Question

The area of the region bounded by $y=x,y=x^3$ is:

• A. $\frac{1}{4}$ sq. units
• B. $\frac{1}{12}$ sq. units
• C. $\frac{1}{3}$ sq. units
• D. $\frac{1}{2}$ sq. units

Answer 1

The correct option is D $\frac{1}{2}$ sq. units

The curve intersect at $x=-1,y=-1$; $x=0,y=0$ and $x=1,y=1$
Therefore area $=-{\int }_{-1}^0(x^3-x)dx+{\int }_0^1(x-x^3)dx$
$=-{[\frac{x^4}{4}-\frac{x^2}{2}]}_{-1}^0+{[\frac{x^2}{2}-\frac{x^4}{4}]}_0^1$
$=-(\frac{1}{4}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{4})=-\frac{1}{2}+1=\frac{1}{2}$