Question

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

• A. $64$
• B. $\frac{64}{3}$
• C. $\frac{32}{3}$
• D. $\frac{16}{3}$

Answer 1

Answer: (D)

Intersecting point of $y=9x^2$ and $y=5x^2+4$ is given by
$9x^2=5x^2+4$

$\Rightarrow 4x^2=4$
$\Rightarrow x^2=1$
$\Rightarrow x=\pm 1$
$\therefore y=9(\pm 1{)}^2=9$
Hence, intersecting points are $(\pm 1,9)$.
$\therefore$ Required area $=2{\int }_0^1\left[\left(5x^2+4\right)-\left(9x^2\right)\right]dx$
$=2{\int }_0^1\left(4-4x^2\right)dx$
$=2{\left[4x-\frac{4x^3}{3}\right]}_0^1$
$=2\left(4-\frac{4}{3}\right)-(0-0)=2\times \frac{8}{3}$
$=\frac{16}{3}$ sq units