# How do you find the arc length of the curve y=lnx over the interval [1,2]?

##### 2 Answers
Jun 28, 2018

Apply the arc length formula.

#### Explanation:

$y = \ln x$

$y ' = \frac{1}{x}$

Arc length is given by:

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

Rearrange:

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

Multiply numerator and denominator by $\sqrt{{x}^{2} + 1}$:

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

Integration is distributive:

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

Apply the substitution $x = \tan \theta$:

$L = {\left[\sqrt{{x}^{2} + 1}\right]}_{1}^{2} + \int \sec \frac{\theta}{\tan} \theta d \theta$

Simplify:

$L = \sqrt{5} - \sqrt{2} + \int \csc \theta d \theta$

Integrate directly:

$L = \sqrt{5} - \sqrt{2} - \left[\ln | \csc \theta + \cot \theta |\right]$

Rewrite in terms of $\tan \theta$ and $\sec \theta$:

$L = \sqrt{5} - \sqrt{2} - \left[\ln | \frac{1 + \sec \theta}{\tan} \theta |\right]$

Reverse the substitution:

$L = \sqrt{5} - \sqrt{2} - {\left[\ln \left(\frac{1 + \sqrt{{x}^{2} + 1}}{x}\right)\right]}_{1}^{2}$

Insert the limits of integration:

$L = \sqrt{5} - \sqrt{2} - \ln \left(\frac{1 + \sqrt{5}}{2 \left(1 + \sqrt{2}\right)}\right)$

Rearrange for clarity:

$L = \sqrt{5} - \sqrt{2} + \ln 2 - \ln \left(\frac{1 + \sqrt{5}}{1 + \sqrt{2}}\right)$

Jun 28, 2018

Length = $\sqrt{5} + \frac{1}{2} \ln \left(\frac{\sqrt{5} - 1}{\sqrt{5} + 1}\right) - \sqrt{2} - \frac{1}{2} \ln \left(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right) \approx 1.222$

#### Explanation:

The arc length of a function $y = f \left(x\right)$ over the interval $\left[a , b\right]$ is given by $L = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$.
$y = \ln x \to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{x}$
$\therefore L = {\int}_{1}^{2} \sqrt{1 + {\left(\frac{1}{x}\right)}^{2}} \mathrm{dx}$
$L = {\int}_{1}^{2} \sqrt{1 + \frac{1}{x} ^ 2} \mathrm{dx}$
I will use this solution $\int \sqrt{1 + \frac{1}{x} ^ 2} \mathrm{dx} = \sqrt{{x}^{2} + 1} + \frac{1}{2} \ln | \frac{\sqrt{{x}^{2} + 1} - 1}{\sqrt{{x}^{2} + 1} + 1} | + C$ courtesy of mason m (https://socratic.org/questions/how-do-you-evaluate-the-integral-int-sqrt-1-1-x-2) to find that
L=sqrt(x^2+1)+1/2ln|(sqrt(x^2+1)-1)/(sqrt(x^2+1)+1)|]_1^2
$\therefore L = \sqrt{5} + \frac{1}{2} \ln \left(\frac{\sqrt{5} - 1}{\sqrt{5} + 1}\right) - \sqrt{2} - \frac{1}{2} \ln \left(\frac{\sqrt{2} - 1}{\sqrt{2} + 1}\right) \approx 1.222$