What is the net area between f(x)=2/x-x in x in[1,2]  and the x-axis?

Apr 7, 2017

$A = \frac{1}{2}$

Explanation:

There is a zero crossing within the interval:

$0 = \frac{2}{x} - x$

${x}^{2} = 2$

$x = \sqrt{2}$

The area will be positive $1 \le x \le \sqrt{2}$ and negative $\sqrt{2} < x \le 2$ Therefore, we subtract the integral in the second region from the first:

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

${\left(2 \ln | x | - {x}^{2} / 2\right]}_{1}^{\sqrt{2}} - {\left(2 \ln | x | - {x}^{2} / 2\right]}_{\sqrt{2}}^{2} =$

${\left(\ln \left({x}^{2}\right) - {x}^{2} / 2\right]}_{1}^{\sqrt{2}} - {\left(\ln \left({x}^{2}\right) - {x}^{2} / 2\right]}_{\sqrt{2}}^{2} =$

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

$\ln \left(2\right) - 1 - 0 + \frac{1}{2} - \ln \left(4\right) + 2 + \ln \left(2\right) - 1 = \frac{1}{2}$