# What is the arclength of f(x)=(x-1)(x+1)  in the interval [0,1]?

Feb 21, 2016

I got $1.478$ units.

#### Explanation:

Remember that arc length is given by ${\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$.

We can see that we need the derivative of the function in question, so let's get to that first:
$f \left(x\right) = \left(x - 1\right) \left(x + 1\right) = {x}^{2} - 1$
$f ' \left(x\right) = 2 x$

Next step is to plug this into the formula:
${\int}_{0}^{1} \sqrt{1 + {\left(2 x\right)}^{2}} \mathrm{dx}$

Before we do anything, we should note that this integral merits a trig substitution. It resembles a right triangle (see below). From the picture, we know that $\tan \left(\theta\right) = \frac{2 x}{1} = 2 x$. That means $\tan \frac{\theta}{2} = x$, and furthermore, $\frac{\mathrm{dx}}{d \theta} = {\sec}^{2} \frac{\theta}{2}$. Multiplying both sides by $d \theta$ tells us $\mathrm{dx} = {\sec}^{2} \frac{\theta}{2} d \theta$.

Making these substitutions into the integral,
${\int}_{0}^{1} \sqrt{1 + {\left(2 \left(\tan \frac{\theta}{2}\right)\right)}^{2}} {\sec}^{2} \frac{\theta}{2} d \theta$

Note that while the stuff inside the integral is in terms of $\theta$ now, the limits of integration (0 and 1) are still in terms of $x$. To fix this issue, recall that $\tan \left(\theta\right) = 2 x$. Taking the inverse tangent of both sides yields $\theta = {\tan}^{- 1} \left(2 x\right)$. Now we can get $x = 0$ and $x = 1$ in terms of $\theta$:
$\theta = {\tan}^{- 1} \left(2 \left(0\right)\right) \to \theta = {\tan}^{- 1} \left(0\right) = 0$
$\theta = {\tan}^{- 1} \left(2 \left(1\right)\right) \to \theta = {\tan}^{- 1} \left(2\right) \approx 1.107$

Unfortunately, there is no exact answer for ${\tan}^{- 1} \left(2\right)$, so we'll have to live with an approximation. Our integral finally becomes
${\int}_{0}^{1.107} \sqrt{1 + {\left(2 \left(\tan \frac{\theta}{2}\right)\right)}^{2}} {\sec}^{2} \frac{\theta}{2} d \theta$

Simplifying:
$\frac{1}{2} {\int}_{0}^{1.107} \sqrt{1 + {\left(\tan \left(\theta\right)\right)}^{2}} {\sec}^{2} \left(\theta\right) d \theta$
$\frac{1}{2} {\int}_{0}^{1.107} \sqrt{1 + {\tan}^{2} \left(\theta\right)} {\sec}^{2} \left(\theta\right) d \theta$

If you remember your trig well, you'll know that $1 + {\tan}^{2} \left(\theta\right) = {\sec}^{2} \left(\theta\right)$.
$\frac{1}{2} {\int}_{0}^{1.107} \sqrt{{\sec}^{2} \left(\theta\right)} {\sec}^{2} \left(\theta\right) d \theta$
$\frac{1}{2} {\int}_{0}^{1.107} {\sec}^{3} \left(\theta\right) d \theta$

Evaluating this integral is unnecessary because we can consult a table of integrals to get a result.
After doing this, we see that $\int {\sec}^{3} \left(\theta\right) d \theta = \frac{1}{2} \sec \left(\theta\right) \tan \left(\theta\right) + \frac{1}{2} \ln \left\mid \sec \left(\theta\right) + \tan \left(\theta\right) \right\mid$. The final step is finding what this is on the interval $\left[0 , 1.107\right]$:
$\frac{1}{2} {\left[\frac{1}{2} \sec \left(\theta\right) \tan \left(\theta\right) + \frac{1}{2} \ln \left\mid \sec \left(\theta\right) + \tan \left(\theta\right) \right\mid\right]}_{0}^{1.107}$
$= \frac{1}{2} \left(\frac{1}{2} \sec \left(1.107\right) \tan \left(1.107\right) + \frac{1}{2} \ln \left\mid \sec \left(1.107\right) + \tan \left(1.107\right) \right\mid\right) - \frac{1}{2} \left(\frac{1}{2} \sec \left(0\right) \tan \left(0\right) + \frac{1}{2} \ln \left\mid \sec \left(0\right) + \tan \left(0\right) \right\mid\right)$
$= 1.478 - 0 = 1.478$