# What is the arclength of f(x)=x+xsqrt(x+3) on x in [-3,0]?

Mar 25, 2018

$s \approx 6.54696$

#### Explanation:

The formula for arc length of a curve defined by $f \left(x\right)$ is:

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

First let's find the derivative $f ' \left(x\right)$:

$f \left(x\right) = x + x {\left(x + 3\right)}^{\frac{1}{2}}$

$f ' \left(x\right) = 1 + {\left(x + 3\right)}^{\frac{1}{2}} + \frac{1}{2} x {\left(x + 3\right)}^{- \frac{1}{2}}$ (by the product rule)

$f ' \left(x\right) = 1 + \sqrt{x + 3} + \frac{x}{2 \sqrt{x + 3}}$

Now let's evaluate $f ' {\left(x\right)}^{2}$

$f ' {\left(x\right)}^{2} = {\left[1 + \sqrt{x + 3} + \frac{x}{2 \sqrt{x + 3}}\right]}^{2}$

$= 1 + \sqrt{x + 3} + \frac{x}{2 \sqrt{x + 3}} + \sqrt{x + 3} + \left(x + 3\right) + \frac{x \cancel{\sqrt{x + 3}}}{2 \cancel{\sqrt{x + 3}}} + \frac{x}{2 \sqrt{x + 3}} + \frac{x \cancel{\sqrt{x + 3}}}{2 \cancel{\sqrt{x + 3}}} + {x}^{2} / \left(4 \left(x + 3\right)\right)$

$= 1 + 2 \sqrt{x + 3} + \cancel{2} \frac{x}{\cancel{2} \sqrt{x + 3}} + \left(x + 3\right) + \cancel{2} \left(\frac{x}{\cancel{2}}\right) + {x}^{2} / \left(4 \left(x + 3\right)\right)$

$= {x}^{2} / \left(4 \left(x + 3\right)\right) + x \left(2 + \frac{1}{\sqrt{x + 3}}\right) + 2 \sqrt{x + 3} + 4$

Now you would need to evaluate this (very nasty) integral:

${\int}_{-} {3}^{0} \sqrt{{x}^{2} / \left(4 \left(x + 3\right)\right) + x \left(2 + \frac{1}{\sqrt{x + 3}}\right) + 2 \sqrt{x + 3} + 5}$

I punched it into Wolfram|Alpha and got the following result:

$s \approx 6.54696$