Question

Find the area of the figure bounded by the following curves
y = $x^3,y^2$ = x.

Answer 1

$y=x^3{y}^2=x$

$x^6=x$

$x^6-x=0$

$x(x^5-1)=0$

$x=0$ or $x=1$

So, $y=0$ at $x=0$ and $y=1$ at $x=1$

$={\int }_0^1(x^3-\sqrt{x})dx$

$=\frac{x^4}{4}-\frac{{\left(x\right)}^{3/2}}{3/2}{]}_0^1$

$=\frac{1}{4}-\frac{2}{3}-0$

$=\frac{3-8}{12}=|-\frac{5}{12}|=\frac{5}{12}$ sq units.