Answer
$\frac{8}{3} $
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
$$f(x)=x^2, \quad g(x)=4 x-x^2$$
First, we find the intersection points
\begin{aligned}
x^2&=4x-x^2\\
2x^2-4x&=0\\
2x(x-2)&=0
\end{aligned}
Then apply $(1)$ with $a=0$ and $b=2$:
$$x=0,\ \ \ \ x=2 $$
Since
$$ 4x-x^2\geq x^2,\ \ \ \ 0\leq x\leq 2 $$Then
\begin{aligned}
\text{Area} &=
\int_0^2\left[\left(4 x-x^2\right)-x^2\right] d x\\
&=\int_0^2\left(4 x-2 x^2\right) d x \\
& =\left[2 x^2-\frac{2}{3} x^3\right]_0^2\\
&=\left[\left(8-\frac{16}{3}\right)-0\right]\\
&=\frac{8}{3}
\end{aligned}
