Question

The area bounded by $y=x^2,$ $y=[x+1],$ $x\le 1$ and the $y$-axis, where $[.]$ represents the greatest integer function, is $A$ sq. unit, then the value of $9A^2$ is

Answer 1

$y=x^2,y=[x+1]=\left[x\right]+1$
For $0\le x<1,y=0+1=1$


Hence, the required area = the shaded area
$A=\underset{0}{\overset{1}{\int }}xdy=\underset{0}{\overset{1}{\int }}\sqrt{y}dy$
$\Rightarrow A=\frac{2}{3}\text{sq. unit}$

$\therefore 9A^2=4$