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

Feb 3, 2017

$\frac{2}{3} \left(2 \sqrt{5} + \sqrt{2}\right) = 3.924$ areal units, nearly. See Socratic graph to see the enclosure.

#### Explanation:

$y = \frac{x}{\sqrt{x + 1}} = \frac{\left(x + 1\right) - 1}{\sqrt{x + 1}} = {\left(x + 1\right)}^{\frac{1}{2}} - {\left(x + 1\right)}^{- \frac{1}{2}}$.-

The area = $\int y \mathrm{dx}$, with $y = {\left(x + 1\right)}^{\frac{1}{2}} - {\left(x + 1\right)}^{- \frac{1}{2}}$ and x from 1 to 4

$= \int \left({\left(x + 1\right)}^{\frac{1}{2}} - {\left(x + 1\right)}^{- \frac{1}{2}}\right) \mathrm{dx}$, for the limits

$= \left[\frac{2}{3} {\left(x + 1\right)}^{\frac{3}{2}} - 2 {\left(x + 1\right)}^{\frac{1}{2}}\right]$, between 1 and 4

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

$= \frac{4}{3} \sqrt{5} + \frac{2}{3} \sqrt{2}$

graph{(y-x/sqrt(x+1))y(x-1+.01y)(x-4)=0 [-10, 10, -5, 5]}