Question

The area of the region bounded by the curves $y=x^2\text{and}y={\sec }^{-1}[-{\cos }^{2}x]$, where [.] represent greatest integer function, is

• A. $\frac{4}{3}{\pi }^{\frac{3}{2}}$ sq.units
• B. $\frac{5}{3}{\pi }^{\frac{3}{2}}$ sq.units
• C. $\frac{1}{3}{\pi }^{\frac{3}{2}}$ sq.units
• D. $\frac{4}{3}{\pi }^{\frac{3}{2}}$ sq.units

Answer 1

The correct option is D $\frac{4}{3}{\pi }^{\frac{3}{2}}$ sq.units

$\therefore 0\le {\cos }^{2}x\le q$
$\Rightarrow 0\ge -{\cos }^{2}x\ge -1$
$\Rightarrow [–{\cos }^{2}x]=–1\text{or}0$
But ${\sec }^{–1}\left(0\right)$ is not defined
$\Rightarrow x\ne (2n+1)\frac{\pi }{2}$


$\therefore y={\sec }^{-1}=\pi$
Solving $y=x^2\text{and}y=\pi \text{we have}\pi =x^2,x=\pm \sqrt{\pi }$
$\therefore$Required area
$=2\underset{0}{\overset{\sqrt{\pi }}{\int }}(\pi -x^2)dx$
$=2{[\pi x-\frac{x^3}{3}]}_0^{\sqrt{\pi }}$
$=2[(\pi \sqrt{\pi }-\frac{\pi \sqrt{\pi }}{3})-(0-0)]$
$=\frac{4}{3}{\pi }^{\frac{3}{2}}$ sq.units