# What is the arclength of f(x)=2-x^2  in the interval [0,1]?

Nov 12, 2015

$\frac{\sqrt{5}}{2} + \ln \frac{\sqrt{5} + 2}{4}$

#### Explanation:

The arc length of a function $f \left(x\right)$ over an interval $\left[a , b\right]$ is given by the definite integral
${\int}_{a}^{b} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$

First, we find $f ' \left(x\right)$
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(2 - {x}^{2}\right) = - 2 x$

So
${\left(f ' \left(x\right)\right)}^{2} = {\left(- 2 x\right)}^{2} = 4 {x}^{2}$

Then the arc length of $2 - {x}^{2}$ on $\left[0 , 1\right]$ is given by
$L = {\int}_{0}^{1} \sqrt{1 + 4 {x}^{2}} \mathrm{dx}$

We can evaluate this integral via trig substitution:
Set $x = \tan \frac{\theta}{2}$ and $\mathrm{dx} = {\sec}^{2} \frac{\theta}{2} d \theta$
We also must change the bounds appropriately.
$x = 0 \implies \theta = 0$ and $x = 1 \implies \theta = {\tan}^{-} 1 \left(2\right)$

So
$L = \frac{1}{2} {\int}_{0}^{{\tan}^{-} 1 \left(2\right)} {\sec}^{3} \left(\theta\right) d \theta$

The indefinite integral of ${\sec}^{3} \left(x\right) \mathrm{dx}$ is
$\int {\sec}^{3} \left(x\right) \mathrm{dx} = \frac{1}{2} \left(\sec \left(x\right) \tan \left(x\right) + \ln | \sec \left(x\right) + \tan \left(x\right) |\right) + C$

(finding the integrals of $\sec \left(x\right)$ and ${\sec}^{3} \left(x\right)$ are interesting exercises on their own, and are frequently solved via a clever $u$ substitution and integration by parts, respectively)

Continuing, this gives us
$L = \frac{1}{4} {\left[\sec \left(x\right) \tan \left(x\right) + \ln | \sec \left(x\right) + \tan \left(x\right) |\right]}_{0}^{{\tan}^{-} 1 \left(2\right)}$

Evaluating this gives us

$L = \frac{1}{4} \left(2 \sec \left({\tan}^{-} 1 \left(2\right)\right) + \ln | \sec \left({\tan}^{-} 1 \left(2\right)\right) + 2 | - 1 \cdot 0 - \ln | 1 + 0 |\right)$
$\implies L = \frac{1}{2} \sec \left({\tan}^{-} 1 \left(2\right)\right) + \frac{1}{4} \ln | \sec \left({\tan}^{-} 1 \left(2\right)\right) + 2 |$

Finally, we can simplify further by noting that
${\tan}^{-} 1 \left(2\right) = {\cos}^{-} 1 \left(\frac{1}{\sqrt{5}}\right)$
(to see this, consider a right triangle with side lengths $1$, $2$, and $\sqrt{5}$)

So, substituting and simplifying (remember $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$), we get the final result

$L = \frac{\sqrt{5}}{2} + \ln \frac{\sqrt{5} + 2}{4}$