## University Calculus: Early Transcendentals (3rd Edition)

$a=-\sqrt 2; b=\sqrt 2$
Following the hint given in the statement, we have $x^4 -2x^2 \le 0$ This implies that $x^2(x^2-2) \le 0$ Thus, we have $x=0$ or, $x^2 \leq 2$ Also, $- \sqrt 2 \leq x^2 \leq \sqrt 2$ Thus, the limits that maximize the integral are: $a=-\sqrt 2; b=\sqrt 2$