# What is the arc length of f(x)=xsqrt(x^2-1)  on x in [3,4] ?

##### 1 Answer
May 20, 2018

$L = 7 + \frac{1}{8} \ln \left(\frac{15}{8}\right) + {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n \left(\frac{1}{8} ^ \left(n - 1\right) - \frac{1}{15} ^ \left(n - 1\right)\right)$

#### Explanation:

$f \left(x\right) = x \sqrt{{x}^{2} - 1}$

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

Arc length is given by:

$L = {\int}_{3}^{4} \sqrt{1 + {\left(2 {x}^{2} - 1\right)}^{2} / \left({x}^{2} - 1\right)} \mathrm{dx}$

Simplify:

$L = {\int}_{3}^{4} x \sqrt{\frac{4 {x}^{2} - 3}{{x}^{2} - 1}} \mathrm{dx}$

Rearrange:

$L = {\int}_{3}^{4} \left(2 x\right) \sqrt{1 + \frac{1}{4 \left({x}^{2} - 1\right)}} \mathrm{dx}$

Apply the substitution ${x}^{2} - 1 = u$:

$L = {\int}_{8}^{15} \sqrt{1 + \frac{1}{4 u}} \mathrm{du}$

For $u \in \left[8 , 15\right]$, $\frac{1}{4 u} < 1$. Take the series expansion of the square root:

$L = {\int}_{8}^{15} {\sum}_{n = 0}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4 u} ^ n \mathrm{du}$

Isolate the $n = 0$ and $n = 1$ terms:

$L = {\int}_{8}^{15} \left(1 + \frac{1}{8 u}\right) \mathrm{du} + {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n {\int}_{8}^{15} \frac{1}{u} ^ n \mathrm{du}$

Integrate term by term:

$L = {\left[u + \frac{1}{8} \ln u\right]}_{8}^{15} + {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n {\left[- \frac{1}{u} ^ \left(n - 1\right)\right]}_{8}^{15}$

Insert the limits of integration:

$L = 7 + \frac{1}{8} \ln \left(\frac{15}{8}\right) + {\sum}_{n = 2}^{\infty} \left(\begin{matrix}\frac{1}{2} \\ n\end{matrix}\right) \frac{1}{4} ^ n \left(\frac{1}{8} ^ \left(n - 1\right) - \frac{1}{15} ^ \left(n - 1\right)\right)$