Answer
$-\frac{2}{3}(2 \sqrt{2}-5)$
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
$$f(x)=\sqrt{x}, \quad g(x)=x^2, \quad x=2$$
find the intersection points
\begin{aligned}
x^2&=\sqrt{x}\\
x^4&=x\\
x^4-x&=0\\
x(x^3-1)&=0
\end{aligned}
Then
$$x=0,\ \ x=1 $$
Since
\begin{aligned}
\sqrt{x}\geq x^2& \ \ \ \ 0\leq x\leq 1 \\
x^2 \geq \sqrt{x}\& \ \ \ \ 1\leq x\leq 2
\end{aligned}
then use $(1)$:
\begin{aligned}
\text{Area}
& = \int_0^1\left(\sqrt{x}-x^2\right) d x+\int_1^2\left(-\sqrt{x}+x^2\right) d x\\
&=\left(\frac{2}{3} x^{3/2} -\frac{1}{3}x^3\right) \bigg|_0^1 +\left(\frac{-2}{3} x^{3/2} +\frac{1}{3}x^3\right)\bigg|_1^2\\
&=-\frac{2}{3}(2 \sqrt{2}-5)
\end{aligned}
