Question

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

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

Answer 1

The correct option is B

$\left(\frac{8}{3}\right)\sqrt{2}-\left(\frac{1}{2}\right)$

Explanation for the correct option:

Finding the area (in sq. units) of the region.

Consider the given region as

$A=\left\{\left(x,y\right):\left(x-1\right)\left[x\right]\le y\le 2\sqrt{x},0\le x\le 2\right\}$

Here, we have

$\left(x-1\right)\left[x\right]\le y\le 2\sqrt{x},0\le x\le 2$

Then,

$$\begin{gathered} y=2\sqrt{x} \\ {y}^2=4x\left[x\ge 0\right] \end{gathered}$$

Also

$y=\left(x-1\right)\left[x\right]=\left\{\begin{array}{ll}0& 0\le x<1\\ x-1& 1\le x<2\\ 2& x=2\end{array}\right.$

The area of a shaded region is given as:

$A={\int }_0^{2}2\sqrt{x}dx-$Area of the triangle

The area of the triangle is $=\frac{1}{2}\times \text{base ×height}$ then the above equation becomes,

$$\begin{gathered} A=2{\int }_0^2\sqrt{x}dx-\frac{1}{2}\times \left(2-1\right)\times 1 \\ A=2{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^2-\frac{1}{2}\times 1\times 1 \\ A=\frac{4}{3}\left[2^{\frac{3}{2}}-0\right]-\frac{1}{2} \\ A=\frac{4}{3}\left[2\times 2^{\frac{1}{2}}\right]-\frac{1}{2} \\ A=\frac{8}{3}\sqrt{2}-\frac{1}{2} \end{gathered}$$

Hence, option (B) is the correct answer.