Question

The area bounded by $y=x^3-x$ and $y=x^2+x$ is

• A. $\frac{37}{12}\text{sq. units}$
• B. $\frac{25}{12}\text{sq. units}$
• C. $\frac{32}{12}\text{sq. units}$
• D. $\frac{43}{12}\text{sq. units}$

Answer 1

The correct option is A $\frac{37}{12}\text{sq. units}$

$y=x^3-x$ and $y=x^2+x$
$\Rightarrow x^3-x=x^2+x$
$\Rightarrow x(x^2-x-2)=0$
$\Rightarrow x(x-2)(x+1)=0$
$\Rightarrow x=-1,0,2$
$\therefore$ Points of intersection are $(-1,0),(0,0),(2,6)$.

From figure required area is given as,
$A=\underset{-1}{\overset{0}{\int }}[(x^3-x)-(x^2+x)]dx+\underset{0}{\overset{2}{\int }}[(x^2+x)-(x^3-x)]dx$
$={[\frac{x^4}{4}-\frac{x^3}{3}-x^2]}_{-1}^0+{[\frac{x^3}{3}+x^2-\frac{x^4}{4}]}_0^2$
$=\frac{37}{12}\text{sq. units}$