How do you find the arc length of the curve f(x)=x^2-1/8lnx over the interval [1,2]?

May 21, 2018

Length of arc is $3 + \frac{1}{8} \ln \left(2\right)$

Explanation:

$f \left(x\right) = {x}^{2} - \frac{1}{8} \ln | x | \therefore {f}^{'} \left(x\right) = 2 x - \frac{1}{8 x}$

$\therefore {\left[{f}^{'} \left(x\right)\right]}^{2} + 1 = {\left(2 x - \frac{1}{8 x}\right)}^{2} + 1$ or

$\therefore {\left[{f}^{'} \left(x\right)\right]}^{2} + 1 = {\left(2 x\right)}^{2} - 2 \cdot 2 \cancel{x} \cdot \frac{1}{8 \cancel{x}} + {\left(\frac{1}{8 x}\right)}^{2} + 1$ or

$\therefore {\left[{f}^{'} \left(x\right)\right]}^{2} + 1 = {\left(2 x\right)}^{2} - \frac{1}{2} + {\left(\frac{1}{8 x}\right)}^{2} + 1$ or

$\therefore {\left[{f}^{'} \left(x\right)\right]}^{2} + 1 = {\left(2 x\right)}^{2} + \frac{1}{2} + {\left(\frac{1}{8 x}\right)}^{2}$ or

$\therefore {\left[{f}^{'} \left(x\right)\right]}^{2} + 1 = {\left(2 x + \frac{1}{8 x}\right)}^{2}$

Length of arc is L=int_1^2 (sqrt [f^'(x)^2+1] dx

$L = {\int}_{1}^{2} \sqrt{{\left(2 x + \frac{1}{8 x}\right)}^{2}} \mathrm{dx}$ or

$L = {\int}_{1}^{2} \left(2 x + \frac{1}{8 x}\right) \mathrm{dx}$ or

$L = {\left[{x}^{2} + \frac{1}{8} \ln \left(x\right)\right]}_{1}^{2}$ or

$L = 4 - 1 + \frac{1}{8} \left[\ln \left(2\right) - \ln \left(1\right)\right]$ or

$L = 3 + \frac{1}{8} \ln \left(2\right) \left[\ln 1 = 0\right]$

Length of arc is $3 + \frac{1}{8} \ln \left(2\right)$ [Ans]