# What is the net area between f(x) = 3-xsqrt(x^2-1)  and the x-axis over x in [2, 3 ]?

Jun 25, 2017

The area $\approx 2.81$

#### Explanation:

Here is a graph of the function $f \left(x\right) = 3 - x \sqrt{{x}^{2} - 1}$:

graph{3-xsqrt(x^2-1) [-1, 5, -7, 3]}

Please observe that the $f \left(x\right)$ is negative over the region $\left[2 , 3\right]$, therefore, to obtain an positive area, we shall evaluate from 3 to 2.

To integrate $- x \sqrt{{x}^{2} - 1} \mathrm{dx}$ let $u = {x}^{2} - 1$ and $\mathrm{du} = 2 x \mathrm{dx}$

$\int x {\left({x}^{2} - 1\right)}^{\frac{1}{2}} \mathrm{dx} = \int \frac{1}{2} {u}^{\frac{1}{2}} \mathrm{du} = \frac{2}{3} \left(\frac{1}{2}\right) {u}^{\frac{3}{2}} = \frac{1}{3} {u}^{\frac{3}{2}}$:

The integration of the constant term is trivial:

${\int}_{3}^{2} 3 - x \sqrt{{x}^{2} - 1} \mathrm{dx} = {\left(3 x - \frac{1}{3} {\left({x}^{2} - 1\right)}^{\frac{3}{2}}\right]}_{3}^{2} =$

$3 \left(2\right) - \frac{1}{3} {\left({2}^{2} - 1\right)}^{\frac{3}{2}} - 3 \left(3\right) + \frac{1}{3} {\left({3}^{2} - 1\right)}^{\frac{3}{2}} \approx 2.81$