Question

The area (in sq. units) of the region $A=\{(x,y):(x-1)[x]\le y\le 2\sqrt{x},0\le x\le 2\}$ where [t] denotes the greatest integer function. is:

• A. $\frac{8}{3}\sqrt{2}-\frac{1}{2}$
• B. $\frac{8}{3}\sqrt{2}-1$
• C. $\frac{4}{3}\sqrt{2}-\frac{1}{2}$
• D. $\frac{4}{3}\sqrt{2}+1$

Answer 1

Answer: (A)

$(x-1)[x]\le y\le 2\sqrt{x},0\le x\le 2$
Draw $y=2\sqrt{x}$
$\Rightarrow y^2=4x\sqrt{x\ge 0}$
$y=(x-1)[x]=\{\begin{array}{ll}0,& 0\le x<1\\ x-1,& 1\le x<2\\ 2,& x=2\end{array}$

$A=\underset{0}{\overset{2}{\int }}2\sqrt{x}dx-\frac{1}{2}1\cdot 1$
$A=2\cdot {\left[\frac{x^{3/2}}{(3/2)}\right]}_0^2-\frac{1}{2}=\frac{8\sqrt{2}}{3}-\frac{1}{2}$