# How do you find the area of the region bounded by the curves y=1+sqrt(x) and y=1+x/3 ?

Sep 22, 2014

Our first step is to find the interval over which we have to integrate. This is accomplished by setting the 2 functions equal to each other. And then solve for x.

$1 + \sqrt{x} = 1 + \frac{x}{3}$, subtract $1$ from both sides

$\sqrt{x} = \frac{x}{3}$, square both sides

$x = \frac{{x}^{2}}{9}$

$x - \frac{{x}^{2}}{9} = 0$

$x \left(1 - \frac{x}{9}\right) = 0$

Set each factor equal to 0.

$x = 0 \to$ lower bound

$1 - \frac{x}{9} = 0$

$- \frac{x}{9} = - 1$

$x = 9 \to$ upper bound

$\left[0 , 9\right] \to$ interval

We now need to figure out which function is greater over this interval. To do this we substitute in a value between 0 and 9. Lets us an $x$ value of $1$.

$y = 1 + \sqrt{1} = 1 + 1 = 2 \to$ Larger function
$y = 1 + \frac{1}{3} = \frac{4}{3} = 1.333$

${\int}_{0}^{9} 1 + \sqrt{x} - \left(1 + \frac{x}{3}\right) \mathrm{dx}$

${\int}_{0}^{9} 1 + \sqrt{x} - 1 - \frac{x}{3} \mathrm{dx}$

${\int}_{0}^{9} \sqrt{x} - \frac{x}{3} \mathrm{dx}$

${\int}_{0}^{9} {x}^{\frac{1}{2}} - \frac{x}{3} \mathrm{dx}$

${\left[{x}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - {x}^{2} / \left(3 \cdot 2\right)\right]}_{0}^{9}$

$\left[{\left(9\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - {\left(9\right)}^{2} / \left(3 \cdot 2\right) - \left({\left(0\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - {\left(0\right)}^{2} / \left(3 \cdot 2\right)\right)\right]$

$\left[{\left(9\right)}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - {\left(9\right)}^{2} / \left(3 \cdot 2\right)\right]$

$\left[\frac{\sqrt{{9}^{3}}}{\frac{3}{2}} - \frac{81}{6}\right]$

$\left[\sqrt{{9}^{3}} \cdot \left(\frac{2}{3}\right) - \frac{81}{6}\right]$

$\left[\sqrt{{\left({3}^{2}\right)}^{3}} \cdot \left(\frac{2}{3}\right) - \frac{81}{6}\right]$

$\left[\sqrt{\left({3}^{6}\right)} \cdot \left(\frac{2}{3}\right) - \frac{81}{6}\right]$

$\left[\left({3}^{3}\right) \cdot \left(\frac{2}{3}\right) - \frac{81}{6}\right]$

$\left[\left({3}^{2}\right) \cdot \left(2\right) - \frac{81}{6}\right]$

$\left[\left(9\right) \cdot \left(2\right) - \frac{81}{6}\right]$

$\left[18 - \frac{27}{2}\right]$

$\left[\frac{36}{2} - \frac{27}{2}\right]$

$\left[\frac{9}{2}\right] = 4.5 \to$ Solution