The area between the curve  $y=2x^4-x^2$, the x-axis, and the ordinates of the two minima of the curve is A then 120A =

The curve is $y=2x^4-x^2=x^2(2x^2-1)$

The curve is symmetrical about the axis of y. Also it is a polynomial of 4 degree having roots 0,0, $\pm$$\frac{1}{\sqrt{2}}$. X = 0 is a repeated root. Hence, graph touches at (0,0).  The curve intersects the axes at O(0,0), $A(\frac{-1}{\sqrt{2}},0)andB(\frac{1}{\sqrt{2}},0)$. Thus the graph of the curve is as shown in the figure.

Here, $y\le 0,asx\mathrm{var}iesfromx=\frac{-1}{2}tox=\frac{1}{2}$

$\therefore$required area =2 Area OCDO

$=2\left|{\int }_0^{\frac{1}{2}}ydx\right|$

$=2\left|{\int }_0^{\frac{1}{2}}(2x^4-x^2)dx\right|$
$$\begin{gathered} =2{\left|\left(\frac{2x^5}{5}-\frac{x^3}{3}\right)\right|}_0^{\frac{1}{2}} \\ =2\left|\left(\frac{1}{80}-\frac{1}{24}\right)\right| \\ =2\left|\frac{-7}{240}\right| \end{gathered}$$

$=\frac{7}{120}sq.units$