The area of the region bounded by the curve \(y^{2}=8 x\) and the line \(y=2 x\) is

Given equation of curve,
\(y^{2}=8 x \text { and line } y=2 x\)
Now, intersecting points of given curve and line.
\(\begin{aligned}
&\quad(2 x)^{2}=8 x \\
&\Rightarrow \quad 4 x^{2}=8 x \\
&\Rightarrow \quad x=0,2 \\
&\text { Putting values of } x \text { in } y=2 x \text { then, we get } y=0,4 \\
&\therefore \text { Required area of bounded region. }
\end{aligned}\)



\(=\int_{0}^{2}(\sqrt{8 x}-2 x) d x\)
\(=\int_{0}^{2}\left(2 \sqrt{2} \cdot x^{1 / 2}-2 x\right) d x\)
\(=2 \int_{0}^{2}\left(\sqrt{2} \cdot x^{1 / 2}-x\right) d x\)
\(=2\left[\sqrt{2} \int_{0}^{2} x^{1 / 2} d x-\int_{0}^{2} x d x\right]\)
\(=2\left[\sqrt{2}\left(\frac{x^{3 / 2}}{3 / 2}\right)_{0}^{2}-\left(\frac{x^{2}}{2}\right)_{0}^{2}\right]\)
\(=2\left[\sqrt{2} \cdot \frac{2}{3}(2)^{3 / 2}-2\right]\)
\(=2\left[\frac{8}{3}-2\right]=2\left(\frac{2}{3}\right)=\frac{4}{3}\) sq. units