Answer
$\frac{5}{3}$
Work Step by Step
Step 1. Graph the functions as shown in the figure.
Step 2. Separate the regions into $x\lt0$ and $x\leq 0$
Step 3. For $x\lt0$, solve the system of equations
$\begin{cases} y=\sqrt {-x} \\ 5y=x+6 \end{cases}$
to get the first intersection point of $(-1,1)$
($x=-36$ is not a solution).
Step 4. For $x\geq0$, solve the system of equations
$\begin{cases} y=\sqrt {x} \\ 5y=x+6 \end{cases}$
to get the second intersection point of $(4,2)$ and the third at $x(9,3)$. Thus there are three intersection points between the two curves.
Step 5. The enclosed area between the two functions can be found as
$A=\int_{-1}^0(\frac{1}{5}x+\frac{6}{5}-\sqrt {-x})dx+\int_0^4(\frac{1}{5}x+\frac{6}{5}-\sqrt {x})dx+\int_4^9(\sqrt {x}-\frac{1}{5}x-\frac{6}{5})dx
=(\frac{1}{10}x^2+\frac{6}{5}x+\frac{2}{3}(-x)^{3/2})|_{-1}^0 + (\frac{1}{10}x^2+\frac{6}{5}x-\frac{2}{3}(x)^{3/2})|_{0}^4 + (\frac{2}{3}(x)^{3/2}-\frac{1}{10}x^2-\frac{6}{5}x)|_{4}^9
=(-\frac{1}{10}(-1)^2-\frac{6}{5}(-1)-\frac{2}{3}(-(-1))^{3/2})+ (\frac{1}{10}(4)^2+\frac{6}{5}(4)-\frac{2}{3}(4)^{3/2})+ (\frac{2}{3}(9)^{3/2}-\frac{1}{10}(9)^2-\frac{6}{5}(9))- (\frac{2}{3}(4)^{3/2}-\frac{1}{10}(4)^2-\frac{6}{5}(4))
=\frac{5}{3}$
(combine fractions with the same denominators and simplify).
