Question

The area bounded by the curves $y=x(x-3{)}^2$ and $y=x$ is (in sq. units) :

• A. $28$
• B. $32$
• C. $4$
• D. $8$

Answer 1

Answer: (D)

$8$

R.E.F image

$y=x(x-3{)}^2$

$x=x(x-3{)}^2\Rightarrow (x\left)\right((x-3{)}^2-1)=0$

$\Rightarrow \left(x\right)(x^2-6x+8)=0$

$\Rightarrow \left(x\right)(x-4)(x-2)=0$

Area bounded by these two curves is

${\int }_0^2\left(\right(x\left)\right(x-3{)}^2-x)dx+{\int }_2^{4}x-x(x-3{)}^{2}dx$

${\int }_0^2{x}^3-6x^2+8xdx+{\int }_2^{4}6x^2-x^5-8xdx$

$[\frac{x^4}{4}-2x^3+4x^2{]}_0^2+[2x^3-\frac{x^4}{4}-4x^2{]}_2^4$

$[4-16+16-0]+[128-64-64-(16-4-16\left)\right]$

$=4+4$

$=8sq.units$