# How do you sketch the region enclosed by y=1+sqrtx, Y=(3+x)/3 and find the area?

Dec 26, 2016

$4.5$ square units

#### Explanation:

Start by finding the intersection points of the two curves.

$1 + \sqrt{x} = \frac{3 + x}{3}$

$3 \left(1 + \sqrt{x}\right) = 3 + x$

$3 + 3 \sqrt{x} = 3 + x$

$3 \sqrt{x} = x + 3 - 3$

${\left(3 \sqrt{x}\right)}^{2} = {x}^{2}$

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

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

$0 = x \left(x - 9\right)$

$x = 0 \mathmr{and} 9$

Next, sketch a rudimentary graph of the curves to see which one lies above the other. You will find the graph of $y = \sqrt{x} + 1$ is above $y = \frac{x + 3}{3}$. You can trace the square root function by drawing the function $y = \sqrt{x}$ and transform it one unit up. $y = \frac{1}{3} \left(x + 3\right)$ is equivalent to $y = \frac{1}{3} x + 1$, which is your run of the mill linear function.

Integrate:

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

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

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

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

$= \frac{1}{3} {\left[2 {x}^{\frac{3}{2}} - \frac{1}{2} {x}^{2}\right]}_{0}^{9}$

$= \frac{1}{3} \left(13.5\right)$

$= 4.5$ square units

Hopefully this helps