Question

Area of region bounded by the curve $y=\frac{16-x^2}{4}$ and $y=se{c}^{-1}[-si{n}^{2}x]$ (where [.] denotes greatest integer function) is

• A. $\frac{1}{3}(4-\pi {)}^{\frac{3}{2}}$ sq. units
• B. $8(4-\pi {)}^{\frac{3}{2}}$ sq. units
• C. $\frac{8}{3}(4-\pi {)}^{\frac{3}{2}}$ sq. units
• D. $\frac{8}{3}(4-\pi {)}^{\frac{1}{2}}$ sq. units

Answer 1

The correct option is C

$\frac{8}{3}(4-\pi {)}^{\frac{3}{2}}$ sq. units

$0\le si{n}^{2}x\le 1\Rightarrow [-si{n}^{2}x]=0or-1$

But $se{c}^{-1}\left(0\right)$ is not defined

$\Rightarrow y=se{c}^{-1}[-si{n}^{2}x]=se{c}^{-1}(-1)=\pi$

Solving $y=\frac{16-x^2}{4}\text{and}y=\pi$

We get $x=\pm 2\sqrt{4-\pi }$

$\therefore$ Required area $=2{\int }_0^{2\sqrt{4-\pi }}(\frac{16-x^2}{4}-\pi )dx=\frac{8}{3}(4-\pi {)}^{\frac{3}{2}}$ sq. units