Question

Find the area of the region bounded by the curve y = x^3, 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

$\therefore Requiredareaofshadedregion={\int }_0^2(x+6-x^3)dx={[\frac{x^2}{2}+6x-\frac{x^4}{4}]}_0^2=[\frac{4}{2}+12-\frac{16}{4}-0]=[2+12-4]=10squnits.$