The area enclosed (in square units) by the curve $y=x^4-x^2$, the $x$-axis and the vertical lines passing through the two minimum points of the curve is

Given that $y=x^4-x^2$

For minimum points of curve:

$\frac{d}{dx}\left(x^4-x^2\right)=0$

$4x^3-2x=0$

$2x\left(2x^2-1\right)=0$

$\Rightarrow 2x=0$ or $2x^2-1=0$

$x=0$ or $x=\frac{\pm 1}{\sqrt{2}}$

So $f$ is minimum at $\frac{\pm 1}{\sqrt{2}}$

The area $=\left|{\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\left(x^4-x^2\right)dx\right|$

$=\left|2{\int }_0^{\frac{1}{\sqrt{2}}}\left(x^4-x^2\right)dx\right| \left[\because f\left(x\right) \text{is even function}\right]$

$=\left|2{{\left[\frac{x^5}{5}-\frac{x^3}{3}\right]}^{\frac{1}{\sqrt{2}}}}_0\right|$

$=\left|2\left({\left(\frac{1}{\sqrt{2}}\right)}^5·\frac{1}{5}-{\left(\frac{1}{\sqrt{2}}\right)}^3·\frac{1}{3}\right)\right|$

$=\left|2\left(\frac{1}{4\sqrt{2}}·\frac{1}{5}-\frac{1}{2\sqrt{2}}·\frac{1}{3}\right)\right|$

$=\left|\frac{2}{2\sqrt{2}}\left(\frac{1}{10}-\frac{1}{3}\right)\right|$

$=\left|\frac{3-10}{30\sqrt{2}}\right|$

$\therefore$ Area $=\frac{7}{30\sqrt{2}}$