Question

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

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

Answer 1

The correct option is B $\frac{8}{3}\sqrt{2}-\frac{1}{2}$

$y=f\left(x\right)=(x-1)\left[x\right]=\{\begin{array}{cc}0& 0\le x<1\\ x-1& 1\le x<2\\ 2(x-1)& x=2\end{array}$
$y^2\le 4x$


$A=\underset{0}{\overset{1}{\int }}(2\sqrt{x}-0)dx+\underset{1}{\overset{2}{\int }}(2\sqrt{x}-(x-1\left)\right)dx$
$={[\frac{2}{3}\times 2x^{3/2}]}_0^1+{[\frac{4}{3}{x}^{3/2}-\frac{x^2}{2}+x]}_1^2$
$=(\frac{4}{3}-0)+\{(\frac{4}{3}\times 2\sqrt{2}-2+2)-(\frac{4}{3}+\frac{1}{2})\}$
$=\frac{4}{3}+\frac{8\sqrt{2}}{3}-\frac{4}{3}-\frac{1}{2}=\frac{8\sqrt{2}}{3}-\frac{1}{2}$
$A=\frac{8\sqrt{2}}{3}-\frac{1}{2}$ sq. units