The area of the region bounded by the curves $y=9 x^2$ and $y=5 x^2+4$ (in sq units) is

• A

  64
• B

  $\frac{64}{3}$
• C

  $\frac{32}{3}$
• Correct answer
  D

  $\frac{16}{3}$

Solution

Intersecting point of $y=9 x^2$ and $y=5 x^2+4$ is given by
$$ \begin{aligned} & \Rightarrow \quad 4 x^2=4 \\ & \Rightarrow \quad x^2=1 \\ & \Rightarrow \quad x= \pm 1 \\ & \therefore \quad y=9( \pm 1)^2=9 \\ & \end{aligned} $$ Hence, intersecting points are $( \pm 1,9)$. $$ \begin{aligned} \therefore \text { Required area } & =2 \int_0^1\left[\left(5 x^2+4\right)-\left(9 x^2\right)\right] d x \\ & =2 \int_0^1\left(4-4 x^2\right) d x \\ & =2\left[4 x-\frac{4 x^3}{3}\right]_0^1 \\ & =2\left(4-\frac{4}{3}\right)-(0-0)=2 \times \frac{8}{3} \\ & =\frac{16}{3} \text { sq units } \end{aligned} $$

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