Question

Find the area of the region bounded by the curve $y=x^3$ and $y=x+6$ and $x=0$.

Answer 1

We have, $y=x^3,y=x+6$ and $x=0$

$\therefore x^3=x+6$

$\Rightarrow x^3-x=6$

$\Rightarrow x^3-x-6=0$

$\Rightarrow x^2(x-2)+2x(x-2)+3(x-2)=0$

$\Rightarrow (x-2)(x^2+2x+3)=0$

$\Rightarrow x=2$, with two imaginary points

$\therefore$ Required area of shaded region $={\int }_0^2(x+6-x^3)dx$

$={[\frac{x^2}{6}+6x-\frac{x^4}{4}]}_0^2$

$=[\frac{4}{2}+12-\frac{16}{4}-0]$

$=[2+12-4]=10squnits$.