Area between the curves $y=2x^4-x^2$, x – axis and the ordinates of two minima of the curve is 

The given curve is $f\left(x\right)=2x^4-x^2$

To get the ordinates where the function has minimum value, find the solutions of the equation $f\text{'}\left(x\right)=0$

it implies that 

   $\begin{array}{rcl}8x^3-2x& =& 0\Rightarrow 2x\left(4x^2-1\right)=0\end{array}$

It implies that $x=0,x=-\frac{1}{2},\frac{1}{2}$

Consider $f\text{'}\text{'}\left(x\right)=24x^2-2$

It has positive values at $x=\pm \frac{1}{2}$

Hence, the function has minimum value at $x=\pm \frac{1}{2}$

The required area is the area of the region bounded by the curve $y=2x^4-x^2$ and the ordinates $x=\pm \frac{1}{2}$

$\begin{array}{rcl}A& =& \left|{\int }_{-\frac{1}{2}}^{\frac{1}{2}}\left(2x^4-x^2\right)dx\right|\\ & =& \left|2{\left(2\frac{x^5}{5}-\frac{x^3}{3}\right)}_0^{\frac{1}{2}}\right|\\ & =& \left|2\left(\frac{1}{80}-\frac{1}{24}\right)\right|\\ & =& \left|\frac{1}{40}-\frac{1}{12}\right|\\ & =& \left|\frac{3-10}{120}\right|\\ & =& \frac{7}{120}\end{array}$

Therefore, the area of required region is $\boxed{\frac{7}{120}}$ square units.