What is the arc length of f(x)= sqrt(x^3+5)  on x in [0,2]?

Dec 28, 2017

The length is around $2.58$.

Explanation:

The formula for arc length of a curve of a function $f \left(x\right)$ on the interval $\left[a , b\right]$ is equal to:
${\int}_{a}^{b} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \setminus \mathrm{dx}$

Let us first work out $f ' \left(x\right)$:
$\frac{d}{\mathrm{dx}} \left(\sqrt{{x}^{3} + 5}\right) =$

I will use the chain rule and let $u = {x}^{3} + 5$
$= \frac{d}{\mathrm{du}} \left(\sqrt{u}\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{3} + 5\right) = \frac{1}{2 \sqrt{u}} \cdot 3 {x}^{2} =$

$= \frac{3 {x}^{2}}{2 \sqrt{{x}^{3} + 5}}$

Now we plug this into our arc length formula:
${\int}_{0}^{2} \sqrt{1 + {\left(\frac{3 {x}^{2}}{2 \sqrt{{x}^{3} + 5}}\right)}^{2}} \setminus \mathrm{dx}$

${\int}_{0}^{2} \sqrt{1 + \frac{9 {x}^{4}}{4 \left({x}^{3} + 5\right)}} \setminus \mathrm{dx}$

This function doesn't have an elementary anti-derivative (that I've been able to find). The best we can really do is use an approximation. Here I did a midpoint Riemann rectangle sum with 25 rectangles to get an answer of about $2.58$: 