What is the arclength of f(x)=x-sqrt(e^x-2lnx) on x in [1,2]?

Mar 3, 2017

$L = {\int}_{1}^{2} \sqrt{1 + {\left(1 - \frac{x {e}^{x} - 2}{2 x \sqrt{{e}^{x} - 2 \ln x}}\right)}^{2}} \mathrm{dx} \approx 1.0630$

Explanation:

The arc length of the curve of $f$ on $x \in \left[a , b\right]$ is given by

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

Here, $f \left(x\right) = x - {\left({e}^{x} - 2 \ln x\right)}^{\frac{1}{2}}$ so

$f ' \left(x\right) = 1 - \frac{1}{2} {\left({e}^{x} - 2 \ln x\right)}^{- \frac{1}{2}} \left({e}^{x} - \frac{2}{x}\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = 1 - \frac{x {e}^{x} - 2}{2 x \sqrt{{e}^{x} - 2 \ln x}}$

Then the arc length is given by

$L = {\int}_{1}^{2} \sqrt{1 + {\left(1 - \frac{x {e}^{x} - 2}{2 x \sqrt{{e}^{x} - 2 \ln x}}\right)}^{2}} \mathrm{dx} \approx 1.0630$