt{1 + \frac{9}{4} x} \cdot \mathrm{dx}$
$L = \frac{3}{2} {\int}_{0}^{3} \sqrt{x + \frac{4}{9}} \cdot \mathrm{dx}$
$L = \cancel{\frac{3}{2}} \cdot \cancel{\frac{2}{3}} {\left(x + \frac{4}{9}\right)}^{\frac{3}{2}} {|}_{0}^{3}$
$L = {\left(3 + \frac{4}{9}\right)}^{\frac{3}{2}} - {\left(\frac{4}{9}\right)}^{\frac{3}{2}} = {\left(31\right)}^{\frac{3}{2}} / 27 - \frac{8}{27}$
$L = \frac{31 \sqrt{31} - 8}{27} \cong 6.0963$