Find the area enclosed between the curves

$y=3x\text{ and }y=6x-x^2$.

Solving the given curves then we get

$y=3x\text{ and }y=6x-x^2\Rightarrow 3x=6x-x^2$

$x^2-3x=0\Rightarrow x=0,3$

$y=6x-x^2\Rightarrow (x-3{)}^2=-(y-9)$

Area  $={\int }_0^3 \left(6 x-x^2-3x\right)dx={\int }_0^3 \left(3x-x^2\right)dx$

${\left.=\frac{3x^2}{2}-\frac{x^3}{3}\right]}_0^3=\frac{27}{2}-9=\frac{9}{2}\text{ sq.unit }$

$\therefore$  The area enclosed between the curves

$y=3x\mathrm{\&}y=6x-x^2\text{ is }\frac{9}{2}\text{ sq.units }$