# Find the exact length of the curve?

## $y = {x}^{3} / 3 + \frac{1}{4 x}$, $1 \setminus \le x \setminus \le 2$

Oct 14, 2017

$\text{Length} = \frac{59}{24} \approx 2.4583$

#### Explanation:

The formula for arclength of a function $f \left(x\right)$ on the interval $\left(a , b\right)$ is color(blue)(int_a^b (sqrt(1 + (f'(x))^2))dx.

In this case, $f \left(x\right) = y = \frac{1}{3} {x}^{3} + \frac{1}{4 x} = \frac{1}{3} {x}^{3} + \frac{1}{4} {x}^{- 1}$.

Find the derivative:
$f ' \left(x\right) = {x}^{2} - \frac{1}{4} {x}^{- 2} = {x}^{2} - \frac{1}{4 {x}^{2}}$

Square $f ' \left(x\right)$:
${\left(f ' \left(x\right)\right)}^{2} = \left({x}^{2} - \frac{1}{4 {x}^{2}}\right) \left({x}^{2} - \frac{1}{4 {x}^{2}}\right)$

$= {x}^{4} - \frac{1}{2} + \frac{1}{16 {x}^{4}}$

$\text{Length} = {\int}_{1}^{2} \sqrt{1 + \left({x}^{4} + \frac{1}{16 {x}^{4}} - \frac{1}{2}\right)} \mathrm{dx}$

$= {\int}_{1}^{2} \sqrt{{x}^{4} + \frac{1}{2} + \frac{1}{16 {x}^{4}}} \mathrm{dx}$

Factor by symmetry:

$= {\int}_{1}^{2} \sqrt{{\left({x}^{2} + \frac{1}{4 {x}^{2}}\right)}^{2}} \mathrm{dx}$

$= {\int}_{1}^{2} \left({x}^{2} + \frac{1}{4} {x}^{- 2}\right) \mathrm{dx}$

$= {\left[\frac{1}{3} {x}^{3} - \frac{1}{4} {x}^{- 1}\right]}_{1}^{2}$

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

$= \frac{8}{3} - \frac{1}{8} - \frac{1}{3} + \frac{1}{4}$

$= \frac{7}{3} + \frac{1}{8}$

$= \frac{56}{24} + \frac{3}{24}$

$= \frac{59}{24}$

$\approx 2.4583$