Question

The area (in sq. units) of the region $A=\{(x, y):(x-1)[x] \leq y \leq 2 \sqrt{x}, 0 \leq x \leq 2\}$ 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)

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

$A=\int\limits_{0}^{2} 2 \sqrt{x} d x-\frac{1}{2} 1 \cdot 1$
$A=2 \cdot\left[\frac{x^{3 / 2}}{(3 / 2)}\right]_{0}^{2}-\frac{1}{2}=\frac{8 \sqrt{2}}{3}-\frac{1}{2}$