# How do you find the area enclosed by the x-axis and the given curve y=(6/x) for x between -4 & -2?

Dec 23, 2016

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

#### Explanation:

Simply integrate the function y=(6/x) between the bounds given, -4 and -2.

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

$\text{ } = 6 {\left[\ln | x |\right]}_{-} {4}^{-} 2$

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