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

Feb 6, 2017

$\left[\left(\frac{1}{2} - \frac{5}{6} \sqrt{5} + \frac{1}{12} \sqrt{5}\right) - \left(8 - \frac{34}{3} \sqrt{17} + \frac{17}{60} \sqrt{17}\right)\right] = 36.38$ areal units, nearly. Please do not attempt to edit this answer. I would review my answer, for bugs, if any, myself, after some time.

#### Explanation:

The integrated area is in y-negative ${Q}_{4}$ and is negative.

The end points for the curved boundary are

A at $\left(1 , - \left(\sqrt{5} - 1\right)\right) = \left(1. - 1.236\right)$ and

B at $\left(4 , = 4 \left(\sqrt{17} - 1\right)\right) = \left(4 , = 12.402\right)$.

The numerical area ( suppressing negative sign )

$= - \int y \mathrm{dx}$, with $y = x \left(1 - \sqrt{4 x + 1}\right)$ and x, from 1 to 4

$= - \int \left(x - x \sqrt{4 x + 1}\right) \mathrm{dx}$, for the limits 1 and 4

$= - \left[{x}^{2} / 2 - \left(\frac{2}{3}\right) \left(\frac{1}{4}\right) \int x d {\left(4 x + 1\right)}^{\frac{3}{2}}\right]$, for x = 1 to 4

#=-[x^2/2-1/6x(4x+1)^(3/2)+1/6int(4x+1)^(3/2) dx], for x = 1 to 4

$= - \left[{x}^{2} / 2 - \frac{1}{6} x {\left(4 x + 1\right)}^{\frac{3}{2}} + \frac{1}{6} \left(\frac{2}{5}\right) \left(\frac{1}{4}\right) {\left(4 x + 1\right)}^{\frac{5}{2}}\right] ,$

between x = 1 and 4

$= - \left[\left(8 - \frac{34}{3} \sqrt{17} + \frac{17}{60} \sqrt{17}\right) - \left(\frac{1}{2} - \frac{5}{6} \sqrt{5} + \frac{1}{12} \sqrt{5}\right)\right]$

graph{x(1-sqrt(4x+1)) [0, 50, -25, 0]}