Question

The area bounded by $y=x^2,y=[x+1],0\le x<1$ and the $y-$axis is ($[.]$ denotes the greatest integer function)

• A. $\frac{1}{3}$sq.unit
• B. $\frac{2}{3}$sq.unit
• C. $1$sq.unit
• D. $\frac{7}{3}$sq.unit

Answer 1

The correct option is B $\frac{2}{3}$sq.unit

$\because 0\le x<1$
$\therefore y=[x+1]=1$
Hence, required area$={\int }_0^1\sqrt{y}dy$
$=\frac{2}{3}{\left[y^{\frac{3}{2}}\right]}_0^1$
$=\frac{2}{3}$.sq.unit.