# What is the net area between f(x)=x/ln(x^2) in x in[3,10]  and the x-axis?

May 6, 2018

The area equal
$A = {\int}_{3}^{10} \frac{x}{\ln} \left({x}^{2}\right) \cdot \mathrm{dx} = {\left[\frac{\Gamma \left(0 , - 2 \cdot \ln \left(x\right)\right)}{2}\right]}_{3}^{10} = \left[\frac{\Gamma \left(0 , - 2 \cdot \ln \left(3\right)\right) - \Gamma \left(0 , - 2 \cdot \ln \left(10\right)\right)}{2}\right] = 12.20245 {\left(u n i t e\right)}^{2}$

#### Explanation:

The sketch of our function $f \left(x\right) = \frac{x}{\ln} \left({x}^{2}\right)$

graph{x/ln(x^2) [-10, 10, -5, 5]}

the interval $x \in \left[3 , 10\right]$

this is the area wanted

$A = {\int}_{a}^{b} y \cdot \mathrm{dx}$

$A = {\int}_{3}^{10} \frac{x}{\ln} \left({x}^{2}\right) \cdot \mathrm{dx} = {\left[\frac{\Gamma \left(0 , - 2 \cdot \ln \left(x\right)\right)}{2}\right]}_{3}^{10} = \left[\frac{\Gamma \left(0 , - 2 \cdot \ln \left(3\right)\right) - \Gamma \left(0 , - 2 \cdot \ln \left(10\right)\right)}{2}\right] = 12.20245 {\left(u n i t e\right)}^{2}$