# What is the net area between f(x) = x^2-sqrt(x+1)  and the x-axis over x in [1, 7 ]?

Dec 14, 2017

See below.

#### Explanation:

$x - \sqrt{x + 1} = 0$

${x}^{2} - x - 1 = 0 \implies x = \frac{1 + \sqrt{5}}{2} \mathmr{and} x = \frac{1 - \sqrt{5}}{2}$

The function is below the x axis in the interval:

$\left(\frac{1 - \sqrt{5}}{2} , \frac{1 + \sqrt{5}}{2}\right)$

So we need two integrals:

$A = - {\int}_{1}^{\frac{1 + \sqrt{5}}{2}} \left({x}^{2} - \sqrt{x + 1}\right) \mathrm{dx} + {\int}_{\frac{1 + \sqrt{5}}{2}}^{7} \left({x}^{2} - \sqrt{x + 1}\right) \mathrm{dx}$

$A = - {\int}_{1}^{\frac{1 + \sqrt{5}}{2}} \left({x}^{2} - \sqrt{x + 1}\right) \mathrm{dx}$

$\to = {\left[\frac{1}{3} {x}^{3} - \frac{2}{3} {\left(x + 1\right)}^{\frac{3}{2}}\right]}^{\left(1 + \sqrt{5}\right)} - {\left[\frac{1}{3} {x}^{3} - \frac{2}{3} {\left(x + 1\right)}^{\frac{3}{2}}\right]}_{1}$

$= - \left[\frac{1}{3} {\left(\frac{1 + \sqrt{5}}{2}\right)}^{3} - \frac{2}{3} {\left(\frac{1 + \sqrt{5}}{2} + 1\right)}^{\frac{3}{2}}\right] - \left[\frac{1}{3} {\left(1\right)}^{3} - \frac{2}{3} {\left(\frac{1 + \sqrt{5}}{2} + 1\right)}^{\frac{3}{2}}\right] = 0.04738$

$A = {\int}_{\frac{1 + \sqrt{5}}{2}}^{7} \left({x}^{2} - \sqrt{x + 1}\right) \mathrm{dx}$

$\to = {\left[\frac{1}{3} {x}^{3} - \frac{2}{3} {\left(x + 1\right)}^{\frac{3}{2}}\right]}^{7} - {\left[\frac{1}{3} {x}^{3} - \frac{2}{3} {\left(x + 1\right)}^{\frac{3}{2}}\right]}_{\frac{1 + \sqrt{5}}{2}}$

$\to = {\left[\frac{1}{3} {\left(7\right)}^{3} - \frac{2}{3} {\left(7 + 1\right)}^{\frac{3}{2}}\right]}^{7} - {\left[\frac{1}{3} {\left(\frac{1 + \sqrt{5}}{2}\right)}^{3} - \frac{2}{3} {\left(\left(\frac{1 + \sqrt{5}}{2}\right) + 1\right)}^{\frac{3}{2}}\right]}_{\frac{1 + \sqrt{5}}{2}} = 100.84806$

Area$= 100.84806 + 0.04738 = 100.89$ units squared.

Graph: