Let I = ∫ a b x 4 - 2 x 2 d x . If I is minimum then the ordered pair ( a ,   b ) is

Given that, $I={\int }_a^b\left(x^4-2x^2\right)dx$

Let $f\left(x\right)=x^4-2x^2$

To draw it's graph, finding the points where it intersects the $x$-axis, put $f\left(x\right)=0$

$\Rightarrow x^4-2x^2=0$

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

So, $x=0$ and $x=\pm \sqrt{2}$.

The graph of $f\left(x\right)$ is,



From the figure, we get minimum area when $-\sqrt{2}\le x\le \sqrt{2}$.

Hence, the ordered pair can be written as $\left(a, b\right)=\left(-\sqrt{2}, \sqrt{2}\right).$