Answer
$\frac{55}{12}$
Work Step by Step
The area between two curves $y=f(x)$ and $y=g(x)$ between $x=a$ and $x=b$ is given by the formula:
$$A=\int_a^b |f(x)-g(x)|dx.\quad\quad\quad(1)$$
Given
$$f(x)=\sqrt{x}, \quad g(x)=-\sqrt[3]{x}, \quad h(x)=x-2$$
First , we find the intersection points
\begin{aligned}
y^2&= y+2\\
y^2-y-2&=0\\
(y-2)(y+1)&=0
\end{aligned}
Then
$$y=2,\ \ \ \ y=-1 $$
and for $y=\sqrt{x},\ \ y=-\sqrt[3]{x}\ \ \to y=0$, since
$$y+2 \geq -y^3,\ \ -1\leq y\leq 0\ \ \text{and}\ \ y+2\geq y^2\ \ , 0\leq y\leq 2 $$Then
\begin{aligned}
\text{Area} &=
\int_{-1}^0\left[\left(y+2\right)+y^3\right] d y+ \int_0^2 [ (y+2)-y^2]dy\\
&= \left(\frac{1}{2}y^2+2y+\frac{1}{4}y^4\right)\bigg|_{-1}^0+ \left(\frac{1}{2}y^2+2y-\frac{1}{3}y^3\right)\bigg|_0^2\\
&= \frac{5}{4} + \frac{10}{3}\\
&= \frac{55}{12}
\end{aligned}
