The area between the curve $\mathrm{y}=2{\mathrm{x}}^4-{\mathrm{x}}^2$, the $\text{x}$-axis and the ordinates of the two minima of the curve is

The equation of the curve is $\mathrm{y}=2{\mathrm{x}}^4-{\mathrm{x}}^2=\left(2{\mathrm{x}}^2-1\right)$${\text{x}}^2$

The curve is symmetrical about the $\text{y}$-axis.

Also, it is a polynomial of degree four having roots $0,0,\pm \frac{1}{\sqrt{2}}.$

$\text{x}=0$ is repeated root. Hence, graph touches $\text{x}$ -axis at $\text{(0,0)}$ and intersects the $\text{x}$ -axis at $\text{A}\left(-\frac{1}{\sqrt{2}}, 0\right)\&\text{B}\left(\frac{1}{\sqrt{2}}, 0\right)$

$\Rightarrow \frac{\text{dy}}{\text{dx}}=8{\text{x}}^3-2\text{x}$

$=2\text{x}(4{\text{x}}^2-1)=0$

$\text{x}=0,\text{x}=\pm \frac{1}{2}$

$\left(\frac{{\text{d}}^2\text{y}}{\text{d}{\text{x}}^2}\right)>0$ at $\text{x}=\frac{1}{2}$ and at $\text{x}=\frac{-1}{2}$

So x = $\pm \frac{1}{2}$ are the points of local minima

Thus, the graph of the curve is shown in diagram.



Here, $\mathrm{y}\le 0$ , as $\text{x}$ varies from $\mathrm{x}=-\frac{1}{2}$ to $\mathrm{x}=\frac{1}{2}$

$\therefore$ The required area

$=2\mathrm{ }\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{O}\mathrm{C}\mathrm{D}\mathrm{O}$

$=2\left|\underset{0}{\overset{\frac{1}{2}}{\int }}\mathrm{y}\mathrm{ }\mathrm{d}\mathrm{x}\right|$

$=2\left|\underset{0}{\overset{\frac{1}{2}}{\int }}\left(2{\mathrm{x}}^4-{\mathrm{x}}^2\right)\mathrm{ }\mathrm{d}\mathrm{x}\right|$

$=\frac{7}{120}\mathrm{ }\mathrm{s}\mathrm{q}.\mathrm{ }\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}$