# What is the net area between f(x) = -sqrt(x+1)  and the x-axis over x in [1, 4 ]?

Jul 21, 2017

$A = \frac{10 \sqrt{5} - 4 \sqrt{2}}{3}$

#### Explanation:

The area is :

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

${\int}_{1}^{4} {\left(x + 1\right)}^{\frac{1}{2}} \mathrm{dx} = {\left[\frac{{\left(x + 1\right)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{1}^{4} =$

$\frac{2}{3} \left(\sqrt{{\left(4 + 1\right)}^{3}} - \sqrt{{\left(1 + 1\right)}^{3}}\right) = \frac{2}{3} \left(\sqrt{125} - \sqrt{8}\right) =$

$\frac{2}{3} \left(5 \sqrt{5} - 2 \sqrt{2}\right) = \frac{10 \sqrt{5} - 4 \sqrt{2}}{3}$