Question

The area of the region $R=\left\{\right(x,y)/x^2\le y\le x\}$is

• A. $\frac{1}{6}$
• B. $\frac{2}{3}$
• C. $\frac{4}{3}$
• D. $2$

Answer 1

The correct option is A $\frac{1}{6}$

$x:(x,y)\{x^2\le y\le x\}$

$y=xy=x^2$

Meeting point $x=0$ and $x=1$

$\therefore area={\int }_0^1(x^2-x)dx$

$\Rightarrow {\frac{x^3}{3}]}_0^1-{\frac{x^2}{2}]}_0^1$

$\Rightarrow |\frac{1}{3}-\frac{1}{2}|$

$\Rightarrow \frac{|2-3|}{6}\Rightarrow \frac{1}{6}sq.unit$

$\therefore Areaofregion‘R^{\prime }is\frac{1}{6}$