⇒Application of Definite Integration

Correct Option is (D)

The required area is shaded.

The point of intersection of the curves are

$\begin{array}{rl}{\mathrm{x}}^2=\mathrm{x}& \Rightarrow \mathrm{x}(\mathrm{x}-1)=0\text{ i.e. }\mathrm{O}(0,0)\text{ and }\mathrm{P}(1,1)\\ \therefore \mathrm{A}& =\underset{0}{\overset{1}{\int }}(\mathrm{x}-{\mathrm{x}}^2)\mathrm{dx}\\ & ={\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^1-{\left[\frac{{\mathrm{x}}^3}{3}\right]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\text{ sq. units }\end{array}$

Correct Option is (B)

Points of intersection of given curves are

${\mathrm{x}}^2-\mathrm{x}=0\Rightarrow \mathrm{x}(\mathrm{x}-1)=0\Rightarrow \mathrm{x}=0,1$

i.e. O(0, 0) and A(1, 1)

Required area is shaded.

$\therefore$ $A=\underset{0}{\overset{1}{\int }}(x-x^2)dx$

$={\left[\frac{x^2}{2}\right]}_0^1-{\left[\frac{x^3}{3}\right]}_0^1=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$ sq. units