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

We have, $y = {x^3}$, $y = x + 6$ and $x = 0$
graph of functions are as shown in the following figure
Solving $y = {x^3}$ and $y = x + 6,$ we get
${x^3} = x + 6$

$\Rightarrow$ ${x^2} - x - 6 = 0$

Clearly $x = 2$ satisfies the above equation.
Also from the figure it is clear that there is only point of intersection

$\therefore$ From the figure, area of shaded region,
$A = \int_0^2 {\left( {x + 6 - {x^3}} \right)} dx$

$= \left[ {\frac{{{x^2}}}{2} + 6x - \frac{{{x^2}}}{4}} \right]_0^2 = \frac{4}{2} + 12 = \frac{{16}}{4} = 10$ sq. units