Answer
$$A=\frac{16}{5}$$
Work Step by Step
Given $$y=x^{3 / 2}, \quad y=2 x$$
First, we get the intersections points of the curves
$y=x^{3 / 2}=2 x \ \Rightarrow x^{3 }=4 x^2\ \Rightarrow x^{2 }(x-4 )=0$
$\Rightarrow x =0\Rightarrow y =0$
and $\Rightarrow x =4\Rightarrow y =8$
Secondly by taking vertical slices, consider the following integration
\begin{aligned}
A &=\int_{x=0}^{x=4} \int_{y=x^{3 / 2}}^{y=2 x} d y d x \\ &=\int_{x=0}^{x=4}[y]_{x^{3 / 2}}^{2 x} d x\\
&=\int_{x=0}^{x=4}\left(2 x-x^{3 / 2}\right) d x \\ &=\left[x^{2}-\frac{2}{5} x^{5 / 2}\right]_{0}^{4}\\
&=16-\frac{2}{5}(32)-0\\
&=\frac{16}{5} \end{aligned}
