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

• A

  $\frac{16}{3}$ sq. units
• Correct answer
  B

  $\frac{4}{3}$ sq. units
• C

  $\frac{3}{4}$ sq. units
• D

  $\frac{8}{3}$ sq. units

Solution

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

Hence, the correct answer is option: 'B'.