# What is the arclength of f(x)=sqrt((x+3)(x/2-1))+5x on x in [6,7]?

May 11, 2017

$L \approx 1.33$

#### Explanation:

The arclength formula is

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

First, find the derivative of the $f \left(x\right)$
$\frac{d}{\mathrm{dx}} \left(\sqrt{\left(x + 3\right) \left(\frac{x}{2} - 1\right)} + 5 x\right)$
$\frac{d}{\mathrm{dx}} \left(\sqrt{\left(x + 3\right) \left(\frac{x}{2} - 1\right)}\right) + \frac{d}{\mathrm{dx}} \left(5 x\right)$
$\frac{d}{\mathrm{dx}} \left(\sqrt{\left(x + 3\right) \left(\frac{x}{2} - 1\right)}\right) + 5$

For the first term, let $u = \left(x + 3\right) \left(\frac{x}{2} - 1\right) = \frac{1}{2} {x}^{2} + \frac{1}{2} x - 3$
$\frac{\mathrm{du}}{\mathrm{dx}} = x + \frac{1}{2}$

Using $u$-substitution, find $\frac{d}{\mathrm{dx}} \left(\sqrt{\left(x + 3\right) \left(\frac{x}{2} - 1\right)}\right)$

$\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \frac{\mathrm{du}}{\mathrm{dx}}$

Replacing these values with $x$, gives
$\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{x + \frac{1}{2}}{2 \sqrt{\frac{1}{2} {x}^{2} + \frac{1}{2} x - 3}}$

Plug this result into the arc-length formula above
$L = {\int}_{a}^{b} \sqrt{1 + f ' \left(x\right)} \mathrm{dx}$
$L = {\int}_{6}^{7} \sqrt{1 + \frac{x + \frac{1}{2}}{2 \sqrt{\frac{1}{2} {x}^{2} + \frac{1}{2} x - 3}}} \mathrm{dx}$

This is a very difficult function to integrate, so using numeric methods is recommended. Using Simpson's Rule (with $n = 10$ and $\Delta x = 1 / 10$), you could calculate

L~~(1//10)/3[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+cdots
cdots+4f(x_10)+f(x_11)]~~1.33