Answer
a=$-\sqrt{2}$
b=$\sqrt{2}$
Work Step by Step
To find where $x^4$-$2x^2$ $\leq0$
let $x^4-2x^2=0$
$x^2(x^2-2)=0$
x=0 or x=+-\sqrt{2}.
By the sign graph
++++++ $0_{-\sqrt{2}}--0_0--0_\sqrt{2}+++++++$,
we can see that $x^4-2x^2\leq0$ on $[-\sqrt{2},\sqrt{2}]$
$=>a=-\sqrt{2}$
and
$b=\sqrt{2}$
minimuze the integrals