Hello, darkvelvet!
From your integral, I conclude that your graph is wrong.
And I don't agree with their answer.
$\displaystyle \text{Find the area of the region bounded by: }\;\begin{Bmatrix}y \:=\:\sqrt{x} \\ y \:=\:2  \sqrt{x} \\ y\:=\:0.\end{Bmatrix}$
The graph of $\displaystyle y = \sqrt{x}$ is the upper half of a "horizontal" parabola,
. . vertex at (0,0), opening to the right.
The graph of $\displaystyle y =\sqrt{x}$ is the above graph reflected over the $\displaystyle x$axis.
The graph of $\displaystyle y \,=\,2  \sqrt{x}$ is the same graph raised 2 units.
And $\displaystyle \,y = 0$ is the $\displaystyle x$axis.
The graph looks like this:
Code:

2 * *
 *
* *
 * *
 *.
 .::::*..
*:::::::::::*...
::::::::::::::::::::*.....
  *  +              *  
 1 4

$\displaystyle \displaystyle \text{Area} \;=\;\int^1_0\!\! x^{\frac{1}{2}}\,dx + \int^4_1\!\!\left(2x^{\frac{1}{2}}\right)\,dx $
My answer is 2.