# How do you find the length of the curve for y= 1/8(4x^2–2ln(x)) for [2, 6]?

Apr 3, 2018

The arc length is $4.46$ units.

#### Explanation:

We have:

$A = {\int}_{2}^{6} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$

And by derivative rules

$y = \frac{1}{2} {x}^{2} - \frac{1}{4} \ln x$

$y ' = x - \frac{1}{4 x}$

Hence:

$A = {\int}_{2}^{6} \sqrt{1 + {\left(x - \frac{1}{4 x}\right)}^{2}} \mathrm{dx}$

$A = {\int}_{2}^{6} \sqrt{1 + {\left(\frac{4 {x}^{2} - 1}{4 x}\right)}^{2}} \mathrm{dx}$

$A = {\int}_{2}^{6} \sqrt{\frac{16 {x}^{2} + 16 {x}^{4} - 8 {x}^{2} + 1}{4 x} ^ 2} \mathrm{dx}$

$A = {\int}_{2}^{6} \frac{\sqrt{16 {x}^{4} + 8 {x}^{2} + 1}}{4 x} \mathrm{dx}$

$A = {\int}_{2}^{6} \frac{\sqrt{{\left(4 {x}^{2} + 1\right)}^{2}}}{4 {x}^{2}} \mathrm{dx}$

$A = {\int}_{2}^{6} \frac{4 {x}^{2} + 1}{4 x} \mathrm{dx}$

$A = {\int}_{2}^{6} x + \frac{1}{4 x} \mathrm{dx}$

$A = {\left[\frac{1}{2} {x}^{2} + \frac{1}{4} \ln | x |\right]}_{2}^{6}$

$A = \frac{1}{2} {\left(6\right)}^{2} + \frac{1}{4} \ln | 6 | - \frac{1}{2} {\left(2\right)}^{2} - \frac{1}{4} \ln | 2 |$

$A = 18 + \frac{1}{4} \ln | 6 | - 2 - \frac{1}{4} \ln | 2 |$

$A = 16.27$

Hopefully this helps!