# How do you find the arc length of the curve f(x)=x^(3/2) over the interval [0,1]?

Sep 6, 2016

$\frac{1}{27} \left(13 \sqrt{13} - 8\right)$

#### Explanation:

The arc length of the curve $f \left(x\right)$ on the interval $\left[a , b\right]$ is defined as:

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

Here, we see that $f \left(x\right) = {x}^{\frac{3}{2}}$ and ${f}^{'} \left(x\right) = \frac{3}{2} {x}^{\frac{1}{2}}$. Thus:

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

$= {\int}_{0}^{1} \sqrt{1 + \frac{9}{4} x} \mathrm{dx}$

$= {\int}_{0}^{1} \sqrt{\frac{4 + 9 x}{4}} \mathrm{dx}$

$= \frac{1}{2} {\int}_{0}^{1} \sqrt{4 + 9 x} \mathrm{dx}$

Substitute with $u = 4 + 9 x$ such that $\mathrm{du} = 9 \mathrm{dx}$.

$= \frac{1}{18} {\int}_{0}^{1} 9 \sqrt{4 + 9 x} \mathrm{dx}$

Substitute $u$ and change the bounds from $x$ to $u$: x=0->u=4+9(0)=4;x=1->u=4+9(1)=13.

$= \frac{1}{18} {\int}_{4}^{13} \sqrt{u} \mathrm{du}$

$= \frac{1}{18} {\int}_{4}^{13} {u}^{\frac{1}{2}} \mathrm{du}$

$= \frac{1}{18} {\left[{u}^{\frac{3}{2}} / \left(\frac{3}{2}\right)\right]}_{4}^{13}$

$= \frac{1}{18} \left(\frac{2}{3}\right) {\left[{u}^{\frac{3}{2}}\right]}_{4}^{13}$

$= \frac{1}{27} {\left[{u}^{\frac{3}{2}}\right]}_{4}^{13}$

$= \frac{1}{27} \left({13}^{\frac{3}{2}} - {4}^{\frac{3}{2}}\right)$

$= \frac{1}{27} \left({13}^{1} \cdot {13}^{\frac{1}{2}} - {\left({2}^{2}\right)}^{\frac{3}{2}}\right)$

$= \frac{1}{27} \left(13 \sqrt{13} - {2}^{3}\right)$

$= \frac{1}{27} \left(13 \sqrt{13} - 8\right)$