Question

The area (in sq. units) of the region $\{x\in R:x\ge ,y\ge 0,y\ge x-2$ and $y\le \sqrt{x}\}$, is

• A. $\frac{13}{3}$
• B. $\frac{8}{3}$
• C. $\frac{16}{3}$
• D. $\frac{5}{3}$

Answer 1

Answer: (C)

$\frac{16}{3}$

The given reqion $\{x\in R:x\ge 0,y\ge 0,y\ge x-2andy\le \sqrt{x}\}$

Here $y\ge x-2$ and $y\le \sqrt{x}$

$\Rightarrow x-2\le y\le \sqrt{x}$

Suppose

$\Rightarrow y=x-2$ …….$\left(1\right)$ and $y=\sqrt{x}$ …………$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\Rightarrow x-2=\sqrt{x}$

$\Rightarrow (x-2{)}^2=(\sqrt{x}{)}^2$ [squaring both sides]

$\Rightarrow x^2+4-4x=x$

$\Rightarrow x^2-4x+4-x=0$

$\Rightarrow x(x-4)-1(x-4)=0$

$\Rightarrow (x-1)(x-4)=0$

$\therefore x=1,4$

At $x=1$, from $\left(1\right)$, $y=1-2=-1$ but $y\ge 0$

$\therefore x=1$ is neglected

At $x=4$, from $\left(1\right)$, $y=4-2=2$

$\therefore$ Area of the region $={\int }_0^4\{\sqrt{x}-(x-2\left)\right\}dx={\int }_0^4(x^{\frac{1}{2}}-x+2)dx$

$={[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{x^2}{\alpha }+2x]}_0^4$

$=\frac{4^{3/2}}{\frac{3}{2}}-\frac{4^2}{2}+2\left(4\right)-0+0-0$

$=\frac{2}{3}\times 8-8+8=\frac{16}{3}$.