Let $I=$$\underset{a}{\overset{b}{\int }}(x^4-2x^2)dx.$ If $I$ is minimum then the ordered pair $\left(a, b\right)$ is 

$\begin{array}{rcl}\mathrm{I}& =& {\int }_{\mathrm{a}}^{\mathrm{b}} \left({\mathrm{x}}^4-2{\mathrm{x}}^2\right) \mathrm{dx}\\ \mathrm{f}\left(\mathrm{x}\right)& =& {\mathrm{x}}^4-2{\mathrm{x}}^2={\mathrm{x}}^2({\mathrm{x}}^2-2)\\ & =& {\mathrm{x}}^2\left(\mathrm{x}+\sqrt{2}\right) \left(\mathrm{x}-\sqrt{2}\right)\end{array}$

As long as $\mathrm{f}\left(\mathrm{x}\right)$ lie below the x-axis, the given integral will remain negative.

So $(\mathrm{a}, \mathrm{b})=\left(-\sqrt{2}, \sqrt{2}\right)$