The area of the region
$\mathbf{R}=\left\{(x, y): 5 x^2 \leq y \leq 2 x^2+9\right\} \text { is : }$

• A

  $11 \sqrt{3}$ square units
• Correct answer
  B

  $12 \sqrt{3}$ square units
• C

  $9 \sqrt{3}$ square units
• D

  $6 \sqrt{3}$ square units

Solution

$\begin{aligned} & \text { Required area }=2 \int_0^{\sqrt{3}}\left(2 x^2+9-5 x^2\right) d x \\ & =2\left[9 x-x^3\right]_0^{\sqrt{3}}=2[9 \sqrt{3}-3 \sqrt{3}]=12 \sqrt{3} .\end{aligned}$

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